# On continuous orbit equivalence rigidity for virtually cyclic group actions

Yongle Jiang  Y. Jiang, School of Mathematical Sciences, Dalian University of
Technology, Dalian, 116024, China<EMAIL_ADDRESS>
###### Abstract.

We prove that for any two continuous minimal (topologically free) actions of
the infinite dihedral group on an infinite compact Hausdorff space, they are
continuously orbit equivalent only if they are conjugate. We also show the
above fails if we replace the infinite dihedral group with certain other
virtually cyclic groups, e.g. the direct product of the integer group with any
non-abelian finite simple group.

###### Key words and phrases:

Cocycle, continuous orbit equivalence, rigidity, infinite dihedral group, skew
product actions

###### 2010 Mathematics Subject Classification:

Primary 37A20, Secondary 37B05, 20F65

##  1\. introduction

Let $G$ be a countable discrete group and $X$ be a compact Hausdorff space.
Denote by $\mathcal{C}$ a class of continuous actions of $G$ on $X$. A natural
and classical topic is to classify members in $\mathcal{C}$ up to certain
equivalence relation. In classical topological dynamical system, people have
studied intensively the classification of ${\mathbb{Z}}$-actions, or amenable
group actions in general, up to conjugacy. Historically speaking, the notion
of entropy was invented to distinguish Bernoulli shifts on finite-bases with
different sizes up to conjugacy (first in the measurable setting). Since one
decade ago, more and more attention has been put on a much wider class of
acting groups, i.e. sofic groups after the pioneering work of L. Bowen on
sofic entropy in the measurable setting [Bow] and D. Kerr and H. Li shortly in
the topological setting [KL1].

In contrast to the conjugacy relation, X. Li initiated a systematic study of
the notion of continuous orbit equivalence (see Definition 2.1) for
topological free actions of a general group $G$ [Li] during his study of
crossed product C∗-algebras. Note that this notion and its weaker versions,
e.g. topological orbit equivalence have been studied in special cases (see
[BT, GPS95, Su]) before Li’s work. In particular, Giordano-Putnam-Skau [GPS95]
and Giordano-Matui-Putnam-Skau [GMPS08, GMPS10] have proved a series of
remarkable results on classification of minimal ${\mathbb{Z}}$-actions on the
Cantor set up to topological orbit equivalence.

Although continuous orbit equivalence relation is by definition, a priori,
weaker than the conjugacy relation, an interesting phenomenon is that these
two notions may coincide in certain cases. To better describe this, let us say
a continuous action $\alpha$ of $G$ on $X$ is _continuous orbit equivalence
rigid for $\mathcal{C}$_ if for any continuous action $\beta\in\mathcal{C}$,
$\alpha$ is continuously orbit equivalent to $\beta$ implies that they are
conjugate, denoted by
$\alpha\overset{coe}{\sim}\beta\Rightarrow\alpha\overset{conj.}{\sim}\beta$.
Note that here, compared with X. Li’s definition [Li, Definition 2.5], we
further assume the acting groups are the same.

Now, we take $\mathcal{C}$ to be the class of minimal topologically free
actions and explain two motivations behind this work.

First, continuous orbit equivalence rigidity has been established for various
group actions, see e.g. [Li, BT, Sc1, Sc2, CM, CJ]. However, as far as we
know, the acting groups are assumed to be torsion-free in all written proofs
(see also Remark 2.5 and Remark 4.2). Thus it is natural to ask whether
continuous orbit equivalence rigidity can be established when the acting
groups contain non-trivial torsion elements. Second, a well-known result due
to Boyle-Tomiyama [BT] says that two topologically transitive, topologically
free ${\mathbb{Z}}$-actions (e.g. minimal free ${\mathbb{Z}}$-actions) are
continuously orbit equivalent only if they are conjugate. This result has been
extended to minimal equicontinuous actions of ${\mathbb{Z}}^{d}$ ($d\geq 2$)
in [CM]. More precisely, it is shown there that for minimal equicontinuous
${\mathbb{Z}}^{d}$-systems, continuous orbit equivalence implies that the
systems are virtually piecewise conjugate. Moreover, actions of virtually
cyclic groups have been considered in the study of dynamic asymptotic
dimension [ALSS] and Matui’s HK conjecture [OS, Sca] recently. Thus, it is
natural to study the analogy of Boyle-Tomiyama’s above mentioned result for
virtually cyclic group actions, i.e. we ask the following question.

######  Question 1.1.

Can we establish continuous orbit equivalence rigidity for minimal
topologically free actions of infinite virtually cyclic (but non-cyclic)
groups?

In general, we could not hope continuous orbit equivalence rigidity holds true
for all infinite virtually cyclic groups. Indeed, we have the following
theorem.

######  Theorem 1.2.

Let $F$ be any non-trivial finite group with trivial center, e.g. a non-
abelian finite simple group. Then there exist two minimal free actions
$\widetilde{\alpha}$ and $\widetilde{\alpha}^{\prime}$ of ${\mathbb{Z}}\times
F$ on an infinite compact Hausdorff space $X$ such that
$\widetilde{\alpha}\overset{coe}{\sim}\widetilde{\alpha}^{\prime}$ but
$\widetilde{\alpha}\overset{conj.}{\not\sim}\widetilde{\alpha}^{\prime}$.

On the positive side, when dealing with the infinite dihedral group, we do
have such a rigidity theorem.

######  Theorem 1.3.

Let $\alpha$ and $\beta$ be any two minimal (topologically free) actions of
the infinite dihedral group
$D_{\infty}={\mathbb{Z}}\rtimes{\mathbb{Z}}/{2{\mathbb{Z}}}$ on an infinite
compact Hausdorff space $X$. Then
$\alpha\overset{coe}{\sim}\beta\Rightarrow\alpha\overset{conj.}{\sim}\beta$.

Note that minimal actions of $D_{\infty}$ on an infinite compact Hausdorff
space is automatically topologically free, see Proposition 2.6. Moreover,
under the assumptions in the above theorem, we know continuous orbit
equivalence implies the two actions have the same topological entropy. This
may be compared with [KT, Theorem 7.5] and [KL2, Theorem D], where it is shown
that many non-virtually cyclic groups admit actions for which topological
entropy is an invariant of continuous orbit equivalence. For possible
generalizations of the theorems to other virtually cyclic groups, see the
discussion in Section 5. Here, we content ourselves with not discussing a
general version of the above theorem. Below, we briefly describe strategy for
the proofs.

For the proof of Theorem 1.2, we use the skew product construction to reduce
it to a problem of constructing minimal actions admitting continuous cocycles
into finite groups which are not coboundaries. On the positive side, one may
compare the proof of Theorem 1.3 with the proof of Boyle-Tomiyama [BT] for
continuous orbit equivalence rigidity for minimal ${\mathbb{Z}}$-actions.
Here, the basic strategy is as follows: by a criterion [Li, Proposition 4.4]
due to X. Li, it suffices to show the orbit cocycle $c$ is cohomologous to a
group isomorphism of $D_{\infty}$. The starting point for our proof is the
characterization of bi-Lipschitz bijections of ${\mathbb{Z}}$ due to
Benjamini-Shamov [BS]. After restricting the orbit cocycle
$c:{\mathbb{Z}}\times X\to D_{\infty}$ on the sub ${\mathbb{Z}}$-action, we
may extract from it a defect cocycle $a$ which attains only finitely many
values in $D_{\infty}$. Here, $a$ is defined to measure the defect between $c$
and a group homomorphism from ${\mathbb{Z}}$ to $D_{\infty}$. By twisting $a$
with a suitable coboundary, we can assume the range of $a$ is contained in
${\mathbb{Z}}$. If the sub ${\mathbb{Z}}$-action is minimal, one can apply the
well-known Gottschalk-Hedlund theorem [GH] for minimal ${\mathbb{Z}}$-actions
to deduce $a$ is actually a coboudary, which implies the original orbit
cocycle $c$ is trivial. For the general case, i.e. the sub
${\mathbb{Z}}$-action is not minimal, we can apply Gottschalk-Hedlund theorem
to each ${\mathbb{Z}}$-component, now the existence of reflection of
${\mathbb{Z}}$ in the group structure of $D_{\infty}$ allows us to patch up
untwisters for each component to get a global one.

The paper is organized into five sections. In Section 2, we explain
terminologies used in this paper, including cocycles, continuous orbit
equivalence, skew product actions and induced actions. We also present
examples of continuous actions. Then we give examples of minimal free actions
admitting continuous cocycles which are not coboundaries in Section 3. The
proof of Theorem 1.2 (resp. Theorem 1.3) is given in Subsection 4.1 (resp.
Subsection 4.2). We conclude the paper with several remarks on the possible
generalization of the theorems in Section 5.

In this paper, $G$ and $H$ usually mean virtually cyclic groups or finitely
generated groups, which is clear from the context. Moreover, we denote a
target group by $K$. Unless otherwise specified, all cocycles are assumed to
be continuous.

##  2\. preliminaries

###  2.1. Cocycles and continuous orbit equivalence

Let $G\curvearrowright X$ be a continuous action on a compact Hausdorff space
and let $K$ be a discrete group. Recall that a continuous map $c:G\times X\to
K$ is called a _$K$ -valued (continuous) cocycle_ if
$c(g_{1}g_{2},x)=c(g_{1},g_{2}x)c(g_{2},x)$ for all $g_{1}$, $g_{2}\in G$ and
all $x\in X$. Two continuous cocycles $c_{1},c_{2}:G\times X\to K$ are
_(continuously) cohomologous/equivalent_ if there exists a continuous map
$b:X\to K$ such that $c_{1}(g,x)=b(gx)^{-1}c_{2}(g,x)b(x)$ for all $g\in G$
and all $x\in X$. A continuous cocycle is called _trivial (resp. a cobundary)_
if it is equivalent to a group homomorphism $\phi:G\to K$ (resp. the trivial
group homomorphism $\phi\equiv e_{K}$ on $G$), which is treated as a constant
cocycle on $X$.

Note that when $G={\mathbb{Z}}$, a $K$-valued cocycle $c$ for
${\mathbb{Z}}\curvearrowright X$ can be represented as a map $f:X\to K$.
Indeed, for one direction, set $f(x)=c(1,x)$; for the other direction, we may
define $c(1,x)=f(x)$ and use the cocycle identity to define $c(N,x)$ for all
$N\in{\mathbb{Z}}$ and hence get a cocycle $c$. Sometimes, we may abuse our
notation by calling $f$ a cocycle.

Given two continuous actions $G\curvearrowright X$ and $K\curvearrowright Y$,
we have the notion of continuous orbit equivalence (written as coe for short)
which is systematically studied by X. Li [Li, Definition 2.5].

######  Definition 2.1 (coe).

$G\curvearrowright X$ and $K\curvearrowright Y$ are continuously orbit
equivalent, denoted by $G\curvearrowright
X\overset{coe}{\sim}K\curvearrowright Y$, if there exists a homeomorphism
$\phi:X\approx Y$ with inverse $\psi=\phi^{-1}$ and continuous maps $a:G\times
X\to K$ and $b:K\times Y\to G$ such that

$\displaystyle\phi(gx)=a(g,x)\phi(x),~{}\psi(hy)=b(h,y)\psi(y)$

for all $g\in G$, $h\in H$, $x\in X$ and $y\in Y$.

If both $a$ and $b$ are further assumed to be group isomorphisms from $G$ to
$K$ and from $K$ to $G$ respectively, then we call these two actions are
_conjugate_ , written as $G\curvearrowright
X\overset{conj.}{\sim}K\curvearrowright Y$. Clearly, conjugacy implies coe.

We remark that if $K\curvearrowright Y$ is _topologically free_ (i.e. for each
$e\neq h\in K$, $\\{y\in Y:hy\neq y\\}$ is dense in $Y$), then $a$ is a
continuous cocycle, see [Li, Lemma 2.8]. When both $a$ and $b$ are cocycles in
the above definition, we simply call $b$ the inverse cocycle of $a$.

The following fact is proved independently in [MST] and [Li_qi], we sketch the
proof for completeness.

######  Proposition 2.2.

Let $G$ and $K$ be finitely generated groups. Let $G\curvearrowright X$ and
$K\curvearrowright Y$ be two topological free actions on compact spaces.
Assume they are continuously orbit equivalent and $a:G\times X\to K$ is the
associated continuous orbit cocycle, then $G\ni g\mapsto a(g,x)\in K$ is a bi-
Lipschitz bijection for each $x\in X$ with respect to the right invariant word
metrics on $G$ and $K$ respectively.

Throughout the paper, a _bi-Lipschitz bijection_ means a bijective map $\phi$
such that both $\phi$ and $\phi^{-1}$ are Lipschitz maps in the usual sense.

###### Proof.

Let $S$ and $T$ be any symmetric generating set for $G$ and $K$ respectively.
Let $|\cdot|_{S}$ and $|\cdot|_{T}$ be associated word lengths, and let
$d_{S}(\cdot,\cdot)$ and $d_{T}(\cdot,\cdot)$ be the corresponding right
invariant word length metrics on $G$ and $K$, i.e.
$d_{S}(g_{1},g_{2})=|g_{2}g_{1}^{-1}|_{S}$ and
$d_{T}(h_{1},h_{2})=|h_{2}h_{1}^{-1}|_{T}$ for all $g_{1}$, $g_{2}\in G$ and
$h_{1}$, $h_{2}\in K$.

Then
$d_{T}(a(g_{1},x),a(g_{2},x))=|a(g_{2},x)a(g_{1},x)^{-1}|_{T}=|a(g_{2}g_{1}^{-1},g_{1}x)|_{T}\leq|g_{2}g_{1}^{-1}|_{S}\cdot\sup_{x\in
X}\max_{s\in S}|a(s,x)|_{T}=d_{S}(g_{1},g_{2})\cdot\sup_{x\in X}\max_{s\in
S}|a(s,x)|_{T}$, where the inequality follows from the cocycle identity. This
shows $g\mapsto a(g,x)$ is a Lipschitz map.

To see $g\mapsto a(g,x)$ is a bijection and its inverse is a Lipschitz map,
apply the similar argument to the inverse cocycle using [Li, Lemma 2.10]. ∎

The following lemma is well-known, but we present its proof for the reader’s
convenience.

######  Lemma 2.3.

Let $\alpha:{\mathbb{Z}}\curvearrowright X$ be a minimal action on a compact
space $X$ and $K_{0}\subset K$ be a closed subgroup in a topological group.
Let $c:{\mathbb{Z}}\times X\to K_{0}$ be a continuous cocycle such that
$c(n,x)=f(\alpha_{n}(x))f(x)^{-1}$ for some continuous map $f:X\to K$ and all
$n\in{\mathbb{Z}}$ and $x\in X$. Then there is a continuous map
$f^{\prime}:X\to K_{0}$ such that
$c(n,x)=f^{\prime}(\alpha_{n}(x))f^{\prime}(x)^{-1}$ holds for all
$n\in{\mathbb{Z}}$ and all $x\in X$.

###### Proof.

Observe that the continuous map

$x\in X\to K_{0}f(x)\in K_{0}\backslash K$

is $\alpha$-invariant. Hence minimality implies there exists some $k\in K$
such that $K_{0}f(x)=K_{0}k$ for all $x\in X$, i.e. $f(x)\in K_{0}k$ for all
$x\in X$. Define $f^{\prime}(x)=f(x)k^{-1}$. We still have
$c(n,x)=f^{\prime}(\alpha_{n}(x))f^{\prime}(x)^{-1}$. ∎

###  2.2. Skew product actions

We recall the following well-known construction of skew product actions. Let
$\sigma:H\to Aut(K)$ be a group homomorphism. Let $c:H\curvearrowright X\to K$
be a _skew cocycle_ , i.e. for all $h_{1}$, $h_{2}\in H$ and all $x\in X$,

$c(h_{1}h_{2},x)=c(h_{1},h_{2}x)\sigma_{h_{1}}(c(h_{2},x)).$

One may check that $c((h_{1}h_{2})h_{3},x)=c(h_{1}(h_{2}h_{3}),x)$ holds for
all $h_{i}\in H$, $1\leq i\leq 3$ and all $x\in X$. Besides, $c$ is a skew
cocycle iff the map $c^{\prime}:H\curvearrowright X\to K\rtimes_{\sigma}H$
given by $c^{\prime}(h,x)=(c(h,x),h)$ is a cocycle in the usual sense.

Given a group homomorphism $\sigma:H\to Aut(K)$, an action $H\curvearrowright
X$ and a skew cocycle $c$ defined as above, we may define the _generalized
skew product action_ :

$\displaystyle G:=K\rtimes_{\sigma}H$ $\displaystyle\curvearrowright X\times
K$ $\displaystyle(k,h)(x,k^{\prime})$
$\displaystyle:=(hx,c(h,x)\sigma_{h}(k^{\prime})k^{-1}).$

Note that when $\sigma$ is trivial, i.e. $\sigma\equiv id_{K}$, skew cocycles
(resp. skew product actions) are reduced to the usual cocycles (resp. usual
skew product actions) for the group $K\times H$. In this paper, we usually
take $H={\mathbb{Z}}$, $K$ be a finite group and $\sigma$ be trivial. To
emphasize the cocycle $c$, we usually denote $X\times K$ by $X\times_{c}K$.

###  2.3. Induced actions

Let $H$ be a subgroup of $G$. Fix any lift map $L:G/H\to G$ such that
$L(gH)H=gH$ for all $g\in G$. Consider the associated cocycle $c:G\times
G/H\to H$ given by $c(g,g^{\prime}H)=L(gg^{\prime}H)^{-1}gL(g^{\prime}H)$ for
all $g,g^{\prime}\in G$. Given any continuous action $H\curvearrowright Y$,
then the _induced action_ $G\curvearrowright G/H\times Y$ is defined as
follows:

$g(g^{\prime}H,y)=(gg^{\prime}H,c(g,g^{\prime}H)y),~{}\mbox{for all
$g,g^{\prime}\in G$ and $y\in Y$.}$

###  2.4. Examples of continuous actions

#### Odometer actions

Fix a sequence of strictly increasing positive integers $(n_{i})_{i\geq 1}$
such that $n_{i}\mid n_{i+1}$ for all $i\geq 1$. Let
$\lim\limits_{\leftarrow}{\mathbb{Z}}/{n_{i}{\mathbb{Z}}}=\\{(x_{i}+n_{i}{\mathbb{Z}})_{i\geq
1}\in\prod_{i\geq
1}{\mathbb{Z}}/{n_{i}{\mathbb{Z}}}:~{}n_{i}\mid(x_{i+1}-x_{i}),~{}\forall~{}i\geq
1\\}$. The _odometer action associated to $(n_{i})$_ is defined as the
continuous action
$\alpha:{\mathbb{Z}}\curvearrowright\lim\limits_{\leftarrow}{\mathbb{Z}}/{n_{i}{\mathbb{Z}}}$
given by

$\alpha_{n}((x_{i}+n_{i}{\mathbb{Z}})_{i})=(n+x_{i}+n_{i}{\mathbb{Z}})_{i},$

where $x_{i}\in{\mathbb{Z}}$.

#### Weakly mixing actions

A continuous action ${\mathbb{Z}}\curvearrowright X$ on a compact Hausdorff
space is called _(topologically) weakly mixing_ [Gl, p.23] if the product
action ${\mathbb{Z}}\curvearrowright X\times X$ is _transitive_ , i.e. for
every pair of non-empty open sets $U$, $V$ in $X\times X$ there exists some
$g\in{\mathbb{Z}}$ with $gU\cap V\neq\emptyset$. Examples of free minimal
weakly mixing actions include the so-called Chacón system [Gl, p. 27].

It is well-known and easy to check that for a continuous action
${\mathbb{Z}}\curvearrowright X$ on a compact Hausdorff space $X$, if $X$ has
no isolated points and there is a point in $X$ with dense orbit, then
${\mathbb{Z}}\curvearrowright X$ is transitive. Conversely, for a compact
metric space $X$, ${\mathbb{Z}}\curvearrowright X$ is transitive implies it
admits a point with dense orbit, see e.g. [KL_book, Proposition 7.9].

###  2.5. Subgroups of $D_{\infty}$

Let $D_{\infty}={\mathbb{Z}}\rtimes\frac{{\mathbb{Z}}}{2{\mathbb{Z}}}=\langle
s\rangle\rtimes\langle t\rangle=\langle s,t~{}|~{}t^{2},tsts\rangle$ be the
infinite dihedral group. We record a simple lemma on the structure of its
subgroups.

######  Lemma 2.4.

If $H$ is a subgroup of $D_{\infty}$, then either $H=\langle
s^{k},s^{i}t\rangle$ for some $k\in{\mathbb{Z}}$ and some $0\leq|i|<k$, or
$H=\langle s^{k}t\rangle$ or $H=\langle s^{k}\rangle$ for some
$k\in{\mathbb{Z}}$. Moreover, all subgroups can be realized as $\phi_{i}(H)$
for some $i\in{\mathbb{Z}}$ and some
$H\in\\{k{\mathbb{Z}}\rtimes\frac{{\mathbb{Z}}}{2{\mathbb{Z}}},~{}k{\mathbb{Z}},~{}\frac{{\mathbb{Z}}}{2{\mathbb{Z}}}:~{}k\in{\mathbb{Z}}\\}$,
where, $\phi_{i}\in Aut(G)$ is defined by $\phi_{i}(s)=s,\phi_{i}(t)=s^{i}t$.

###### Proof.

Write $H\cap{\mathbb{Z}}=\langle s^{k}\rangle$ for some $k\in{\mathbb{Z}}$.
Note that the set of torsion elements of $G$ is precisely
$\\{s^{i}t:i\in{\mathbb{Z}}\\}$.

If $H$ contains no non-trivial torsion elements, then $H\subset{\mathbb{Z}}$,
i.e. $H=H\cap{\mathbb{Z}}=\langle s^{k}\rangle$.

If $H$ contains some non-trivial torsion element, say $s^{i}t\in H$ for some
$i\in{\mathbb{Z}}$. W.L.O.G., we may assume $|i|$ is smallest among all non-
trivial torsion elements in $H$.

Case 1: $i=0$. Then clearly, $H=\langle s^{k},t\rangle$.

Case 2: $|i|>0$.

If $k=0$, then $H=\langle s^{i}t\rangle$. Indeed, assume not, then $s^{j}t\in
H$ for some $j\neq i$, which implies that $(s^{i}t)(s^{j}t)=s^{i-j}\in
H\cap{\mathbb{Z}}$, a contradiction.

We may assume $k>0$. Clearly, $0<|i|<k$, then $H=\langle s^{k},s^{i}t\rangle$.
Indeed, assume $s^{j}t\in H$, then $(s^{j}t)(s^{i}t)^{-1}=s^{j-i}\in
H\cap{\mathbb{Z}}=\langle s^{k}\rangle$, hence $s^{j}t\in\langle
s^{k},s^{i}t\rangle$.

The moreover part is now clear. ∎

######  Remark 2.5.

Note that this lemma implies that for two odometer actions of $D_{\infty}$,
the notion of structurally conjugacy between them, as introduced in [CM,
Definition 3.1], coincides with the usual conjugacy. Thus [CM, Theorem 3.3]
implies that for two free odometer actions of $D_{\infty}$, continuous orbit
equivalence between them implies they are conjugate.

Next, let us observe the following hold, see also [OS, Proposition 2.8].

######  Proposition 2.6.

If $\alpha:D_{\infty}\curvearrowright X$ is a minimal action on an infinite
compact Hausdorff space $X$, then it is topologically free.

###### Proof.

Without loss of generality, we may assume $\alpha$ is not free. Write
$D_{\infty}={\mathbb{Z}}\rtimes\frac{{\mathbb{Z}}}{2{\mathbb{Z}}}=\langle
s,t|~{}t^{2},tsts\rangle=\langle s\rangle\rtimes\langle t\rangle$. Then
observe that the sub ${\mathbb{Z}}=\langle s\rangle$-action, i.e.
$\alpha|_{{\mathbb{Z}}}:{\mathbb{Z}}\curvearrowright X$ is minimal.

Indeed, note that the number of minimal ${\mathbb{Z}}$-components in $X$ is
either one or two (See the discussion before Case I in the proof of Theorem
1.3 for details). If we have two minimal ${\mathbb{Z}}$-components, say
$X_{0}$ and $tX_{0}$, then $\alpha|_{{\mathbb{Z}}}$ on each component is free
and so is $\alpha$, contradicting to our assumption.

Now, as the action is minimal and $X$ is infinite, every stabilizer subgroup
must be of infinite index in $D_{\infty}$. By Lemma 2.4, we deduce that for
each $x\in X$, if the stabilizer group $Stab(x)$ is not trivial, then
$Stab(x)=\\{e,s^{n}t\\}$ for some $n\in{\mathbb{Z}}$. To check topological
freeness of $\alpha$, we only need to show that for each $n\in{\mathbb{Z}}$,
the closed set $X_{n}:=\\{x\in X:s^{n}tx=x\\}$ is nowhere dense.

Note that $x\in X_{n}$ iff $Stab(x)=\\{e,s^{n}t\\}$. Now, assume that
$\exists~{}\emptyset\neq U\subsetneq X_{n}$, where $U$ is an open set. Then we
may pick any $y\in X\setminus U$ and use ${\mathbb{Z}}\curvearrowright X$ is
minimal to deduce that for each $N$, $\exists~{}|m|>N$ such that $s^{m}y\in
U$. Hence, $\\{e,s^{n}t\\}=Stab(s^{m}y)=s^{m}Stab(y)s^{-m}$. Since
$Stab(y)=\\{e\\}$ or $\\{e,s^{k}t\\}$ for some $k\in{\mathbb{Z}}$, we deduce
that $2m=n-k$, which gives us a contradiction since $m$ can be arbitrarily
large. ∎

###  2.6. Bi-Lipschitz bijections of ${\mathbb{Z}}$

We need the following result in [BS].

######  Theorem 2.7 (Benjamini-Shamov).

Let $f:{\mathbb{Z}}\to{\mathbb{Z}}$ be a bi-Lipschitz bijection
(${\mathbb{Z}}$ is equipped with its usual metric, namely $\rho(x,y):=|x-y|$).
Then either $\sup_{x\in{\mathbb{Z}}}|f(x)-x|<+\infty$ or
$\sup_{x\in{\mathbb{Z}}}|f(x)+x|<+\infty$. More precisely, $f(x)=\pm
x+\mbox{const}+r(x)$, where $\sup_{x\in{\mathbb{Z}}}|r(x)|<\infty$.

###  2.7. Gottschalk-Hedlund theorem

Another ingredient we need is the classical Gottschalk-Hedlund theorem [GH,
Theorem 14.11]. We record the following version as presented in [KH, P. 102,
Theorem 2.9.4] and explain how to use it in our setting. Note that the proof
of [KH, P. 102, Theorem 2.9.4] still works without the metrizability
assumption on the space.

######  Theorem 2.8 (Gottschalk-Hedlund).

Let $X$ be a compact Hausdorff space, $f:X\to X$ a minimal continuous map and
$g:X\to\mathbb{R}$ continuous such that
$\sup_{n\in{\mathbb{N}}}|\sum_{i=0}^{n}g\circ f^{i}(x_{0})|<\infty$ for some
$x_{0}\in X$. Then there is a continuous $\phi:X\to\mathbb{R}$ such that
$\phi\circ f-\phi=g$.

Now, let $f:X\to X$ be a minimal continuous action and $c:{\mathbb{Z}}\times
X\to\mathbb{R}$ be a continuous cocycle for this action. Set $g(x)=c(1,x)$,
then a calculation shows that $\sum_{i=0}^{n}g\circ
f^{i}(x_{0})=c(n+1,x_{0})$. Thus, Gottschalk-Hedlund theorem actually shows
that if $c$ is a bounded cocycle, i.e. $\sup_{(n,x)\in{\mathbb{Z}}\times
X}|c(n,x)|<\infty$, then it is a continuous coboundary map.

##  3\. existence of cocycles that are not coboundaries

For the proof of Theorem 1.2, we need to construct free minimal actions such
that they admit continuous cocycles which are not coboundaries with values
into finite groups. We collect some results along this direction.

######  Proposition 3.1.

Let ${\mathbb{Z}}\curvearrowright X$ be a minimal weakly mixing and continuous
action on a compact metric space $X$. Then for each non-trivial group
homomorphism $\phi$ from ${\mathbb{Z}}$ to $F$, where $F$ is a finite abelian
group, $\phi$ is not a coboundary.

###### Proof.

Assume that $\phi(s)=L(sx)-L(x)$ for some continuous map $L:X\to F$ and all
$s\in{\mathbb{Z}}$ and all $x\in X$. Then $L(sx)-L(x)=L(sy)-L(y)$ for all
$x,y\in X$. Then $L^{\prime}(x,y):=L(x)-L(y)$ is a continuous map which is
constant on each orbit of any point in $X\times X$. By assumption, we know
that $L^{\prime}$ is constant since the diagonal action
${\mathbb{Z}}\curvearrowright X\times X$ admits a point with dense orbit. Thus
$L^{\prime}(x,y)=L^{\prime}(x,x)=0$ for all $x,y\in X$, i.e. $L$ is a constant
function on $X$, hence $\phi(s)=0$ for all $s\in{\mathbb{Z}}$, i.e. $\phi$ is
a trivial homomorphism, a contradiction. ∎

For the next result, we need the following known characterization of
coboundaries using essential values for continuous cocycles in [LM], which was
adapted from Schmidt’s notion for measurable cocycles in [Sc].

Let $K$ be a locally compact group with the unit element $e$. By $K_{\infty}$
we denote the one point compactification of $K$, i.e.
$K_{\infty}=K\cup\\{\infty\\}$. We may extend the group operation form the
group $K$ onto the set $K_{\infty}$ by $g\cdot\infty=\infty$ for all $g\in
K_{\infty}$. One can check that the operation $K_{\infty}\times
K_{\infty}\ni(g_{1},g_{2})\to g_{1}g_{2}\in K_{\infty}$ is continuous.

######  Definition 3.2 (Definition 3.1 in [LM]).

Let $T:{\mathbb{Z}}\curvearrowright X$ be a continuous action on a compact
space $X$. Let $c:{\mathbb{Z}}\times X\to K$ be a continuous cocycle. We say
that $r\in K_{\infty}$ is an _essential value of $\phi$_ if for each nonempty
open $U\subset X$ and each neighborhood $V$ of $r$ there exists
$N\in{\mathbb{Z}}$ such that

$U\cap T^{-N}U\cap\\{x\in X:c(N,x)\in V\\}\neq\emptyset.$

The set of all essential values of $c$ is denote by $E_{\infty}(c)$. Put also
$E(c)=E_{\infty}(c)\cap K$.

######  Proposition 3.3.

Let ${\mathbb{Z}}\curvearrowright X$ be a minimal action and $K$ be a compact
abelain group, e.g. a finite abelian group. Let $c:{\mathbb{Z}}\times X\to K$
be a continuous cocycle. Then $E(c)=\\{e\\}$ iff $c$ is a coboundary, where
$e$ denotes the neutral element in $K$.

###### Proof.

$\Rightarrow$: this is Lemma 3.1 (ii) in [LM]. $\Leftarrow$: one can check
this directly using the definition of $E(c)$. ∎

The following result might be known to experts, since we do not find a
reference, we include the proof.

######  Proposition 3.4.

Let $({\mathbb{Z}},X)$ be a minimal odometer action, say
$X=\lim\limits_{\leftarrow}{\mathbb{Z}}/{n_{i}{\mathbb{Z}}}$, where $n_{i}\mid
n_{j}$ for all $i\leq j$ and $\lim_{j\to\infty}n_{j}=\infty$. Let $p$ be a
prime number. Then the following statements are equivalent.
(1) Every continuous map $f:X\to{\mathbb{Z}}/{p{\mathbb{Z}}}$ gives rise to a
continuous coboundary.
(2) $\sup_{i}ord(p,n_{i})=\infty$, where $ord(p,n_{i}):=\max_{p^{k}\mid
n_{i}}k$.

###### Proof.

(1)$\Rightarrow$(2): By Proposition 3.3, we know that for all continuous
$f:X\to{\mathbb{Z}}/{p{\mathbb{Z}}}$, $E(f)=\\{0\\}$, i.e. for all $0\neq
c\in{\mathbb{Z}}/{p{\mathbb{Z}}}$, there exists some non-empty open set
$U\subset X$ such that for all $n\in{\mathbb{Z}}$, we have $x\in U\cap
T^{-n}U\Rightarrow f(x)+f(Tx)+\cdots+f(T^{n-1}x)\neq c$.

Now, denote by $\pi_{i}:X\to{\mathbb{Z}}/{n_{i}{\mathbb{Z}}}$ the natural
projection map, i.e.
$\pi_{i}((x_{i}+n_{i}{\mathbb{Z}})_{i})=x_{i}+n_{i}{\mathbb{Z}}$. Assume that
$\sup_{i}ord(p,n_{i})$ is bounded, let us take some $j$ such that
$ord(p,n_{j})=\sup_{i}ord(p,n_{i})$ and we may assume $n_{j}>1$, then we
define a map $f:X\to{\mathbb{Z}}/{p{\mathbb{Z}}}$ as the composition
$X\overset{\pi_{j}}{\to}{\mathbb{Z}}/{n_{j}{\mathbb{Z}}}\overset{f^{\prime}}{\to}{\mathbb{Z}}/{p{\mathbb{Z}}}$,
where
$f^{\prime}(i+n_{j}{\mathbb{Z}})=\delta_{i,1}\in{\mathbb{Z}}/{p{\mathbb{Z}}}$
for all $0\leq i\leq n_{1}-1$. Take $c=1\in{\mathbb{Z}}/{p{\mathbb{Z}}}$, then
we find some $U\subset X$ as above. We may shrink $U$ if necessary to assume
$U=\pi_{k}^{-1}(i_{0}+n_{k}{\mathbb{Z}})$ for some $k>j$ and $i_{0}$. Now,
take any $x=(x_{i}+n_{i}{\mathbb{Z}})_{i}\in U$ and $n=n_{k}\ell$ for some
integer $\ell$ such that $\frac{n_{k}}{n_{j}}\ell\equiv 1~{}mod~{}p$ (This is
possible since $p\nmid\frac{n_{k}}{n_{j}}$ by our choice of $j$ and $p$ is a
prime number). Note that $n_{k}\mid n$ implies that $T^{n}x\in U$. Now, a
calculation shows that

$\displaystyle f(x)+f(Tx)+\cdots+f(T^{n-1}x)$
$\displaystyle=\sum_{s=0}^{n-1}f^{\prime}(x_{j}+s+n_{j}{\mathbb{Z}})$
$\displaystyle=1\cdot(\frac{n}{n_{j}})=\frac{n_{k}\ell}{n_{j}}=1=c\in{\mathbb{Z}}/{p{\mathbb{Z}}}.$

This is a contradiction.

(2)$\Rightarrow$(1): the proof is similar as above. By Proposition 3.3, we aim
to show that for all continuous $f:X\to{\mathbb{Z}}/{p{\mathbb{Z}}}$ and all
$0\neq c\in{\mathbb{Z}}/{p{\mathbb{Z}}}$, there exists some non-empty set
$U\subset X$ such that for all $n\in{\mathbb{Z}}$, $x\in U\cap
T^{-n}U\Rightarrow f(x)+f(Tx)+\cdots+f(T^{n-1}x)\neq c$.

Pick $j$ large enough such that $f$ factors through
$\pi_{j}:X\to{\mathbb{Z}}/{n_{j}{\mathbb{Z}}}$, i.e. $f=\pi_{j}\circ
f^{\prime}$ for some map
$f^{\prime}:{\mathbb{Z}}/{n_{j}{\mathbb{Z}}}\to{\mathbb{Z}}/{p{\mathbb{Z}}}$.
Since $\sup_{i}ord(p,n_{i})=\infty$, we may find $k>j$ such that
$ord(p,n_{k})>ord(p,n_{j})$. Then let $U=\pi_{k}^{-1}(0+n_{k}{\mathbb{Z}})$.
Now, for each $n\in{\mathbb{Z}}$, if $x\in U\cap T^{-n}U$, then
$x_{k}=0=x_{k}+n$ in ${\mathbb{Z}}/{n_{k}{\mathbb{Z}}}$, so $n_{k}\mid n$.
Now, a calculation shows that

$\displaystyle f(x)+f(Tx)+\cdots+f(T^{n-1}x)$
$\displaystyle=\sum_{s=0}^{n-1}f^{\prime}(x_{j}+s+n_{j}{\mathbb{Z}})$
$\displaystyle=(\sum_{s=0}^{n_{j}-1}f^{\prime}(s+n_{j}{\mathbb{Z}}))\frac{n}{n_{j}}$
$\displaystyle=[(\sum_{s=0}^{n_{j}-1}f^{\prime}(s+n_{j}{\mathbb{Z}}))\frac{n}{n_{k}}]\frac{n_{k}}{n_{j}}\in
p{\mathbb{Z}}~{}\mbox{(since $ord(p,n_{k})>ord(p,n_{j})$)}.$

Thus, $0=f(x)+f(Tx)+\cdots+f(T^{n-1}x)\neq c$ in
${\mathbb{Z}}/{p{\mathbb{Z}}}$. ∎

From the above proposition, we can deduce that for the free minimal odometer
action
${\mathbb{Z}}\curvearrowright\lim\limits_{\leftarrow}\frac{{\mathbb{Z}}}{2^{i}{\mathbb{Z}}}:=X$,
it admits a continuous ${\mathbb{Z}}/{3{\mathbb{Z}}}$-valued cocycle that is
not a coboundary.

##  4\. proofs

###  4.1. Non-rigidity part

Theorem 1.2 is a direct corollary of the following proposition.

######  Proposition 4.1.

Let $\alpha:{\mathbb{Z}}\curvearrowright X$ be a minimal free action on a
compact Hausdorff space. Let $c,c^{\prime}:{\mathbb{Z}}\times X\to F$ be two
continuous cocycles (w.r.t. the action $\alpha$) into a finite non-trivial
group $F$. Consider the associated skew product actions
$\widetilde{\alpha}:F\times{\mathbb{Z}}\curvearrowright X\times_{c}F$ and
$\widetilde{\alpha}^{\prime}:F\times{\mathbb{Z}}\curvearrowright
X\times_{c^{\prime}}F$ as defined in Subsection 2.2. Then the following hold:
(i) Both $\widetilde{\alpha}$ and $\widetilde{\alpha}^{\prime}$ are minimal
free actions.
(ii) $\widetilde{\alpha}\overset{coe}{\sim}\widetilde{\alpha}^{\prime}$.
(iii) If $c^{\prime}\equiv e\in F$, the neutral element in $F$, then
$\widetilde{\alpha}\overset{conj.}{\sim}\widetilde{\alpha}^{\prime}$ implies
that $c$ is cohomologous to a group homomorphism from ${\mathbb{Z}}$ into
$C(F)$.
(iv) For each $F$ with trivial center, there exist $\alpha$, $c$ and
$c^{\prime}$ such that
$\widetilde{\alpha}\overset{conj.}{\not\sim}\widetilde{\alpha}^{\prime}$.

###### Proof.

(i) We leave it as an exercise.
(ii) Recall that $\widetilde{\alpha}:F\times{\mathbb{Z}}\curvearrowright
X\times_{c}F$ is defined as follows:

$\widetilde{\alpha}_{(f,n)}(x,f^{\prime})=(\alpha_{n}(x),c(n,x)f^{\prime}f^{-1})~{}\mbox{for
all $f$, $f^{\prime}\in F$, $n\in{\mathbb{Z}}$ and $x\in X$}.$

Define $\theta:(F\times{\mathbb{Z}})\times(X\times_{c}F)\to
F\times{\mathbb{Z}}$ by setting

$\theta((t,n),(x,f))=(tf^{-1}c(n,x)^{-1}c^{\prime}(n,x)f,n)~{}\mbox{for all
$n\in{\mathbb{Z}}$, $t\in F$ and $x\in X$}.$

One can check it is a continuous cocycle w.r.t. the skew product action
$\widetilde{\alpha}$. Indeed,

$\displaystyle\theta((t_{1},n_{1})(t_{2},n_{2}),(x,f))$
$\displaystyle=\theta((t_{1}t_{2},n_{1}+n_{2}),(x,f))$
$\displaystyle=(t_{1}t_{2}f^{-1}c(n_{1}+n_{2},x)^{-1}c^{\prime}(n_{1}+n_{2},x)f,n_{1}+n_{2})$
$\displaystyle\theta((t_{1},n_{1}),\widetilde{\alpha}_{(t_{2},n_{2})}(x,f))\theta((t_{2},n_{2}),(x,f))$
$\displaystyle=\theta((t_{1},n_{1}),(\alpha_{n_{2}}(x),c(n_{2},x)ft_{2}^{-1}))\theta((t_{2},n_{2}),(x,f))$
$\displaystyle=(t_{1}t_{2}f^{-1}c(n_{2},x)^{-1}c(n_{1},\alpha_{n_{2}}(x))^{-1}c^{\prime}(n_{1},\alpha_{n_{2}}(x))c(n_{2},x)ft_{2}^{-1},n_{1})\cdot$
$\displaystyle\quad\quad(t_{2}f^{-1}c(n_{2},x)^{-1}c^{\prime}(n_{2},x)f,n_{2})$
$\displaystyle=(t_{1}t_{2}f^{-1}c(n_{2},x)^{-1}c(n_{1},\alpha_{n_{2}}(x))^{-1}c^{\prime}(n_{1},\alpha_{n_{2}}(x))c^{\prime}(n_{2},x)f,n_{1}+n_{2})$
$\displaystyle=(t_{1}t_{2}f^{-1}c(n_{1}+n_{2},x)^{-1}c^{\prime}(n_{1}+n_{2},x)f,n_{1}+n_{2}).$

Thus,

$\displaystyle\theta((t_{1},n_{1})(t_{2},n_{2}),(x,f))=\theta((t_{1},n_{1}),\widetilde{\alpha}_{(t_{2},n_{2})}(x,f))\theta((t_{2},n_{2}),(x,f)).$

Moreover, one can verify that

$\widetilde{\alpha}_{(t,n)}(x,f)=\widetilde{\alpha}^{\prime}_{\theta((t,n),(x,f))}(x,f).$

By symmetry, one can also define the inverse cocycle for $\theta$, hence
$\widetilde{\alpha}\overset{coe}{\sim}\widetilde{\alpha}^{\prime}$.

(iii) First, we observe that for each $\Phi\in Aut(F\times{\mathbb{Z}})$, we
have $\Phi(t,1)=(\epsilon(t)g,\pm 1)$ for some $\epsilon\in Aut(F)$ and $g\in
C(F)$, the center of $F$. Indeed, since $F$ is finite,
$\Phi(F\times\\{0\\})=F\times\\{0\\}$, thus we define
$\epsilon=\Phi|_{F\times\\{0\\}}$. Next, we may write $\Phi(e,1)=(g,\pm 1)$
for some $g\in F$. From $\Phi((e,1)(t,0))=\Phi((t,0)(e,1))$ for all $t\in F$,
we deduce that $g\in C(F)$. Note that $\Phi(t,n)=(\epsilon(t)g^{n},\pm n)$ for
all $n\in{\mathbb{Z}}$ and $t\in F$. For such $\Phi$, we simply write
$\Phi=(\epsilon(\cdot)g,\pm)$.

Next, assume $\widetilde{\alpha}$ is conjugate to
$\widetilde{\alpha}^{\prime}$, say via a homeomorphism $\tau:X\times_{c}F\to
X\times_{c^{\prime}}F$ and a group isomorphism
$(\epsilon(\cdot)g,\pm):F\times{\mathbb{Z}}\cong F\times{\mathbb{Z}}$ defined
in the last paragraph. Then we may write

$\tau((x,f^{\prime}))=(\phi(x,f^{\prime}),\psi(x,f^{\prime}))$

for some continuous maps $\phi:X\times_{c}F\to X$ and $\psi:X\times_{c}F\to
F$.

Fix any $(x,f^{\prime})\in X\times_{c}F$ and $(f,n)\in F\times{\mathbb{Z}}$,
the $F\times{\mathbb{Z}}$-equivariance of $\tau$ shows that for all $x\in X$,
$f$, $f^{\prime}\in F$ and all $n\in{\mathbb{Z}}$, we have

(1) $\displaystyle\begin{split}\alpha_{\pm
n}(\phi(x,f^{\prime}))&=\phi(\alpha_{n}(x),c(n,x)f^{\prime}f^{-1})\\\
c^{\prime}(\pm
n,\phi(x,f^{\prime}))\psi(x,f^{\prime})g^{-n}\epsilon(f)^{-1}&=\psi(\alpha_{n}(x),c(n,x)f^{\prime}f^{-1}).\end{split}$

Since $f$ is arbitrary in the 1st expression in (1), we deduce that $\phi$
only depends on its 1st coordinate, so we may directly write
$\phi(x,-)=\phi(x)$ and hence $\alpha_{\pm n}(\phi(x))=\phi(\alpha_{n}(x))$.
Using $\tau$ is a homeomorphism, we can check that $\phi|_{X}$ is also a
homeomorphism.

Now, set $f=c(n,x)$ and $f^{\prime}=e$ into the 2nd line in (1), we get that

$c^{\prime}(\pm
n,\phi(x))\psi(x,e)g^{-n}\epsilon(c(n,x))^{-1}=\psi(\alpha_{n}(x),e).$

Thus,

(2) $\displaystyle c^{\prime}(\pm
n,\phi(x))=\psi(\alpha_{n}(x),e)\epsilon(c(n,x))g^{n}\psi(x,e)^{-1}.$

Since $\alpha_{\pm n}(\phi(x))=\phi(\alpha_{n}(x))$, (2) may be rewritten as

$\displaystyle c^{\prime}(\pm n,x)=\psi(\phi^{-1}(\alpha_{\pm
n}(x)),e)\epsilon(c(n,\phi^{-1}(x)))g^{n}\psi(\phi^{-1}(x),e)^{-1}.$

or simply

(3) $\displaystyle c^{\prime}(n,x)=\psi^{\prime}(\alpha_{n}(x))\epsilon(c(\pm
n,\phi^{-1}(x)))g^{\pm n}\psi^{\prime}(x)^{-1}~{}\mbox{for all
$n\in{\mathbb{Z}}$ and $x\in X$,}$

where $\psi^{\prime}(x):=\psi(\phi^{-1}(x),e)$.

Now, let $c\equiv e\in F$, the neutral element in $F$, then (3) is the same as
saying $c^{\prime}$ is cohomologous to a group homomorphism ${\mathbb{Z}}\ni
n\mapsto g^{\pm n}\in C(F)$.

(iv) Set $c\equiv e$. We are left to construct $\alpha$ and $c$ such that
$\widetilde{\alpha}\overset{conj.}{\not\sim}\widetilde{\alpha}^{\prime}$. It
suffices to make sure the above identity (3) fails.

Since $C(F)$ is trivial, we deduce $g=e$. It suffices to find some $\alpha$
and a cocycle $c^{\prime}:{\mathbb{Z}}\times X\to F$ (w.r.t. $\alpha$) which
is not a coboundary.

By Cauchy theorem, we may find some non-trivial $t\in F$ such that
${\mathbb{Z}}/{p{\mathbb{Z}}}\cong\langle t\rangle\subset F$ for some prime
number $p$. By Lemma 2.3, it suffices to find some $\alpha$ and a continuous
cocycle $c^{\prime}:{\mathbb{Z}}\times X\to{\mathbb{Z}}/{p{\mathbb{Z}}}$ which
is not a coboundary.

We can apply Proposition 3.1 or Proposition 3.4 to construct a suitable free
minimal and weakly mixing or free minimal odometer action
${\mathbb{Z}}\curvearrowright X$ and a continuous cocycle $c$ which is not a
coboundary. ∎

###  4.2. Rigidity part

We are ready to prove Theorem 1.3.

###### Proof of Theorem 1.3.

Let $D_{\infty}={\mathbb{Z}}\rtimes\frac{{\mathbb{Z}}}{2{\mathbb{Z}}}=\langle
s\rangle\rtimes\langle t\rangle$. Denote by $S$ the symmetric generating set
$\\{s^{\pm},t^{\pm}\\}$ for $D_{\infty}$. Denote by $|\cdot|_{S}$ the word
length on $G$ with respect to $S$. Note that $|s^{n}t|_{S}=n+1$ and
$|s^{n}|_{S}=n$. Let $d(\cdot,\cdot)$ be the right-invariant word metric on
$D_{\infty}$ with respect to $S$, i.e.
$d(g_{1},g_{2}):=|g_{2}g_{1}^{-1}|_{S}$.

Let $c:D_{\infty}\times X\to D_{\infty}$ be the continuous orbit cocycle
associated to a given continuous orbit equivalence.

Claim 1: for all $x\in X$, either
$\sup_{n\in{\mathbb{Z}}}d(c(s^{n},x),s^{n})<\infty$ or
$\sup_{n\in{\mathbb{Z}}}d(c(s^{n},x),s^{-n})<\infty$.

###### Proof of Claim 1.

Define $\pi:{\mathbb{Z}}\to D_{\infty}$ by setting $\pi(s^{2n})=s^{n}$ and
$\pi(s^{2n+1})=ts^{n}$ for all $n\in{\mathbb{Z}}$. One can verify that $\pi$
is a bi-Lipschitz bijection. In fact, $\frac{|n-m|}{2}\leq
d(\pi(s^{n}),\pi(s^{m}))\leq 2|n-m|$ for all $n$, $m\in{\mathbb{Z}}$.

Fix any $x\in X$, write $\phi_{x}(g)=c(g,x)$ for all $g\in D_{\infty}$. Define
$\phi:{\mathbb{Z}}\to{\mathbb{Z}}$ by setting
$\phi=\pi^{-1}\circ\phi_{x}\circ\pi$. Clearly, $\phi$ is again a bi-Lipschitz
bijection by Proposition 2.2.

From Theorem 2.7, we deduce that either
$\sup_{n\in{\mathbb{Z}}}|\phi(n)-n|<\infty$ or
$\sup_{n\in{\mathbb{Z}}}|\phi(n)+n|<\infty$. This implies Claim 1 holds. ∎

Define $X_{+}=\\{x\in X:\sup_{n\in{\mathbb{Z}}}d(c(s^{n},x),s^{n})<\infty\\}$
and $X_{-}=\\{x\in X:\sup_{n\in{\mathbb{Z}}}d(c(s^{n},x),s^{-n})<\infty\\}$.
Clearly, $X_{+}\cap X_{-}=\emptyset$ since $s$ has infinite order. From Claim
1, we know that $X=X_{+}\sqcup X_{-}$.

Claim 2: Both $X_{+}$ and $X_{-}$ are clopen subsets of $X$.

###### Proof of Claim 2.

We follow the idea while dealing with ${\mathbb{Z}}$-actions in [BT]. For each
$r\geq 1$, define $B(r):=\\{ts^{i},s^{i}:~{}|i|\leq r\\}\subset D_{\infty}$.
Note that $B(r)\nearrow D_{\infty}$ as $r\to\infty$.

First, for all $N>0$, there exists $K>0$ such that $B(N)\subset c(B(K),x)$ for
all $x\in X$. Indeed, since $D_{\infty}\ni g\mapsto c(g,x)\in D_{\infty}$ is a
bijection, we may find $K_{x}>0$ such that $B(N)\subset c(B(K_{x}),x)$ for all
$x\in X$. Moreover, by continuity of $c$, we may find a small open
neighborhood $V_{x}\ni x$ such that $B(N)\subset c(B(K_{x}),y)$ for all $y\in
V_{x}$. Since $X=\cup_{x\in X}V_{x}$, compactness of $X$ implies we may find a
finite subcover $X=\cup_{i=1}^{n}V_{x_{i}}$, set $K=\max_{1\leq i\leq
n}K_{x_{i}}$.

Let $N:=\max_{x\in X,g\in S}|c(g,x)|$, and $K$ as above for this $N$. Then for
all $x\in X$, either

$c(s^{>K},x)\subseteq\\{s^{>0},ts^{>0}\\}~{}\mbox{and}~{}c(s^{<-K},x)\subseteq\\{s^{<0},ts^{<0}\\}~{}\mbox{or}$
$c(s^{>K},x)\subseteq\\{s^{<0},ts^{<0}\\}~{}\mbox{and}~{}c(s^{<-K},x)\subseteq\\{s^{>0},ts^{>0}\\}.$

Here, $c(s^{>K},x):=\\{c(s^{i},x):i>K\\}$, and $c(s^{-K},x)$ is similarly
defined; $\\{s^{>0},ts^{>0}\\}:=\\{s^{i},ts^{j}:i>0,~{}j>0\\}$ and
$\\{s^{<0},ts^{<0}\\}$ is similarly defined.

To see this, for each $|k|>K$, we have $c(s^{k},x)\not\in B(N)$ as $c(-,x)$ is
bijective. Since $|c(s^{\pm 1},s^{k}x)|\leq N$, from $c(s^{\pm
1+k},x)=c(s^{\pm 1},s^{k}x)c(s^{k},x)$, we deduce that $c(s^{\pm 1+k},x)$ and
$c(s^{k},x)$ must have the same sign for the $s$-exponent when writing them as
elements in $t{\mathbb{Z}}=t\langle s\rangle$ or ${\mathbb{Z}}=\langle
s\rangle$. Then apply Claim 1 to see that if
$c(s^{>K},x)\subseteq\\{s^{>0},ts^{>0}\\}$ holds, then $x\in X_{+}$ and thus
$c(s^{-K},x)\subseteq\\{s^{<0},ts^{<0}\\}$ is automatic.

Now, denote by $X_{+}^{\prime}:=\\{x\in
X:c(s^{>K},x)\subseteq\\{s^{>0},ts^{>0}\\}~{}\mbox{and}~{}c(s^{<-K},x)\subseteq\\{s^{<0},ts^{<0}\\}\\}$
and $X_{-}^{\prime}:=\\{x\in
X:c(s^{>K},x)\subseteq\\{s^{<0},ts^{<0}\\}~{}\mbox{and}~{}c(s^{<-K},x)\subseteq\\{s^{>0},ts^{>0}\\}\\}$.

From above, we have shown $X_{+}^{\prime}\sqcup X_{-}^{\prime}=X=X_{+}\sqcup
X_{-}$. It is easy to see that $X_{+}^{\prime}=X_{+}$ and
$X_{-}^{\prime}=X_{-}$. Hence, to finish the proof, we just need to observe
that both $X_{+}^{\prime}$ and $X_{-}^{\prime}$ are closed as $c$ is
continuous and $D_{\infty}$ is countable and discrete. ∎

Define $a:{\mathbb{Z}}\times X\to D_{\infty}$ by setting

$\displaystyle a(s^{n},x)=\begin{cases}c(s^{n},x)s^{-n},&~{}\mbox{if $x\in
X_{+}$}\\\ c(s^{n},x)s^{n},&~{}\mbox{if $x\in X_{-}$}.\end{cases}$

Claim 3: $a$ is a continuous cocycle taking finitely many values as $n$
changes.

###### Proof of Claim 3.

From Claim 1, we know $a$ takes only finitely many values.

Fix any $x\in X_{+}$ and $n\in{\mathbb{Z}}$. We have

$\displaystyle
a(s^{m+n},x)=c(s^{m+n},x)s^{-m-n}=c(s^{m},s^{n}x)c(s^{n},x)s^{-n}s^{-m}$
$\displaystyle=\begin{cases}a(s^{m},s^{n}x)s^{m}a(s^{n},x)s^{-m},~{}&\mbox{if
$s^{n}x\in X_{+}$}\\\ a(s^{m},s^{n}x)s^{-m}a(s^{n},x)s^{-m},~{}&\mbox{if
$s^{n}x\in X_{-}$}.\end{cases}$

Case 1: $s^{n}x\in X_{+}$.

Since both $a(s^{m+n},x)$ and $a(s^{m},s^{n}x)$ lie in a finite set as $m$
changes, we deduce that $\\{s^{m}a(s^{n},x)s^{-m}:m\in{\mathbb{Z}}\\}$ is a
finite set. Thus, we have $a(s^{n},x)\in{\mathbb{Z}}$ and thus
$a(s^{m+n},x)=a(s^{m},s^{n}x)a(s^{n},x)$.

Case 2: $s^{n}x\in X_{-}$.

Since both $a(s^{m+n},x)$ and $a(s^{m},s^{n}x)$ lie in a finite set as $m$
changes, we deduce that $\\{s^{-m}a(s^{n},x)s^{-m}:m\in{\mathbb{Z}}\\}$ is a
finite set. Thus, $a(s^{n},x)\in{\mathbb{Z}}t$. Once again, this implies that
$a(s^{m+n},x)=a(s^{m},s^{n}x)a(s^{n},x)$. To sum up, we have shown that for
all $x\in X_{+}$, $a(s^{m+n},x)=a(s^{m},s^{n}x)a(s^{n},x)$ holds for all $m$,
$n\in{\mathbb{Z}}$.

Similarly, one can show this also holds for all $x\in X_{-}$. We include the
details for completeness.

Fix any $x\in X_{-}$ and $n\in{\mathbb{Z}}$. We have

$\displaystyle
a(s^{m+n},x)=c(s^{m+n},x)s^{m+n}=c(s^{m},s^{n}x)c(s^{n},x)s^{m+n}$
$\displaystyle=\begin{cases}a(s^{m},s^{n}x)s^{m}a(s^{n},x)s^{m},~{}&\mbox{if
$s^{n}x\in X_{+}$}\\\ a(s^{m},s^{n}x)s^{-m}a(s^{n},x)s^{m},~{}&\mbox{if
$s^{n}x\in X_{-}$}.\end{cases}$

Case i: $s^{n}x\in X_{+}$.

Since $a(s^{m+n},x)$ and $a(s^{m},s^{n}x)$ take finitely many values as $m$
changes, we get that $\\{s^{m}a(s^{n},x)s^{m}:m\in{\mathbb{Z}}\\}$ is a finite
set. Therefore, $a(s^{n},x)\in{\mathbb{Z}}t$, which implies that
$a(s^{m+n},x)=a(s^{m},s^{n}x)a(s^{n},x)$.

Case ii: $s^{n}x\in X_{-}$.

This case can be checked similarly. ∎

Define $D:X\to D_{\infty}$ by

$\displaystyle D(x)=\begin{cases}t,&~{}\mbox{if $x\in X_{-}$}\\\
e,&~{}\mbox{if $x\in X_{+}$.}\end{cases}$

From Claim 2, we know $D$ is a continuous map. Moreover, one can verify that
$D(s^{n}x)a(s^{n},x)D(x)^{-1}\in{\mathbb{Z}}$ for all $x\in X$ and
$n\in{\mathbb{Z}}$.

Indeed, one can check this by discussing four cases depending on $x\in
X_{\pm}$ and $s^{n}x\in X_{\pm}$. We explain the proof for the case $x\in
X_{+}$ and $s^{n}x\in X_{-}$ below, the other three cases can be checked
similarly.

By assumption, there exists a finite subset $F\subset D_{\infty}$ such that
$c(s^{k},x)s^{-k}\in F\ni c(s^{k},s^{n}x)s^{k}$ for all $k\in{\mathbb{Z}}$.
From the cocycle identity $c(s^{n},x)=c(s^{k},s^{n}x)^{-1}c(s^{k+n},x)$, we
deduce that $c(s^{n},x)=s^{k}g_{k}s^{k+n}$ for some $g_{k}\in F^{-1}F$ for all
$k\in{\mathbb{Z}}$. Taking a sufficiently large $k$, we deduce that $g_{k}\in
t{\mathbb{Z}}=t\langle s\rangle$ and thus $c(s^{n},x)\in t\langle s\rangle$.
Hence $D(s^{n}x)a(s^{n},x)D(x)^{-1}=tc(s^{n},x)s^{-n}\in{\mathbb{Z}}=\langle
s\rangle$.

Thus, the map $a^{\prime}:{\mathbb{Z}}\times X\to{\mathbb{Z}}\subset
D_{\infty}$ defined by

$a^{\prime}(s^{n},x):=D(s^{n}x)a(s^{n},x)D(x)^{-1}$

is a continuous cocycle taking finitely many values in ${\mathbb{Z}}$.

Take a ${\mathbb{Z}}$-minimal component, say $X_{0}$, i.e.
$X_{0}=\overline{Orb({\mathbb{Z}},x_{0})}$ for some $x_{0}\in X$ is minimal
w.r.t. the sub ${\mathbb{Z}}$-action. Then since ${\mathbb{Z}}\lhd
D_{\infty}$, we know $tX_{0}=\overline{Orb({\mathbb{Z}},tx_{0})}$ is also a
minimal ${\mathbb{Z}}$-component. Moreover, $X_{0}\cup tX_{0}$ is a
$G$-invariant closed subset, hence $X=X_{0}\cup tX_{0}$ by minimality
assumption. Clearly, either $X_{0}\cap tX_{0}=\emptyset$ or $X_{0}=tX_{0}$.

At this stage, we need to split the proof into two cases.

Case I: $X_{0}=tX_{0}$, i.e. the sub ${\mathbb{Z}}$-action
${\mathbb{Z}}\curvearrowright X$ is still minimal.

We may think of $a^{\prime}$ as a cocycle taking values in ${\mathbb{R}}$ via
the natural inclusion ${\mathbb{Z}}\hookrightarrow{\mathbb{R}}$. As we assume
${\mathbb{Z}}\curvearrowright X$ is still minimal, we may apply Gottschalk-
Hedlund theorem, i.e. Theorem 2.8 and Lemma 2.3 to deduce there exists some
continuous map $L:X\to{\mathbb{Z}}$ such that
$a^{\prime}(s^{n},x)=L(s^{n}x)^{-1}L(x)$. Thus, letting
$L^{\prime}(x)=L(x)D(x)$, we deduce that

$a(s^{n},x)=L^{\prime}(s^{n}x)^{-1}L^{\prime}(x).$

Therefore,

$\displaystyle c(s^{n},x)$
$\displaystyle=\begin{cases}L^{\prime}(s^{n}x)^{-1}L^{\prime}(x)s^{n},~{}&\mbox{if
$x\in X_{+}$}\\\ L^{\prime}(s^{n}x)^{-1}L^{\prime}(x)s^{-n},~{}&\mbox{if $x\in
X_{-}$}\end{cases}$
$\displaystyle=L^{\prime}(s^{n}x)^{-1}s^{n}L^{\prime}(x)~{}\mbox{for all $x\in
X$ and $n\in{\mathbb{Z}}$.}$

Here, the last equality can be checked by observing that $L(x)\in{\mathbb{Z}}$
which is abelian and $t$ acts on ${\mathbb{Z}}$ as a reflection.

Claim 4: there exists some $k\in{\mathbb{Z}}$ such that
$c(t,x)=L^{\prime}(tx)^{-1}(s^{k}t)L^{\prime}(x)$ for all $x\in X$.

###### Proof of Claim 4.

Notice that on the one hand,
$c(s^{-n},x)=L^{\prime}(s^{-n}x)^{-1}s^{-n}L^{\prime}(x)$; on the other hand,

$\displaystyle
c(s^{-n},x)=c(ts^{n}t^{-1},x)=c(t,s^{n}t^{-1}x)c(s^{n},t^{-1}x)c(t^{-1},x)$
$\displaystyle=c(t,s^{n}t^{-1}x)L^{\prime}(s^{n}t^{-1}x)^{-1}s^{n}L^{\prime}(t^{-1}x)c(t^{-1},x).$

Thus,

(4) $\displaystyle
s^{-n}(L^{\prime}(x)c(t^{-1},x)^{-1}L^{\prime}(t^{-1}x)^{-1})s^{-n}=L^{\prime}(s^{-n}x)c(t,s^{n}t^{-1}x)L^{\prime}(s^{n}t^{-1}x)^{-1}.$

Since the right hand side of the above takes only finitely many values as $n$
changes, we deduce that
$L^{\prime}(x)c(t^{-1},x)^{-1}L^{\prime}(t^{-1}x)^{-1}\in{\mathbb{Z}}t$.

Write $L^{\prime}(x)c(t^{-1},x)^{-1}L^{\prime}(t^{-1}x)^{-1}=s^{k(x)}t$ for
some $k(x)\in{\mathbb{Z}}$. Clearly, $X\ni x\mapsto k(x)\in{\mathbb{Z}}$ is
continuous from definition. Next, from (4), we may deduce that
$c(t^{-1},x)^{-1}=L^{\prime}(x)^{-1}s^{k(x)}tL^{\prime}(t^{-1}x)$ and
$c(t,s^{n}t^{-1}x)=L^{\prime}(s^{-n}x)^{-1}s^{k(x)}tL^{\prime}(s^{n}t^{-1}x)$.

Since $t^{2}=e$, the above implies that $k(x)=k(s^{n}x)$ for all
$n\in{\mathbb{Z}}$ and $x\in X$. Since the sub ${\mathbb{Z}}$-action is
minimal and $k(\cdot)$ is continuous, we deduce that $k(x)\equiv k$ for all
$x\in X$. ∎

To sum up, we have shown that

$\displaystyle
c(s^{n},x)=L^{\prime}(s^{n}x)^{-1}s^{n}L^{\prime}(x),~{}c(t,x)=L^{\prime}(tx)^{-1}s^{k}tL^{\prime}(x)~{}\mbox{for
all}~{}n\in{\mathbb{Z}}~{}\mbox{and all}~{}x\in X.$

Clearly, this is equivalent to say $c$ is cohomologous to $\phi\in
Aut(D_{\infty})$, where $\phi$ is given by $\phi(s)=s$ and $\phi(t)=s^{k}t$ as
mentioned in Lemma 2.4.

Finally, we can apply Proposition 4.4 in [Li] to conclude that the two actions
are conjugate.

Case II: $X_{0}\cap tX_{0}=\emptyset$, i.e. $X=X_{0}\sqcup tX_{0}$ and hence
$X_{0}$ is a clopen subset.

We apply Gottschalk-Hedlund theorem to the minimal subactions
${\mathbb{Z}}\curvearrowright X_{0}$ and ${\mathbb{Z}}\curvearrowright tX_{0}$
respectively. Argue similarly as in Case I, we find two continuous maps
$L^{\prime\prime}:X_{0}\to D_{\infty}$ and $L^{\prime}:tX_{0}\to D_{\infty}$
such that

(5) $\displaystyle
c(s^{n},x)=\begin{cases}L^{\prime\prime}(s^{n}x)^{-1}s^{n}L^{\prime\prime}(x),~{}\mbox{for
all $x\in X_{0}$ and $n\in{\mathbb{Z}}$},\\\
L^{\prime}(s^{n}x)^{-1}s^{n}L^{\prime}(x),~{}\mbox{for all $x\in tX_{0}$ and
$n\in{\mathbb{Z}}$}.\end{cases}$

Fix any $x\in X_{0}$, we compute as follows:

$\displaystyle c(s^{-n},x)$
$\displaystyle=L^{\prime\prime}(s^{-n}x)^{-1}s^{-n}L^{\prime\prime}(x)$
$\displaystyle c(s^{-n},x)$ $\displaystyle=c(ts^{n}t^{-1},x)$
$\displaystyle=c(t,s^{n}t^{-1}x)c(s^{n},t^{-1}x)c(t^{-1},x)$
$\displaystyle=c(t,s^{n}t^{-1}x)[L^{\prime}(s^{n}t^{-1}x)^{-1}s^{n}L^{\prime}(t^{-1}x)]c(t^{-1},x).$

Thus, we deduce

$L^{\prime\prime}(s^{-n}x)c(t,s^{n}t^{-1}x)L^{\prime}(s^{n}t^{-1}x)^{-1}=s^{-n}[L^{\prime\prime}(x)c(t^{-1},x)^{-1}L^{\prime}(t^{-1}x)^{-1}]s^{-n}.$

As
$\\{L^{\prime\prime}(s^{-n}x)c(t,s^{n}t^{-1}x)L^{\prime}(s^{n}t^{-1}x)^{-1}:n\in{\mathbb{Z}}\\}$
is a finite set, we deduce that

$\displaystyle
L^{\prime\prime}(x)c(t^{-1},x)^{-1}L^{\prime}(t^{-1}x)^{-1}=L^{\prime\prime}(s^{-n}x)c(t,s^{n}t^{-1}x)L^{\prime}(s^{n}t^{-1}x)^{-1}\in{\mathbb{Z}}t~{}\mbox{for
all $n\in{\mathbb{Z}}$}.$

Let us write

(6) $\displaystyle
L^{\prime\prime}(x)c(t^{-1},x)^{-1}L^{\prime}(t^{-1}x)^{-1}=s^{k(x)}t$

for some map $k:X_{0}\to{\mathbb{Z}}$. Clearly, $k(\cdot)$ is a continuous
map. Moreover, since $X_{0}$ is ${\mathbb{Z}}=\langle s\rangle$-invariant,
from the above identity, we deduce that

$\displaystyle
L^{\prime\prime}(s^{-n}x)c(t^{-1},s^{-n}x)^{-1}L^{\prime}(t^{-1}s^{-n}x)^{-1}$
$\displaystyle=s^{k(s^{-n}x)}t$
$\displaystyle[L^{\prime\prime}(s^{-n}x)c(t,s^{n}t^{-1}x)L^{\prime}(s^{n}t^{-1}x)^{-1}]^{-1}$
$\displaystyle=[s^{k(x)}t]^{-1}.$

Multiply the above two expressions and use $t^{2}=e$, we deduce that
$e=s^{k(s^{-n}x)-k(x)}$, i.e. $k(s^{-n}x)=k(x)$ for all $n\in{\mathbb{Z}}$.
Since $X_{0}$ is ${\mathbb{Z}}=\langle s\rangle$-minimal, we deduce that $k$
is constant, say $k(x)=k$ for all $x\in X_{0}$. Then, (6) can be simplified to

$\displaystyle
L^{\prime}(tx)=L^{\prime}(t^{-1}x)=t^{-1}s^{-k}L^{\prime\prime}(x)c(t^{-1},x)^{-1}.$

From (5), a calculation shows that for each $x\in X_{0}$,

$\displaystyle c(s^{n},tx)$
$\displaystyle=L^{\prime}(s^{n}tx)^{-1}s^{n}L^{\prime}(tx)=L^{\prime}(ts^{-n}x)^{-1}s^{n}L^{\prime}(tx)$
$\displaystyle=[t^{-1}s^{-k}L^{\prime\prime}(s^{-n}x)c(t^{-1},s^{-n}x)^{-1}]^{-1}s^{n}[t^{-1}s^{-k}L^{\prime\prime}(x)c(t^{-1},x)^{-1}]$
$\displaystyle=c(t^{-1},s^{-n}x)L^{\prime\prime}(s^{-n}x)^{-1}s^{-n}L^{\prime\prime}(x)c(t^{-1},x)^{-1}.$

In fact, using cocycle identity, we may compute the full expression for $c$.
More precisely, for all $x\in X_{0}$, we have

(7)
$\displaystyle\begin{split}c(s^{n},x)&=L^{\prime\prime}(s^{n}x)^{-1}s^{n}L^{\prime\prime}(x),\\\
c(s^{n}t,x)&=c(t^{-1},s^{-n}x)L^{\prime\prime}(s^{-n}x)^{-1}s^{-n}L^{\prime\prime}(x),\\\
c(s^{n},tx)&=c(t^{-1},s^{-n}x)L^{\prime\prime}(s^{-n}x)^{-1}s^{-n}L^{\prime\prime}(x)c(t^{-1},x)^{-1},\\\
c(s^{n}t,tx)&=L^{\prime\prime}(s^{n}x)^{-1}s^{n}L^{\prime\prime}(x)c(t,tx).\end{split}$

Now, to avoid confusion, let us write $\beta$ for the 2nd action, i.e.
$gx=\beta_{c(g,x)}(x)$ for all $g\in D_{\infty}$ and $x\in X$. Define a map

$\displaystyle X=X_{0}\sqcup tX_{0}$
$\displaystyle\overset{\phi}{\longrightarrow}X$ $\displaystyle\phi(x)$
$\displaystyle=\beta_{L^{\prime\prime}(x)}(x)$ $\displaystyle\phi(tx)$
$\displaystyle=\beta_{t^{-1}L^{\prime\prime}(x)}(x)~{}\mbox{for all}~{}x\in
X_{0}.$

Clearly, $\phi$ is continuous as $X_{0}$ is clopen and both $c$ and
$L^{\prime\prime}:X_{0}\to D_{\infty}$ are continuous.

Claim 5: $\phi$ is the desired conjugacy between the two actions, i.e. $\phi$
is a homeomorphism and $\phi(gx)=\beta_{g}(\phi(x))$ for all $g\in D_{\infty}$
and $x\in X$.

###### Proof of Claim 5.

The conjugacy identity can be checked by computation using
$\beta_{c(g,x)}(x)=gx$ and the above expression for $c(g,x)$ in (7). We are
left to show that $\phi$ is a bijection.

(a) $\phi$ is injective.

First, we check $\phi|_{X_{0}}$ is injective. Indeed, take any $x,y\in X_{0}$,
suppose $\phi(x)=\phi(y)$, that is,
$\beta_{L^{\prime\prime}(x)}(x)=\beta_{L^{\prime\prime}(y)}(y)$, i.e.
$\beta_{L^{\prime\prime}(y)^{-1}L^{\prime\prime}(x)}(x)=y$.

Since $D_{\infty}\ni g\mapsto c(g,x)\in D_{\infty}$ is a bijection by
Proposition 2.2, we may find some $g\in D_{\infty}$ such that
$c(g,x)=L^{\prime\prime}(y)^{-1}L^{\prime\prime}(x)$. Thus,
$y=\beta_{c(g,x)}(x)=gx$. Since $x,y\in X_{0}$ and $X_{0}\cap
tX_{0}=\emptyset$, we deduce that $g\in{\mathbb{Z}}$, say $g=s^{n}$. Hence,
$c(s^{n},x)=L^{\prime\prime}(s^{n}x)^{-1}L^{\prime\prime}(x)$. In view of (7),
we deduce $s^{n}=e$, i.e. $x=y$.

Clearly, this also shows that $\phi|_{tX_{0}}$ is injective.

We are left to show that $\phi(x)\neq\phi(ty)$ for any $x,y\in X_{0}$.

Assume not, then
$\beta_{L^{\prime\prime}(x)}(x)=\beta_{t^{-1}L^{\prime\prime}(y)}(y)$, i.e.
$\beta_{L^{\prime\prime}(y)^{-1}tL^{\prime\prime}(x)}(x)=y$. We may find some
$g\in D_{\infty}$ such that
$c(g,x)=L^{\prime\prime}(y)^{-1}tL^{\prime\prime}(x)$. Hence,
$y=\beta_{c(g,x)}(x)=gx$. So $g\in{\mathbb{Z}}$, say $g=s^{n}$. Thus,
$c(s^{n},x)=L^{\prime\prime}(s^{n}x)^{-1}tL^{\prime\prime}(x)$. In view of
(7), this implies $s^{n}=t$, a contradiction.

(b) $\phi$ is surjective.

First, we observe that the 2nd action also has two distinct minimal
${\mathbb{Z}}$-components. Suppose not, we may apply the proof of Case I to
the 2nd action to see the first action is conjugate to the 2nd one. Thus,
these two actions must have the same number of minimal
${\mathbb{Z}}$-components (as each automorphism of $D_{\infty}$ must fix the
subgroup ${\mathbb{Z}}=\langle s\rangle$ globally), but this contradicts our
assumption in Case II.

Let us write the two minimal ${\mathbb{Z}}$-components of the 2nd action as
$Y_{0}$ and $\beta_{t}(Y_{0})$, i.e. $X=Y_{0}\sqcup\beta_{t}(Y_{0})$.

Now, since $\phi(gx)=\beta_{g}(\phi(x))$ for all $g\in D_{\infty}$ and all
$x\in X$, we know $\phi(X_{0}),\phi(tX_{0})\in\\{Y_{0},\beta_{t}(Y_{0})\\}$.
As $\phi$ is injective and $X_{0}\cap tX_{0}=\emptyset$, we deduce that
$\\{\phi(X_{0}),\phi(tX_{0})\\}=\\{Y_{0},\beta_{t}(Y_{0})\\}$, so $\phi$ is
surjective. ∎

∎

######  Remark 4.2.

By [CJ, Co], we know that for every finitely generated one-ended group $G$,
its full shifts $G\curvearrowright A^{G}$ for finite $A$ are continuous
cocycle superrigid actions with respect to any countable target groups. It was
mentioned in [CJ, Corollary 5] that this can be combined with [Li, Theorem
1.6] to deduce the full shifts as above are continuous orbit equivalence rigid
actions if $G$ is further assumed to be torsion free and amenable. In fact,
this further assumption is unnecessary. Indeed, we just observe that from
continuous cocycle superrigidity, we deduce $c(g,x)=L(gx)^{-1}\phi(g)L(x)$.
Then take $x$ to be any fixed point for this full shift action and use
$g\mapsto c(g,x)$ is a bijection for all $x\in X$, we deduce that
$g\mapsto\phi(g)$ is automatically a group isomorphism of $G$. Hence, we may
just apply [Li, Proposition 4.4] instead to get the conclusion.

##  5\. concluding remarks

Here are several remarks on the theorems and possible generalizations.

(1) Let $F$ be any non-trivial finite abelian group. From the proof of
Proposition 4.1, we know that if ${\mathbb{Z}}\curvearrowright X$ is a minimal
action admitting a continuous cocycle $c:{\mathbb{Z}}\times X\to F$ which is
not cohomologous to a group homomorphism from ${\mathbb{Z}}$ to $C(F)$, the
center of $F$, then we can construct two continuously orbit equivalent but not
conjugate actions.

In fact, for each non-trivial finite abelian group $F$, such a minimal action
${\mathbb{Z}}\curvearrowright X$ and a cocycle as above do exist (by taking
${\mathbb{Z}}\curvearrowright X$ to be a well-chosen Toeplitz system [Gl,
Example 1.20], as shown to me by Prof. Lemańczyk). By combining this with
Lemma 2.3, we can deduce that for each finite non-trivial group $F$,
$F\times{\mathbb{Z}}$ admits two continuously orbit equivalent but not
conjugate minimal actions.

(2) The construction in the proof of item (ii) Proposition 4.1 can be modified
to deal with a general semi-direct product group
$G=F\rtimes_{\sigma}{\mathbb{Z}}$ for a finite group $F$. Since
$Aut(F\rtimes_{\sigma}{\mathbb{Z}})$ becomes more involved and the existence
of minimal actions that admit skew cocycles that are not trivial into finite
groups is not clear to us, we do not study the full generality here.

(3) Let $G=D_{\infty}$. Then we take a ${\mathbb{Z}}$-minimal component, say
$X_{0}$, i.e. $X_{0}=\overline{Orb({\mathbb{Z}},x_{0})}$ for some $x_{0}\in X$
is minimal w.r.t. the sub ${\mathbb{Z}}$-action. Then since ${\mathbb{Z}}\lhd
G$, we know $tX_{0}=\overline{Orb({\mathbb{Z}},tx_{0})}$ is also a minimal
${\mathbb{Z}}$-component. Moreover, $X_{0}\cup tX_{0}$ is a $G$-invariant
closed subset, hence $X=X_{0}\cup tX_{0}$ if we assume $G\curvearrowright X$
is minimal.

We observe below that once we have two minimal ${\mathbb{Z}}$-components, then
the $G$-action is an induced action.

######  Proposition 5.1.

With the above notations and assumptions, if $X_{0}\cap tX_{0}=\emptyset$,
then the action $G\curvearrowright X=X_{0}\sqcup tX_{0}$ is conjugate to the
induced action $G\curvearrowright G/{{\mathbb{Z}}}\times X_{0}$ associated to
a cocycle $\delta:G\times G/{{\mathbb{Z}}}\to{\mathbb{Z}}$.

###### Proof.

Indeed, consider the natural lift map $L:G/{{\mathbb{Z}}}\to G$ given by
$L({\mathbb{Z}})=e$ and $L(t{\mathbb{Z}})=t$. Then for the associated cocycle
$\delta:G\times G/{\mathbb{Z}}\to{\mathbb{Z}}$ given by
$\delta(g,g^{\prime}{\mathbb{Z}})=L(gg^{\prime}{\mathbb{Z}})^{-1}gL(g^{\prime}{\mathbb{Z}})$
satisfies that

$\displaystyle\delta(s^{n},{\mathbb{Z}})=s^{n},~{}\delta(s^{n},t{\mathbb{Z}})=s^{-n},~{}\delta(s^{n}t,{\mathbb{Z}})=s^{-n}~{}\mbox{and}~{}\delta(s^{n}t,t{\mathbb{Z}})=s^{n}~{}\mbox{for
all $n\in{\mathbb{Z}}$.}$

Now, define $\psi:X\to G/{{\mathbb{Z}}}\times X_{0}$ by setting

$\displaystyle\psi(x)=\begin{cases}({\mathbb{Z}},x)&\mbox{if $x\in X_{0}$}\\\
(t{\mathbb{Z}},tx)&\mbox{if $x\in tX_{0}$.}\end{cases}$

Clearly, $\psi$ is a bijection and continuous (as $X_{0}$ is clopen). One can
check that $\psi(gx)=g\psi(x)$ for all $g\in G$ and $x\in X$.

Hence, $\psi$ is a conjugacy between $G\curvearrowright X=X_{0}\sqcup tX_{0}$
and the induced action $G\curvearrowright G/{{\mathbb{Z}}}\times X_{0}$ for a
free minimal action ${\mathbb{Z}}\curvearrowright X_{0}$ (i.e. the sub
${\mathbb{Z}}$-action) w.r.t. the above chosen cocycle $c$. ∎

Next, consider any two induced actions $G\curvearrowright
G/{{\mathbb{Z}}}\times X_{0}$ from two minimal topologically free
${\mathbb{Z}}$-actions on $X_{0}$ defined above and let $\phi$ be a
homeomorphism which induces a continuous orbit equivalence between these two
actions. Suppose we know that $\phi=id$, then one can show the two actions are
conjugate directly. This shows the main difficulty for dealing with Case II in
the proof of Theorem 1.3 is that we do not know, a priori, how the
homeomorphism behaves on the two minimal ${\mathbb{Z}}$-components.

Before proving the above assertion, let us fix some notation for the induced
actions. Fix any lift $L:G/{{\mathbb{Z}}}\to G$, i.e.
$L(g{\mathbb{Z}}){\mathbb{Z}}=g{\mathbb{Z}}$ for all $g\in G$. We may further
assume $L({\mathbb{Z}})=e$ and $L(t{\mathbb{Z}})=t$. Then let $\delta:G\times
G/{{\mathbb{Z}}}\to{\mathbb{Z}}$ be the associated cocycle
$\delta(g,s{\mathbb{Z}})=L(gs{\mathbb{Z}})^{-1}gL({\mathbb{Z}})$. Observe that
$\delta(s,{\mathbb{Z}})=L(s{\mathbb{Z}})^{-1}sL({\mathbb{Z}})=s$ for all
$s\in{\mathbb{Z}}$.

Fix a minimal topologically free action $\alpha:{\mathbb{Z}}\curvearrowright
X_{0}$, then the induced action $\tilde{\alpha}:G\curvearrowright
X=G/{{\mathbb{Z}}}\times X_{0}$ is defined as follows:
$\tilde{\alpha}_{s}(g{\mathbb{Z}},x):=(sg{\mathbb{Z}},\alpha_{\delta(s,g{\mathbb{Z}})}(x))$
for all $s,g\in G$ and $x\in X_{0}$.

Now it suffices to show the following holds.

######  Proposition 5.2.

Let $\alpha,\beta:{\mathbb{Z}}\curvearrowright X_{0}$ be two minimal
topologically free actions. Let
$\tilde{\alpha},\tilde{\beta}:G\curvearrowright X:=G/{{\mathbb{Z}}}\times
X_{0}$ be the associated induced actions as recalled above. Then the following
hold.
(1) $\tilde{\alpha}\overset{coe}{\sim}\tilde{\beta}~{}\mbox{via the identity
homeomorphism}\Rightarrow\alpha\overset{coe}{\sim}\beta$.
(2)
$\alpha\overset{coe}{\sim}\beta\Rightarrow\tilde{\alpha}\overset{conj.}{\sim}\tilde{\beta}$.

###### Proof.

(1) Let $c:G\times X\to G$ be the continuous orbit cocycle w.r.t.
$\tilde{\alpha}$, i.e.
$\tilde{\alpha}_{g}(\tilde{x})=\tilde{\beta}_{c(g,\tilde{x})}(\tilde{x})$,
where $\tilde{x}=(s{\mathbb{Z}},x)\in X$ is any point. Then a calculation
using the definition of the induced actions shows that

$\displaystyle
gs{\mathbb{Z}}=c(g,\tilde{x})s{\mathbb{Z}},~{}~{}\alpha_{\delta(g,s{\mathbb{Z}})}(x)=\beta_{\delta(c(g,\tilde{x}),s{\mathbb{Z}})}(x),~{}\forall~{}g,s\in
G,~{}\forall~{}x\in X_{0}.$

Notice that this implies that $c(g,\tilde{x}){\mathbb{Z}}=g{\mathbb{Z}}$ as
${\mathbb{Z}}\lhd G$. In particular, $c(s,\tilde{x})\in{\mathbb{Z}}$ for all
$s\in{\mathbb{Z}}$.

Now, we define the following map $\theta:{\mathbb{Z}}\times
X_{0}\to{\mathbb{Z}}$ given by $\theta(s,x)=c(s,\tilde{x})$, where
$\tilde{x}:=({\mathbb{Z}},x)$.

Clearly, $\theta$ is well-defined and continuous. Then, we verify that
$\theta$ is a cocycle w.r.t. $\alpha$.

Take any $s_{1},s_{2}\in{\mathbb{Z}}$, and $x\in X_{0}$, we have

$\displaystyle\theta(s_{1}s_{2},x)=c(s_{1}s_{2},\tilde{x})=c(s_{1},\tilde{\alpha}_{s_{2}}(\tilde{x}))c(s_{2},\tilde{x})$
$\displaystyle\theta(s_{1},\alpha_{s_{2}}(x))\theta(s_{2},x)=c(s_{1},\widetilde{\alpha_{s_{2}}(x)})c(s_{2},\tilde{x}).$

It suffices to check that
$\tilde{\alpha}_{s_{2}}(\tilde{x})=\widetilde{\alpha_{s_{2}}(x)}$. Recall that
$\tilde{x}=({\mathbb{Z}},x)$ and
$\tilde{\alpha}_{s_{2}}(\tilde{x})=(s_{2}{\mathbb{Z}},\alpha_{\delta(s_{2},{\mathbb{Z}})}x)=({\mathbb{Z}},\alpha_{s_{2}}(x))$.
Meanwhile, $\widetilde{\alpha_{s_{2}}(x)}=({\mathbb{Z}},\alpha_{s_{2}}(x))$.

Now, let us check that $\theta$ gives us a coe between $\alpha$ and $\beta$.

Indeed, we have

$\displaystyle\alpha_{s}(x)$
$\displaystyle=\alpha_{\delta(s,{\mathbb{Z}})}(x)=\beta_{\delta(c(s,\tilde{x}),{\mathbb{Z}})}(x)$
$\displaystyle=\beta_{L(c(s,\tilde{x}){\mathbb{Z}})^{-1}c(s,\tilde{x})L({\mathbb{Z}})}(x)$
$\displaystyle=\beta_{L(s{\mathbb{Z}})^{-1}\theta(s,x)}(x)$
$\displaystyle=\beta_{\theta(s,x)}(x),~{}\forall~{}s\in{\mathbb{Z}},~{}\forall~{}x\in
X_{0}.$

(2) Let $\theta:{\mathbb{Z}}\times X_{0}\to{\mathbb{Z}}$ be the cocycle from
the coe. By [BT], we know that $\theta(s,x)=f(\alpha_{s}(x))^{-1}s^{\pm}f(x)$
for some continuous map $f:X\to{\mathbb{Z}}$. One can check $X_{0}\ni
x\mapsto\beta_{f(x)}(x)\in X_{0}$ intertwines $\alpha_{s}$ with
$\beta_{s^{\pm}}$, which implies
$Y\ni(g{\mathbb{Z}},x)\mapsto(g{\mathbb{Z}},\beta_{f(x)}(x))\in X$ intertwines
$\tilde{\alpha_{g}}$ with $\tilde{\beta}_{\tau(g)}$. Here, $\tau\in Aut(G)$ is
defined as follows:

$\displaystyle\tau=\begin{cases}id,~{}\mbox{ if
$\theta(s,x)=f(\alpha_{s}(x))^{-1}sf(x)$ for all $s\in{\mathbb{Z}}$}\\\
\phi,~{}\mbox{ if $\theta(s,x)=f(\alpha_{s}(x))^{-1}s^{-1}f(x)$ for all
$s\in{\mathbb{Z}}$}\end{cases}$

Here, $\tau\in Aut(G)$ is determined by $\tau(s)=s^{-1}$ for all
$s\in{\mathbb{Z}}$ and $\tau(t)=t$ for the reflection $t\in G$. ∎

### Acknowledgements

This work is partially supported by NSFC grant no. 12001081. He is grateful to
Prof. Nhan-Phu Chung and Prof. Xin Li for useful discussion, to Prof. Xin Li
for sharing his unpublished note on coe rigidity dating back to 2016. He also
thanks Prof. Mariusz Lemańczyk and Prof. Zhengxing Lian for very helpful
correspondence on cocycles. He also thanks the referee for his/her helpful
suggestion which improves the presentation.

## References


and consider heterogeneous data: for two opposite nodes in the graph ($v=0$
and $v=25$ _e.g._), $a_{v}$ are $\mathcal{N}(0,1)$ and $b_{v}$ uniform in
$[0,1]$ (and the functions are re-normalized), while they are taken equal to
$0$ in the rest of the graph. In Figure 1(a), the graph is well connected and
Assumption 1 is verified for a small enough $\bar{\sigma}^{2}$, so that all
algorithms perform comparatively well; MC-SGD because of the homogeneity, the
gossip-based ones and Walkman thanks to the connectivity. However, decreasing
the connectivity and increasing data heterogeneity in a pathological way, we
obtain Figure 1(b): MC-SGD fails due to function-heterogeneity
($\bar{\sigma}^{2}$ too big), while the three others are slowed down by their
communication inefficiency, illustrating how depending only on $\tau_{\rm
hit}$, MC-SAGA outperforms the other algorithms, that rather depend on
$n\tau_{\rm mix}$ or $n\tau_{\rm mix}^{2}$.

(a) Geometric graph, homogeneous functions

(b) Cycle graph, heterogeneous functions

Figure 1: Comparison of MC-SAGA and MC-SGD with existing algorithms. In
abscissa are the total number of communications, and in ordinate
$f(\mathbf{x})-f(\mathbf{x}^{\star})$.