# A dynamical classification for crossed products of fiberwise essentially
minimal zero-dimensional dynamical systems

Paul Herstedt

###### Abstract.

We prove that crossed products of fiberwise essentially minimal zero-
dimensional dynamical systems have isomorphic $K$-theory if and only if the
dynamical systems are strong orbit equivalent. Under the additional assumption
that the dynamical systems have no periodic points, this gives a
classification theorem including isomorphism of the $C^{*}$-algebras as well.
We additionally explore the $K$-theory of such crossed products and the
Bratteli diagrams associated to the dynamical systems.

This research was supported by the Israel Science Foundation grant no. 476/16.

## Introduction

In 1990, Ian Putnam proved in [14] that the crossed product $C^{*}$-algebras
associated to minimal Cantor systems are A$\mathbb{T}$-algebras of real rank
zero. Using the classification results of Geroge Elliott in [5] and Dadarlat-
Gong in [3], one sees that such $C^{*}$-algebras are classifiable by their
$K$-theory. In 1995, Putnam, along with Thierry Giordano and Christian Skau,
expanded this classification theorem to include dynamics; in [6], they showed
that there is a condition on the dynamical systems, called “strong orbit
equivalence”, that is equivalent to isomorphism of the crossed product
$C^{*}$-algebras. This dynamical classification was motivated by Krieger’s
theorem (see [9] and [10]), which says that for ergodic non-singular systems,
the associated von Neumann crossed product factors are isomorphic if and only
if the systems are orbit equivalent. The goal of this paper is to expand these
results in the $C^{*}$-setting.

In our previous paper [8], we determined a condition on a zero-dimensional
dynamical system called “fiberwise essentially minimal” (see Definition 1.2
below) that guarantees that the associated crossed product is an
A$\mathbb{T}$-algebra. As its name suggests, this class is a broadening of
minimal (and also essentially minimal). If we further restrict the system to
have no periodic points, this crossed product has real rank zero and is
therefore classifiable by $K$-theory (due to the work of Elliott in [5] and
Dadarlat-Gong in [3]). This was an expansion of work done on the minimal
Cantor case in 1990 by Ian Putnam (see [13] and [14]) in which the crossed
products are simple, and work done on the essentially minimal case in 1992 by
Putnam and Skau along with Richard Herman (see [7]) in which the crossed
products are not necessarily simple. Our result from [8] includes many more
non-simple crossed products.

This paper expands on the work in [8] in two major ways. The first of which is
what we explore in Section 2, where we discuss some specifics about the
$K$-theory of the crossed products. We define “large subalgebras” of our
crossed products (see Definition 2.6) which are AF-subalgebras (see Theorem
2.7) that have the same $K_{0}$ group as the crossed product (see Theorem
2.9). This mirrors the result of large subalgebras in the minimal case by
Putnam in [14]. We also give a simple description of the $K_{1}$ group of the
crossed product in Theorem 2.10.

The second major aspect of this paper is expanding the dynamical
classification of minimal Cantor systems that coincides with the $K$-theoretic
classification, introduced by Giordano, Putnam, and Skau in [6]. They
introduce the notion of “strong orbit equivalence”, which we expand to the
fiberwise essentially minimal case in Definition 1.2. In Section 3, we discuss
how the circle algebra direct system that gives the A$\mathbb{T}$-algebra of
the crossed product gives us an ordered Bratteli diagram whose Vershik system
is conjugate to the original dynamical system. Using this Bratteli diagram
along with our $K$-theory results, we then prove Theorem 4.2, which tells us
that for fiberwise essentially minimal zero-dimensional systems, $K$-theory
isomorphism of the crossed products is equivalent to strong orbit equivalence
of the dynamical systems. This, combined with the classification result of [5]
and [3], gives us Theorem 4.3, which tells us that if the dynamical systems
have no periodic points, this is also equivalent to isomorphism of the crossed
products.

Acknowledgements. This work was done at the Ben Gurion University of the
Negev, funded by the Israel Science Foundation (grant no.476/16).

##  1\. Preliminaries

This section introduces terms that will be used to prove the main theorems of
the paper, along with some examples and some previous relevant results.

Let $X$ be a totally disconnected compact metrizable space and let $h:X\to X$
be a homeomorphism of $X$. We call $(X,h)$ a _zero-dimensional system_. Let
$\alpha$ be the automorphism of $C(X)$ induced by $h$; that is, $\alpha$ is
defined by $\alpha(f)(x)=f(h^{-1}(x))$ for all $f\in C(X)$ and all $x\in X$.
Then we denote the crossed product of $C(X)$ by $\alpha$ by
$C^{*}(\mathbb{Z},X,h)$ (or, less commonly, $C^{*}(\mathbb{Z},C(X),\alpha)$).
We denote the “standard unitary” of $C^{*}(\mathbb{Z},X,h)$ by $u$, which is a
unitary element of $C^{*}(\mathbb{Z},X,h)$ that satisfies $ufu^{*}=\alpha(f)$
for all $f\in C(X)$.

We will use the disjoint union symbol $\bigsqcup$ to denote unions of disjoint
sets. We will not always say explicitly that the sets in this union are
disjoint, as this will be implied by the notation. By a partition
$\mathcal{P}$ of $X$, we mean a finite set of mutually disjoint compact open
subsets of $X$ whose union is $X$.

We say that a nonempty closed subset $Y$ of $X$ is a _minimal set_ if it is
$h$-invariant and has no nonempty $h$-invariant proper closed subsets. By
Zorn’s lemma, minimal sets exist for every zero-dimensional system. We say
that a dynamical system $(X,h)$ is an _essentially minimal system_ if it has a
unique minimal set.

The following definition is introduced as Definition 1.10 in [8].

######  Definition 1.1.

Let $(X,h)$ be a zero-dimensional system and let $\mathcal{P}$ be a partition
of $X$. We define a _system of finite first return time maps subordinate to
$\mathcal{P}$_ to be a tuple

$\mathcal{S}=(T,(X_{t})_{t=1,\ldots,T},(K_{t})_{t=1,\ldots,T},(Y_{t,k})_{t=1,\ldots,T;k=1,\ldots,K_{t}},(J_{t,k})_{t=1,\ldots,T;k=1,\ldots,K_{t}})$

such that:

  1. (a)

We have $T\in\mathbb{Z}_{>0}$.

  2. (b)

For each $t\in\\{1,\ldots,T\\}$, $X_{t}$ is a compact open subset of $X$. That
$\mathcal{S}$ is subordinate to $\mathcal{P}$ means that, for each
$t\in\\{1,\ldots,T\\}$, $X_{t}$ is contained in an element of $\mathcal{P}$.

  3. (c)

For each $t\in\\{1,\ldots,T\\}$, $K_{t}\in\mathbb{Z}_{>0}$.

  4. (d)

For each $t\in\\{1,\ldots,T\\}$ and each $k\in\\{1,\ldots,K_{t}\\}$, $Y_{t,k}$
is a compact open subset of $X_{t}$. Moreover, for each
$t\in\\{1,\ldots,T\\}$, $\\{Y_{t,1},\ldots,Y_{t,K_{t}}\\}$ is a partition of
$X_{t}$; that is,

$\bigsqcup_{k=1}^{K_{t}}Y_{t,k}=X_{t}.$

  5. (e)

For each $t\in\\{1,\ldots,T\\}$ and each $k\in\\{1,\ldots,K_{t}\\}$,
$J_{t,k}\in\mathbb{Z}_{>0}$. Using Definition LABEL:defnLambdaU, we can write
$\\{J_{t,k}\\}=\lambda_{X_{t}}(Y_{t,k})$. Moreover, for each
$t\in\\{1,\ldots,T\\}$,
$\\{h^{J_{t,1}}(Y_{t,1}),\ldots,h^{J_{t,K_{t}}}(Y_{t,K_{t}})\\}$ is a
partition of $X_{t}$; that is,

$\bigsqcup_{k=1}^{K_{t}}h^{J_{t,k}}(Y_{t,k})=X_{t}.$

  6. (f)

The set

$\mathcal{P}_{1}(\mathcal{S})=\big{\\{}h^{j}(Y_{t,k})\,\big{|}\,\mbox{$t\in\\{1,\ldots,T\\}$,
$k\in\\{1,\ldots,K_{t}\\}$, and $j\in\\{0,\ldots,J_{t,k}-1\\}$}\big{\\}}$

is a partition of $X$. Note that this combined with condition (e) also implies

$\mathcal{P}_{2}(\mathcal{S})=\big{\\{}h^{j}(Y_{t,k})\,\big{|}\,\mbox{$t\in\\{1,\ldots,T\\}$,
$k\in\\{1,\ldots,K_{t}\\}$, and $j\in\\{1,\ldots,J_{t,k}\\}$}\big{\\}}$

is a partition of $X$.

The following definition is introduced as Definition 1.18 in [8].

######  Definition 1.2.

Let $(X,h)$ be a zero-dimensional system and let $Z\subset X$ be a closed
subset. We say that the triple $(X,h,Z)$ is a _fiberwise essentially minimal
zero-dimensional system_ if there is a quotient map $\psi:X\to Z$ such that

  1. (a)

$\psi|_{Z}:Z\to Z$ is the identity map.

  2. (b)

$\psi\circ h=\psi$.

  3. (c)

For each $z\in Z$, $(\psi^{-1}(z),h|_{\psi^{-1}(z)})$ is an essentially
minimal system and $z$ is in its minimal set.

For examples of fiberwise essentially minimal zero-dimensional systems, see
Examples 1.18 in [8]. The connection between Definition 1.1 and Definition 1.2
is the following theorem, which appears as Theorem 2.1 of [8].

######  Theorem 1.3.

Let $(X,h)$ be a zero-dimensional system. Then there exists some closed
$Z\subset X$ such that $(X,h,Z)$ is fiberwise essentially minimal if and only
if for any partition $\mathcal{P}$ of $X$, $(X,h)$ admits a system of finite
first return time maps subordinate to $\mathcal{P}$.

The following is Theorem 2.2 in [8].

######  Theorem 1.4.

Let $(X,h,Z)$ be a fiberwise essentially minimal zero-dimensional system. Then
$C^{*}(\mathbb{Z},X,h)$ is an A$\mathbb{T}$-algebra.

The following is a consequence of the proof of Theorem 1.4. By “circle
algebra”, we mean an algebra isomorphic to a finite direct sum of matrices and
matrices over $C(S^{1})$.

######  Corollary 1.5.

Let $(X,h,Z)$ be a fiberwise essentially minimal zero-dimensional system, let
$\mathcal{P}$ be a partition of $X$, let $a_{1},\ldots,a_{n}\in
C^{*}(\mathbb{Z},X,h)$, and let $\varepsilon>0$. Then there is a circle
algebra $A\subset C^{*}(\mathbb{Z},X,h)$ and a partition
$\mathcal{P}^{\prime}$ of $X$ that is finer than $\mathcal{P}$ such that

  1. (a)

The diagonal matrices of $A$ are $C(\mathcal{P}^{\prime})$.

  2. (b)

For each $k\in\\{1,\ldots,n\\}$, there is a $b_{k}\in A$ such that
$\|a_{k}-b_{k}\|<\varepsilon$.

We now introduce the concepts important to the dynamical side of the
discussion in this paper. Let $(X_{1},h_{1})$ and $(X_{2},h_{2})$ be dynamical
systems. By an _orbit map_ , we mean a homeomorphism $F:X_{1}\to X_{2}$ such
that for all $x\in X_{1}$, we have
$F(\text{orb}_{h_{1}}(x))=\text{orb}_{h_{2}}(F(x))$, where
$\text{orb}_{h_{1}}(x)$ denotes the $h_{1}$-orbit of $x$ (and likewise for
$\text{orb}_{h_{2}}$). We say that $(X_{1},h_{1})$ and $(X_{2},h_{2})$ are
_orbit equivalent_ if there exists such an $F$. If the orbit map satisfies
$F\circ h_{1}=h_{2}\circ F$, we say that $(X_{1},h_{1})$ and $(X_{2},h_{2})$
are _conjugate_. When we consider orbit maps between $(X_{1},h_{1},Z_{1})$ and
$(X_{2},h_{2},Z_{2})$ for closed sets $Z_{1}\subset X_{1}$ and $Z_{2}\subset
X_{2}$, we require that $F(Z_{1})=Z_{2})$.

######  Definition 1.6.

Let $(X_{1},h_{1})$ and $(X_{2},h_{2})$ be dynamical systems and let
$F:X_{1}\to X_{2}$ be an orbit map. Then there are functions
$\beta,\gamma:X_{1}\to\mathbb{Z}$, called _orbit cocycles_ , that satisfy
$(F\circ h_{1}^{\beta(x)})(x)=(h_{2}\circ F)(x)$ and $(h_{2}^{\gamma(x)}\circ
F)(x)=(F\circ h_{1})(x)$.

The following is a generalization of Definition 1.3 of [6] from the minimal
case to the fiberwise essentially minimal case.

######  Definition 1.7.

Let $(X_{1},h_{1},Z_{1})$ and $(X_{2},h_{2},Z_{2})$ be fiberwise essentially
minimal zero-dimensional systems. We say that $(X_{1},h_{1},Z_{1})$ and
$(X_{2},h_{2},Z_{2})$ are _strong orbit equivalent_ if there is an orbit map
$F:X_{1}\to X_{2}$ such that the orbit cocycles
$\beta,\gamma:X_{1}\to\mathbb{Z}$ are continuous on $X_{1}\setminus Z_{1}$.

##  2\. $K$-Theory

In this section we discuss the $K$-theory of the crossed products associated
to fiberwise essentially minimal zero-dimensional systems. For some references
on work done in the minimal case, see [14] and [6]. For a reference on work
done in the essentially minimal case, see [7].

######  Definition 2.1.

An _ordered group_ is a pair $(G,G^{+})$, where $G$ is a countable abelian
group and $G^{+}$ is a subset of $G$, called the _positive cone_ , that
satisfies the following:

  1. (a)

For all $g_{1},g_{2}\in G^{+}$, we have $g_{1}+g_{2}\in G^{+}$.

  2. (b)

For all $g\in G$, there are $g_{1},g_{2}\in G^{+}$ such that $g=g_{1}-g_{2}$.

  3. (c)

The identity of $G$ is the only element in both $G^{+}$ and $-G^{+}$.

We call $e\in G^{+}$ an _order_ unit if for all $g\in G^{+}$, there is some
$n\in\mathbb{Z}_{>0}$ such that $ne-g\in G^{+}$.

Given an ordered group $(G,G^{+})$, we may write $g\geq 0$ to denote $g\in
G^{+}$. The notation $g_{1}\geq g_{2}$ means that $g_{1}-g_{2}\in G^{+}$. By a
homomorphism of ordered groups $(G_{1},G_{1}^{+})$ and $(G_{2},G_{2}^{+})$, we
mean a homomorphism of groups $\varphi:G_{1}\to G_{2}$ such that
$\varphi(G_{1}^{+})\subset G_{2}^{+}$.

When we fix a particular order unit $e\in G^{+}$, we may write the triple
$(G,G^{+},e)$ and call this an ordered group with distinguished order unit. By
a homomorphism of ordered groups with distinguished order units
$(G_{1},G_{1}^{+},e_{1})$ and $(G_{2},G_{2}^{+},e_{2})$, we mean a
homomorphism of ordered groups $\varphi:G_{1}\to G_{2}$ such that
$\varphi(e_{1})=e_{2}$.

We introduce notation important to the following proposition, which is
Proposition 2.2 of [8], and a direct consequence of Theorem 2.4 of [11]. Let
$\mathcal{T}$ denote the Toeplitz algebra, the universal $C^{*}$-algebra
generated by a single isometry $s$. Let $A$ be a unital $C^{*}$-algebra and
let $\alpha$ be an automorphism of $A$, and let $u$ be the standard unitary of
$C^{*}(\mathbb{Z},A,\alpha)$. We denote by $\mathcal{T}(A,\alpha)$ the
Toeplitz extension of $A$ by $\alpha$, which is the subalgebra of
$C^{*}(\mathbb{Z},A,\alpha)\otimes\mathcal{T}$ generated by $A\otimes 1$ and
$u\otimes s$. The ideal generated by $A\otimes(1-ss^{*})$ is isomorphic to
$A\otimes\mathcal{K}$, and the quotient by this ideal is isomorphic to
$C^{*}(\mathbb{Z},A,\alpha)$.

######  Proposition 2.2.

Let $(X,h)$ be a zero-dimensional system. Let $\alpha$ be the automorphism of
$C(X)$ induced by $h$; that is, $\alpha$ is defined by
$\alpha(f)(x)=f(h^{-1}(x))$ for all $f\in C(X)$ and all $x\in X$. Let $\delta$
be the connecting map obtained from the exact sequence

${0}$${C(X)\otimes\mathcal{K}}$${\mathcal{T}(C(X),\alpha)}$${C^{*}(\mathbb{Z},A,\alpha)}$${0,}$

where $K_{0}(C(X)\otimes\mathcal{K})$ is identified with $K_{0}(C(X))$ in the
standard way. Let $i:C(X)\to C^{*}(\mathbb{Z},X,h)$ be the natural inclusion.
Then there is an exact sequence

${0}$${K_{1}(C^{*}(\mathbb{Z},X,h))}$${K_{0}(C(X))}$${K_{0}(C(X))}$${K_{0}(C^{*}(\mathbb{Z},X,h))}$${0.}$$\scriptstyle{\delta}$$\scriptstyle{\text{id}-\alpha_{*}}$$\scriptstyle{i_{*}}$

###### Proof.

Since $K_{1}(C(X))=0$, this follows immediately from Theorem 2.4 of [11]. ∎

Note that $K_{0}(C(X))\cong C(X,\mathbb{Z})$; we will use this identification
throughout the paper. Let $C(X,\mathbb{Z})^{+}$ denote the subset of
$C(X,\mathbb{Z})$ consisting of $f$ such that $f(x)\geq 0$ for all $x\in X$.
Then it is easy to check that $(C(X,\mathbb{Z}),C(X,\mathbb{Z})^{+})$ is an
ordered group and the function $\chi_{X}$ is an order unit.

The following is closely related to Proposition 5.1 of [7]; although the
hypotheses are broadened, the proof is essentially the same. Adopting the
notation of Proposition 2.2, we denote
$K_{0}(C(X))/\text{im}(\text{id}-\alpha_{*})$ by $K^{0}(X,h)$.

######  Proposition 2.3.

Let $(X,h,Z)$ be a fiberwise essentially minimal zero-dimensional system and
adopt the notation of Proposition 2.2. Let $\pi:C(X,\mathbb{Z})\to K^{0}(X,h)$
denote the quotient map. Define $K^{0}(X,h)^{+}=\pi(C(X,\mathbb{Z})^{+})$.
Then $(K^{0}(X,h),K^{0}(X,h)^{+},\pi(1))$ is an ordered group with
distinguished order unit.

###### Proof.

We check the conditions of Definition 2.1. Conditions (a) and (b) follow from
surjectivity of $\pi$. For condition (c), let $g\in
K^{0}(X,h)^{+}\cap-K^{0}(X,h)^{+}$. This means that there is $f_{1}\in
C(X,\mathbb{Z})^{+}$ such that $\pi(f_{1})=g$ and $f_{2}\in
C(X,\mathbb{Z})^{+}$ such that $\pi(-f_{2})=g$. But then $\pi(f_{1}+f_{2})=0$
and so $f_{1}+f_{2}\in\text{im}(\text{id}-\alpha_{*})$. Let $f\in
C(X,\mathbb{Z})$ satisfy $f-\alpha_{*}(f)=f_{1}+f_{2}$. Let
$E=f^{-1}(\max_{x\in X}f(x))$. Since $f-\alpha_{*}(f)\geq 0$, we must have
$h(E)\subset E$. Let $\psi$ be as in Definition 1.2 and let $z\in\psi(E)$ and
define $E_{z}=E\cap\psi^{-1}(z)$. Since $h(E_{z})\subset E_{z}$, $E_{z}$ is
invariant so must intersect the minimal set. But then by Theorem 1.1 of [7],
since $E_{z}\neq\varnothing$, we have $\bigcup_{n\in\mathbb{Z}_{\geq
0}}h^{n}(E_{z})=\psi^{-1}(z)$, and so $E_{z}=\psi^{-1}(z)$. Since this holds
for all $z\in\psi(E)$, we see that $f$ is constant on $\psi^{-1}(z)$ for all
$z\in Z$. Since $\psi^{-1}(z)$ is invariant for all $z\in Z$, we must have
$f=\alpha_{*}(f)$, and so $f_{1}+f_{2}=0$, and since $f_{1},f_{2}\geq 0$, we
see $f_{1}=f_{2}=0$, and finally we see $g=0$. This proves condition (c).

Finally, the fact that $\pi(1)$ is an order unit is also clear from the
surjectivity of $\pi$. ∎

######  Theorem 2.4.

Let $(X,h)$ be a zero-dimensional system. Then adopting the notation of
Proposition 2.2, we have
$(K_{0}(C^{*}(\mathbb{Z},X,h)),K_{0}(C^{*}(\mathbb{Z},X,h))^{+},1)\cong(K^{0}(X,h),K^{0}(X,h)^{+},1)$.

###### Proof.

Since $\ker(i_{*})=\text{im}(\text{id}-\alpha_{*})$, and since $i:C(X)\to
C^{*}(\mathbb{Z},X,h)$ is the natural inclusion, the map $i_{*}$ induces a map
$\varphi:K^{0}(X,h)\to K_{0}(C^{*}(\mathbb{Z},X,h))$ which is an isomorphism
of groups and satisfies $K^{0}(X,h)^{+}\subset
K_{0}(C^{*}(\mathbb{Z},X,h))^{+}$.

Let $p\in C^{*}(\mathbb{Z},X,h)$ be a projection. By applying Corollary 1.5
with $a_{1}=p$ and $\varepsilon=1/2$, $p$ is unitarily equivalent to
$\chi_{U}$ for some compact open $U\subset X$. Let $q$ be the image of
$\chi_{U}$ under the quotient map $C(X)\to C(X)/\text{im}(\text{id}-\alpha)$.
Then $[\varphi(q)]=[\chi_{U}]=[p]$. Repeating this argument for
$M_{n}(C^{*}(\mathbb{Z},X,h))$, we see that
$K_{0}(C^{*}(\mathbb{Z},X,h))^{+}\subset K^{0}(X,h)^{+}$.

Finally, that $\varphi(1)=1$ is clear, proving the theorem. ∎

What we have also shown in the previous proof is the following.

######  Corollary 2.5.

Let $(X,h,Z)$ be a fiberwise essentially minimal zero-dimensional system. Let
$i:C(X)\to C^{*}(\mathbb{Z},X,h)$ denote the canonical inclusion. Then the
induced map $i_{*}:K_{0}(C(X))\to K_{0}(C^{*}(\mathbb{Z},X,h))$ is surjective
as a map between ordered groups.

The following definition is from Section 2 of [12], later studied in the
minimal case in [13]. These have been referred to as “large subalgebras” in
the literature.

######  Definition 2.6.

Let $(X,h,Z)$ be a fiberwise essentially minimal zero-dimensional system. We
define $A_{Z}$ to be the $C^{*}$-algebra generated by $C(X)$ and
$uC_{0}(X\setminus Z)$.

The following theorem is contained in Theorem 2.3 of [12]. We provide a direct
proof in our context for the reader, which helps give an idea of the AF
structure of the large subalgebra.

######  Theorem 2.7.

Let $(X,h,Z)$ be a fiberwise essentially minimal zero-dimensional system. Then
$A_{Z}$ is an AF-algebra.

###### Proof.

Let $(\mathcal{P}^{(n)})$ be a generating sequence of partitions of $X$. For
each $n\in\mathbb{Z}_{>0}$, we inductively define systems
$\mathcal{S}^{(n)}=(T^{(n)},(X_{t}^{(n)}),(K_{t}^{(n)}),(Y_{t,k}^{(n)}),(J_{t,k}^{(n)}))$
of finite first return time maps. First, let
$\mathcal{S}^{(1)}=(T^{(1)},(X_{t}^{(1)}),(K_{t}^{(1)}),(Y_{t,k}^{(1)}),(J_{t,k}^{(1)}))$
be any system of finite first return time maps subordinate to
$\mathcal{P}^{(n)}$ such that $\mathcal{P}_{1}(\mathcal{S}^{(1)})$ is finer
than $\mathcal{P}^{(1)}$ and such that
$\bigsqcup_{t=1}^{T^{(1)}}X_{t}^{(1)}\supset Z$ (the former is possible by
Proposition 1.14 of [8] and the latter is possible by Lemma 4.12 of [8]). Now,
let $n\in\mathbb{Z}_{>0}$ and suppose we have chosen
$\mathcal{S}^{(n)}=(T^{(n)},(X_{t}^{(n)}),(K_{t}^{(n)}),(Y_{t,k}^{(n)}),(J_{t,k}^{(n)}))$.
Let
$\mathcal{S}^{(n+1)}=(T^{(n+1)},(X_{t}^{(n+1)}),(K_{t}^{(n+1)}),(Y_{t,k}^{(n+1)}),(J_{t,k}^{(n+1)}))$
be any system of finite first return time maps subordinate to
$\mathcal{P}^{(n+1)}$ such that $\mathcal{P}_{1}(\mathcal{S}^{(n+1)})$ is
finer than $\mathcal{P}^{(n+1)}$ and finer than
$\mathcal{P}_{1}(\mathcal{S}^{(n)})$ and such that
$\bigsqcup_{t=1}^{T^{(n+1)}}X_{t}^{(n+1)}\supset Z$.

Let $n\in\mathbb{Z}_{>0}$. Let $A^{(n)}$ be the finite dimensional
$C^{*}$-subalgebra of $C^{*}(\mathbb{Z},X,h)$ spanned by the matrix units
$u^{i-j}\chi_{h^{j}(Y_{t,k}^{(n)})}$ for $t\in\\{1,\ldots,T^{(n)}\\}$,
$k\in\\{1,\ldots,K_{t}^{(n)}\\}$, and $i,j\in\\{0,\ldots,J_{t,k}^{(n)}-1\\}$.
We see
$A^{(n)}\cong\bigoplus_{t=1}^{T^{(n)}}\bigoplus_{k=1}^{K_{t}^{(n)}}M_{J_{t,k}^{(n)}}$.
Notice that $C(\mathcal{P}_{1}(\mathcal{S}^{(n)}))\subset A^{(n)}$ as the
diagonal matrices. Set $Z^{(n)}=\bigsqcup_{t=1}^{T^{(n+1)}}X_{t}^{(n)}$ and
then notice that $uC(X\setminus Z^{(n)})\subset A^{(n)}$ as the superdiagonal
matrices, so $A^{(n)}$ is generated by $C(\mathcal{P}_{1}(\mathcal{S}^{(n)}))$
and $uC(X\setminus Z^{(n)})$.

Notice that $A^{(n)}\subset A^{(n+1)}$, so we get a directed system of finite
dimensional $C^{*}$-algebras, whose limit $A^{(\infty)}$ contains $C(X)$ since
$(\mathcal{P}_{1}(\mathcal{S}^{(n)}))$ is a generating sequence of partitions
and since $\bigcap_{n=1}^{\infty}Z^{(n)}=Z$, we have $uC(X\setminus
Z^{(n)})\to uC_{0}(X\setminus Z)\subset A_{Z}$. It now clear that
$A^{(\infty)}$ is generated by $C(X)$ and $uC_{0}(X\setminus Z)$, and is
therefore equal to $A_{Z}$. ∎

The following is Lemma 4.2 from [13].

######  Lemma 2.8.

Adopt the notation of Theorem 2.7 and its proof. Let $p$ be a projection in
$C(X)\cap A^{(n)}$ and suppose that $p=0$ on $Z^{(n)}$. Then $\alpha(p)\in
C(X)\cap A^{(n)}$ and $[\alpha(p)]=[p]$ in $K_{0}(A^{(n)})$.

We finally have the following theorem, which tells us enough about the $K_{0}$
structure of the crossed product to be able to prove Theorems 4.2 and Theorem
4.3. The proof follows that of Theorem 4.1 in [13].

######  Theorem 2.9.

Let $(X,h,Z)$ be a fiberwise essentially minimal zero-dimensional system. Then
$K_{0}(A_{Z})\cong K_{0}(C^{*}(\mathbb{Z},X,h))$ as ordered groups.

###### Proof.

Let $i:A_{Z}\to C^{*}(\mathbb{Z},X,h)$ denote the inclusion map, and let
$i_{*}:K_{0}(A_{Z})\to K_{0}(C^{*}(\mathbb{Z},X,h))$ denote map induced by $i$
on $K_{0}$. Let $i_{1}:C(X)\to A_{Z}$ denote the canonical inclusion, let
$i_{2}:C(X)\to C^{*}(\mathbb{Z},X,h)$ denote the canonical inclusion, and let
$(i_{1})_{*}$ and $(i_{2})_{*}$ denote the induced maps on $K_{0}$. We then
clearly have $i\circ i_{1}=i_{2}$. By Corollary 2.5,
$(i_{2})_{*}:K_{0}(C(X))\to K_{0}(C^{*}(\mathbb{Z},X,h))$ is a surjective map
between ordered groups, and therefore so is $i^{*}$.

By Proposition 2.2, $\ker((i_{2})_{*})=\text{ran}(\text{id}-\alpha_{*})$.
Thus, since $(i_{2})_{*}=(i_{1})_{*}\circ i_{*}$, we have
$(i_{1})_{*}(\text{ran}(\text{id}-\alpha_{*}))\subset\ker(i_{*})$. Now suppose
that $a\in\ker(i_{*})$. Because $(i_{1})_{*}$ is surjective, we can find $g\in
C(X,\mathbb{Z})$ such that $(i_{1})_{*}(g)=a$. Then
$(i_{2})_{*}(g)=i_{*}(a)=0$, so
$g\in\ker((i_{2})_{*})=\text{ran}(\text{id}-\alpha_{*})$, so
$a\in(i_{1})_{*}(\text{ran}(\text{id}-\alpha_{*}))$. Altogether, we have

(1) $(i_{1})_{*}(\text{ran}(\text{id}-\alpha_{*}))=\ker(i_{*}).$

We adopt the notation of the proof of Theorem 2.7. That is, we have a
generating sequence of partitions $(\mathcal{P}^{(n)})$ and a sequence of
systems of finite first return time maps $(\mathcal{S}^{(n)})$ and a sequence
of subalgebras $(A^{(n)})$ of $C^{*}(\mathbb{Z},X,h)$ as in the proof. We now
claim that $(i_{1})_{*}(\text{ran}(\text{id}-\alpha_{*}))=0$. Suppose
$g_{1},g_{2}\in C(X,\mathbb{Z})$ satisfy $g_{1}|_{Z}=g_{2}|_{Z}$. Since
$(\mathcal{P}^{(n)})$ is a generating sequence of partitions and for each
$n\in\mathbb{Z}_{>0}$, we have $\mathcal{P}_{1}(\mathcal{S}^{(n)})$ is finer
than $\mathcal{P}^{(n)}$, there is some $n\in\mathbb{Z}_{>0}$ such that
$g_{1},g_{2}\in A^{(n)}$ and such that $g_{1}|_{Z^{(n)}}=g_{2}|_{Z^{(n)}}$
(where $Z^{(n)}$ is defined in the proof of Theorem 2.7). So $g_{1}-g_{2}$ is
$0$ on $Z^{(n)}$, so we can write $g_{1}-g_{2}$ as a linear combination of
projections in $C(X)\cap A^{(n)}$ each of which is zero on $Z^{(n)}$. So by
Lemma 2.8, we have $[\alpha(g_{1}-g_{2})]=[g_{1}-g_{2}]$ in $K_{0}(A^{(n)})$,
and so $(i_{1})_{*}(g_{1}-\alpha(g_{1}))=(i_{1})_{*}(g_{2}-\alpha(g_{2}))$ in
$K_{0}(A_{Z})$.

So let $g\in C(X,\mathbb{Z})$. Define $f\in C(X,\mathbb{Z})$ by
$f=g\circ\psi$. Then $f|_{Z}=g|_{Z}$, and so by the above paragraph, we have
$(i_{1})_{*}(g-\alpha(g))=(i_{1})_{*}(f-\alpha(f))$. But then notice that
$\alpha(f)=g\circ\psi\circ h^{-1}=g\circ\psi=f$, and so
$(i_{1})_{*}(f-\alpha(f))=0$. Thus, we have
$(i_{1})_{*}(\text{ran}(\text{id}-\alpha_{*}))=0$. Combining this with (1), we
see $\ker(i_{*})=0$. Altogether, we have shown that $i_{*}$ is an isomorphism
of ordered groups. ∎

######  Theorem 2.10.

Let $(X,h,Z)$ be a fiberwise essentially minimal zero-dimensional system. Then
$K_{1}(C^{*}(\mathbb{Z},X,h))\cong C(Z,\mathbb{Z})$.

###### Proof.

Adopt the notation of Proposition 2.2. Then
$K_{1}(C^{*}(\mathbb{Z},X,h))\cong\ker(\text{id}-\alpha_{*})$. Identifying
$K_{0}(C(X))$ with $C(X,\mathbb{Z})$, we may replace $\alpha_{*}$ with
$\alpha$.

Let $f\in\ker(\text{id}-\alpha)$ and let $z\in Z$. Suppose $f|_{\psi^{-1}(z)}$
is not constant. Then there is some $x\in\psi^{-1}(z)$ such that $f(z)\neq
f(x)$. Let $U$ be a compact open subset of $\psi^{-1}(z)$ such that
$f(U)=f(z)$. Since $(\psi^{-1}(z),h|_{\psi^{-1}(z)})$ is an essentially
minimal zero-dimensional system, by Theorem 1.1 of [7], there is an
$n\in\mathbb{Z}_{>0}$ such that $x\in h^{-n}(U)$. Let $x^{\prime}=h^{n}(x)\in
U$. Then $f(x^{\prime})\neq
f(x)=f(h^{-n}(x^{\prime}))=\alpha^{n}(f)(x^{\prime})$, and so
$f\neq\alpha^{n}(f)$, and so $f\neq\alpha(f)$, a contradiction to
$x\in\ker(\text{id}-\alpha)$. Therefore, $f|_{\psi^{-1}(z)}$ is constant.

Now suppose $f\in C(X,\mathbb{Z})$ and suppose $f|_{\psi^{-1}(z)}$ is constant
for each $z\in Z$. Then for each $x\in X$, we have
$\alpha(f)(x)=f(h^{-1}(x))=f(\psi(x))=f(x)$, and so $f=\alpha(f)$, and so
$f\in\ker(\text{id}-\alpha)$.

Thus, we have

$\ker(\text{id}-\alpha)=\\{f\in
C(X,\mathbb{Z})\,\big{|}\,\mbox{$f|_{\psi^{-1}(z)}$ is constant for each $z\in
Z$}\\}\cong C(Z,\mathbb{Z})$

as desired. ∎

######  Corollary 2.11.

Let $(X_{1},h_{1},Z_{1})$ and $(X_{2},h_{2},Z_{2})$ be fiberwise essentially
minimal zero-dimensional systems such that $C^{*}(\mathbb{Z},X_{1},h_{1})\cong
C^{*}(\mathbb{Z},X_{2},h_{2})$. Then $Z_{1}\cong Z_{2}$.

##  3\. Bratelli Diagrams

In this section, we explore the construction of ordered Bratelli diagrams
associated to fiberwise essentially minimal zero-dimensional systems. This
correspondence is used to prove Theorem 4.2. For work done in the minimal
case, see [2] and [6]. For work done in the essentially minimal case, see [1]
and [7].

######  Definition 3.1.

A _Bratteli diagram_ $B$ is a pair of sets $(V,E)$ such that:

  1. (a)

The set $V$ is called _the set of vertices of $B$_. We can write
$V=\bigsqcup_{n=0}^{\infty}V_{n}$ where $V_{0}$ contains a single point
$v_{0}$ and $V_{n}$ is finite and nonempty for all $n\in\mathbb{Z}_{\geq 0}$.
For each $n\in\mathbb{Z}_{\geq 0}$, we call $V_{n}$ _the set of vertices of
$B$ at level $n$_.

  2. (b)

The set $E$ is called _the set of edges of $B$_. We can write
$E=\bigsqcup_{n=1}^{\infty}E_{n}$ where $E_{n}$ is finite and nonempty for all
$n\in\mathbb{Z}_{>0}$. For each $n\in\mathbb{Z}_{>0}$, we call $E_{n}$ _the
set of edges at level $n$_.

  3. (c)

There are maps $r,s:E\to V$ such that for $n\in\mathbb{Z}_{>0}$ and $e\in
E_{n}$, $s(e)\in V_{n-1}$ and $r(e)\in V_{n}$. Moreover, $r^{-1}(v)$ is
nonempty for all $v\in V$ and $s^{-1}(v)$ is nonempty for all $v\in V\setminus
V_{0}$. The map $r$ is called the _range map of $B$_ and the map $s$ is called
_the source map of $B$_.

######  Notation 3.2.

Let $B=(V,E)$ be a Bratteli diagram. For each $v\in V$, we denote
$R(v)=r(s^{-1}(v))$, and for each $v\in V\setminus V_{0}$, we denote
$S(v)=s(r^{-1}(v))$. If $v\in V_{n}$, then $R(v)$ is the set of all vertices
in $V_{n+1}$ that are connected to $v$ by an edge, and $S(v)$ is the set of
all vertices in $V_{n-1}$ connected to $v$ by an edge. In a reasonable sense,
this gives us range and source maps for vertices.

######  Definition 3.3.

An _ordered Bratteli diagram_ $B$ is a Bratteli diagram $(V,E)$ together with
a partial order $\leq$ on $E$ such that $e,e^{\prime}\in E$ are comparable if
and only if $r(e)=r(e^{\prime})$. We write $B=(V,E,\leq)$.

Let $B=(V,E,\leq)$ be an ordered Bratteli diagram. We define $E_{\min}$
($E_{\max}$) to be the set of all edges that are minimal (maximal, resp.) with
respect to $\leq$. We define $V_{\min}$ ($V_{\max}$) to be the set of all
$v\in V$ such that there is an $e$ in $E_{\min}$ ($E_{\max}$, resp.) with
$s(e)=v$.

######  Definition 3.4.

Let $B=(V,E,\leq)$ be an ordered Bratteli diagram. We define a _partial
Vershik transformation_ $\widetilde{h}_{B}:(X_{B}\setminus X_{B,\max})\cup
X_{B,\min}\to(X_{B}\setminus X_{B,\min})\cup X_{B,\max}$ in the following way.
If $x\in X_{B,\max}\cap X_{B,\min}$, we define $\widetilde{h}_{B}(x)=x$. If
$x=(x_{1},x_{2},\ldots)\in X_{B}\setminus X_{B,\max}$, then there is some
smallest $k\in\mathbb{Z}_{>0}$ such that $x_{k}\notin E_{\max}$. Let $y_{k}$
denote the successor of $x_{k}$ in $E$ and let $(y_{1},y_{2},\ldots,y_{k-1})$
be the unique path from $v_{0}$ to $s(y_{k})$ such that $y_{j}\in E_{\min}$
for all $j\in\\{1,\ldots,k-1\\}$. We define
$\widetilde{h}_{B}(p)=(y_{1},y_{2},\ldots,y_{k},x_{k+1},x_{k+2},\ldots)$.

Let $B=(V,E,\leq)$ be an ordered Bratteli diagram. For each
$k,k^{\prime}\in\mathbb{Z}_{>0}$ with $k<k^{\prime}$, we define
$P_{k,k^{\prime}}$ to be the set of all paths from $V_{k}$ to
$V_{k^{\prime}}$. Formally, $P_{k,k^{\prime}}$ is the set of
$(e_{k+1},\ldots,e_{k^{\prime}})$ such that for all
$j\in\\{k+1,\ldots,k^{\prime}\\}$, $e_{j}\in E_{j}$ and for all
$j\in\\{k+1,\ldots,k^{\prime}-1\\}$, we have $r(e_{j})=s(e_{j+1})$.

######  Definition 3.5.

Let $B=(V,E,\leq)$ be an ordered Bratteli diagram. We define the _infinite
path space_ $X_{B}$ to be the set of all sequences $x=(x_{1},x_{2},\ldots)$
where $x_{n}\in E_{n}$ and $r(x_{n})=s(x_{n+1})$ for all $n\in\mathbb{Z}_{>0}$
together with the topology generated by sets of the form
$U(e_{1},\ldots,e_{k})$, which is the set of all $x=(x_{1},x_{2},\ldots)$ with
$x_{j}=e_{j}$ for all $j\in\\{1,\ldots,k\\}$.

Let $B=(V,E,\leq)$ be an ordered Bratteli diagram. It is easy to see that the
infinite path space is a zero-dimensional space. We define $X_{B,\min}$
($X_{B,\max}$) to be the set of all $x=(x_{1},x_{2},\ldots)\in X_{B}$ such
that $x_{j}$ is in $E_{\min}$ ($E_{\max}$, resp.) for all
$j\in\mathbb{Z}_{>0}$.

The following terminology appears in Definition 2.18 of
[BezuglyiKwiatkowskiYassawi13], although we restate it to give more clarity as
to when the definition applies.

######  Definition 3.6.

Let $B=(V,E,\leq)$ be an ordered Bratteli diagram and let $\widetilde{h}_{B}$
be its partial Vershik transformation. We say that the ordering on $B$ is
_perfect_ if for every $e\in X_{B,\min}$,
$\overline{\text{orb}_{\widetilde{h}_{B}(e)}}\cap X_{B,\max}$ contains a
single element, and if for every $e\in X_{B,\max}$,
$\overline{\text{orb}_{\widetilde{h}_{B}(e)}}\cap X_{B,\min}$ contains a
single element. In this case, we define the _Vershik transformation_ of
$X_{B}$, denoted by $h_{B}$, to be the extension of $\widetilde{h}_{B}$ which,
for each $e\in X_{B,\min}$, sends the unique element of
$\overline{\text{orb}_{\widetilde{h}_{B}(e)}}\cap X_{B,\max}$ to $e$.

Thus, given an ordered Bratteli diagram $B=(V,E,\leq)$ with a perfect
ordering, the system $(X_{B},h_{B})$ is a zero-dimensional system.

######  Lemma 3.7.

Let $(X,h)$ be a fiberwise essentially minimal zero-dimensional system, let
$\mathcal{P}$ and $\mathcal{P}^{\prime}$ be partitions of $X$, and let
$\mathcal{S}=(T,(X_{t}),(K_{t}),(Y_{t,k}),(J_{t,k}))$ be a system of finite
first return time maps subordinate to $\mathcal{P}$ such that for each
$t\in\\{1,\ldots,T\\}$, we have $\psi(X_{t})\subset X_{t}$. Then there is a
system
$\mathcal{S}^{0}=(T^{0},(X_{t}^{0}),(K_{t}^{0}),(Y_{t,k}^{0}),(J_{t,k}^{0}))$
of finite first return time maps subordinate to $\mathcal{P}$ and a system
$\mathcal{S}^{\prime}=(T^{\prime},(X_{t}^{\prime}),(K_{t}^{\prime}),(Y_{t,k}^{\prime}),(J_{t,k}^{\prime}))$
of finite first return time maps subordinate to $\mathcal{P}^{\prime}$ such
that:

  1. (a)

The partition $\mathcal{P}_{1}(\mathcal{S}^{\prime})$ is finer than
$\mathcal{P}^{\prime}$ and $\mathcal{P}_{1}(\mathcal{S}^{0})$.

  2. (b)

The partition $\mathcal{P}_{1}(\mathcal{S}^{0})$ is finer than
$\mathcal{P}_{1}(\mathcal{S})$.

  3. (c)

We have $T^{0}=T$ and for each $t\in\\{1,\ldots,T\\}$, we have
$X_{t}^{0}=X_{t}$.

  4. (d)

For each $t^{\prime}\in\\{1,\ldots,T^{\prime}\\}$, there is a
$t\in\\{1,\ldots,T^{0}\\}$ and $k\in\\{1,\ldots,K_{t}^{0}\\}$ such that
$X_{t^{\prime}}^{\prime}\subset Y_{t,k}^{0}$.

  5. (e)

For each $t^{\prime}\in\\{1,\ldots,T^{\prime}\\}$, there is a
$t\in\\{1,\ldots,T^{0}\\}$ and $k\in\\{1,\ldots,K_{t}^{0}\\}$ such that
$X_{t^{\prime}}^{\prime}\subset h^{J_{t,k}^{0}}(Y_{t,k}^{0})$.

  6. (f)

For each $t\in\\{1,\ldots,T^{0}\\}$ and each $k\in\\{1,\ldots,K^{0}\\}$, there
is a $t^{\prime}\in\\{1,\ldots,T^{\prime}\\}$ such that
$Y_{t,k}^{0}\subset\bigcup_{j\in\mathbb{Z}}h^{j}(X_{t^{\prime}}^{\prime})$.

###### Proof.

We first construct $\mathcal{S}^{\prime}$ and then use it to modify
$\mathcal{S}$ to obtain $\mathcal{S}^{0}$. By applying Lemma 4.13 of [8], we
may assume that $\mathcal{S}$ satisfies the conclusions of the lemma; in
particular, for all $t\in\\{1,\ldots,T\\}$, we have $\psi(X_{t})\subset
Y_{t,1}$, and the partitions $\mathcal{P}_{1}(\mathcal{S})$ and
$\mathcal{P}_{2}(\mathcal{S})$ are finer than $\mathcal{P}$. Let
$\mathcal{P}^{\prime\prime}$ be a partition finer than $\mathcal{P}^{\prime}$,
$\mathcal{P}_{1}(\mathcal{S})$, and $\mathcal{P}_{2}(\mathcal{S})$. Then apply
Lemma 3.2 of [8] to obtain a system
$\mathcal{S}^{\prime}=(T^{\prime},(X_{t}^{\prime}),(K_{t}^{\prime}),(Y_{t,k}^{\prime}),(J_{t,k}^{\prime}))$
of finite first return time maps subordinate to $\mathcal{P}^{\prime\prime}$
such that for all $t^{\prime}\in\\{1,\ldots,T^{\prime}\\}$, there is a
$t\in\\{1,\ldots,T\\}$ such that $X_{t^{\prime}}^{\prime}\subset X_{t}$. Since
$\mathcal{P}^{\prime\prime}$ is finer than $\mathcal{P}^{\prime}$,
$\mathcal{S}^{\prime}$ is also subordinate to $\mathcal{P}^{\prime}$. By
applying Proposition 1.9 of [8], we may assume that
$\mathcal{P}_{1}(\mathcal{S}^{\prime})$ is finer than
$\mathcal{P}^{\prime\prime}$.

Let $t^{\prime}\in\\{1,\ldots,T^{\prime}\\}$ and let $t\in\\{1,\ldots,T\\}$ be
such that $X_{t^{\prime}}^{\prime}\subset X_{t}$. Since
$\mathcal{P}^{\prime\prime}$ is finer than $\mathcal{P}_{1}(\mathcal{S})$,
there is some $k\in\\{1,\ldots,K_{t}\\}$ such that

(2) $X_{t^{\prime}}^{\prime}\subset Y_{t,k}.$

Since $\mathcal{P}^{\prime\prime}$ is finer than
$\mathcal{P}_{2}(\mathcal{S})$, there is some $l\in\\{1,\ldots,K_{t}\\}$ such
that

(3) $X_{t^{\prime}}^{\prime}\subset h^{J_{t,l}}(Y_{t,l}).$

Define $T^{0}=T$ and for each $t\in\\{1,\ldots,T\\}$, define $X_{t}^{0}=X_{t}$
(note that after finishing this construction this verifies conclusion (c) of
the lemma). Let $t\in\\{1,\ldots,T\\}$ and let $\\{s(1),\ldots,s(N_{t})\\}$ be
the set of all $t^{\prime}\in\\{1,\ldots,T^{\prime}\\}$ such that
$X_{t^{\prime}}\subset X_{t}$. For each $n\in\\{1,\ldots,N_{t}\\}$, let
$\\{a(n,1),\ldots,a(n,C_{n})\\}$ be the set of all $k\in\\{1,\ldots,K_{t}\\}$
such that
$Y_{t,k}\cap\bigcup_{j\in\mathbb{Z}}h^{j}(X_{s(n)}^{\prime})\neq\varnothing$.
Define $K_{t}^{0}=\sum_{n=1}^{N_{t}}C_{n}$ and define $C_{0}=0$. Let
$k\in\\{1,\ldots,K_{t}^{0}\\}$ and let $n\in\\{1,\ldots,N_{t}\\}$ and
$c\in\\{1,\ldots,C_{n}\\}$ be such that $k=C_{n-1}+c$. Then define

(4)
$Y_{t,k}^{0}=Y_{t,a(n,c)}\cap\bigcup_{j\in\mathbb{Z}}h^{j}(X_{s(n)}^{\prime})$

and define $J_{t,k}^{0}=J_{t,a(n,c)}$. It is routine to verify that
$\mathcal{S}^{0}=(T^{0},(X_{t}^{0}),(K_{t}^{0}),(Y_{t,k}^{0}),(J_{t,k}^{0}))$
is a system of finite first return time maps subordinate to $\mathcal{P}$. By
applying Proposition 1.9 of [8], we may assume that
$\mathcal{P}_{1}(\mathcal{S}^{0})$ is finer than $\mathcal{P}$ (note that this
proves conclusion (b) of the lemma).

We now verify the conclusions of the lemma. Conclusions (a), (b), and (c) have
already been verified. Conclusion (d) follows from (2) and from the fact that
$\mathcal{P}_{1}(\mathcal{S}^{0})$ is finer than
$\mathcal{P}_{1}(\mathcal{S})$. Conclusion (e) follows from (2) and the fact
that that $\mathcal{P}_{2}(\mathcal{S}^{0})$ is finer than
$\mathcal{P}_{2}(\mathcal{S})$. Conclusion (f) is shown by (4). This proves
the lemma. ∎

######  Proposition 3.8.

Let $(X,h,Z)$ be a fiberwise essentially minimal zero-dimensional system.
There is an ordered Bratteli diagram $B=(V,E,\leq)$ with a perfect ordering
such that

  1. (a)

The system $(X_{B},h_{B},X_{B,\min})$ is conjugate to $(X,h,Z)$.

  2. (b)

For each $v$ in $V_{\min}$ (or $V_{\max}$) there is a $v^{\prime}$ in
$V_{\min}$ ($V_{\max}$, resp.) and an edge $e$ in $E_{\min}$ ($E_{\max}$,
resp.) such that $s(e)=v$ and $r(e)=v^{\prime}$.

  3. (c)

For each $v$ in $V_{\min}$ (or $V_{\max}$) and each $e$ in $E_{\min}$
($E_{\max}$, resp.) with $r(e)\in R(v)$ satisfies $s(e)=v$.

  4. (d)

For each $v$ in $V_{\min}$ (or $V_{\max}$) and each $m\in\mathbb{Z}_{>0}$,
$R^{m}(v)=(R^{m}\circ S^{m}\circ R^{m})(v)$.

###### Proof.

Let $Z$ and $\psi$ correspond to $(X,h)$ as in Definition 1.2, and let
$(\mathcal{P}^{(n)})$ be a generating sequence of partitions of $X$. We will
construct an ordered Bratteli diagram $B=(V,E,\leq)$ such that $(X_{B},h_{B})$
is conjugate to $(X,h)$ via a map $F:X\to X_{B}$ that satisfies
$F(Z)=X_{B,\min}$.

First, we construct a sequence $(\mathcal{S}^{(n)})$ of finite first return
time maps subordinate to $(\mathcal{P}^{(n)})$. First, let
$\mathcal{S}^{(1)\prime}$ be any system of finite first return time maps
subordinate to $\mathcal{P}^{(1)}$ such that
$\mathcal{P}_{1}(\mathcal{S}^{(1)\prime})$ is finer than $\mathcal{P}^{(1)}$
(such a system exists by Proposition 1.9 of [8]). We construct the other
systems inductively. For each $n\in\mathbb{Z}_{>0}$, we apply Lemma 3.7 with
$\mathcal{S}^{(n)\prime}$ in place of $\mathcal{S}$, $\mathcal{P}^{(n)}$ in
place of $\mathcal{P}$, $\mathcal{P}^{(n+1)}$ in place of
$\mathcal{P}^{\prime}$, and get $\mathcal{S}^{(n)}$ (that is $\mathcal{S}^{0}$
in the lemma) and $\mathcal{S}^{(n+1)\prime}$ (that is $\mathcal{S}^{\prime}$
in the lemma) satisfying the conclusions of the lemma. Thus, to construct the
sequence of systems, we only need to define $\mathcal{S}^{(0)}$ by $T^{(0)}=1$
, $X_{1}^{(0)}=X$, $K_{1}^{(0)}=1$, $Y_{1,1}^{(0)}=1$, and $J_{1,1}^{(0)}=1$.

Now we begin to define $B$. For each $n\in\mathbb{Z}_{\geq 0}$, define

$V_{n}=\\{(n,t,k)\,\big{|}\,\mbox{$t\in\\{1,\ldots,T^{(n)}\\}$ and
$k\in\\{1,\ldots,K_{t}^{(n)}\\}$}\\}$

The set of edges from $(n,t,k)\in V_{n}$ to $(n+1,t^{\prime},k^{\prime})\in
V_{n+1}$ is the set of all $(n+1,t^{\prime},k^{\prime},j)$ such that
$h^{j}(Y_{t^{\prime},k^{\prime}}^{(n+1)})\subset Y_{t,k}^{(n)}$. Note that
this is well defined; by assumption,
$h^{j}(Y_{t^{\prime},k^{\prime}}^{(n+1)})$ is a subset of an element of
$\mathcal{P}_{1}(\mathcal{S}^{(n)})$, so we do not need to include $t$ and $k$
in the tuple defining this edge. We define an order on the edges
$r^{-1}((n,t,k))$ by $(n,t,k,j_{1})\leq(n,t,k,j_{2})$ if $j_{1}<j_{2}$.

We now construct the orbit map $F:X\to X_{B}$. Let $x\in X$. Then for each
$n\in\mathbb{Z}_{>0}$, there is precisely one $t\in\\{1,\ldots,T^{(n)}\\}$,
one $k\in\\{1,\ldots,K_{t}^{(n)}\\}$, and one
$j\in\\{0,\ldots,J_{t,k}^{(n)}\\}$ such that $x\in h^{j}(Y_{t,k}^{(n)})$. If
$x\in h^{j}(Y_{t,k}^{(n)})\cap
h^{j^{\prime}}(Y_{t^{\prime},k^{\prime}}^{(n+1)})$, then $j^{\prime}\geq j$,
since otherwise we would have $h^{j-j^{\prime}}(Y_{t,k}^{(n)})\subset
X_{t}^{(n+1)}$ (this follows from Lemma 3.7(c) and (d)), which is not possible
since by definition $h^{i}(Y_{t,k}^{(n)})\cap X_{t}^{(n+1)}=\varnothing$ for
$i\in\\{1,\ldots,J_{t,k}^{(n)}-1\\}$. This, combined with the fact that
$\mathcal{P}_{1}(\mathcal{S}^{(n+1)})$ is finer than
$\mathcal{P}_{1}(\mathcal{S}^{(n)})$, tells us that
$h^{j^{\prime}-j}(Y_{t^{\prime},k^{\prime}}^{(n+1)})\subset Y_{t,k}^{(n)}$,
and therefore there is an edge from $(n,t,k)$ to
$(n+1,t^{\prime},k^{\prime})$; namely,
$(n+1,t^{\prime},k^{\prime},j^{\prime}-j)$. Thus, this gives us an infinite
path in $X_{B}$ associated to $x$. We define $F$ by sending $x$ to this
infinite path.

We now show that $F$ is injective. Suppose $x,x^{\prime}\in X$ and
$F(x)=F(x^{\prime})=(e_{1},e_{2},\ldots)$ where we write
$e_{n}=(n,t_{n},k_{n},j_{n})$ for $n\in\mathbb{Z}_{>0}$. First, by definition
is it clear that for each $n\in\mathbb{Z}_{>0}$, there are
$i_{n},i_{n}^{\prime}\in\\{0,\ldots,J_{t_{n},k_{n}}^{(n)}-1\\}$ such that
$x\in h^{i_{n}}(Y_{t_{n},k_{n}}^{(n)})$ and $x^{\prime}\in
h^{i_{n}^{\prime}}(Y_{t_{n},k_{n}}^{(n)})$. First, notice that $i_{1}=j_{1}$
and $i_{1}^{\prime}=j_{1}$ by definition of $F$. Then, by definition of $F$ we
have $j_{2}=i_{2}-j_{1}$ and $j_{2}=i_{2}^{\prime}-j_{2}$; in particular,
$i_{2}=i_{2}^{\prime}$. Proceeding like this, we see that
$i_{n}=i_{n}^{\prime}$ for all $n\in\mathbb{Z}_{>0}$. Since
$(\mathcal{P}^{(n)})$ is a generating sequence of partitions, so is
$(\mathcal{P}_{1}(\mathcal{S}^{(n)}))$, and therefore
$\bigcap_{n=0}^{\infty}h^{i_{n}}(Y_{t_{n},k_{n}}^{(n)})$ contains at most one
element of $X$. Thus, $x=x^{\prime}$, and therefore $F$ is injective.

Next, we show that $F$ is surjective. Let $e=(e_{1},e_{2},\ldots)\in X_{B}$
and write $e_{n}=(n,t_{n},k_{n},j_{n})$ for $n\in\mathbb{Z}_{>0}$. We
construct a sequence $(i_{n})$ with
$i_{n}\in\\{0,\ldots,J_{t_{n},k_{n}}^{(n)}-1\\}$ for all $n\in\mathbb{Z}_{>0}$
such that $\bigcap_{n=0}^{\infty}h^{i_{n}}(Y_{t_{n},k_{n}}^{(n)})$ is nonempty
and contains the element of $X$ that $F$ maps to $e$. First, let
$i_{1}=j_{1}$. Then, for all $n\in\mathbb{Z}_{>1}$, let $i_{n}=j_{n}+i_{n-1}$
(note that this can be rewritten as $i_{n}=\sum_{k=1}^{n}j_{k}$). The claim
now follows from the definition of $B$, and $F$ is therefore surjective.

So far we have shown that $F$ is bijective. We now show that $F$ is a
homeomorphism. Let $U(e_{1},\ldots,e_{N})$ be an element of the basis of the
topology of $X_{B}$. For each $n\in\\{1,\ldots,N\\}$, write
$e_{n}=(n,t_{n},k_{n},j_{n})$. For each $n\in\\{1,\ldots,N\\}$, write
$j_{n}^{\prime}=\sum_{k=1}^{n}j_{k}$. We claim that if $x\in
h^{j_{N}^{\prime}}(Y_{t_{N},k_{N}}^{(N)})$, then $F(x)\in
U(e_{1},\ldots,e_{N})$. So let $x\in
h^{j_{N}^{\prime}}(Y_{t_{N},k_{N}}^{(N)})$. First notice that by definition of
$e_{N}$,

$h^{j_{N}}(Y_{t_{N},k_{N}}^{(N)})\subset Y_{t_{N-1},k_{N-1}}^{(N-1)}.$

Thus,

$h^{j_{N}^{\prime}}(Y_{t_{N},k_{N}}^{(N)})\subset
h^{j_{N-1}^{\prime}}(Y_{t_{N-1},k_{N-1}}^{(N-1)}),$

since $j_{N}+j_{N-1}^{\prime}=j_{N}^{\prime}$. Similarly, we have

$h^{j_{N-1}}(Y_{t_{N-1},k_{N-1}}^{(N-1)})\subset Y_{t_{N-2},k_{N-2}}^{(N-2)}.$

Thus, for every $n\in\\{1,\ldots,N\\}$, we have

$x\in h^{j_{n}^{\prime}}(Y_{t_{n},k_{n}}^{(N)}),$

and so since $j_{n}^{\prime}-j_{n-1}^{\prime}=j_{n}$, the $n$th edge of $F(x)$
is indeed $e_{n}$, and $F(x)\in U(e_{1},\ldots,e_{N})$ as desired. Next, we
claim that if $x\in X$ satisfies $F(x)\in U(e_{1},\ldots,e_{N})$, then $x\in
h^{j_{N}^{\prime}}(Y_{t_{N},k_{N}}^{(N)})$. So let $x\in X$ satisfy $F(x)\in
U(e_{1},\ldots,e_{N})$. Then for each $n\in\\{1,\ldots,N\\}$, $x\in
h^{i_{n}}(J_{t_{n},k_{n}}^{(n)})$ for some
$i_{n}\in\\{0,\ldots,J_{t_{n},k_{n}}^{(n)}-1\\}$. It is clear that
$i_{1}=j_{1}$. Then, notice that $i_{2}$ is such that $j_{2}=i_{2}-j_{1}$, and
so $i_{2}=j_{1}+j_{2}=j_{2}^{\prime}$. Repeating this process inductively, we
see that $i_{N}=j_{N}^{\prime}$, and so $x\in
h^{j_{N}^{\prime}}(Y_{t_{N},k_{N}}^{(N)})$ as desired. Altogether, this shows
that $F$ is a homeomorphism.

If $x\in Z$, then there are sequences of integers $(t_{n})$ and $(k_{n})$ such
that $x\in\bigcap_{n=0}^{\infty}Y_{t_{n},k_{n}}^{(n)}$. By definition of the
order on $B$, this means that $F(x)\in X_{B,\min}$. Conversely, suppose $x\in
X$ satisfies $F(x)\in X_{B,\min}$. Write $F(x)=(e_{1},e_{2},\ldots)$ and for
$n\in\mathbb{Z}_{>0}$ write $e_{n}=(n,t_{n},k_{n},i_{n})$. Since $e_{n}$ is
minimal, $i_{n}$ is the minimal element of
$\\{0,\ldots,J_{t_{n},k_{n}}^{(n)}-1\\}$ such that
$h^{i_{n}}(Y_{t_{n},k_{n}}^{(n)})\subset X_{t_{n-1}}^{(n-1)}$. But since
$X_{t_{n}}^{(n)}\subset X_{t_{n-1}}^{(n-1)}$, we have $i_{n}=0$. Hence,
$x\in\bigcap_{n=0}^{\infty}Y_{t_{n},k_{n}}^{(n)}$. Thus, $F(Z)=X_{B,\min}$.
Also notice that if $x\in h^{-1}(Z)$, there are sequences of integers
$(t_{n})$ and $(k_{n})$ such that
$x\in\bigcap_{n=0}^{\infty}h^{J_{t_{n},k_{n}}^{(n)}-1}(Y_{t_{n},k_{n}}^{(n)})$.
By definition of the order on $B$, this means that $F(x)\in X_{B,\max}$.
Similarly, the converse holds, and so $F(h^{-1}(Z))=X_{B,\max}$.

We now show that $(F\circ h)|_{X\setminus h^{-1}(Z)}=(\widetilde{h}_{B}\circ
F)|_{X\setminus h^{-1}(Z)}$. Let $x\in X\setminus h^{-1}(Z)$. For each
$n\in\mathbb{Z}_{>0}$, let $t_{n}\in\\{1,\ldots,T^{(n)}\\}$, let
$k_{n}\in\\{1,\ldots,K_{t_{n}}^{(n)}\\}$, and let
$j_{n}\in\\{0,\ldots,K_{t_{n}}^{(n)}-1\\}$ satisfy $x\in
h^{j_{n}}(Y_{t_{n},k_{n}}^{(n)})$. Since $x\notin h^{-1}(Z)$, there is some
smallest $N\in\mathbb{Z}_{>0}$ such that $j_{N}\neq J_{t_{N},k_{N}}^{(N)}-1$.
We have

$h^{j_{N}+1}(Y_{t_{N},k_{N}}^{(N)})\subset
J_{t_{N-1},k_{N-1}}^{(N-1)}(Y_{t_{N-1},k_{N-1}}^{(N-1)})$

and so by Lemma 3.7(d), there is a
$k_{N-1}^{\prime}\in\\{1,\ldots,K_{t_{N-1}}^{(N-1)}\\}$ such that

(5) $h^{j_{N}+1}(Y_{t_{N},k_{N}}^{(N)})\subset
Y_{t_{N-1},k_{N-1}^{\prime}}^{(N-1)}.$

Inductively, we can find for each $n\in\\{1,\ldots,N-2\\}$ a
$k_{n}^{\prime}\in\\{1,\ldots,K_{t_{n}}^{(n)}\\}$ such that
$Y_{t_{n+1}^{\prime},k_{n+1}^{\prime}}^{(n+1)}\subset
Y_{t_{n}^{\prime},k_{n}^{\prime}}^{(n)}$. Since
$j_{N}+1<J_{t_{N},k_{N}}^{(N)}$ and since for each $n\in\mathbb{Z}_{>N}$ we
have $X_{t_{n}}^{(n)}\subset X_{t_{N}}^{(N)}$, it follows that
$j_{n}+1<J_{t_{N},k_{N}}^{(N)}$ as well. So let $k_{n}^{\prime}=k_{n}$ and let
$j_{n}^{\prime}=0$ for $n\in\\{1,\ldots,N-1\\}$, and let
$j_{n}^{\prime}=j_{n}+1$ for $n\in\mathbb{Z}_{\geq N}$. Then for each
$n\in\mathbb{Z}_{>0}$, $h(x)\in
h^{j_{n}^{\prime}}(Y_{t_{n},k_{n}^{\prime}}^{(n)})$.

Now write $F(x)=(e_{1},e_{2},\ldots)$ and
$\widetilde{h}_{B}(F(x))=(e_{1}^{\prime},e_{2}^{\prime},\ldots)$ and for each
$n\in\mathbb{Z}_{>0}$ write $e_{n}=(n,s_{n},l_{n},i_{n})$ and
$e_{n}^{\prime}=(n,s_{n}^{\prime},l_{n}^{\prime},i_{n}^{\prime})$. By
definition of $F$, for all $n\in\mathbb{Z}_{>0}$ we have $s_{n}=t_{n}$,
$l_{n}=k_{n}$, and $i_{n}=j_{n}-j_{n-1}$ where $j_{0}=0$. We also see that $N$
is the smallest element of $\mathbb{Z}_{>0}$ such that $e_{N}\notin E_{\max}$,
so $e_{N}^{\prime}$ is the successor of $e_{N}$,
$(e_{1}^{\prime},\ldots,e_{N-1}^{\prime})$ is the minimal path such that
$r(e_{N-1}^{\prime})=s(e_{N}^{\prime})$, and $e_{n}^{\prime}=e_{n}$ for all
$n\in\mathbb{Z}_{>N}$. In particular we see that $s_{n}^{\prime}=t_{n}$ for
all $n\in\mathbb{Z}_{>0}$, $l_{n}^{\prime}=k_{n}$ for all
$n\in\mathbb{Z}_{\geq N}$, $i_{n}^{\prime}=0$ for $n\in\\{1,\ldots,N-1\\}$ and
$i_{n}^{\prime}=i_{n}=j_{n}-j_{n-1}$ for all $n\in\mathbb{Z}_{>N}$. Observe
that $i_{N}^{\prime}$ is the smallest integer greater than $i_{N}$ such that
$h^{i_{N}^{\prime}}(Y_{t_{N},k_{N}}^{(N)})\subset
Y_{t_{N-1},l_{N-1}^{\prime}}^{(N-1)}$. Thus, this combined with

(6) $h^{i_{N}}(Y_{t_{N},k_{N}}^{(N)})\subset Y_{t_{N-1},k_{N-1}}^{(N-1)}$

tells us

$\displaystyle i_{N}^{\prime}$
$\displaystyle=i_{N}+J_{t_{N-1},k_{N-1}}^{(N-1)}$
$\displaystyle=j_{N}-j_{N-1}+J_{t_{N-1},k_{N-1}}^{(N-1)}$
$\displaystyle=j_{N}-(J_{t_{N-1},k_{N-1}}^{(N-1)}-1)+J_{t_{N-1},k_{N-1}}^{(N-1)}$
$\displaystyle=j_{N}+1$ $\displaystyle=j_{N}^{\prime}.$

From (5) and (6) we see $l_{N-1}^{\prime}=k_{N-1}^{\prime}$. Similarly, for
$n\in\\{2,\ldots,N-1\\}$ since $i_{n}^{\prime}=j_{n}^{\prime}=0$, we have
$l_{n-1}^{\prime}=k_{n-1}^{\prime}$. Write
$F(h(x))=(e_{1}^{\prime\prime},e_{2}^{\prime\prime},\ldots)$ and for each
$n\in\mathbb{Z}_{>0}$ write
$e_{n}^{\prime\prime}=(n,s_{n}^{\prime\prime},l_{n}^{\prime\prime},i_{n}^{\prime\prime})$.
For each $n\in\mathbb{Z}_{>0}$, we have
$s_{n}^{\prime\prime}=t_{n}^{\prime}=t_{n}$ and
$l_{n}^{\prime\prime}=k_{n}^{\prime}$. For $n\in\\{1,\ldots,N-1\\}$ , we have
$i_{n}^{\prime\prime}=i_{n}^{\prime}=0$. We also have
$i_{N}^{\prime\prime}=j_{N}^{\prime}-j_{N-1}^{\prime}=j_{N}^{\prime}-0=i_{N}^{\prime}$.
For $n\in\mathbb{Z}_{>N}$, we have

$\displaystyle i_{n}^{\prime\prime}=j_{n}^{\prime}-j_{n-1}^{\prime}$
$\displaystyle=j_{n}+1-(j_{n-1}-1)$ $\displaystyle=j_{n}-j_{n-1}$
$\displaystyle=i_{n}$ $\displaystyle=i_{n}^{\prime}.$

Altogether, we see $F(h(x))=\widetilde{h}_{B}(F(x))$, and so

(7) $(F\circ h)|_{X\setminus h^{-1}(Z)}=(\widetilde{h}_{B}\circ
F)|_{X\setminus h^{-1}(Z)}.$

We now show that the order on $B$ is perfect. For each $x\in X$, $z$ is in the
minimal set of the essentially minimal zero-dimensional system
$(\psi^{-1}(Z),h|_{\psi^{-1}(Z)})$, $\overline{\text{orb}}(x)$ contains
exactly one element of $Z$ and exactly one element of $h^{-1}(Z)$. Now, (7)
combined with $F(Z)=X_{B,\min}$ and $F(h^{-1}(Z))=X_{B,\max}$ tells us that
the ordering on $B$ is perfect. It is now clear that $F\circ h=h_{B}\circ F$.
This proves conclusion (a) of the proposition.

Before we prove the rest, we first prove two claims that will be used a few
times in the proof.

Claim ($\ast$): Let $n\in\mathbb{Z}_{>0}$, $t_{n}\in\\{1,\ldots,T^{(n)}\\}$,
$k_{n}\in\\{1,\ldots,K_{t_{n}}^{(n)}\\}$, and
$t_{n+1}\in\\{1,\ldots,T^{(n+1)}\\}$. If $X_{t_{n+1}}^{(n+1)}\subset
h^{J_{t_{n},k_{n}}^{(n)}}(Y_{t_{n},k_{n}}^{(n)})$, then for any
$k_{n+1}\in\\{1,\ldots,K_{t_{n+1}}^{(n+1)}\\}$, there is an
$i_{n+1}\in\\{1,\ldots,J_{t_{n+1},k_{n+1}}^{(n+1)}\\}$ such that
$e=(n+1,t_{n+1},k_{n+1},i_{n+1})$ is a maximal edge with
$s(e)=(n,t_{n},k_{n})$.

We now prove ($\ast$). Let $k_{n+1}\in\\{1,\ldots,K_{t_{n+1}}^{(n+1)}\\}$.
Then

$h^{J_{t_{n+1},k_{n+1}}^{(n+1)}}(Y_{t_{n+1},k_{n+1}})\subset
X_{t_{n+1}^{(n+1)}}\subset h^{J_{t_{n},k_{n}}^{(n)}}(Y_{t_{n},k_{n}}^{(n)}),$

so we have

$h^{J_{t_{n+1},k_{n+1}}^{(n+1)}-J_{t_{n},k_{n}}^{(n)}}(Y_{t_{n+1},k_{n+1}})\subset
Y_{t_{n},k_{n}}^{(n)}$

Set $i_{n+1}=J_{t_{n+1},k_{n+1}}^{(n+1)}-J_{t_{n},k_{n}}^{(n)}$. Let
$j\in\\{1,\ldots,J_{t_{n},k_{n}}^{(n)}-1\\}$. Then

$h^{i_{n+1}+j}(Y_{t_{n+1},k_{n+1}}^{(n+1)})\subset
h^{j}(Y_{t_{n},k_{n}}^{(n)}),$

and since
$h^{j}(Y_{t_{n},k_{n}}^{(n)})\cap\left(\bigsqcup_{t=1}^{T^{(n)}}X_{t}^{(n)}\right)=\varnothing$,
we indeed see that $e=(n+1,t_{n+1},k_{n+1},i_{n+1})\in E_{\max}$. This proves
($\ast$).

Claim ($\ast\ast$): If there is a
$k_{n+1}\in\\{1,\ldots,K_{t_{n+1}}^{(n+1)}\\}$ and a
$i\in\\{1,\ldots,J_{t_{n+1},k_{n+1}}^{(n+1)}\\}$ such that
$e=(n+1,t_{n+1},k_{n+1},i_{n+1})$ is a maximal edge with
$s(e)=(n,t_{n},k_{n})$, then $X_{t_{n+1}}^{(n+1)}\subset
h^{J_{t_{n},k_{n}}^{(n)}}(Y_{t_{n},k_{n}}^{(n)})$.

We now prove ($\ast\ast$). Since $s(e)=(n,t_{n},k_{n})$, we have

$h^{i_{n+1}}(Y_{t_{n+1},k_{n+1}}^{(n+1)})\subset Y_{t_{n},k_{n}}^{(n)}$

Since $e$ is maximal, there is no
$j\in\\{i_{n+1}+1,\ldots,J_{t_{n+1},k_{n+1}}^{(n+1)}-1\\}$ with
$h^{j}(Y_{t_{n+1},k_{n+1}}^{(n+1)})\subset X_{t_{n},k_{n}}^{(n)}$. However,
notice that

$h^{i_{n+1}+J_{t_{n},k_{n}}^{(n)}}(Y_{t_{n+1},k_{n+1}}^{(n+1)})\subset
h^{J_{t_{n},k_{n}}^{(n)}}(Y_{t_{n},k_{n}}^{(n)})\subset
X_{t_{n},k_{n}}^{(n)}.$

Thus, we must have
$i_{n+1}+J_{t_{n},k_{n}}^{(n)}=J_{t_{n+1},k_{n+1}}^{(n+1)}$. Thus,
$X_{t_{n+1}}^{(n+1)}\cap
h^{J_{t_{n},k_{n}}^{(n)}}(Y_{t_{n},k_{n}}^{(n)})\neq\varnothing$, and so by
Lemma 3.7(e), we actually have $X_{t_{n+1}}^{(n+1)}\subset
h^{J_{t_{n},k_{n}}^{(n)}}(Y_{t_{n},k_{n}}^{(n)})$. This proves ($\ast\ast$).

We now prove conclusion (b) of the proposition. Let $v\in V_{\min}$. Write
$v=(n,t_{n},k_{n})$. Since $v\in V_{\min}$, there is some $v^{\prime\prime}\in
V_{n+1}$ and some $e^{\prime}\in E_{\min}$ with $s(e^{\prime})=v$ and
$r(e^{\prime})=v^{\prime\prime}$. Write
$v^{\prime\prime}=(n+1,t_{n+1},k_{n+1})$ and then
$e^{\prime}=(n+1,t_{n+1},k_{n+1},i_{n+1})$. By Lemma 3.7(d), since
$e^{\prime}\in E_{\min}$, we have $i_{n+1}=0$, so
$Y_{t_{n+1},k_{n+1}}^{(n+1)}\subset Y_{t_{n},k_{n}}^{(n)}$. Again by Lemma
3.7(d), we have

(8) $X_{t_{n+1}}^{(n+1)}\subset Y_{t_{n},k_{n}}^{(n)}.$

Now, let $t_{n+2}\in\\{1,\ldots,T^{(n+2)}\\}$ satisfy

$X_{t_{n+2}}^{(n+2)}\cap\left(\bigcup_{j\in\mathbb{Z}}h^{j}(X_{t_{n+1}}^{(n+1)})\right)\neq\varnothing.$

By Lemma 3.7(d), there is actually a
$k_{n+1}^{\prime}\in\\{1,\ldots,K_{t_{n+1}}^{(n+1)}\\}$ such that

$X_{t_{n+2}}^{(n+2)}\subset Y_{t_{n+1},k_{n+1}^{\prime}}^{(n+1)}.$

So for any $k_{n+2}\in\\{1,\ldots,K_{t_{n+2}}^{(n+2)}\\}$, we have a minimal
edge $e^{\prime\prime}=(n+2,t_{n+2},k_{n+2},0)$ with
$s(e^{\prime\prime})=v^{\prime}=(n+1,t_{n+1},k_{n+1}^{\prime})$. Thus,
$v^{\prime}\in V_{\min}$. By (8), $Y_{t_{n+1},k_{n+1}^{\prime}}^{(n+1)}\subset
Y_{t_{n},k_{n}}^{(n)}$, and so there is a minimal edge
$e=(n+1,t_{n+1},k_{n+1}^{\prime},0)$ with $s(e)=v$ and $r(e)=v^{\prime}$.

Now let $v\in V_{\max}$ and write $v=(n,t_{n},k_{n})$. Since $v\in V_{\max}$,
there is some $v^{\prime\prime}\in V_{n+1}$ and some $e^{\prime}\in E_{\max}$
with $s(e^{\prime})=V$ and $r(e^{\prime})=v^{\prime\prime}$. Write
$v^{\prime\prime}=(n+1,t_{n+1},k_{n+1})$ and then
$e^{\prime}=(n+1,t_{n+1},k_{n+1},i_{n+1})$. By $(\ast\ast)$, we have

$X_{t_{n+1}}^{(n+1)}\subset h^{J_{t_{n},k_{n}}^{(n)}}(Y_{t_{n},k_{n}}^{(n)})$

Now, let $t_{n+2}\in\\{1,\ldots,T^{(n+2)}\\}$ satisfy

$X_{t_{n+2}}^{(n+2)}\cap\left(\bigcup_{j\in\mathbb{Z}}h^{j}(X_{t_{n+1}}^{(n+1)})\right)\neq\varnothing.$

By Lemma 3.7(e), there is actually a
$k_{n+1}^{\prime}\in\\{1,\ldots,K_{t_{n+1}}^{(n+1)}\\}$ such that

$X_{t_{n+2}}^{(n+2)}\subset
h^{J_{t_{n+1},k_{n+1}^{\prime}}^{(n+1)}}(Y_{t_{n+1},k_{n+1}^{\prime}}^{(n+1)}).$

By $(\ast)$, $v^{\prime}=(n+1,t_{n+1},k_{n+1}^{\prime})\in V_{\max}$. By
$(\ast)$, there is $e\in E_{\max}$ with $s(e)=v$ and $r(e)=v^{\prime}$. This
completes the proof of conclusion (b).

We now prove conclusion (c) of the proposition. Let $v\in V_{\min}$ and $e\in
E_{\min}$ satisfy $r(e)\in R(v)$. Write $v=(n,t_{n},k_{n})$ and
$e=(n+1,t_{n+1},k_{n+1},i_{n+1})$. Since $r(e)\in R(v)$, there is an edge
$e^{\prime}=(n+1,t_{n+1},k_{n+1},i_{n+1}^{\prime})$ with $s(e^{\prime})=v$,
which tells us there is
$j_{n+1}\in\\{0,\ldots,J_{t_{n+1},k_{n+1}}^{(n+1)}-1\\}$ such that
$h^{j_{n+1}}(Y_{t_{n+1},k_{n+1}})\subset Y_{t_{n},k_{n}}^{(n)}$. By Lemma
3.7(f),
$Y_{t_{n},k_{n}}^{(n)}\subset\bigcup_{j\in\mathbb{Z}}h^{j}(X_{t_{n+1}}^{(n+1)})$.
Since $v\in V_{\min}$, there is some edge
$e^{\prime\prime}=(n+1,t_{n+1},k_{n+1}^{\prime\prime},i_{n+1}^{\prime\prime})\in
E_{\min}$ with $s(e)=v$. By Lemma 3.7(d), since $e^{\prime\prime}\in
E_{\min}$, we have $i_{n+1}^{\prime\prime}=0$, so
$Y_{t_{n+1},k_{n+1}^{\prime\prime}}^{(n+1)}\subset Y_{t_{n},k_{n}}^{(n)}$.
Again by Lemma 3.7(d), we actually have

(9) $X_{t_{n+1}}^{(n+1)}\subset Y_{t_{n},k_{n}}^{(n)}.$

Now, since $e\in E_{\min}$, by Lemma 3.7(d), we have $i_{n+1}=0$. But by (9),
we must have $Y_{t_{n+1},k_{n+1}}^{(n+1)}\subset Y_{t_{n},k_{n}}^{(n)}$, which
means that $s(e)=v$.

Let $v\in V_{\max}$ and $e\in E_{\max}$ satisfy $r(e)\in R(v)$. Write
$v=(n,t_{n},k_{n})$ and $e=(n+1,t_{n+1},k_{n+1},i_{n+1})$. Since $r(e)\in
R(v)$, there is an edge $e^{\prime}=(n+1,t_{n+1},k_{n+1},i_{n+1}^{\prime})$
with $s(e^{\prime})=v$, which tells us there is
$j_{n+1}\in\\{0,\ldots,J_{t_{n+1},k_{n+1}}^{(n+1)}-1\\}$ such that
$h^{j_{n+1}}(Y_{t_{n+1},k_{n+1}}^{(n+1)})\subset Y_{t_{n},k_{n}}^{(n)}$. By
Lemma 3.7(f),
$Y_{t_{n},k_{n}}^{(n)}\subset\bigcup_{j\in\mathbb{Z}}h^{j}(X_{t_{n+1}}^{(n+1)})$.
Since $v\in V_{\max}$, there is some edge
$e^{\prime\prime}=(n+1,t_{n+1},k_{n+1}^{\prime\prime},i_{n+1}^{\prime\prime})\in
E_{\max}$ with $s(e)=v$. By $(\ast\ast)$, we have

$X_{t_{n+1}}^{(n+1)}\subset h^{J_{t_{n},k_{n}}^{(n)}}(Y_{t_{n},k_{n}}^{(n)}).$

By ($\ast$), $s(e)=v$. This completes the proof of conclusion (c).

Now we prove conclusion (d) of this lemma. Let $v\in V_{\min}$, let
$m\in\mathbb{Z}_{>0}$, and write $v=(n,t_{n},k_{n})$. It is clear that
$v\in(S^{m}\circ R^{m})(v)$, and so

$R^{m}(v)\subset(R^{m}\circ S^{m}\circ R^{m})(v).$

Now, let $w=(n+m,t_{n+m},k_{n+m})\in R^{m}(v)$. By Lemma 3.7(d) (applied $m$
times), we have

$\bigcup_{j\in\mathbb{Z}}h^{j}(X_{t_{n+m}}^{(n+m)})\subset\bigcup_{j\in\mathbb{Z}}h^{j}(X_{t_{n}}^{(n)})$

So every element of $(S^{m}\circ R^{m})(v)$ has the form
$(n,t_{n},k_{n}^{\prime})$ for some
$k_{n}^{\prime}\in\\{1,\ldots,K_{t_{n}}^{(n)}\\}$. Let
$w^{\prime}=(n+m,t_{n+m}^{\prime},k_{n+m}^{\prime})\in(R^{m}\circ S^{m}\circ
R^{m})(v)$, meaning there is an path $(e_{1},\ldots,e_{m})$ with
$s(e_{1})=(n,t_{n},k_{n}^{\prime})$ for some
$k_{n}^{\prime}\in\\{1,\ldots,K_{t_{n}}^{(n)}\\}$ and $r(e_{m})=w^{\prime}$.
Write $e_{m}=(n+m,t_{n+m}^{\prime},k_{n+m}^{\prime},i_{n+m}^{\prime})$. This
means that
$h^{i_{n+m}^{\prime}}(Y_{t_{n+m}^{\prime},k_{n+m}^{\prime}}^{(n+m)})\subset
Y_{t_{n},k_{n}^{\prime}}^{(n)}$. But then by Lemma 3.7(d) (applied $m$ times),
we must have $Y_{t_{n+m}^{\prime},k_{n+m}^{\prime}}^{(n+m)}\subset
Y_{t_{n},k_{n}}^{(n)}$. Therefore $w^{\prime}\in R^{m}(v)$ (by a minimal
path), as desired. An identical argument using Lemma 3.7(e) in place of (d)
shows that this equation also holds when $v\in V_{\max}$. This proves
conclusion (d) of this proposition and therefore finishes the proof of the
proposition. ∎

######  Proposition 3.9.

Let $(X,h)$ be a fiberwise essentially minimal zero-dimensional system and let
$B=(V,E,\leq)$ be an ordered Bratteli diagram that satisfies the conclusions
of Proposition 3.8. If $B^{\prime}=(V^{\prime},E^{\prime},\leq^{\prime})$ is a
telescoping of $B$, then $B^{\prime}$ also satisfies the conclusions of
Proposition 3.8.

###### Proof.

Let $(k_{n})$ be the telescoping sequence corresponding to $B^{\prime}$, so
that $V_{n}^{\prime}=V_{k_{n}}$ and $E_{n}^{\prime}=P_{k_{n}+1,k_{n+1}}$ for
all $n\in\mathbb{Z}_{>0}$ (where $k_{0}=0$). This identification induces a map
$\varphi:X_{B}\to X_{B^{\prime}}$ by sending $e=(e_{1},e_{2},\ldots)$ to
$\varphi(e)=((e_{1},\ldots,e_{k_{1}}),(e_{k_{1}+1},\ldots,e_{k_{2}}),\ldots)$.
By definition of the induced order on a telescoped Bratteli diagram, $\varphi$
is a conjugation, and therefore $B^{\prime}$ satisfies conclusion (a) of
Proposition 3.8.

We now show that conclusion (b) holds. Let $v\in V_{n,\min}^{\prime}$. What we
are looking for is a $v^{\prime}\in V_{n+1,\min}^{\prime}$ and $e\in
E_{n+1,\min}^{\prime}$ such that $s(e)=v$ and $r(e)=v^{\prime}$. We can regard
all of this as happening in $B$ instead of $B^{\prime}$, so that we have $v\in
V_{k_{n},\min}$, and we are looking for $v^{\prime}\in V_{k_{n+1},\min}$ and a
minimal path $(e_{k_{n}+1},\ldots,e_{k_{n+1}})\in P_{k_{n}+1,k_{n+1}}$. By
Proposition 3.8(b), there is a $e_{k_{n}+1}\in E_{k_{n}+1,\min}$ and a
$v_{k_{n}+1}\in V_{k_{n}+1,\min}$ with $s(e)=v$ and $r(e)=v_{k_{n}+1}$.
Proceeding inductively, we construct the desired result, with
$v^{\prime}=v_{k_{n+1}}$. The same argument works for $V_{\max}^{\prime}$ in
place of $V_{\min}^{\prime}$. This proves that conclusion (b) holds.

Next, we prove that conclusion (c) holds. Let $v\in V_{n,\min}^{\prime}$ and
let $e\in E_{n+1,\min}^{\prime}$ satisfy $r(e)\in R(v)$. We want to show that
$s(e)=v$. We once again regard all of this as happening in $B$ instead of
$B^{\prime}$. What this means is that we have $v\in V_{k_{n},\min}^{\prime}$
and a minimal path $(e_{k_{n}+1},\ldots,e_{k_{n+1}})\in P_{k_{n}+1,k_{n+1}}$
such that there is some path
$(e_{k_{n}+1}^{\prime},\ldots,e_{k_{n+1}}^{\prime})\in P_{k_{n}+1,k_{n+1}}$
such that $s(e_{k_{n}+1}^{\prime})=v$ and
$r(e_{k_{n+1}}^{\prime})=r(e_{k_{n+1}})$, and we want to show that
$s(e_{k_{n}+1})=v$. Suppose not. Then by Proposition 3.8(c), we have
$r(e_{k_{n}+1})\notin R(v)$; in particular, we have $r(e_{k_{n}+1})\neq
r(e_{k_{n}+1}^{\prime})$. Proceeding like this, we eventually see
$r(e_{k_{n+1}})\neq r(e_{k_{n+1}}^{\prime})$, a contradiction. Thus,
$s(e_{k_{n}+1})=v$. The proof for $V_{\max}^{\prime}$ in place of
$V_{\min}^{\prime}$ is analogous. This proves that conclusion (c) holds.

That conclusion (d) holds is immediate; replace $m$ with $k_{n+m}-k_{n}$. This
completes the proof of the proposition. ∎

##  4\. The Dynamical Classification Theorem

We now prove our main theorems, Theorem 4.2 and Theorem 4.3.

######  Lemma 4.1.

Let $B=(V,E,\leq)$ be an ordered Bratteli diagram with a perfect ordering. Let
$e,f\in X_{B}$ and suppose $e$ and $f$ pass through the same vertex $v$ at
level $k$. Then there is some $N\in\mathbb{Z}$ such that $h_{B}^{N}(e)=f$.

###### Proof.

Write $e=(e_{1},e_{2},\ldots)$ and write $f=(f_{1},f_{2},\ldots)$. Let $n_{1}$
be the largest element of $\\{1,\ldots,k\\}$ such that $e_{n_{1}}\neq
f_{n_{1}}$.

Suppose $e_{n_{1}}<f_{n_{1}}$. Then there is an $N_{1}\in\mathbb{Z}_{>0}$ such
that

$h_{B}^{N_{1}}(e)=(e_{1}^{\prime},\ldots,e_{n_{1}-1}^{\prime},f_{n_{1}},e_{n_{1}+1},e_{n_{1}+2},\ldots),$

where $(e_{1}^{\prime},\ldots,e_{n_{1}-1}^{\prime})$ is the minimal path from
$v_{0}\in V_{0}$ to $s(f_{n_{1}})$. Now, $h_{B}^{M}(e)$ and $f$ pass through
the same vertex at level $n_{1}-1$. So let $n_{2}$ be the largest element of
$\\{1,\ldots,n_{1}-1\\}$ such that $e_{n_{2}}^{\prime}\neq f_{n_{2}}$. Clearly
$e_{n_{2}}^{\prime}<f_{n_{2}}$. So repeating the above process, we find an
integer $N_{2}\in\mathbb{Z}_{>0}$ such that
$h_{B}^{N_{1}+N_{2}}(e)=(e_{1}^{\prime\prime},\ldots,e_{n_{2}-1}^{\prime\prime},f_{n_{2}},\ldots,f_{n_{1}},e_{n_{1}+1},\ldots)$.
Repeating this process inductively, we arrive at an integer $N$ such that
$h_{B}^{N}(e)=f$. ∎

######  Theorem 4.2.

Let $(X_{1},h_{1},Z_{1})$ and $(X_{2},h_{2},Z_{2})$ be fiberwise essentially
minimal zero-dimensional systems. Then $(X_{1},h_{1},Z_{1})$ and
$(X_{2},h_{2},Z_{2})$ are strong orbit equivalent if and only if

$(K_{0}(C^{*}(\mathbb{Z},X_{1},h_{1})),K_{0}(C^{*}(\mathbb{Z},X_{1},h_{1}))^{+},1)\cong(K_{0}(C^{*}(\mathbb{Z},X_{2},h_{2})),K_{0}(C^{*}(\mathbb{Z},X_{2},h_{2}))^{+},1)$

and

$K_{1}(C^{*}(\mathbb{Z},X_{1},h_{1}))\cong
K_{1}(C^{*}(\mathbb{Z},X_{2},h_{2})).$

###### Proof.

($\Leftarrow$). Let $B_{1}$ and $B_{2}$ be the Bratteli diagrams satisfying
the conclusions of Proposition 3.8 for $(X_{1},h_{1},Z_{1})$ and
$(X_{2},h_{2},Z_{2})$, respectively. By Theorem 2.4, we have
$K^{0}(X_{1},h_{1})\cong K^{0}(X_{2},h_{2})$. By Proposition 3.8(a), we have
$K^{0}(X_{B_{1}},h_{B_{1}})\cong K^{0}(X_{B_{2}},h_{B_{2}})$. By a slight but
trivial extension of Theorem 5.4 in [7], we have $K_{0}(B_{1})\cong
K_{0}(B_{2})$. By [4], we have $B_{1}\sim B_{2}$ (in the equivalence class of
Bratteli diagrams generated by telescoping and isomorphism), which tells us
that there is a (non-ordered) Bratteli diagram $B$ such that telescoping $B$
to odd levels yields a telescoping of $B_{1}$ and telescoping $B$ to even
levels yields a telescoping of $B_{2}$. By replacing $B_{1}$ and $B_{2}$ with
their telescopings (which can be done without changing the above due to
Proposition 3.9), we may assume that telescoping $B$ to odd levels yields
$B_{1}$ and telescoping $B_{2}$ to even levels yields $B_{2}$. Let
$B^{\prime}$ be the telescoping of $B$ by the sequence $(3n-2)$, so that
telescoping $B$ to odd levels yields a telescoping of $B_{1}$ by $(3n-2)$ and
telescoping $B$ to even levels yields a telescoping of $B_{2}$ by $(3n-1)$.
Note that by Proposition 3.9, these telescopings of $B_{1}$ and $B_{2}$ also
satisfy the conclusions of Proposition 3.8.

We denote by $V_{\min}^{\prime}$ ($V_{\max}^{\prime}$) the minimal (resp.
maximal) vertices as inherited by $B_{1}$ and $B_{2}$. We claim that
$B^{\prime}$ has the following properties:

  1. (i)

Let $v\in V_{\min}^{\prime}$. There is at least one $v^{\prime}\in
V_{\min}^{\prime}$ with $v^{\prime}\in R(v)$.

  2. (ii)

Let $v\in V_{\min}^{\prime}$. There is precisely one $v^{\prime}\in
V_{\min}^{\prime}$ with $v^{\prime}\in S(v)$.

We first prove property (i). Let $v\in V_{\min,n}^{\prime}$. The statement is
trivially true for $n=0$ so suppose $n>0$ and without loss of generality
suppose that $n$ is odd. View $v$ as a vertex in $V_{\min,3n-2}$, so that the
statement we are trying to prove becomes $R^{3}(v)\cap
V_{\min}\neq\varnothing$. Notice that for every vertex $w\in V^{(2)}\setminus
V^{(2)}_{0}$, $S(w)\cap V^{(2)}_{\min}\neq\varnothing$, since there is always
an $e\in E^{(2)}_{\min}$ with $s(e)=w$. Thus, it follows that $(S^{2}\circ
R^{3})(v)\cap V_{\min}\neq\varnothing$. By Proposition 3.8(b), we have
$(R^{2}\circ S^{2}\circ R^{3})(v)\cap V_{\min}\neq\varnothing$. By Proposition
3.8(d), we have $R^{2}\circ S^{2}\circ R^{2}=R^{2}$. Thus, $R^{2}\circ
S^{2}\circ R^{3}=R^{3}$, so $R^{3}(v)\cap V_{\min}\neq\varnothing$. The proof
for $v\in V_{\max}^{\prime}$ is analogous. Altogether, this proves property
(i).

We now prove property (ii). Let $v\in V_{\min,n}^{\prime}$ and without loss of
generality suppose that $n$ is odd. The case $n=1$ is trivial so suppose
$n>1$. View $v$ as a vertex in $V_{\min,3n-2}$, so that the statement we are
trying to prove is that $S^{3}(v)$ contains precisely one element. Let $w\in
S(v)$ and let $(e_{1},e_{2})$ be a minimal path with $r(e_{2})=w$. Since
$s(e_{1})\in V_{\min}$, we have $S^{3}(v)\cap V_{\min}\neq\varnothing$. Now,
suppose $S^{3}(v)\cap V_{\min}$ contains two elements, $v_{1}$ and $v_{2}$. By
Proposition 3.8(c), we have $R^{2}(v_{1})\cap R^{2}(v_{2})=\varnothing$. By
Proposition 3.8(b), $R^{2}(v_{1})\cap V_{\min}\neq\varnothing$ and
$R^{2}(v_{2})\cap V_{\min}\neq\varnothing$. By Proposition 3.8(c) again,

(10) $R^{2}(R^{2}(v_{1})\cap V_{\min})\cap R^{2}(R^{2}(v_{2})\cap
V_{\min})=\varnothing$

Since $v\in R^{3}(v_{1})$ and $v\in R^{3}(v_{2})$, we have

(11) $R^{4}(v_{1})\cap R^{4}(v_{2})\neq\varnothing.$

But now notice that by Proposition 3.8(d), we have
$R^{4}(v_{1})=R^{2}(R^{2}(v_{1})\cap V_{\min})$ and
$R^{4}(v_{2})=R^{2}(R^{2}(v_{2})\cap V_{\min})$. Thus, we see that (10) and
(11) yield a contradiction. Therefore, $S^{3}(v)\cap V_{\min}$ contains
precisely one element. The proof for $v\in V_{\max}^{\prime}$ is analogous.
This proves property (ii).

For convenience, replace $B^{(1)}$ with its telescoping by $(3n-2)$ and
replace $B^{(2)}$ with its telescoping by $(3n-1)$. Now, let
$e=(e_{1},e_{2},\ldots)\in X_{B^{(1)},\min}$ and let $(v_{1},v_{2},\ldots)$ be
its associated vertices. Let $n\in\mathbb{Z}_{>1}$. We view $v_{n}$ as a
vertex in $V_{2n-1}^{\prime}$. By property (ii), there is a unique vertex
$v_{n-1}^{\prime}\in V_{\min,2n-2}^{\prime}$ such that $v_{n-1}^{\prime}\in
S(v_{n})$. We claim that $v_{n-1}\in S(v_{n-1}^{\prime})$. If not, then by
property (ii) there is some $w\in S(v_{n-1}^{\prime})$ such that $w\neq
v_{n-1}$. But then $R^{2}(w)\cap R^{2}(v_{n-1})\neq\varnothing$, which is a
contradiction again by Proposition 3.8(c). Thus, for each
$n\in\mathbb{Z}_{>1}$, $v_{n-1}^{\prime}$ is connected by a path to
$v_{n}^{\prime}$, there is an $e_{n}^{\prime}\in E_{\min,n}^{(2)}$ with
$s(e_{n}^{\prime})=v_{n-1}^{\prime}$ and $r(e_{n}^{\prime})=v_{n}^{\prime}$ by
Proposition 3.8(c). Thus, this gives us
$e^{\prime}=(e_{1}^{\prime},e_{2}^{\prime},\ldots)\in X_{B^{(2)},\min}$ such
that for each $n\in\mathbb{Z}_{>0}$, we have
$s(e_{n}^{\prime})=v_{n-1}^{\prime}$ and $r(e_{n}^{\prime})=v_{n}^{\prime}$.

Let $f=(f_{1},f_{2},\ldots)\in X_{B^{\prime}}$ be any path with
$r(f_{n})=s(f_{n+1})=v_{2n-1}$ for all odd $n\in\mathbb{Z}_{>0}$ and
$r(f_{n})=s(f_{n+1})=v_{2n}^{\prime}$ for all even $n\in\mathbb{Z}_{>0}$.
Since this respects the vertices and the range and source maps, we are free to
define $F_{1}(e)=f$ and $F_{2}(e^{\prime})=f$.

We now show that the above pairing is a bijection between $X_{B,\min}^{(1)}$
and $X_{B,\min}^{(2)}$. Let
$e^{\prime}=(e_{1}^{\prime},e_{2}^{\prime},\ldots)\in X_{B^{(2)},\min}$ and
$(v_{n}^{\prime})$ be as above. By property (ii), for each
$n\in\mathbb{Z}_{>0}$ there is precisely one $w_{n}\in V_{\min}^{\prime}$ with
$w_{n}\in S(v_{n}^{\prime})$. Since $v_{n}\in S(v_{n}^{\prime})$, we must have
$w_{n}=v_{n}$. Thus, the bijection is established, and so
$F_{1}(X_{B^{(1)},\min})=F_{2}(X_{B^{(2)},\min})$. We now repeat the process
for maximal vertices, choosing edges which are not minimal if the number of
edges between the vertices is more than one. This extends $F_{1}$ and $F_{2}$
so that $F_{1}(X_{B^{(1)},\max})=F_{2}(X_{B^{(2)},\max})$.

We now extend $F_{1}$ and $F_{2}$ by any bijection between $E_{n}^{(1)}$ and
$E_{2n-1}^{\prime}$ and any bijection between $E_{n}^{(2)}$ and $E_{2n}$ that
respect the range and source maps. In this way, we get homeomorphisms
$F_{1}:X_{B^{(1)}}\to X_{B^{\prime}}$ and $F_{2}:X_{B^{(2)}}\to
X_{B^{\prime}}$. Define $F=F_{2}^{-1}\circ F_{1}:X_{B^{(1)}}\to X_{B^{(2)}}$.

Let $\alpha,\beta:X_{B^{(1)}}\to\mathbb{Z}$ be the orbit cocyles of $F$. We
will show that $\alpha$ and $\beta$ are continuous on $X_{B^{(1)}}\setminus
X_{B^{(1)},\max}$. So let $e=(e_{1},e_{2},\ldots)\in X_{B^{(1)}}\setminus
X_{B^{(1)},\max}$. Let $k$ be the smallest element of $\mathbb{Z}_{>0}$ such
that $e_{k}\notin E^{(1)}_{\max}$. Then $e$ and $h_{B^{(1)}}(e)$ are confinal
from level $k$. This means that $F_{1}(e)$ and $F_{1}(h_{B^{(1)}}(e))$ are
cofinal from level $2k-1$, and so $F(e)$ and $F(h_{B^{(1)}}(e))$ are cofinal
from level $k$. In particular, $F(e)$ and $F(h_{B^{(1)}}(e))$ pass through the
same vertex $v$ at level $k$. By Lemma 4.1, there is an integer $N$ such that

$h_{B^{(1)}}^{N}(e)=F(h_{B^{(1)}}(e))$

Let $f\in U(e_{1},\ldots,e_{k+1})$. Then $y$ and $h_{B^{(1)}}(f)$ are confinal
from level $k$. This means that $F_{1}(f)$ and $F_{1}(h_{B^{(1)}}(f))$ are
cofinal from level $2k-1$, and so $F(y)$ and $F(h_{B^{(1)}}(f))$ are cofinal
from level $k$. Since $e,f\in U(e_{1},\ldots,e_{k+1})$, $h_{B^{(1)}}(x)$ and
$h_{B^{(1)}}(f)$ have the same initial segment from level $0$ to level $k+1$,
and so $F(h_{B^{(1)}}(e))$ and $F(h_{B^{(1)}}(f))$ have the same initial
segment from level $0$ to level $k$. Similarly, $F(e)$ and $F(f)$ have the
same initial segment from level $0$ to level $k$. Thus, the integer $N$ from
above satisfies

$h_{B^{(1)}}^{N}(f)=F(h_{B^{(1)}}(f))$

Since $f\in U(e_{1},\ldots,e_{k+1})$ was arbitrary, this shows that $\alpha$
is continuous at $e$.

The argument for $\beta$ is analogous to $\alpha$. Thus,
$(X_{B^{(1)}},h_{B^{(1)}},X_{B^{(1)},\max})$ and
$(X_{B^{(2)}},h_{B^{(2)}},X_{B^{(2)},\max})$ are strong orbit equivalent. By
replacing $F$ with $h_{B^{(1)}}^{-1}\circ F$, we see that
$(X_{B^{(1)}},h_{B^{(1)}},X_{B^{(1)},\min})$ and
$(X_{B^{(2)}},h_{B^{(2)}},X_{B^{(2)},\min})$ are strong orbit equivalent.
Therefore, by Proposition 3.8(a), $(X_{1},h_{1}.Z_{1})$ and
$(X_{2},h_{2},Z_{2})$ are strong orbit equivalent.

($\Rightarrow$). Assume $(X_{1},h_{1},Z_{1})$ and $(X_{2},h_{2},Z_{2})$ are
strong orbit equivalent. By the definition of strong orbit equivalence, we
have $Z_{1}\cong Z_{2}$ and so by Theorem 2.10, we have
$K_{1}(C^{*}(\mathbb{Z},X_{1},h_{1}))\cong K_{1}(C^{*}(X_{2},h_{2},Z_{2}))$.

Let $\beta,\gamma:X_{1}\to\mathbb{Z}$ be the associated orbit cocyles defined
by $F\circ h_{1}=h_{2}^{\beta}\circ F$ and $F\circ h_{1}^{\gamma}=h_{2}\circ
F$. Recall that strong orbit equivalence tells us that $\beta$ and $\gamma$
are continuous on $X_{1}\setminus Z_{1}$ and $F(Z_{1})=Z_{2}$.

Let $\widetilde{h}_{2}=F^{-1}\circ h_{2}\circ F:X_{1}\to X_{1}$. Then
$\widetilde{h}_{2}$ is conjugate to $h_{2}$, has the same orbits as $h_{1}$,
and

$h_{1}=\widetilde{h}_{2}^{\beta},\widetilde{h}_{2}=h_{1}^{\gamma}.$

We also see that $(X_{1},\widetilde{h}_{2},Z_{1})$ is a fiberwise essentially
minimal zero-dimensional system that is conjugate to $(X_{2},h_{2},Z_{2})$, so
we work with the former for the remainder of the proof.

Let $\alpha_{1}$ be the automorphism of $C(X_{1})$ induced by $h_{1}$ and let
$\alpha_{2}$ be the automorphism of $C(X_{1})$ induced by $\widetilde{h}_{2}$.
We now show that
$\text{ran}(\text{id}-(\alpha_{1})_{*})\subset\text{ran}(\text{id}-(\alpha_{2})_{*})$.
It is enough to show that for any compact open set $E\subset X_{1}$, we have
$(\text{id}-(\alpha_{1})_{*})(\chi_{E})\in\text{ran}(\text{id}-(\alpha_{2})_{*})$.

Given $f\in C(X_{1},\mathbb{Z})$, we denote the image of $f$ in
$K_{0}(C^{*}(\mathbb{Z},X_{1},h_{i}))$ by $[f]_{1}$ and denote the image of
$f$ in $K_{0}(C^{*}(\mathbb{Z},X_{1},\widetilde{h}_{2}))$ by $[f]_{2}$. Since
$h_{1}$ and $\widetilde{h}_{2}$ have the same orbits, there is a probably
measure $\mu$ that is both $h_{1}$\- and $\widetilde{h}_{2}$-invariant. Note
that $C^{*}(\mathbb{Z},X_{1},h_{1})$ is the $C^{*}$-subalgebra of
$L(L^{2}(X_{1},\mu))$ generated by $C(X)$ and the unitary $u_{1}:g\to g\circ
h_{1}^{-1}$. Also note that $C^{*}(\mathbb{Z},X_{1},\widetilde{h}_{2})$ is the
$C^{*}$-subalgebra of $L(L^{2}(X_{1},\mu))$ generated by $C(X)$ and the
unitary $u_{2}:g\to g\circ\widetilde{h}_{2}^{-1}$. For the remainder of the
proof, we identify these crossed products with these corresponding subalgebras
of $L(L^{2}(X_{1},\mu))$.

We claim that map $\varphi:K_{0}(C^{*}(\mathbb{Z},X_{1},h_{1}))\to
K_{0}(C^{*}(\mathbb{Z},X_{1},\widetilde{h}_{2}))$ defined by
$\varphi([\chi_{U}]_{1})=[\chi_{U}]_{2}$ is an isomorphism of ordered groups.
Since the positive cone and distinguished order units agree via this map, the
only thing to check is that
$\text{im}(\text{id}-(\alpha_{1})_{*})=\text{im}(\text{id}-(\alpha_{2})_{*})$.

Let $E$ be a compact open subset of $X_{1}$ such that $E\cap
Z_{1}=\varnothing$. Then since $\beta$ is continuous on $E$,
$\text{ran}(\beta|_{E})=\\{k_{1},\ldots,k_{N}\\}$. Then

$u_{1}\chi_{E}=\sum_{n=1}^{N}u_{2}^{k_{n}}\chi_{E\cap\beta^{-1}(k_{n})}.$

Thus, $u_{1}\chi_{E}\in C^{*}(\mathbb{Z},X_{1},\widetilde{h}_{2})$, and,
letting $A_{Z_{1}}^{(1)}$ correspond to $C^{*}(\mathbb{Z},X_{1},h_{1})$ as in
Definition 2.6, we have $A_{Z_{1}}^{(1)}\subset
C^{*}(\mathbb{Z},X_{1},\widetilde{h}_{2})$. Since $E\cap Z_{1}=\varnothing$,
there is a unitary $v\in A_{Z_{1}}^{(1)}$ such that
$v\chi_{E}v^{*}=\chi_{h_{1}(E)}$. Since we also have $v\in
C^{*}(\mathbb{Z},X_{1},\widetilde{h}_{2})$, we have
$[\chi_{E}]_{2}=[\chi_{h_{1}(E)}]_{2}$, so
$\chi_{E}-\chi_{h_{1}(E)}\in\text{im}(\text{id}-(\alpha_{2})_{*})$. Theorem
2.9 (and the proof) tells us that $K_{0}(A_{Z_{1}})^{(1)}\cong
K_{0}(\mathbb{Z},X_{1},h_{1})$ by the injection map. Thus,
$\text{im}(\text{id}-(\alpha_{1})_{*})\subset\text{im}(\text{id}-(\alpha_{2})_{*})$.
By repeating the above process with $\gamma$ instead of $\beta$, we can
similarly show that
$\text{im}(\text{id}-(\alpha_{2})_{*})\subset\text{im}(\text{id}-(\alpha_{1})_{*})$.
This completes the proof. ∎

######  Theorem 4.3.

Let $(X_{1},h_{1},Z_{1})$ and $(X_{2},h_{2},Z_{2})$ be fiberwise essentially
minimal zero-dimensional systems with no periodic points. The following are
equivalent:

  1. (a)

$C^{*}(\mathbb{Z},X_{1},h_{1})\cong C^{*}(\mathbb{Z},X_{2},h_{2})$.

  2. (b)

$(K_{0}(C^{*}(\mathbb{Z},X_{1},h_{1})),K_{0}(C^{*}(\mathbb{Z},X_{1},h_{1}))^{+},1)\cong(K_{0}(C^{*}(\mathbb{Z},X_{2},h_{2})),K_{0}(C^{*}(\mathbb{Z},X_{2},h_{2}))^{+},1)$
and

$K_{1}(C^{*}(\mathbb{Z},X_{1},h_{1}))\cong
K_{1}(C^{*}(\mathbb{Z},X_{2},h_{2}))$.

  3. (c)

$(X_{1},h_{1},Z_{1})$ and $(X_{2},h_{2},Z_{2})$ are strong orbit equivalent.

###### Proof.

(a) $\Longleftrightarrow$ (b). By Theorem 2.2 and 2.3 of [8],
$C^{*}(\mathbb{Z},X_{1},h_{1})$ and $C^{*}(\mathbb{Z},X_{2},h_{2})$ are
A$\mathbb{T}$-algebras of real rank zero, so this result follows from [3].

(b) $\Longleftrightarrow$ (c). This is implied by Theorem 4.2.

∎

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