University of Wrocław

Doctoral Thesis

Ellis groups in model theory and strongly generic sets

Adam Malinowski

Advisor: prof. Ludomir Newelski

Wrocław 2023

## Abstract

Assume $G$ is a group and $\mathcal{A}$ is an algebra of subsets of $G$ closed
under left translation. We study various ways to understand the Ellis group of
the $G$-flow $S(\mathcal{A})$ (the Stone space of $\mathcal{A}$), with
particular interest in the model-theoretic setting where $G$ is definable in a
first order structure $M$ and $\mathcal{A}$ consists of externally definable
subsets of $G$.

In one part of the thesis we explore strongly generic sets. Maximal algebras
of such sets are shown to carry enough information to retrieve the Ellis
group. A subset of $G$ is strongly generic if each non-empty Boolean
combination of its translates is generic. Trivial examples include what we
call _periodic_ sets, which are unions of cosets of finite index subgroups of
$G$. We give several characterizations of strongly generic sets, in
particular, we relate them to almost periodic points of the flow $2^{G}$. For
groups without a smallest finite index subgroup we show how to construct non-
periodic strongly generic subsets in a systematic way. When $G$ is definable
in a model $M$, a definable, strongly generic subset of $G$ will remain as
such in any elementary extension of $M$ only if it is strongly generic in $G$
in an adequately uniform way. Sets satisfying this condition are called
_uniformly strongly generic_. We analyse a few examples of these sets in
different groups.

In the second part we assume that $G$ is a topological group and consider a
particular algebra of its subsets denoted $\mathcal{SBP}$. It consists of
subsets of $G$ that have the _strong Baire property_ , meaning nowhere dense
boundary. We explicitly describe the Ellis group of $S(\mathcal{A})$ for an
arbitrary subalgebra $\mathcal{A}$ of $\mathcal{SBP}$ under varying
assumptions on the group $G$, including the case when $G$ is a compact
topological group. We use this description to relate the Ellis groups computed
for a model and its elementary extension in particular scenarios. Under some
of those assumptions we also decide whether the obvious inclusions between the
families of strongly generic, uniformly strongly generic and periodic sets can
be reversed. These results can be applied in o-minimal structures, in which
externally definable subsets are proved to have the strong Baire property.
Finally, we propose a procedure of finding a maximal generic algebra in a
given subalgebra of $\mathcal{SBP}$ given that we succeeded in doing so while
neglecting nowhere dense sets.

### Acknowledgements

I would like to express deep gratitude to my advisor, Ludomir Newelski. His
expert guidance, limitless engagement and incredibly dry wit have earned my
truest appreciation.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Topological dynamics
    2. 2.2 Boolean algebras and Stone spaces
    3. 2.3 Alternative representation of the Ellis semigroup
    4. 2.4 Image algebras
  3. 3 Strongly generic sets
    1. 3.1 Basic properties
    2. 3.2 Concrete examples
    3. 3.3 Uniformly strongly generic sets
  4. 4 The Ellis group across models
    1. 4.1 Groups with profinite topology
    2. 4.2 Compact groups
    3. 4.3 Precompact groups
    4. 4.4 Types at infinity
    5. 4.5 Regular open sets
  5. A Explicit description of an Ellis group
    1. A.1 A variant
  6. B USG families

##  1 Introduction

The idea of applying topological dynamics in model theory is due to Newelski
[New09, New12b], who suggested it could serve to extend the results from
stable group theory to unstable context. It was later explored and broadened
by several authors, see e.g. [Pil13, GPP14, Jag15, CS18, KNS19]. One area of
study is to investigate dynamical objects, such as the Ellis group,111The
phrase _Ellis group_ is now a common name in model theory for an object
defined in Section 2. associated with any definable group. From a model-
theoretic perspective an important question arises: to what extent, if any,
are these objects preserved when computed for different models of the same
theory? This dissertation aims to examine some constructions and techniques
related to the issue, among which are the strongly generic sets.

Assume $G$ is a group definable in a first-order structure $M$. The space
$S_{G}(M)$, the Stone space of the algebra of $M$-definable subsets of $G$, is
naturally a $G$-flow. In case $M$ is stable, $S_{G}(M)$ is canonically
isomorphic to the enveloping (Ellis) semigroup of itself and there is a unique
minimal ideal $\mathcal{M}$ of $S_{G}(M)$, consisting of the generic types of
$G$. $\mathcal{M}$ is also a group isomorphic to $G/G^{0}$, where $G^{0}$ is
the connected component of $G$. However, when $M$ is not stable, $S_{G}(M)$ is
typically not isomorphic to its Ellis semigroup and we need to consider
$S_{\mathrm{ext},G}(M)$ instead, which is the Stone space of the algebra of
_externally_ definable subsets of $G$. As there is usually little relation
between subsets externally definable in a model $M$ and its elementary
extension $M^{*}$, it remains unclear whether any properties of the Ellis
groups defined with respect to the corresponding $G$-flows,
$S_{\mathrm{ext},G}(M)$ and $S_{\mathrm{ext},G}(M^{*})$, are preserved.

An idea to overcome this is to express the Ellis group in terms of objects
that are closer to definability, so that their properties will carry over
between models. An algebra of subsets of $G$ closed under left translation is
called a _$G$ -algebra_. It was observed in [New12a] that to any Ellis group
we can assign an _image algebra_ , which is a particular $G$-algebra of
externally definable subsets of $G$. Its unique property is that it consists
of generic sets, i.e. subsets of $G$ for which we can find finitely many
translates covering $G$. It is easy to find examples of generic sets, but it
is quite unusual for them to form non-trivial $G$-algebras. A set generating a
$G$-algebra consisting only of generic sets (except the empty set) is thereby
called _strongly generic_.

Newelski proved that any maximal $G$-algebra consisting of externally
definable generic sets is an image algebra and any minimal ideal is determined
by image algebras up to homeomorphism. With a little bit more work, also the
algebraic structure of the Ellis group can be retrieved from strongly generic
sets. We present the details in Section 2. On the other hand, strongly generic
sets feature a certain kind of regularity which ideally could reveal itself
uniformly across many models of the same theory. For instance, Newelski proved
in [New12a] that a strongly generic definable subset of a stable group must be
_periodic_ , i.e. a union of cosets of a finite index subgroup. These
properties lead us to believe that strongly generic sets are worth a study on
their own. Ultimately this may result in a discovery of some interesting
connections between Ellis groups of a model and its elementary extension.

We also consider a strenghtening of the notion of strongly generic sets, which
are the uniformly strongly generic sets. These are precisely the sets that
remain strongly generic in every elementary extension, taking it one step
closer to the realization of the idea from the previous paragraph. However,
not every strongly generic set is uniformly strongly generic. In fact, we
prove that certain kinds of groups, including the compact groups, are
guaranteed to have a strongly generic subset that is not uniformly strongly
generic.

Our second idea is localization. Given a large Boolean algebra $\mathcal{A}$,
any ultrafilter $p\in S(\mathcal{A})$ is determined by the family of its
restrictions $p\cap\mathcal{A}_{i}$ to subalgebras $\mathcal{A}_{i}$ of
$\mathcal{A}$ that jointly generate it. For example, it is common in stability
theory to view a type $p\in S(M)$ as a consistent collection of
$\Delta$-types, where $\Delta$ is a finite set of formulas. This
representation is sufficient to capture vital properties of types, such as
forking or genericity, by means of their local ranks. It is easy to show, as
we do in Section 4, that the Ellis group of $S_{\mathrm{ext},G}(M)$ can be
similarly decomposed into simpler, local parts. Thereby a potentially fruitful
approach to understanding the Ellis group is to identify and study its
tractable fragments with respect to this decomposition. Assuming that $G$ is a
topological group, we explore one such fragment, corresponding to the algebra
$\mathcal{SBP}$ of all sets having what we name the _strong Baire property_
(abbreviated SBP), in Section 4.

The main contribution of the thesis consists in:

  * –

Establishing a correspondence between strongly generic sets in an arbitrary
group $G$ and almost periodic points in the flow $2^{G}$; see Theorem 3.12.

  * –

Providing a wide range of examples of sets that are strongly generic but not
periodic; see Proposition 3.17 together with Proposition 3.21.

  * –

Providing and analysing a few examples of uniformly strongly generic sets that
are not periodic; see Examples 3.31, 3.32, 3.34.

  * –

Discovering the algebra $\mathcal{SBP}$, which allows of a full description of
the Ellis group in multiple scenarios; see Theorems 4.6, 4.18, 4.29.
Furthermore, identifying the boundary of the techniques used to give that
description; see Subsection 4.4.

  * –

Proving that every infinite compact Hausdorff group has a strongly generic
subset with SBP that is not periodic; see Theorem 4.25.

  * –

Showing that externally definable subsets of densely ordered o-minimal
structures have SBP, making the previous results applicable to groups
definable in such structures; see Proposition 4.33 and Corollary 4.34.

  * –

Revealing a correspondence between maximal generic subalgebras of
$\mathcal{SBP}$ and maximal generic subalgebras of the algebra of regular open
subsets of $G$; see Theorem 4.57.

The dissertation is organized as follows. Section 2 serves as a reminder of
known theory that is the starting point of our study. We review classical
topics such as topological dynamics and Stone duality, then proceed to recount
the work from [New12a, New12b] that motivates our interest in strongly generic
sets. The next two sections contain our original results. Section 3 explores
the basic properties and characterizations of strongly generic sets and
describes an organized way to construct them. It also introduces uniformly
strongly generic sets and gives a few interesting examples. In Section 4 we
compute the Ellis group of an arbitrary subalgebra of $\mathcal{SBP}$ under
varying assumptions on the group $G$. Next we explain why our approach fails
for other groups. Finally, we propose a technique of obtaining maximal generic
algebras in $\mathcal{SBP}$ from like algebras consisting of regular open
subsets of $G$.

##  2 Preliminaries

###  2.1 Topological dynamics

In this subsection we recall the relevant notions and results from topological
dynamics. For a comprehensive study on the subject the reader is directed to
[Aus88]. Throughout the subsection $G$ is a fixed (discrete) group.

A $G$_-flow_ is a compact (Hausdorff) topological space $X$ on which $G$ acts
by homeomorphisms.222The classical definition of a $G$-flow is more general,
assuming that $G$ is a topological group and the action need be jointly
continuous. However, in our usage of the notion $G$ will always be treated as
discrete. A _subflow_ is a subset $Y\subseteq X$ closed both topologically and
under the action of $G$. For any point $p\in X$ there is a smallest subflow of
$X$ containing $p$, namely $\operatorname{cl}(G\cdot p)$. A minimal subflow is
a non-empty subflow without a non-empty proper subflow. By compactness and the
Zorn’s lemma, any non-empty subflow of $X$ contains a minimal subflow. A point
$p\in X$ is called _almost periodic_ if it belongs to any minmal subflow, or
equivalently, if $\operatorname{cl}(G\cdot p)$ is a minimal subflow. A
$G$_-flow morphism_ is a continuous function between $G$-flows preserving the
action of $G$. It is a $G$_-flow isomorphism_ if it has an inverse which is a
morphism.

Below we state some basic facts about $G$-flows to be used later.

######  Remark 2.1.

Let $f_{1},f_{2}:X\to Y$ be $G$-flow morphisms. Then the set

$X_{0}=\\{x\in X:f_{1}(x)=f_{2}(x)\\}$

is a subflow of $X$.

######  Fact 2.2.

Assume $f:X\to Y$ is a $G$-flow morphism.

  1. (i)

If $X_{0}\subseteq X$ is a subflow, then $f[X_{0}]\subseteq Y$ is a subflow.

  2. (ii)

If $Y_{0}\subseteq Y$ is a subflow, then $f^{-1}[Y_{0}]\subseteq X$ is a
subflow.

  3. (iii)

If $X_{0}\subseteq X$ is a minimal subflow, then $f[X_{0}]\subseteq Y$ is a
minimal subflow.

  4. (iv)

If $p\in X$ is almost periodic, then $f(p)\in Y$ is almost periodic.

  5. (v)

If $q\in Y$ is almost periodic and $f$ is onto (or just $q\in f[X]$), then
$q=f(p)$ for some almost periodic $p\in X$.

###### Proof.

(i) – (iv) are easy. To prove (v), let $Y_{0}=\operatorname{cl}(G\cdot q)$ and
$X_{0}=f^{-1}[Y_{0}]$. Then $X_{0}\subseteq X$ is a non-empty subflow, so it
contains a minimal subflow $X_{1}$. We have that $f[X_{1}]\subseteq Y_{0}$ is
a subflow, so in fact $f[X_{1}]=Y_{0}$ by minimality of $Y_{0}$. In
particular, there is $p\in X_{1}$ such that $f(p)=q$. ∎

######  Remark 2.3.

A non-empty subflow $X_{0}\subseteq X$ is minimal if and only if for each open
$U\subseteq X$ satisfying $U\cap X_{0}\neq\varnothing$ there exist
$g_{1},\ldots,g_{n}\in G$ such that $X_{0}\subseteq g_{1}U\cup\ldots\cup
g_{n}U$.

###### Proof.

$({\implies})$ The set $X_{0}\setminus G\cdot U$ is a proper subflow of
$X_{0}$, so it is empty, i.e. $X_{0}\subseteq G\cdot U$. The conclusion
follows from compactness.

$({\impliedby})$ Assume for contradiction that there is a non-empty proper
subflow $X_{1}\subseteq X_{0}$ and take an open set $U\subseteq X$ such that
$U\cap X_{0}\neq\varnothing$ and $U\cap X_{1}=\varnothing$. Then $gU\cap
X_{1}=\varnothing$ for each $g\in G$, which contradicts the assumption. ∎

######  Definition 2.4.

Assume $X$ is a $G$-flow.

  1. (i)

A subset $U\subseteq X$ is _generic_ if $X=\bigcup_{i=1}^{n}g_{i}U$ for some
$g_{1},\ldots,g_{n}\in G$.

  2. (ii)

A point $p\in X$ is _generic_ if every open neighbourhood of $p$ is generic.

######  Fact 2.5 ([New09, Lemma 1.7, Corollary 1.9]).

A generic point $p\in X$ exists if and only if there is a unique minimal
subflow $X_{0}\subseteq X$. In this case $X_{0}$ consists precisely of the
generic points in $X$.

###### Proof.

Since distinct minimal subflows are clearly disjoint, it suffices to show that
a point $p\in X$ is generic if and only if $p$ belongs to every minimal
subflow of $X$. For one direction, assume that $p\in X$ is generic and
$p\notin X_{0}$ for some minimal subflow $X_{0}\subseteq X$. The set
$U=X\setminus X_{0}$ is an open neighbourhood of $p$ closed under the action
of $G$, so $p$ is not generic, which is a contradiction.

For the other direction, assume a non-generic point $p$ belongs to every
minimal subflow and take an open neighbourhood $U$ of $p$ that is not generic.
By compactness, $X_{1}:=X\setminus G\cdot U$ is a non-empty subflow of $X$, so
it contains a minimal subflow $X_{0}\subseteq X_{1}$. Clearly $p\notin X_{0}$,
a contradiction. ∎

Assume $X$ is a $G$-flow. For any $g\in G$, we denote by $\pi_{g}:X\to X$ the
homemorphism given by the action of $G$ on $X$. Regarding $\\{\pi_{g}:g\in
G\\}$ as a subset of $X^{X}$ endowed with the product topology, the
_enveloping semigroup_ or _Ellis semigroup_ $E(X)$ is the closure of this
subset. $E(X)$ equipped with function composition is a semigroup with identity
$\pi_{e}$. It is moreover a _left topological semigroup_ , meaning that the
semigroup operation is continuous in the first coordinate. $E(X)$ is also a
$G$-flow itself with the action $g\cdot f=\pi_{g}\circ f$, where $g\in G,f\in
E(X)$.

More generally, assume $S$ is a left topological semigroup with identity. A
subset $I\subseteq S$ is a (left) ideal, written $I\trianglelefteqslant S$,
provided that $S\cdot I\subseteq I$. It is a minimal ideal, denoted
$I\trianglelefteqslant_{m}S$, if it is minimal among the non-empty ideals. For
every element $a\in S$ there is a smallest ideal containing $a$, namely
$S\cdot a$. Moreover if $I\trianglelefteqslant_{m}S$, then $I\cdot a=I$ for
each $a\in I$. When $S$ is compact, it is routine to show that any ideal
contains a minimal ideal. An element $u\in S$ is called an idempotent if it
satisfies $u^{2}=u$.

######  Remark 2.6.

  1. (i)

For $f\in E(X)$, the smallest subflow of $E(X)$ containing $f$ equals the
smallest ideal containing $f$.

  2. (ii)

In $E(X)$, the minimal subflows and minimal ideals coincide.

###### Proof.

(i) The function $r_{f}:E(X)\to E(X)$ defined by $r_{f}(h)=h\circ f$ is
continuous, so by compactness of $E(X)$,

$\operatorname{cl}(G\cdot f)=\operatorname{cl}r_{f}[\\{\pi_{g}:g\in
G\\}]=r_{f}[\operatorname{cl}\\{\pi_{g}:g\in G\\}]=r_{f}[E(X)]=E(X)\circ f.$

(ii) For $Y\subseteq E(X)$, we have that

$\displaystyle Y\text{ is a minimal subflow}$ $\displaystyle\iff(\forall y\in
Y)\,Y=\operatorname{cl}(G\cdot y),$ $\displaystyle Y\text{ is a minimal
ideal}$ $\displaystyle\iff(\forall y\in Y)\,Y=E(X)\circ y.$

The conclusion follows from (i). ∎

Below we formulate a fundamental theorem of Ellis:

######  Theorem 2.7 ([Ell69]).

Assume $S$ is a compact left topological semigroup.

  1. (i)

Given $I\trianglelefteqslant_{m}S$, the set $J(I)$ of idempotents in $I$ is
non-empty.

  2. (ii)

For every $I\trianglelefteqslant_{m}S$ and $u\in J(I)$, the set $uI$ is a
group and $I$ is a disjoint union of the groups $uI$, where $u\in J(I)$.

  3. (iii)

The groups $uI$, where $I\trianglelefteqslant_{m}S$ and $u\in J(I)$, are all
isomorphic.

  4. (iv)

For each $I,J\trianglelefteqslant_{m}S$ and $u\in J(I)$ there is $v\in J(J)$
such that $uv=v$ and $vu=u$.

In particular, the theorem applies to $E(X)$. In the context of model theory
the groups $uI$, into which any minimal ideal splits, are called _ideal
subgroups_ , and their common isomorphism type is the _Ellis group_ of the
$G$-flow $X$.

The following facts will be used for explicit Ellis group computation:

######  Lemma 2.8.

Assume $\mathcal{G}$ is a group, $I\trianglelefteqslant_{m}E(X)$ and
$\varphi:I\to\mathcal{G}$ is a semigroup epimorphism such that
$\varphi^{-1}[\\{e\\}]=J(I)$. Then the Ellis group of $X$ is isomorphic to
$\mathcal{G}$.

###### Proof.

Take any $u\in J(I)$. We will check that $\varphi\restriction uI$ is an
isomorphism of groups $uI\to\mathcal{G}$. It is clearly injective because
$\ker(\varphi\restriction uI)=uI\cap J(I)=\\{u\\}$. For the proof of
surjectivity, fix $g\in\mathcal{G}$ and pick $q\in I$ such that
$\varphi(q)=g$. Then $uq\in uI$ and $\varphi(uq)=\varphi(u)\varphi(q)=g$. ∎

######  Corollary 2.9.

Assume $\mathcal{G}$ is a group, $I\trianglelefteqslant_{m}E(X)$ and
$\varphi:E(X)\to\mathcal{G}$ is a semigroup epimorphism such that
$I\cap\varphi^{-1}[\\{e\\}]=J(I)$. Then the Ellis group of $X$ is isomorphic
to $\mathcal{G}$.

###### Proof.

By Lemma 2.8 it suffices to show that $\varphi[I]=\mathcal{G}$. Take any $u\in
J(I)$, fix $g\in G$ and pick $f\in E(X)$ satisfying $\varphi(f)=g$. Then
$fu\in I$ and

$\varphi(fu)=\varphi(f)\varphi(u)=\varphi(f)=g.\qed$

In the last lemma and corollary the respective assumptions
$\varphi^{-1}[\\{e\\}]=J(I)$ and $I\cap\varphi^{-1}[\\{e\\}]=J(I)$ each can be
replaced with a left-to-right inclusion ($\subseteq$), since the reverse
inclusion holds automatically.

###  2.2 Boolean algebras and Stone spaces

Here we recall the correspondence (formally: dual equivalence of categories)
between Boolean algebras and Stone spaces, which we will generally refer to as
_Stone duality_. A _Stone space_ is a topological space which is compact
Hausdorff and totally disconnected, meaning it has a basis of sets that are
both closed and open.333We call such sets _clopen_. To any Boolean algebra
$\mathcal{A}$ we assign the space of ultrafilters on $\mathcal{A}$, called the
_Stone space of_ $\mathcal{A}$ and denoted as $S(\mathcal{A})$. For
$A\in\mathcal{A}$, define

$[A]=\\{p\in S(\mathcal{A}):A\in p\\}.$

The set $S(\mathcal{A})$ equipped with the topology generated by
$\\{[A]:A\in\mathcal{A}\\}$ is a Stone space. On the other hand, when $X$ is a
topological space, we assign to it the family $\mathcal{CO}(X)$ of clopen
subsets of $X$, which is a Boolean algebra.

When $S$ is a set and $\mathcal{A}$ is an algebra of subsets of $S$, for $s\in
S$, let

$\hat{s}=\\{A\in\mathcal{A}:s\in A\\}.$

It is an ultrafilter on $\mathcal{A}$ which we call the _principal ultrafilter
corresponding to_ $s$. If additionally $S=X$ is a Stone space and
$\mathcal{A}=\mathcal{CO}(X)$, the map $x\mapsto\hat{x}$ is a homemorphism
between $X$ and $S(\mathcal{CO}(X))$. On the other hand, if $\mathcal{A}$ is a
Boolean algebra, the map $A\mapsto[A]$ is an isomorphism between $\mathcal{A}$
and $\mathcal{CO}(S(\mathcal{A}))$.

Let $\mathcal{A},\mathcal{B}$ be Boolean algebras. To a homomorphism
$\varphi:\mathcal{A}\to\mathcal{B}$ we assign the continuous function
$\varphi^{*}:S(\mathcal{B})\to S(\mathcal{A})$, called the _dual map of_
$\varphi$, given by the condition $A\in\varphi^{*}(q)\iff\varphi(A)\in q$.
Furthermore, to a continuous function $f:X\to Y$ between topological spaces
$X,Y$ we assign the homomorphism $f^{*}:\mathcal{CO}(Y)\to\mathcal{CO}(X)$,
also called the dual map of $f$, defined as $f^{*}(B)=f^{-1}[B]$. These
assignments make a pair of contravariant functors between the category of
Boolean algebras and the category of Stone spaces. So we have
$(\varphi\circ\psi)^{*}=\psi^{*}\circ\varphi^{*}$ etc.

When $\mathcal{CO}(S(\mathcal{A}))$ is identified with $\mathcal{A}$ via the
isomorphism defined above, the map
$(\varphi^{*})^{*}:\mathcal{CO}(S(\mathcal{A}))\to\mathcal{CO}(S(\mathcal{B}))$
is identical to $\varphi:\mathcal{A}\to\mathcal{B}$. If $X$ and $Y$ are Stone
spaces, the same holds the other way: for $f:X\to Y$, $(f^{*})^{*}$ is
identical to $f$ after the identification of $S(\mathcal{CO}(X))$ with $X$.
Hence the pair of functors establishes dual equivalence of the categories of
Boolean algebras and Stone spaces.

######  Fact 2.10.

Assume $\varphi:\mathcal{A}\to\mathcal{B}$ is a homomorphism of Boolean
algebras and consider the dual map $\varphi^{*}:S(\mathcal{B})\to
S(\mathcal{A})$.

  1. (i)

If $\varphi$ is injective, then $\varphi^{*}$ is surjective.

  2. (ii)

If $\varphi$ is surjective, then $\varphi^{*}$ is injective.

The same fact holds for any continuous function $f:X\to Y$ between Stone
spaces and its dual. It follows that both implications are actually
equivalences.

######  Fact 2.11.

Assume $\varphi:\mathcal{A}\to\mathcal{B}$ is a homomorphism of Boolean
algebras. Then the image of $\varphi^{*}:S(\mathcal{B})\to S(\mathcal{A})$
consists precisely of all $p\in S(\mathcal{A})$ such that
$p\cap\operatorname{Ker}\varphi=\varnothing$. ∎

Assume $G$ is a group. A Boolean algebra on which $G$ acts by automorphisms
will be called a $G$_-algebra_. A $G$_-algebra homomorphism_ is a Boolean
algebra homomorphism preserving the group action. When $\mathcal{A}$ is a
$G$-algebra, there is a natural structure of a $G$-flow on $S(\mathcal{A})$,
namely $g\cdot q=\\{gA:A\in q\\}$ for $g\in G,q\in S(\mathcal{A})$.
Conversely, when $X$ is a topological space on which $G$ acts by
homeomorphisms (such as a $G$-flow), the action $g\cdot A=\\{g\cdot p:p\in
A\\}$, where $g\in G,A\in\mathcal{CO}(X)$, defines the structure of a
$G$-algebra on $\mathcal{CO}(X)$. All facts related to Stone duality naturally
extend to $G$-flows and $G$-algebras.

Assume $\mathcal{A}$ is a $G$-algebra. When a subset
$\mathcal{B}\subseteq\mathcal{A}$ is closed under all $G$-algebra operations,
we call it a _$G$ -subalgebra_, denoted as $\mathcal{B}\leqslant\mathcal{A}$,
and regard it as a $G$-algebra with the induced structure. In particular,
given $\mathcal{B}\subseteq\mathcal{P}(G)$, we write
$\mathcal{B}\leqslant\mathcal{P}(G)$ when $\mathcal{B}$ is closed under union,
complement and left translation, and regard it as a $G$-algebra equipped with
these operations. On the other hand, sometimes $\mathcal{B}$ already has a
structure of a $G$-algebra and $\mathcal{B}\subseteq\mathcal{A}$, in which
case we write $\mathcal{B}\leqslant\mathcal{A}$ to indicate that the
structures coincide, or $\mathcal{B}\not\leqslant\mathcal{A}$ when they are
different.444This happens in Subsection 4.5:
$\mathbb{RO}\subseteq\mathcal{SBP}$ are $G$-algebras, but
$\mathbb{RO}\not\leqslant\mathcal{SBP}$.

Sometimes we simply write _algebra_ instead of $G$-algebra. Whenever we speak
of an algebra without the action of $G$, we use the full term _Boolean
algebra_.

The $G$-algebras have a natural variant of the first isomorphism theorem:

######  Fact 2.12.

Assume $\varphi:\mathcal{A}\to\mathcal{B}$ is a $G$-algebra homomorphism.

  1. (i)

$\operatorname{Ker}\varphi=\\{A\in\mathcal{A}:\varphi(A)=0\\}$ is a $G$-ideal
in $\mathcal{A}$, meaning an ideal closed under the action of $G$.

  2. (ii)

$\operatorname{Im}\varphi=\\{\varphi(A):A\in\mathcal{A}\\}$ is a
$G$-subalgebra of $\mathcal{B}$.

  3. (iii)

If $I\trianglelefteqslant\mathcal{A}$ is any $G$-ideal, there is a natural
$G$-algebra structure on the quotient $\mathcal{A}/I$ and a natural quotient
epimorphism $\pi:\mathcal{A}\to\mathcal{A}/I$.

  4. (iv)

There is a unique isomorphism
$\overline{\varphi}:\mathcal{A}/\operatorname{Ker}\varphi\to\operatorname{Im}\varphi$
such that $\varphi$ decomposes as $\mathcal{A}\xrightarrow{\ \pi\
}\mathcal{A}/\operatorname{Ker}\varphi\xrightarrow{\ \overline{\varphi}\
}\operatorname{Im}\varphi$.

In the following remark and corollary, $\mathcal{A}\leqslant\mathcal{P}(G)$ is
arbitrary.

######  Remark 2.13.

The orbit of $\hat{e}$ is dense in $S(\mathcal{A})$, i.e.
$S(\mathcal{A})=\operatorname{cl}(G\cdot\hat{e})$.

###### Proof.

We have that $G\cdot\hat{e}=\\{\hat{g}:g\in G\\}$, which is clearly dense in
$S(\mathcal{A})$. ∎

######  Corollary 2.14.

$G$-flow morphisms $f_{1},f_{2}:S(\mathcal{A})\to Y$ are equal if and only if
$f_{1}(\hat{e})=f_{2}(\hat{e})$.

###### Proof.

Follows directly from Remarks 2.1 and 2.13. ∎

###  2.3 Alternative representation of the Ellis semigroup

In the next two subsections we present the relevant portions of [New12a,
New12b] together with some thus far unpublished folklore. Several proofs have
been reworked with the use of Stone duality, which we think offers a valuable
perspective on the subject.

Assume that $G$ is a group and $\mathcal{A}\leqslant\mathcal{P}(G)$. Thus
$X=S(\mathcal{A})$ is a Stone space and a $G$-flow.

######  Definition 2.15 ([New12b], before Lemma 1.2).

For $A\in\mathcal{A}$ and $q\in S(\mathcal{A})$, let

$d_{q}(A):=\\{h\in G:A\in hq\\}=\\{h\in G:h^{-1}A\in q\\}.$ (1)

It is easy to check that for a fixed $q\in S(\mathcal{A})$, the function
$d_{q}:\mathcal{A}\to\mathcal{P}(G)$ is a $G$-algebra homomorphism. It need
not always be the case that $d_{q}(A)\in\mathcal{A}$. If, however,
$d_{q}(A)\in\mathcal{A}$ for all $q\in S(\mathcal{A})$ and $A\in\mathcal{A}$,
we say that $\mathcal{A}$ is $d$_-closed_.

######  Remark 2.16.

The definition $(\ref{eq:dop})$ also makes sense (and defines a $G$-algebra
homomorphism) whenever $q\in S(\mathcal{B})$, where $\mathcal{B}$ is a
$G$-algebra containing $\mathcal{A}$, particularly for $q\in\beta
G=S(\mathcal{P}(G))$. Fix such $\mathcal{B}$ and let
$r=r_{\mathcal{B},\mathcal{A}}:S(\mathcal{B})\to S(\mathcal{A})$ be the dual
map of the inclusion $\mathcal{A}\to\mathcal{B}$, which we call the
_restriction_. Then for any $A\in\mathcal{A},q\in S(\mathcal{B})$ we have that

$d_{q}(A)=d_{r(q)}(A),$

which implies that the function $d_{q}:\mathcal{A}\to\mathcal{A}$ depends only
on $r(q)=q\cap\mathcal{A}$.

More generally:

######  Fact 2.17.

Assume $\varphi:\mathcal{A}\to\mathcal{B}$ is a homomorphism of $G$-algebras,
$A\in\mathcal{A}$ and $p\in S(\mathcal{B})$. Then
$d_{p}(\varphi(A))=d_{\varphi^{*}(p)}(A)$. ∎

######  Definition 2.18.

Assume $\mathcal{A}$ is $d$-closed and $p,q\in S(\mathcal{A})$. The
multiplication on $S(\mathcal{A})$ is defined by

$p\ast q:=\\{A\in\mathcal{A}:d_{q}(A)\in p\\}.$ (2)

Again, it can be checked directly (see [New12b]) that $p\ast q\in
S(\mathcal{A})$ and $*$ is associative and continuous in the first coordinate,
so $(S(\mathcal{A}),\ast)$ is a left topological semigroup. Below we prove it
using the Stone duality.

The operations $d_{q}$ and $\ast$ can be understood as follows. For $q\in
S(\mathcal{A})$, denote as $r_{q}:G\to S(\mathcal{A})$ the function
$r_{q}(g)=g\cdot q$. When $G$ is given the discrete topology and the action of
itself by left translation, $r_{q}$ becomes a morphism between topological
spaces on which $G$ acts by homeomorphisms. Thus from Stone duality we get the
$G$-algebra homomorphism
$r_{q}^{*}:\mathcal{CO}(S(\mathcal{A}))\to\mathcal{P}(G)$, defined by
$r_{q}^{*}(C)=r_{q}^{-1}[C]$. After the standard identification $r_{q}^{*}$
can be thought of as a map $\mathcal{A}\to\mathcal{P}(G)$, whereafter it is
identical to $d_{q}$. It follows that $d_{q}$ is a $G$-algebra homomorphism.

In case $\mathcal{A}$ is $d$-closed, we have that
$d_{q}:\mathcal{A}\to\mathcal{A}$ and we denote the dual $G$-flow morphism
$d_{q}^{*}:S(\mathcal{A})\to S(\mathcal{A})$ as $\hat{r}_{q}$. The maps
$r_{q}$ and $\hat{r}_{q}$ agree on $G$ in the sense that
$\hat{r}_{q}(\hat{g})=r_{q}(g)$ for $g\in G$, thus from now we write $r_{q}$
for both.555We may think of $G$ as a subset of $S(\mathcal{A})$ via the map
$G\to S(\mathcal{A}),g\mapsto\hat{g}$, even though it need not be injective.
We also have that $p\ast q=r_{q}(p)$, hence $\ast$ is continuous in the first
coordinate. Moreover, $r_{q}(r_{p}(\hat{e}))=r_{q}(p)=p\ast q=r_{p\ast
q}(\hat{e})$, where both $r_{q}\circ r_{p}$ and $r_{p\ast q}$ are $G$-flow
morphisms. From Corollary 2.14 we get $r_{p\ast q}=r_{q}\circ r_{p}$, which
means precisely that $\ast$ is associative.

######  Lemma 2.19.

Assume that $\mathcal{A}$ is d-closed, $p,q\in S(\mathcal{A})$ and $A\in
p,B\in q$. Moreover, assume that the set

$AB:=\\{ab:a\in A,b\in B\\}$

belongs to $\mathcal{A}$. Then $AB\in p\ast q$.

###### Proof.

We have that $A\subseteq d_{q}(AB)$ because for any $a\in A$ we have
$B\subseteq a^{-1}AB$, so $a^{-1}AB\in q$. It follows that $d_{q}(AB)\in p$,
i.e. $AB\in p\ast q$. ∎

There is another description of the $d_{q}$ operation:

######  Remark 2.20.

Consider $Y=\mathcal{P}(G)$ as a topological space with topology induced from
the product topology on $2^{G}$ via the natural bijection. It is a Stone space
and a $G$-flow with the action $\varphi_{g}(A):=Ag^{-1}=\\{ag^{-1}:a\in A\\}$,
which we denote here as $\varphi_{g}$ to avoid confusion with ordinary left
translation.

  1. (i)

For $g\in G,A\in\mathcal{A}$ we have that $d_{\hat{g}}(A)=Ag^{-1}$.

  2. (ii)

For a fixed $A\in\mathcal{A}$, the map $f:S(\mathcal{A})\to\mathcal{P}(G)$
defined by $f(q)=d_{q}(A)$ is a $G$-flow morphism.

  3. (iii)

For $A\in\mathcal{A}$ and $q\in S(\mathcal{A})$,

$d_{q}(A)=\lim_{\hat{g}\to q}Ag^{-1}.$

  4. (iv)

For $A\in\mathcal{A}$, we have that $\\{d_{q}A:q\in
S(\mathcal{A})\\}=\operatorname{cl}_{\mathcal{P}(G)}\\{Ag^{-1}:g\in G\\}$.

###### Proof.

(i) By definition

$h\in d_{\hat{g}}(A)\iff h^{-1}A\in\hat{g}\iff g\in h^{-1}A\iff h\in Ag^{-1}.$

(ii) The family $\\{V_{b}:b\in G\\}\cup\\{\mathcal{P}(G)\setminus V_{b}:b\in
G\\}$ is a subbasis of $\mathcal{P}(G)$, where
$V_{b}=\\{B\in\mathcal{P}(G):b\in B\\}$. The map $f$ is continuous as for
$b\in G$:

$f(q)\in V_{b}\iff b\in f(q)\iff b\in d_{q}(A)\iff b^{-1}A\in q\iff
q\in[b^{-1}A],$

so the preimage $f^{-1}[V_{b}]=[b^{-1}A]$ is a clopen set. Moreover, for $g\in
G$,

$\displaystyle h\in f(gq)$ $\displaystyle\iff h\in d_{gq}(A)\iff A\in hgq$
$\displaystyle\iff hg\in d_{q}(A)\iff h\in f(q)g^{-1},$

so $f(gq)=\varphi_{g}\big{(}f(q)\big{)}$.

(iii) Follows from (i) and (ii).

(iv) By Remark 2.13 and the compactness of $S(\mathcal{A})$:

$\displaystyle\\{d_{q}A:q\in S(\mathcal{A})\\}$
$\displaystyle=f[S(\mathcal{A})]=f[\operatorname{cl}\\{\hat{g}:g\in G\\}]$
$\displaystyle=\operatorname{cl}\\{f(\hat{g}):g\in
G\\}=\operatorname{cl}\\{Ag^{-1}:g\in G\\}.\qed$

######  Corollary 2.21.

If $\mathcal{A}$ is d-closed, then it is closed under right translation.

Since by Remark 2.16 the $G$-algebra $\mathcal{A}$ is $d$-closed iff
$d_{q}(A)\in\mathcal{A}$ for each $A\in\mathcal{A},q\in\beta G$, i.e. it is
closed under operations $\\{d_{q}:q\in\beta G\\}$ which do not depend on
$\mathcal{A}$, any intersection of $d$-closed $G$-subalgebras of
$\mathcal{P}(G)$ is again $d$-closed. Thus the $G$-algebra $\mathcal{A}$ has a
$d$_-closure_ $\mathcal{A}^{d}$, the smallest $d$-closed $G$-algebra
containing it, which is the intersection of all such $G$-algebras.

Using this notion we define a useful generalization of (2): the same formula
defines a function $\ast:S(\mathcal{B})\times S(\mathcal{A})\to
S(\mathcal{A})$ whenever $\mathcal{A}$ is a $G$-algebra and $\mathcal{B}$ is a
$G$-algebra containing $\mathcal{A}^{d}$. Again, for $p\in S(\mathcal{B}),q\in
S(\mathcal{A})$, we have that $p\ast q=r_{q}(p)$, where
$r_{q}:S(\mathcal{B})\to S(\mathcal{A})$ is the dual map of
$d_{q}:\mathcal{A}\to\mathcal{B}$, hence $\ast$ is continuous in the first
coordinate. By the same argument as before it is also associative in the sense
that if $\mathcal{C}$ is a $G$-algebra containing $\mathcal{B}^{d}$, then
$(p\ast q)\ast s=p\ast(q\ast s)$ for $p\in S(\mathcal{C}),q\in
S(\mathcal{B}),s\in S(\mathcal{A})$.

The following fact is an explicit description of $d$-closure.

######  Fact 2.22.

$\mathcal{A}^{d}$ is the Boolean algebra generated by the set

$\mathcal{C}=\bigcup_{q\in
S(\mathcal{A})}d_{q}[\mathcal{A}]=\bigcup_{q\in\beta G}d_{q}[\mathcal{A}].$

###### Proof.

Let $\langle\mathcal{C}\rangle$ be the Boolean algebra generated by
$\mathcal{C}$. Clearly $\langle\mathcal{C}\rangle\subseteq\mathcal{A}^{d}$ and
$\mathcal{A}\subseteq\langle\mathcal{C}\rangle$ because
$d_{\hat{e}}=\mathrm{id}$ by Remark 2.20, so it suffices to show that
$\langle\mathcal{C}\rangle\leqslant\mathcal{P}(G)$ and it is d-closed. The set
$\mathcal{C}$ is closed under left translation since
$d_{q}[\mathcal{A}]\leqslant\mathcal{P}(G)$ for any $q\in\beta G$. Thus
$\langle\mathcal{C}\rangle\leqslant\mathcal{P}(G)$. To show that it is
$d$-closed, it suffices to check that $\mathcal{C}$ is closed under $d_{p}$
for $p\in\beta G$, as then $d_{p}[\langle\mathcal{C}\rangle]=\langle
d_{p}[\mathcal{C}]\rangle\subseteq\langle\mathcal{C}\rangle$. Fix $p\in\beta
G$ and $C\in\mathcal{C}$, so that $C=d_{q}(A)$ for some $q\in
S(\mathcal{A}),A\in\mathcal{A}$. By Fact 2.17,
$d_{p}(C)=d_{p}(d_{q}(A))=d_{d_{q}^{*}(p)}(A)=d_{p\ast q}(A)\in\mathcal{C}$,
where $d_{q}:\mathcal{A}\to\mathcal{P}(G)$. ∎

######  Remark 2.23.

Assume $\mathcal{A}\leqslant\mathcal{B}\leqslant\mathcal{P}(G)$ are d-closed
and consider $S(\mathcal{A})$, $S(\mathcal{B})$ as semigroups with $\ast$.
Then the restriction $\pi:S(\mathcal{B})\to S(\mathcal{A})$ is an epimorphism.

###### Proof.

Take $q\in S(\mathcal{B})$. The diagram on the left commutes, hence by Stone
duality, the diagram on the right also commutes:

${\mathcal{B}}$${\mathcal{B}}$${\mathcal{A}}$${\mathcal{A}}$$\scriptstyle{d_{q}}$$\scriptstyle{i}$$\scriptstyle{d_{\pi(q)}}$$\scriptstyle{i}$
${S(\mathcal{B})}$${S(\mathcal{B})}$${S(\mathcal{A})}$${S(\mathcal{A})}$$\scriptstyle{\pi}$$\scriptstyle{r_{q}}$$\scriptstyle{\pi}$$\scriptstyle{r_{\pi(q)}}$

It follows that $\pi(r_{q}(p))=r_{\pi(q)}(\pi(p))$ for $p\in S(\mathcal{B})$,
so $\pi(p\ast q)=\pi(p)\ast\pi(q)$. ∎

The next result, due to Newelski, has not been published before.

######  Theorem 2.24.

The following are isomorphic as semigroups:

  1. (i)

$E(S(\mathcal{A}))$ with the structure described before;

  2. (ii)

$S(\mathcal{A}^{d})$ with $\ast$;

  3. (iii)

$\operatorname{End}_{G}(\mathcal{A}^{d})$, the set of all $G$-algebra
endomorphisms of $\mathcal{A}^{d}$ with function composition;

  4. (iv)

$\operatorname{End}_{G}(S(\mathcal{A}^{d}))^{\mathrm{op}}$, the opposite
semigroup to the set of all $G$-flow endomorphisms of $S(\mathcal{A}^{d})$
with function composition.

Moreover, the first two are isomorphic as $G$-flows via the same isomorphism.

###### Proof.

(i) $\cong$ (ii) We will check that the function $\ell:S(\mathcal{A}^{d})\to
E(S(\mathcal{A}))$ defined by $\ell(p)=\ell_{p}$, where $\ell_{p}(q)=p\ast q$
for $q\in S(\mathcal{A})$, is an isomorphism of semigroups and $G$-flows. For
$g\in G$, we have that $\ell_{\hat{g}}(q)=\hat{g}\ast q=g\cdot q$, so
$\ell(\hat{g})=\pi_{g}$. Since $\\{\hat{g}:g\in G\\}$ is dense in
$S(\mathcal{A}^{d})$, it follows that

$\ell[S(\mathcal{A}^{d})]=\ell\big{[}\operatorname{cl}\\{\hat{g}:g\in
G\\}\big{]}=\operatorname{cl}\\{\ell(\hat{g}):g\in
G\\}=\operatorname{cl}\\{\pi_{g}:g\in G\\}=E(S(\mathcal{A})),$

so $\ell$ is both well defined and surjective. It is also injective since

$\ell(p)(\hat{e})=p\ast\hat{e}=p.$

Since $\ast$ is associative, $\ell$ is a semigroup homomorphism. Furthermore,

$\ell(g\cdot q)=\ell(\hat{g}\ast
q)=\ell(\hat{g})\circ\ell(q)=\pi_{g}\circ\ell(q),$

so $\ell(g\cdot q)=g\cdot\ell(q)$. It remains to prove that $\ell$ is a
homemorphism.

A subbasis of $E(S(\mathcal{A}))$ consists of sets of the form

$V_{q,A}=\\{f\in E(S(\mathcal{A})):f(q)\in[A]\\},$

where $q\in S(\mathcal{A})$ and $A\in\mathcal{A}$. We have that

$\displaystyle\ell(p)\in V_{q,A}\iff\ell_{p}(q)\in[A]\iff A\in p\ast q\iff
d_{q}(A)\in p.$

Hence $\ell^{-1}[V_{q,A}]=[d_{q}(A)]$ is clopen in $S(\mathcal{A}^{d})$, which
implies the continuity of $\ell$. By Fact 2.22, the sets $\\{d_{q}(A):q\in
S(\mathcal{A}),A\in\mathcal{A}\\}$ generate $\mathcal{A}^{d}$ as a Boolean
algebra. It follows that the family $\\{[d_{q}(A)]:q\in
S(\mathcal{A}),A\in\mathcal{A}\\}$ is a subbasis of $S(\mathcal{A}^{d})$,
meaning that $\ell^{-1}$ is also continuous.

(ii) $\cong$ (iv) The function
$r:S(\mathcal{A}^{d})\to\operatorname{End}_{G}(S(\mathcal{A}^{d}))$,
$r(q)=r_{q}=(d_{q})^{*}$, is clearly a well defined semigroup
antihomomorphism. It is injective because if $r(p)=r(q)$, then
$p=r_{p}(\hat{e})=r_{q}(\hat{e})=q$. It remains to show the surjectivity of
$r$. Let $f:S(\mathcal{A}^{d})\to S(\mathcal{A}^{d})$ be a $G$-flow
endomorphism and take $q=f(\hat{e})$. Then $r_{q}:S(\mathcal{A}^{d})\to
S(\mathcal{A}^{d})$ is also a $G$-flow morphism such that $r_{q}(\hat{e})=q$,
so by Corollary 2.14, we have $f=r_{q}$.

(iii) $\cong$ (iv) Follows directly from Stone duality. ∎

###  2.4 Image algebras

In this subsection we continue to assume that $G$ is a group and
$\mathcal{A}\leqslant\mathcal{P}(G)$.

Theorem 2.24 allows us to better understand the structure of ideal subgroups
in $E(S(\mathcal{A}))$. For any $p\in S(\mathcal{A}^{d})$, we consider

$\displaystyle\operatorname{Ker}d_{p}=\\{B\in\mathcal{A}^{d}:d_{p}(B)=\varnothing\\}$
and $\displaystyle\operatorname{Im}d_{p}=\\{d_{p}(B):B\in\mathcal{A}^{d}\\}.$

######  Lemma 2.25.

If $p\in S(\mathcal{A}^{d})$, then

$\operatorname{cl}(G\cdot
p)=\bigcap\\{[B]^{c}:B\in\operatorname{Ker}d_{p}\\}.$

###### Proof.

For $B\in\mathcal{A}^{d}$, we have that

$B\in\operatorname{Ker}d_{p}\iff(\forall g\in
G)\,gp\notin[B]\iff[B]\cap\operatorname{cl}(G\cdot p)=\varnothing.$

Given that for any closed subset $F\subseteq S(\mathcal{A}^{d})$ we have

$F=\bigcap\\{[B]^{c}:[B]\cap F=\varnothing\\},$

the conclusion follows. ∎

######  Lemma 2.26.

For any set $X$, assume $S\subseteq X^{X}$ is a semigroup with function
composition and let $I\trianglelefteqslant_{m}S$, $u\in J(I)$. Then

  1. (i)

$u(y)=y$ for all $y\in\operatorname{Im}u$;

  2. (ii)

$qu=q$ for any $q\in I$.

###### Proof.

(i) Follows directly from $u\circ u=u$.

(ii) Since $q\in I=I\circ u$, there is $\alpha\in I$ such that $q=\alpha\circ
u$. Therefore

$qu=\alpha uu=\alpha u=q.\qed$

######  Theorem 2.27.

Assume $I\trianglelefteqslant_{m}\operatorname{End}_{G}(\mathcal{A}^{d})$.

  1. (i)

All $\varphi\in I$ share a common kernel $K_{I}$, which uniquely determines
$I$.

  2. (ii)

Endomorphisms $\varphi,\psi\in I$ have the same image if and only if they
belong to the same ideal subgroup $uI$, $u\in J(I)$.

  3. (iii)

Let $\mathcal{R}_{I}=\\{\operatorname{Im}\varphi:\varphi\in I\\}$. Then
$\mathcal{R}_{I}=\mathcal{R}_{J}$ for any
$J\trianglelefteqslant_{m}\operatorname{End}_{G}(\mathcal{A}^{d})$.

  4. (iv)

Let
$\mathcal{K}=\\{K_{I}:I\trianglelefteqslant_{m}\operatorname{End}_{G}(\mathcal{A}^{d})\\}$
and write $\mathcal{R}$ for the common $\mathcal{R}_{I}$ for all
$I\trianglelefteqslant_{m}\operatorname{End}_{G}(\mathcal{A}^{d})$. Then the
function

$\\{uI:I\trianglelefteqslant_{m}\operatorname{End}_{G}(\mathcal{A}^{d}),u\in
J(I)\\}\to\mathcal{K}\times\mathcal{R}$

assigning to any $uI$ the common
$(\operatorname{Ker}\varphi,\operatorname{Im}\varphi)$ for all $\varphi\in
uI$, is a bijection.

###### Proof.

(i) By Theorem 2.24, for any
$J\trianglelefteqslant_{m}\operatorname{End}_{G}(\mathcal{A}^{d})$ we can
write $I=\\{d_{p}:p\in I^{\prime}\\}$ and $J=\\{d_{q}:q\in J^{\prime}\\}$,
where $I^{\prime},J^{\prime}\trianglelefteqslant_{m}S(\mathcal{A}^{d})$. It
suffices to show that for $p\in I^{\prime},q\in J^{\prime}$ we have that
$\operatorname{Ker}d_{p}=\operatorname{Ker}d_{q}$ if and only if
$I^{\prime}=J^{\prime}$. Indeed, if
$\operatorname{Ker}d_{p}=\operatorname{Ker}d_{q}$, then by Lemma 2.25 we have
$I^{\prime}=\operatorname{cl}(G\cdot p)=\operatorname{cl}(G\cdot
q)=J^{\prime}$. On the other hand, if $I^{\prime}=J^{\prime}$, then as in
Lemma 2.25,

$\operatorname{Ker}d_{p}=\\{B\in\mathcal{A}^{d}:[B]\cap
I^{\prime}=\varnothing\\}=\operatorname{Ker}d_{q}.$

(ii) Let $u,v\in J(I)$ and fix $\varphi\in uI,\psi\in vI$. We want to show
that $\operatorname{Im}\varphi=\operatorname{Im}\psi$ if and only if $uI=vI$.
On the one hand, if $uI=vI$, then by simple group properties there is
$\alpha\in uI$ such that $\varphi=\psi\circ\alpha$. Consequently,
$\operatorname{Im}\varphi\subseteq\operatorname{Im}\psi$ and
$\operatorname{Im}\varphi=\operatorname{Im}\psi$ by symmetry. On the other
hand, assume that $\operatorname{Im}\varphi=\operatorname{Im}\psi$. Using the
already proved implication we get
$\operatorname{Im}u=\operatorname{Im}\varphi=\operatorname{Im}\psi=\operatorname{Im}v$.
From Lemma 2.26 (i) it follows that $uv=v$ (because $u$ equals identity on
$\operatorname{Im}v=\operatorname{Im}u$) and from Lemma 2.26 (ii) that $uv=u$,
hence $u=uv=v$ and $uI=vI$.

(iii) Take $R\in\mathcal{R}_{I}$ and pick $u\in J(I)$ so that
$R=\operatorname{Im}u$. By Theorem 2.7, there is $v\in J(J)$ such that $uv=v$
and $vu=u$. It follows that $\operatorname{Im}v=\operatorname{Im}u=R$, so
$R\in\mathcal{R}_{J}$.

(iv) Follows from (i) – (iii). ∎

The elements of $\mathcal{R}$ are called _image algebras_. By Theorem 2.24,
they are exactly the algebras of the form $R=d_{q}[\mathcal{A}^{d}]$, where
$q\in S(\mathcal{A}^{d})$ is almost periodic. The image algebras determine any
minimal ideal (or subflow) of $S(\mathcal{A}^{d})$ up to $G$-flow isomorphism.

######  Fact 2.28.

Assume $R\in\mathcal{R}$ and $I\trianglelefteqslant_{m}S(\mathcal{A}^{d})$.
Then $I\cong S(R)$ as $G$-flows. Moreover, one such $G$-flow isomorphism
between $I$ and $S(R)$ is $\pi\restriction I$, where
$\pi:S(\mathcal{A}^{d})\to S(R)$ denotes the restriction.

###### Proof.

Take $u\in J(I)$ such that $R=\operatorname{Im}d_{u}$ and consider the dual
diagrams

${\mathcal{A}^{d}}$${R}$${R}$$\scriptstyle{d_{u}}$$\scriptstyle{\iota}$$\scriptstyle{\mathrm{id}}$
${S(\mathcal{A}^{d})}$${S(R)}$${S(R)}$$\scriptstyle{\pi}$$\scriptstyle{\mathrm{id}}$$\scriptstyle{r_{u}}$

Since $d_{u}\circ d_{u}=d_{u\ast u}=d_{u}$, the diagram on the left is
commutative, hence so is the one on the right. Moreover, $d_{u}$ is an
epimorphism of $G$-algebras, so by Fact 2.10, $r_{u}$ is an injective $G$-flow
morphism. Therefore it is an isomorphism onto its image (the continuity of
$r_{u}^{-1}$ follows from compactness). By Fact 2.11 and Lemma 2.25, that
image is equal to

$\\{p\in
S(\mathcal{A}^{d}):p\cap\operatorname{Ker}d_{u}=\varnothing\\}=\operatorname{cl}(G\cdot
u)=I.\qed$

Hence $I\cong S(R)$. Finally, from the commutativity of the right hand side
diagram we get that $\pi\restriction I$ is the inverse of $r_{u}$, so it is
also an isomorphism.

Now we will explain the relation between image algebras and strongly generic
sets. We should note that some similar ideas have already been studied in
[BF97] for an arbitrary semigroup $G$, but the study only applies to the
universal minimal $G$-flow and has no model-theoretic connotations.

######  Definition 2.29.

  1. (a)

A subset $B\subseteq G$ is _generic_ if finitely many (left) translates of $B$
cover $G$, i.e.

$G=\bigcup_{i=1}^{n}g_{i}\cdot B\quad\text{for some }g_{1},\ldots,g_{n}\in G.$

  2. (b)

A $G$-algebra $\mathcal{B}\leqslant\mathcal{P}(G)$ is _generic_ if every set
in $\mathcal{B}\setminus\\{\varnothing\\}$ is generic.

  3. (c)

A subset $B\subseteq G$ is _strongly generic_ if the $G$-algebra generated by
$B$ is generic.

######  Lemma 2.30.

Assume $\mathcal{B}\leqslant\mathcal{P}(G)$ is a $G$-algebra. Then

$\mathcal{B}\text{ is generic}\iff S(\mathcal{B})\text{ is a minimal subflow
of itself.}$

###### Proof.

$({\implies})$ Assume for contradiction there is a proper non-empty subflow
$Y\subseteq S(\mathcal{B})$. Pick $p\in S(\mathcal{B})\setminus Y$ and find
$B\in\mathcal{B}$ such that $p\in[B]$ and $[B]\cap Y=\varnothing$. Then
$B\neq\varnothing$ and $g[B]\cap Y=g([B]\cap Y)=\varnothing$ for any $g\in G$.
On the other hand, for some $g_{1},\ldots,g_{n}\in G$ we have that
$G=\bigcup_{i=1}^{n}g_{i}B$, which implies that
$S(\mathcal{B})=[G]=\bigcup_{i=1}^{n}g_{i}[B]$, a contradiction.

$({\impliedby})$ Take any $B\in\mathcal{B}\setminus\\{\varnothing\\}$ and
consider the set

$G\cdot[B]:=\bigcup_{g\in G}g[B].$

Its complement is a proper $G$-subflow of $S(\mathcal{B})$, so by assumption
it must be empty, i.e. $G\cdot[B]=S(\mathcal{B})=[G]$. By compactness,
$[G]=\bigcup_{i=1}^{n}g_{i}[B]$ for some $g_{1},\ldots,g_{n}\in G$, which
means that $G=\bigcup_{i=1}^{n}g_{i}B$, so $B$ is generic. ∎

######  Corollary 2.31.

Image algebras are generic.

###### Proof.

This follows directly from Fact 2.28 and Lemma 2.30. ∎

######  Corollary 2.32.

If $p\in S(\mathcal{A}^{d})$ is almost periodic and $B\in\mathcal{A}^{d}$,
then $d_{p}(B)$ is strongly generic. ∎

######  Lemma 2.33.

Assume $R_{1},R_{2}\in\mathcal{R}$ and $R_{1}\subseteq R_{2}$. Then
$R_{1}=R_{2}$.

###### Proof.

Take $I{\trianglelefteqslant_{m}}S(\mathcal{A}^{d})$ and $u,v\in J(I)$ such
that $R_{1}=d_{u}[\mathcal{A}^{d}]$, $R_{2}=d_{v}[\mathcal{A}^{d}]$. It
suffices to show that $u=v$. Take any $B\in\mathcal{A}^{d}$. By assumption we
can find $B^{\prime}\in\mathcal{A}^{d}$ such that $d_{u}B=d_{v}B^{\prime}$. By
Lemma 2.26, $v\ast u=v$, so

$d_{u}B=d_{v}B^{\prime}=d_{v}(d_{v}B^{\prime})=d_{v}(d_{u}B)=d_{v\ast
u}B=d_{v}B.$

In particular,

$B\in u\iff e\in d_{u}B\iff e\in d_{v}B\iff B\in v,$

hence $u=v$, as desired. ∎

######  Lemma 2.34.

Every generic $G$-subalgebra $\mathcal{B}\leqslant\mathcal{A}^{d}$ can be
extended to an image algebra $R\in\mathcal{R}$.

###### Proof.

By Fact 2.28, the flow $S(\mathcal{B})$ is minimal, so the point

$q=\hat{e}=\\{B\in\mathcal{B}:e\in B\\}\in S(\mathcal{B})$

is almost periodic. Applying Fact 2.2 (v) to the restriction
$r:S(\mathcal{A}^{d})\to S(\mathcal{B})$, we can find an almost periodic point
$p\in S(\mathcal{A}^{d})$ such that $r(p)=q$. Given $B\in\mathcal{B}$, Remark
2.20 gives

$d_{p}(B)=d_{q}(B)=d_{\hat{e}}(B)=Be^{-1}=B.$

Therefore $\mathcal{B}\subseteq d_{p}[\mathcal{A}^{d}]$ and
$R:=d_{p}[\mathcal{A}^{d}]$ is an image algebra, since $p\in
S(\mathcal{A}^{d})$ is almost periodic. ∎

######  Corollary 2.35.

Image algebras are precisely maximal generic $G$-subalgebras of
$\mathcal{A}^{d}$.

###### Proof.

First assume $R$ is an image algebra. By Corollary 2.31, $R$ is generic, so it
extends to a maximal generic $G$-algebra
$R\subseteq\mathcal{B}\subseteq\mathcal{A}^{d}$. By Lemma 2.34, $\mathcal{B}$
further extends to an image algebra $R_{2}\supseteq\mathcal{B}$. It follows
that $R\subseteq R_{2}$, so by Lemma 2.33, $R=R_{2}=\mathcal{B}$. Thus $R$ is
maximal generic.

Now assume $\mathcal{B}\subseteq\mathcal{A}^{d}$ is a maximal generic
$G$-algebra. By Lemma 2.34, $\mathcal{B}$ can be extended to an image algebra
$R\supseteq\mathcal{B}$, which is generic by Corollary 2.31. Hence
$\mathcal{B}=R$ and so $\mathcal{B}$ is an image algebra. ∎

The next proposition is a joint result of Newelski and the author. It shows
how to recover the Ellis group of $S(\mathcal{A})$ from any image algebra.

######  Proposition 2.36.

Assume $\mathcal{B}\leqslant\mathcal{A}^{d}$ is an image algebra and define

$\mathcal{E}=\\{q\in
S(\mathcal{B}):d_{q}[\mathcal{B}]\subseteq\mathcal{B}\\}.$

  1. (i)

The operation

$p\ast q=\\{B\in\mathcal{B}:d_{q}B\in p\\}$

is well defined as an operation $\ast:\mathcal{E}\times\mathcal{E}\to
S(\mathcal{B})$.

  2. (ii)

$\mathcal{E}$ is closed under $\ast$.

  3. (iii)

$(\mathcal{E},\ast)$ is isomorphic to the Ellis group of $S(\mathcal{A})$.

###### Proof.

(i) If $q\in\mathcal{E}$, then by definition
$d_{q}:\mathcal{B}\to\mathcal{B}$, so we have the dual map
$r_{q}:S(\mathcal{B})\to S(\mathcal{B})$. As usual, for any $p\in
S(\mathcal{B})$ we have that $p\ast q=r_{q}(p)$.

(ii) Fix $p,q\in\mathcal{E}$ and $B\in\mathcal{B}$. Applying Fact 2.17 to
$d_{q}:\mathcal{B}\to\mathcal{B}$, we get

$d_{p\ast q}(B)=d_{r_{q}(p)}(B)=d_{p}(d_{q}(B))\in\mathcal{B}.$

Hence $p\ast q\in\mathcal{E}$.

(iii) Take $I\trianglelefteqslant_{m}S(\mathcal{A}^{d})$ and pick $u\in J(I)$
such that $\mathcal{B}=d_{u}[\mathcal{A}^{d}]$. By Fact 2.28, the restriction
$\pi:I\to S(\mathcal{B})$ is a $G$-flow isomorphism. We first show that
$\pi[uI]=\mathcal{E}$. If $q\in uI$, then
$d_{q}[\mathcal{A}^{d}]=d_{u}[\mathcal{A}^{d}]=\mathcal{B}$, so in particular
$d_{\pi(q)}[\mathcal{B}]=d_{q}[\mathcal{B}]\subseteq\mathcal{B}$, hence
$\pi(q)\in\mathcal{E}$. On the other hand, assume that $q_{0}\in\mathcal{E}$
and take $q\in I$ such that $\pi(q)=q_{0}$. Then $q\ast u=q$ by Lemma 2.26, so

$d_{q}[\mathcal{A}^{d}]=d_{q\ast
u}[\mathcal{A}^{d}]=d_{q}[d_{u}[\mathcal{A}^{d}]]=d_{q}[\mathcal{B}]=d_{q_{0}}[\mathcal{B}]\subseteq\mathcal{B}=d_{u}[\mathcal{A}^{d}].$

By Lemma 2.33, these algebras are equal. Hence $q\in uI$ and so
$q_{0}\in\pi[uI]$.

It follows that $\pi$ is a bijection between $uI$ and $\mathcal{E}$. It is
also a homomorphism since for $p,q\in uI$ and $B\in\mathcal{B}$,

$\displaystyle B\in\pi(p\ast q)$ $\displaystyle\iff B\in p\ast q\iff d_{q}B\in
p\iff d_{\pi(q)}B\in p$ $\displaystyle\iff d_{\pi(q)}B\in\pi(p)\iff
B\in\pi(p)\ast\pi(q).$

Hence $\mathcal{E}$ is isomorphic to the Ellis group $uI$. ∎

The results from this section lead to an interesting way of studying the Ellis
group of $S(\mathcal{A})$ that only refers to the notion of a strongly generic
set. Take any strongly generic set $B\in\mathcal{A}^{d}$ (e.g. the empty set)
so that the $G$-algebra $\mathcal{B}\leqslant\mathcal{A}^{d}$ generated by $B$
is generic. Extend $\mathcal{B}$ to a maximal generic $G$-subalgebra $R$ of
$\mathcal{A}^{d}$, which is also an image algebra. Every minimal ideal
$I\trianglelefteqslant_{m}E(S(\mathcal{A}))$ is isomorphic to $S(R)$ as a
$G$-flow and the Ellis group of $S(\mathcal{A})$ is isomorphic to
$\mathcal{E}$ as defined in Proposition 2.36. Thus if the strongly generic
subsets of $G$ can be understood to the point of characterizing the maximal
generic $G$-subalgebras of $\mathcal{A}^{d}$, a complete description of the
Ellis group of $S(\mathcal{A})$ will follow.

######  Corollary 2.37.

Assume $R\in\mathcal{R}$ and $R\leqslant\mathcal{B}\leqslant\mathcal{A}^{d}$.
Then the Ellis groups of $S(\mathcal{A})$ and $S(\mathcal{B})$ are isomorphic.

###### Proof.

By Corollary 2.35, $R$ is a maximal generic $G$-subalgebra of
$\mathcal{A}^{d}$. Since $R\leqslant\mathcal{B}^{d}\leqslant\mathcal{A}^{d}$,
$R$ is also a maximal generic $G$-subalgebra of $\mathcal{B}^{d}$, hence it is
an image algebra in $\mathcal{B}^{d}$. By Proposition 2.36, the Ellis groups
of $S(\mathcal{A})$ and $S(\mathcal{B})$ can be computed directly from $R$,
giving the same result up to isomorphism. ∎

##  3 Strongly generic sets

In this section we study abstract properties of strongly generic sets. We also
provide a way to construct non-trivial examples of such sets in a wide class
of groups, namely those that do not have the smallest subgroup of finite
index.

###  3.1 Basic properties

Throughout the section assume that $G$ is a group. We regard $2^{G}$ as a
$G$-flow, where for $g\in G,f\in 2^{G}$, the action is defined as

$(g\odot f)(x)=f(xg)\text{ for }x\in G.$

Equivalently, $g\odot\chi_{A}=\chi_{Ag^{-1}}$ for $A\subseteq G$, where
$\chi_{A}$ denotes the characteristic function of $A$.

The simplest examples of strongly generic sets are periodic sets, defined
below:

######  Definition 3.1.

Assume $f\in 2^{G}$.

  1. (i)

$t\in G$ is a _period_ of $f$ if $(\forall x\in G)\,f(x\cdot t)=f(x)$.

  2. (ii)

$\operatorname{Per}(f)$ is the set of periods of $f$:

$\operatorname{Per}(f)=\\{t\in G:(\forall x\in G)\,f(x\cdot t)=f(x)\\}.$

By the next remark, $\operatorname{Per}(f)$ is a subgroup of $G$.

  3. (iii)

$f$ is _periodic_ if $\operatorname{Per}(f)$ has finite index in $G$.

  4. (iv)

A subset $A\subseteq G$ is _periodic_ if $\chi_{A}\in 2^{G}$ is periodic.

######  Remark 3.2.

$\operatorname{Per}(f)$ is the stabilizer of $f$ in the $G$-flow $2^{G}$. In
particular, it is a subgroup of $G$.

$\operatorname{Per}(f)$ can also be understood as follows: assume $A\subseteq
G$ and $H\leqslant G$. Clearly the following are equivalent:

$\displaystyle H\subseteq\operatorname{Per}(\chi_{A})\iff(\forall h\in
H)\,Ah=A\iff AH=A$ $\displaystyle\iff A\text{ is a union of some left cosets
of }H.$

As a consequence:

######  Remark 3.3.

$\operatorname{Per}(\chi_{A})$ is the greatest subgroup $H\leqslant G$ such
that $A$ can be written as a union of left cosets of $H$. ∎

The following remark is a useful characterization of periodic sets.

######  Remark 3.4.

For $A\subseteq G$ the following are equivalent:

  1. (i)

$A$ is periodic;

  2. (ii)

$A$ is a union of left cosets of some subgroup $H\leqslant G$ of finite index;

  3. (iii)

The $G$-algebra $\langle A\rangle\subseteq\mathcal{P}(G)$ generated by $A$ is
finite;

  4. (iv)

$A^{-1}:=\\{a^{-1}:a\in A\\}$ is periodic.

###### Proof.

(i)$\implies$(ii) Follows directly from Remark 3.3.

(ii)$\implies$(iii) Assume $A$ is a union of left cosets of a subgroup
$H\leqslant G$ of finite index. All such unions form a finite $G$-algebra of
size $2^{[G:H]}$ in which $\langle A\rangle$ is contained, hence it is also
finite.

(iii) $\implies$ (iv) From the assumption the set $\\{gA:g\in G\\}$ is finite,
hence also finite is the set

$\\{\chi_{A^{-1}g^{-1}}:g\in G\\}=\\{g\cdot\chi_{A^{-1}}:g\in G\\},$

which is the orbit of $\chi_{A^{-1}}$ in $2^{G}$. It follows that the
stabilizer of $\chi_{A^{-1}}$ has finite index in $G$, so $A^{-1}$ is
periodic.

(iv)$\implies$(i) Follows from the converse implication by symmetry. ∎

######  Corollary 3.5.

Assume $A\subseteq G$ is periodic. Then $\operatorname{Per}(\chi_{A})$ can be
expressed as an intersection of finitely many left translates of $A$ and
$A^{c}$.

###### Proof.

By definition

$\operatorname{Per}(\chi_{A})=\bigcap_{a\in A}a^{-1}A\cap\bigcap_{b\in
G\setminus A}b^{-1}A^{c}.$

By Remark 3.4 there are finitely many sets of the form $gA$ where $g\in G$, so
the conclusion follows. ∎

######  Corollary 3.6.

Assume $A\subseteq G$ is periodic. Then there is a normal subgroup
$N\trianglelefteqslant G$ of finite index such that $N\in\langle A\rangle^{d}$
and $A$ is a union of cosets of $N$.

###### Proof.

Let

$N:=\bigcap_{g\in G}g\operatorname{Per}(\chi_{A})g^{-1}\trianglelefteqslant
G.$

Since $\operatorname{Per}(\chi_{A})$ has finite index, there are only finitely
many distinct sets in this intersection. Each of them belongs to $\langle
A\rangle^{d}$, since by Corollary 3.5 we have
$\operatorname{Per}(\chi_{A})\in\langle A\rangle$ and so
$g\operatorname{Per}(\chi_{A})g^{-1}\in\langle A\rangle^{d}$ by Corollary
2.21. Hence $N\in\langle A\rangle^{d}$ and $N$ has finite index. Also
$N\subseteq\operatorname{Per}(\chi_{A})$, so by Remark 3.3, $A$ is a union of
cosets of $N$. ∎

It is now easy to see that every periodic subset $A\subseteq G$ is strongly
generic. Indeed, if $A$ is periodic, then by Remark 3.4 any non-empty element
of $\langle A\rangle$ contains at least one coset of some subgroup $H\leqslant
G$ of finite index (e.g. $H=\operatorname{Per}(\chi_{A})$), hence it is
generic. It was shown in [New12a, Proposition 2.8] that in the context of
stable groups these are the only definable strongly generic subsets of $G$.
Below is a known natural generalization of that result.

######  Theorem 3.7.

If $A\subseteq G$ is strongly generic and the formula $\varphi(x;y)\equiv x\in
yA$ is stable in $(G,\cdot,A)$, then $A$ is periodic.

###### Proof.

Note that the $G$-algebra $\mathcal{A}\leqslant\mathcal{P}(G)$ generated by
$A$ contains an atom. Indeed, otherwise we can find $\langle b_{\eta}:\eta\in
2^{<\omega}\rangle$ such that for each $\alpha\in 2^{\omega}$ the family
$\\{b_{\alpha\restriction n}A^{\alpha(n)}:n\in\omega\\}$ has the finite
intersection property. It follows that there are $2^{\aleph_{0}}$-many
$\varphi$-types over the countable parameter set $\\{b_{\eta}:\eta\in
2^{<\omega}\\}$, which contradicts the stability of $\varphi$.

Choose an atom $B\in\mathcal{A}$. Since $A$ is strongly generic, $B$ is
generic, so $G$ is a union of finitely many left translates of $B$, which are
also atoms. Therefore $\mathcal{A}$ is finite and the conclusion follows from
Remark 3.4. ∎

The stability assumption is essential – as we will show in the next
subsection, strongly generic non-periodic sets exist in general. However, the
property of being a strongly generic set turns out to be equivalent to a
specific weaker version of periodicity.

######  Definition 3.8.

Assume $f\in 2^{G}$.

  1. (i)

For any finite $U\subseteq G$, an element $t\in G$ is a $U$_-period_ of $f$ if

$(\forall x\in U)\,f(xt)=f(x).$

  2. (ii)

$\operatorname{Per}_{U}(f)$ is the set of $U$-periods of $f$.

  3. (iii)

$f$ is _locally periodic_ if for each finite $U\subseteq G$ the set
$\operatorname{Per}_{U}(f)$ is generic.

  4. (iv)

A subset $A\subseteq G$ is _locally periodic_ if $\chi_{A}$ is locally
periodic.

Note that $\operatorname{Per}_{U}(f)$ is typically not a subgroup of $G$, so
in (iii) we can no longer expect it to be a subgroup of finite index; the
correct condition here is for it to be a generic set.

We will also utilize the notion of a _self-replicating function_. Although it
was originally defined for functions $f\in 2^{N}$ where $N$ is a
$\mathbb{Q}$-vector space (Definition 3.12 in [New09]), it can be naturally
interpreted when $N=G$ is an arbitrary group:

######  Definition 3.9.

We say that $f\in 2^{G}$ is self-replicating if

$(\forall\underset{\text{finite}}{U\subseteq
G})(\exists\underset{\text{finite}}{V\subseteq G})(\forall g\in G)(\exists
h\in G)\,h\odot(f\restriction U)\subseteq f\restriction Vg.$

The expression $h\odot(f\restriction U)$ here means the partial function
defined by the same formula as in the beginning of the section:

$\big{(}h\odot(f\restriction U)\big{)}(x)=(f\restriction U)(xh)\quad\text{for
}x\in Uh^{-1}.$

In [New09] it was proved that self-replication is a sufficient condition for
almost periodicity. The following proposition states that in fact these
conditions are equivalent to each other, and also to the property of being
locally periodic.

######  Proposition 3.10.

For $f\in 2^{G}$ the following are equivalent:

  1. (i)

$f$ is self-replicating;

  2. (ii)

$f$ is an almost periodic point of $2^{G}$;

  3. (iii)

$f$ is locally periodic.

###### Proof.

(i)$\implies$(ii) Take any basic open neighbourhood of $f$. We can write it as

$[\sigma]=\\{f^{\prime}\in 2^{G}:\sigma\subseteq f^{\prime}\\},$

where $\sigma=f\restriction U$ for some finite $U\subseteq G$. By Remark 2.3,
it suffices to show that $\operatorname{cl}(G\odot f)\subseteq
W^{-1}\odot[\sigma]$ for some finite $W\subseteq G$. From the assumption there
is a finite $V\subseteq G$ such that

$(\forall g\in G)(\exists h\in G)\,h\odot(f\restriction U)\subseteq
f\restriction Vg.$

Let $W=\\{w\in G:Uw\subseteq V\\}$ and take any
$f^{\prime}\in\operatorname{cl}(G\odot f)$. We have that
$f^{\prime}\in[f^{\prime}\restriction V]\cap\operatorname{cl}(G\odot f)$, so
there is $g\in G$ such that $g\odot f\in[f^{\prime}\restriction V]$, or
equivalently $g\odot(f\restriction Vg)=f^{\prime}\restriction V$. Take $h\in
H$ such that $h\odot(f\restriction U)\subseteq f\restriction Vg$. It follows
that

$gh\odot(f\restriction U)\subseteq g\odot(f\restriction
Vg)=f^{\prime}\restriction V.$

Let $w:=(gh)^{-1}$. By the above, $Uw\subseteq V$, hence $w\in W$ and $w\odot
f^{\prime}\in[\sigma]$, as needed.

(ii)$\implies$(iii) Take any finite $U\subseteq G$. By Remark 2.3, we can find
a finite set $W\subseteq G$ such that $\operatorname{cl}(G\odot f)\subseteq
W^{-1}\odot[f\restriction U]$. It suffices to show that
$W^{-1}\cdot\operatorname{Per}_{U}(f)=G$. For any $g\in G$ we can find $w\in
W$ such that $g\odot f\in w^{-1}\odot[f\restriction U]$, i.e. $f\restriction
U\subseteq wg\odot f$. It follows that $wg\in\operatorname{Per}_{U}(f)$.

(iii)$\implies$(i) Fix any finite $U\subseteq G$. By assumption
$\operatorname{Per}_{U}(f)$ is generic, so there is a finite $W\subseteq G$
such that $W^{-1}\cdot\operatorname{Per}_{U}(f)=G$. We will show that the set
$V=U\cdot W$ has the required property. Take any $g\in G$ and write
$g=w^{-1}h$, where $w\in W,h\in\operatorname{Per}_{U}(f)$. Then
$Uh=Uwg\subseteq Vg$ and $h^{-1}\odot(f\restriction U)\subseteq f$, so
$h^{-1}\odot(f\restriction U)\subseteq f\restriction Vg$, as required. ∎

######  Proposition 3.11.

A subset $A\subseteq G$ is strongly generic if and only if it is locally
periodic.

###### Proof.

Write $A^{0}:=G\setminus A$ and $A^{1}:=A$. The following are equivalent:

  * –

$A$ is strongly generic;

  * –

For any Boolean term $\tau(x_{1},\ldots,x_{n})$ and any $u_{1},\ldots,u_{n}\in
G$, if the set $\tau(u_{1}^{-1}A,\ldots,u_{n}^{-1}A)$ is non-empty, then it is
generic;

  * –

For any $\varepsilon_{1},\ldots,\varepsilon_{n}\in\\{0,1\\}$ and
$u_{1},\ldots,u_{n}\in G$, if the set
$\bigcap_{j=1}^{n}u_{j}^{-1}A^{\varepsilon_{j}}$ is non-empty, then it is
generic;

  * –

For any $\varepsilon_{1},\ldots,\varepsilon_{n}\in\\{0,1\\}$ and
$u_{1},\ldots,u_{n}\in G$, if
$e\in\bigcap_{j=1}^{n}u_{j}^{-1}A^{\varepsilon_{j}}$, then this set is
generic;

  * –

For any $u_{1},\ldots,u_{n}\in G$ the set
$\bigcap_{j=1}^{n}u_{j}^{-1}A^{\chi_{A}(u_{j})}$ is generic.

Let $U=\\{u_{1},\ldots,u_{n}\\}\subseteq G$ and $t\in G$. We have that

$\displaystyle t\in\bigcap_{j=1}^{n}u_{j}^{-1}A^{\chi_{A}(u_{j})}$
$\displaystyle\iff\bigwedge_{j=1}^{n}u_{j}t\in A^{\chi_{A}(u_{j})}$
$\displaystyle\iff\bigwedge_{j=1}^{n}\chi_{A}(u_{j}t)=\chi_{A}(u_{j})$
$\displaystyle\iff(\forall u\in U)\,\chi_{A}(ut)=\chi_{A}(u)$
$\displaystyle\iff t\in\operatorname{Per}_{U}(\chi_{A}),$

hence

$\bigcap_{j=1}^{n}u_{j}^{-1}A^{\chi_{A}(u_{j})}=\operatorname{Per}_{U}(\chi_{A}).$

By the previous equivalences, the set $A$ is strongly generic if and only if
for every finite $U\subseteq G$, the set $\operatorname{Per}_{U}(\chi_{A})$ is
generic, i.e. $A$ is locally periodic. ∎

######  Theorem 3.12.

For $A\subseteq G$,

$A\text{ is strongly generic}\iff\chi_{A}\text{ is an almost periodic point of
}2^{G}.$

###### Proof.

Follows directly from the last two propositions. ∎

Using these results we give an alternative proof of Corollary 2.31.

###### Proof of Corollary 2.31.

Take any almost periodic point $q\in S(\mathcal{A}^{d})$ and let
$B=d_{q}(A)\in\operatorname{Im}d_{q}$. By Fact 2.2 (iv) and Remark 2.20 (ii),
the point $\chi_{B}$ is almost periodic in $2^{G}$. Therefore by Theorem 3.12,
the set $B$ is strongly generic. ∎

Consider the additive group of integers, $G=(\mathbb{Z},+)$. It is well known
that there are almost periodic points in $2^{\mathbb{Z}}$ that are not
periodic functions. It follows from Theorem 3.12 that there is a strongly
generic subset $A\subseteq\mathbb{Z}$ that is not periodic. In the next
subsection we explicitly describe a whole class of such subsets.

###  3.2 Concrete examples

Consider a finitely branching tree $T\subseteq\omega^{<\omega}$ with leaves
$T_{L}\subseteq T$. Pick indexed families $\mathcal{H}=\langle
H_{\eta}:\eta\in T\rangle$ of subgroups of $G$ and $d=\langle d_{\eta}:\eta\in
T\rangle$ of elements of $G$, satisfying for any $\eta\in T$:

  * •

$H_{\varnothing}=G$;

  * •

$H_{\eta^{\frown}i}=H_{\eta^{\frown}j}$ whenever
$\eta^{\frown}i,\eta^{\frown}j\in T$;

  * •

$H_{\eta^{\frown}i}$ is a proper subgroup of $H_{\eta}$ of finite index
whenever $\eta^{\frown}i\in T$;

  * •

$d_{\eta}H_{\eta}$ is a disjoint union of
$\\{d_{\eta^{\frown}i}H_{\eta^{\frown}i}:\eta^{\frown}i\in T\\}$ if
$\eta\notin T_{L}$.

Assume that each $g\in G$ belongs to some (necessarily unique) coset
$d_{\eta}H_{\eta}$, where $\eta\in T_{L}$. Equivalently, the intersection of
cosets $d_{\eta}H_{\eta}$ along any infinite branch of $T$ is empty. Thus we
can define $f_{T}:G\to T_{L}$ so that $f_{T}(g)=\eta$ for the unique $\eta\in
T_{L}$ such that $g\in d_{\eta}H_{\eta}$.

######  Definition 3.13.

  1. (i)

A _tree of cosets_ of $G$ is a tuple $(T,\mathcal{H},d)$ satisfying the above
assumptions.

  2. (ii)

A _valued tree of cosets_ of $G$ is a tuple $(T,\mathcal{H},d,v)$, where
$(T,\mathcal{H},d)$ is a tree of cosets of $G$ and $v:T_{L}\to\\{0,1\\}$.

  3. (iii)

A function $f:G\to\\{0,1\\}$ is _founded_ on the valued tree of cosets
$(T,\mathcal{H},d,v)$ if $f=v\circ f_{T}$. It is _founded_ on the tree of
cosets $(T,\mathcal{H},d)$ if it is founded on $(T,\mathcal{H},d,v)$ for some
$v$.

  4. (iv)

A function $f:G\to\\{0,1\\}$ is _tree-founded_ if it is founded on some tree
of cosets of $G$.

  5. (v)

A subset $A\subseteq G$ is _tree-founded_ if $\chi_{A}:G\to\\{0,1\\}$ is tree-
founded.

The generality of the definition is convenient, but it is possible to put a
simplifying restriction without narrowing down the class of tree-founded
functions. Namely, assume $f$ is a function founded on the tree of cosets
$(T,\mathcal{H},d)$. Then without loss of generality we can assume that each
$H_{\eta}$ depends only on $|\eta|$, i.e. $H_{\eta}=A_{|\eta|}$ for some
decreasing sequence of subgroups of finite index $G=A_{0}\geqslant
A_{1}\geqslant A_{2}\geqslant\ldots$. Moreover, we can assume that each
$A_{n}$ is a normal subgroup of $G$ so that the left and right cosets
coincide. When $(T,\mathcal{H},d)$ satisfies these additional assumptions, we
shall call it a _linear tree of cosets_.

To see this, let $A_{n}$ be the intersection of all conjugates of $H_{\eta}$
with $|\eta|\leqslant n$. It is easy to prove that $A_{n}$ is a decreasing
sequence of normal subgroups of finite index in $G$. The node of $T$
corresponding to each coset of $H_{\eta}$ can be split into finitely many
nodes corresponding to all cosets of $A_{|\eta|}$ contained in it and the
function $v$ can be modified accordingly.

######  Example 3.14.

Take any $2$-adic integer $\alpha\in\mathbb{Z}_{2}\setminus\mathbb{Z}$. Let
$H_{n}=2^{n}\mathbb{Z}\leqslant\mathbb{Z}$ and $d_{n}=\alpha\bmod{2^{n}}$ so
that $\langle d_{n}+H_{n}:n\in\mathbb{N}\rangle$ is a descending sequence of
cosets in $\mathbb{Z}$ with empty intersection. Define
$f:\mathbb{Z}\to\\{0,1\\}$ by

$\begin{cases}f\restriction\mathbb{Z}\setminus(d_{1}+H_{1})=0,\\\
f\restriction(d_{1}+H_{1})\setminus(d_{2}+H_{2})=1,\\\
f\restriction(d_{2}+H_{2})\setminus(d_{3}+H_{3})=0,\\\
f\restriction(d_{3}+H_{3})\setminus(d_{4}+H_{4})=1,\\\ \vdots\end{cases}$

and so on. Then $f$ is tree-founded. Figure 1 illustrates the construction for

$\alpha=1\cdot 2^{0}+0\cdot 2^{1}+1\cdot 2^{2}+1\cdot
2^{3}+\ldots=1+0+4+8+\ldots\in\mathbb{Z}_{2}\setminus\mathbb{Z}.$

$\mathbb{Z}$$2\mathbb{Z}$$0$$2\mathbb{Z}+1$$4\mathbb{Z}+3$$1$$4\mathbb{Z}+1$$8\mathbb{Z}+1$$0$$8\mathbb{Z}+5$$16\mathbb{Z}+5$$1$$16\mathbb{Z}+13$$\ddots$
Figure 1: A valued tree of cosets defining a tree-founded function.

Now we set to prove that every tree-founded set is strongly generic. We will
also give a natural sufficient condition for a tree-founded set to be non-
periodic. Assuming $G$ has sufficiently many subgroups of finite index, this
describes a large class of explicitly definable strongly generic, non-periodic
subsets of $G$.

Consider the topology on $G$ generated by cosets of subgroups of $G$ of finite
index, which is called _the profinite topology_ on $G$. Contrary to what the
name can suggest, it need be neither compact nor Hausdorff. For instance, when
$G=(\mathbb{Q},+)$, the topology consists of just two sets, $\varnothing$ and
$\mathbb{Q}$. On the other hand, when $G=(\mathbb{Z},+)$, the topology is
Hausdorff but not compact – as demonstrated in the example above, where
infinitely many disjoint cosets cover $\mathbb{Z}$. Finally, let
$G=(\hat{\mathbb{Z}},+)$ be the profinite completion of $(\mathbb{Z},+)$. Then
the profinite topology on $G$ is the usual topology on $\hat{\mathbb{Z}}$
treated as a profinite group, hence it is both Hausdorff and compact. In this
case every tree-founded subset of $G$ is founded on a finite tree, as an
infinite tree would have an infinite branch, which by compactness would have a
non-empty intersection. Therefore in $\hat{\mathbb{Z}}$, tree-founded sets are
precisely periodic sets.

We will say that a subset $A\subseteq G$ is _pf-clopen_ if it is clopen in the
profinite topology. We will also say that a function $f:G\to\\{0,1\\}$ is _pf-
continuous_ if it is continuous with respect to the profinite topology on $G$.
Clearly $A\subseteq G$ is pf-clopen if and only if $\chi_{A}:G\to\\{0,1\\}$ is
pf-continuous.

######  Remark 3.15.

Every tree-founded set $A\subseteq G$ is pf-clopen.

###### Proof.

Take the tree of cosets $(T,\mathcal{H},d)$ on which $\chi_{A}$ is founded.
Let $a\in A$ and take $\eta=f_{T}(a)$ so that $\eta$ is a leaf in $T$ and
$a\in d_{\eta}H_{\eta}$. Then $\chi_{A}$ is constant on $d_{\eta}H_{\eta}$, so
$d_{\eta}H_{\eta}\subseteq A$, hence $A$ is open. By the same reasoning if
$b\in G\setminus A$, there is a coset $d_{\eta}H_{\eta}$ such that $b\in
d_{\eta}H_{\eta}\subseteq G\setminus A$, so $G\setminus A$ is open as well. ∎

The converse of Remark 3.15 is also true provided that the family of finite
index subgroups of $G$ ordered by inverse inclusion has countable cofinality,
i.e. if there is a decreasing sequence $(H_{n})$ of finite index subgroups of
$G$ such that each finite index subgroup of $G$ has some $H_{n}$ as a
subgroup.

######  Proposition 3.16.

Every pf-clopen set is strongly generic.

###### Proof.

All pf-clopen sets form a $G$-algebra of subsets of $G$. The algebra is
generic, since if a pf-clopen set is non-empty, then it contains a coset of a
subgroup of finite index, hence it is generic. The conclusion follows. ∎

######  Proposition 3.17.

Every tree-founded subset of $G$ is strongly generic. ∎

Some valued trees are needlessly complicated with respect to the tree-founded
function they define. Figure 2 shows tree simplifications that do not change
the functions founded on them.

$G$$1$$1$$1$$0$$0$$G$$1$$0$$G$$0$$1$$1$$1$$1$$1$$\vdots$$G$$0$$1$ Figure 2:
Reducible trees and their reductions.

######  Definition 3.18.

Assume $(T,\mathcal{H},d,v)$ is a valued tree of cosets of $G$.

  1. (i)

A node $\eta\in T$ is _homogeneous_ if $v$ is constant on the set of leaves
extending $\eta$, or equivalently, if $v\circ f_{T}$ is constant on
$d_{\eta}H_{\eta}$.

  2. (ii)

$(T,\mathcal{H},d,v)$ is _irreducible_ if every homogeneous $\eta\in T$ is a
leaf. Otherwise it is _reducible_.

######  Remark 3.19.

Assume $f:G\to\\{0,1\\}$ is founded on a valued tree of cosets
$(T,\mathcal{H},d,v)$. Then $f$ is founded on some irreducible valued tree of
cosets.

###### Proof.

Let $T^{\prime}$ be the tree obtained by removing from $T$ all proper
extensions of $\eta$ for each minimal homogeneous $\eta\in T$. Thus each such
$\eta$ will become a leaf in $T^{\prime}$. Define $v^{\prime}(\eta)$ for such
$\eta$ as the common value of $v$ on the leaves extending $\eta$ in $T$ and
$v^{\prime}=v$ elsewhere. Then $(T^{\prime},\mathcal{H}\restriction
T^{\prime},d\restriction T^{\prime},v^{\prime})$ is an irreducible valued tree
of cosets of $G$ and $f=v^{\prime}\circ f_{T^{\prime}}$. ∎

For the purpose of stating the next lemma, we introduce some auxiliary
notation.

######  Notation.

Assume $f:G\to\\{0,1\\}$.

  1. (i)

Denote by $\mathcal{F}$ the family of subgroups of $G$ of finite index;

  2. (ii)

For $H\in\mathcal{F}$, let $\chi(f,H)=\bigcup\\{A\in G/H:f\text{ is constant
on }A\\}$;

  3. (iii)

For $H,H^{\prime}\in\mathcal{F}$, write
$H\mathbin{\sqsubseteq\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-2.84526pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to15.649pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{/}}}}}}\hss}\hskip
0.0pt}}}}}_{f}H^{\prime}$ if $H\leqslant H^{\prime}$ and
$\chi(f,H)\supsetneq\chi(f,H^{\prime})$.

######  Lemma 3.20.

Assume $f:G\to\\{0,1\\}$ is pf-continuous. The following are equivalent:

  1. (i)

$f$ is periodic;

  2. (ii)

There is $N\in\mathbb{N}$ such that each sequence
$G\mathbin{\sqsupseteq\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-2.84526pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to17.07164pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{/}}}}}}\hss}\hskip
0.0pt}}}}}_{f}H_{1}\mathbin{\sqsupseteq\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-2.84526pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to17.07164pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{/}}}}}}\hss}\hskip
0.0pt}}}}}_{f}H_{2}\mathbin{\sqsupseteq\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-2.84526pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to17.07164pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{/}}}}}}\hss}\hskip
0.0pt}}}}}_{f}\ldots\mathbin{\sqsupseteq\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-2.84526pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to17.07164pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{/}}}}}}\hss}\hskip
0.0pt}}}}}_{f}H_{n}$ has length $n\leqslant N$;

  3. (iii)

There is no infinite sequence
$G\mathbin{\sqsupseteq\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-2.84526pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to17.07164pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{/}}}}}}\hss}\hskip
0.0pt}}}}}_{f}H_{1}\mathbin{\sqsupseteq\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-2.84526pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to17.07164pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{/}}}}}}\hss}\hskip
0.0pt}}}}}_{f}H_{2}\mathbin{\sqsupseteq\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-2.84526pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to17.07164pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{/}}}}}}\hss}\hskip
0.0pt}}}}}_{f}\ldots$, i.e. the strict partial order
$(\mathcal{F},\mathbin{\sqsubseteq\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-2.84526pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to15.649pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{/}}}}}}\hss}\hskip
0.0pt}}}}}_{f})$ is well-founded.

###### Proof.

(i)$\implies$(ii) Take a normal subgroup $T\trianglelefteqslant G$ of finite
findex such that $T\leqslant\operatorname{Per}(f)$. Consider any subgroup
$H\leqslant G$ such that $f$ is constant on some coset $aH$, $a\in G$. Then
$HT\leqslant G$ and $f$ is constant on $aHT$ since for any $h,h^{\prime}\in
H,t,t^{\prime}\in T$: $f(aht)=f(ah)=f(ah^{\prime})=f(ah^{\prime}t^{\prime})$.
Since $aH\subseteq aHT$ and $aHT$ is a union of left cosets of $T$, it follows
that $\chi(f,H)=\bigcup\mathcal{T}$ for some family $\mathcal{T}\subseteq
G/T$. Therefore any sequence
$G\mathbin{\sqsupseteq\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-2.84526pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to17.07164pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{/}}}}}}\hss}\hskip
0.0pt}}}}}_{f}H_{1}\mathbin{\sqsupseteq\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-2.84526pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to17.07164pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{/}}}}}}\hss}\hskip
0.0pt}}}}}_{f}H_{2}\mathbin{\sqsupseteq\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-2.84526pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to17.07164pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{/}}}}}}\hss}\hskip
0.0pt}}}}}_{f}\ldots\mathbin{\sqsupseteq\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-2.84526pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to17.07164pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{/}}}}}}\hss}\hskip
0.0pt}}}}}_{f}H_{n}$ has length $n\leqslant N$, where $N=[G:T]$.

(ii)$\implies$(iii) is obvious.

(iii)$\implies$(i) Assume for contradiction that $f$ is not periodic. It
suffices to show that there is no
$\mathbin{\sqsubseteq\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-2.84526pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to15.649pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{/}}}}}}\hss}\hskip
0.0pt}}}}}_{f}$-minimal element in $\mathcal{F}$. Take any $H\in\mathcal{F}$.
Since $f$ is not periodic, we have that $\chi(f,H)\neq G$, so we can find
$a,b\in G\setminus\chi(f,H)$ such that $f(a)=0$ and $f(b)=1$. By pf-continuity
of $f$ there are subgroups $A,B\in\mathcal{F}$ satisfying $f\restriction
aA\equiv 0$ and $f\restriction bB\equiv 1$. It follows that $H_{0}:=H\cap
A\cap
B\mathbin{\sqsubseteq\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-2.84526pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to15.649pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{/}}}}}}\hss}\hskip
0.0pt}}}}}_{f}H$, as required. ∎

######  Proposition 3.21.

Assume $f:G\to\\{0,1\\}$ is founded on a valued tree of cosets
$(T,\mathcal{H},d,v)$ which is linear, irreducible and infinite. Then $f$ is
not periodic.

###### Proof.

Recall that linearity means that $H_{\theta}=H_{\eta}$ whenever
$|\theta|=|\eta|$. Take an infinite branch $t\in\omega^{\omega}$ of $T$. For
each $n\in\mathbb{N}$ we have that $t\restriction n$ is not a leaf in $T$, so
by irreducibility it is not homogeneous. Hence $f$ is not constant on
$d_{t\restriction n}H_{t\restriction n}$ and $\chi(f,H_{t\restriction n})\cap
d_{t\restriction n}H_{t\restriction n}=\varnothing$. Take any $a\in
d_{t\restriction n}H_{t\restriction n}$ and find a leaf $\eta$ extending
$t\restriction n$ such that $a\in d_{\eta}H_{\eta}$. Then $f$ is constant on
$d_{\eta}H_{\eta}$, so
$d_{\eta}H_{\eta}\subseteq\chi(f,H_{\eta})\setminus\chi(f,H_{t\restriction
n})$ and therefore
$H_{t\restriction|\eta|}=H_{\eta}\mathbin{\sqsubseteq\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-2.84526pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to15.649pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{/}}}}}}\hss}\hskip
0.0pt}}}}}_{f}H_{t\restriction n}$. Thus for any $n\in\mathbb{N}$ there is
$m\geqslant n$ such that $H_{t\restriction
m}\mathbin{\sqsubseteq\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-2.84526pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to15.649pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{/}}}}}}\hss}\hskip
0.0pt}}}}}_{f}H_{t\restriction n}$, so by Lemma 3.20, $f$ is not periodic. ∎

$\alpha=1+3+9+27+\ldots=-\frac{1}{2}\in\mathbb{Z}_{3}\setminus\mathbb{Z}$$\mathbb{Z}$$3\mathbb{Z}$$3\mathbb{Z}+2$$6\mathbb{Z}$$0$$6\mathbb{Z}+3$$1$$6\mathbb{Z}+2$$0$$6\mathbb{Z}+5$$1$$3\mathbb{Z}+1$$9\mathbb{Z}+1$$9\mathbb{Z}+7$$18\mathbb{Z}+10$$0$$18\mathbb{Z}+1$$1$$18\mathbb{Z}+16$$0$$18\mathbb{Z}+7$$1$$9\mathbb{Z}+4$$\begin{subarray}{c}\vdots\\\\[2.15277pt]
\displaystyle\downarrow\\\\[2.15277pt] \displaystyle\alpha\end{subarray}$
Figure 3: A periodic function founded on a non-linear infinite irreducible
tree.

The linearity assumption above is essential. Indeed, consider the function
$f:\mathbb{Z}\to\\{0,1\\}$ founded on the tree shown in Figure 3. Then $f$ is
periodic as $f\restriction 2\mathbb{Z}\equiv 0$ and
$f\restriction(2\mathbb{Z}+1)\equiv 1$, even though the tree is infinite and
irreducible.

This concludes our description of a class of strongly generic, non-periodic
subsets of an arbitrary group $G$. However, this is not a complete
characterization even in $(\mathbb{Z},+)$, as there are strongly generic
subsets which are not even pf-clopen.

######  Example 3.22.

Let $v_{2}:\mathbb{Z}\to\mathbb{N}\cup\\{\infty\\}$ denote the $2$-adic
valuation, i.e. $v_{2}(0)=\infty$ and $v_{2}(k)$ is the highest
$n\in\mathbb{N}$ such that $2^{n}\mid k$ for $k\neq 0$. Define
$A_{\varepsilon}\subseteq\mathbb{Z}$, $\varepsilon\in\\{0,1\\}$, as follows:

$\chi_{A_{\varepsilon}}(k)=\begin{cases}\varepsilon&\text{if }\ k=0,\\\
v_{2}(k)\bmod 2&\text{if }\ k\neq 0.\end{cases}$

Clearly neither $A_{\varepsilon}$ is pf-clopen, because $\chi_{A_{0}}$ and
$\chi_{A_{1}}$ are not pf-continuous at $k=0$. However, both $A_{0}$ and
$A_{1}$ are strongly generic. To see this for e.g. $A_{0}$, by Proposition
3.11 it suffices to check it is locally periodic. Fix any finite
$U\subseteq\mathbb{Z}$ and take an even natural number $N>v_{2}(u)$ for all
$u\in U\setminus\\{0\\}$. We claim that

$2^{N}+2^{N+1}\mathbb{Z}=\\{(2k+1)\cdot
2^{N}:k\in\mathbb{Z}\\}\subseteq\operatorname{Per}_{U}(\chi_{A_{0}}).$

Indeed, for any $k\in\mathbb{Z}$ and $u\in U\setminus\\{0\\}$ we have that
$v_{2}\big{(}(2k+1)\cdot 2^{N}\big{)}=N>v_{2}(u)$, so
$v_{2}\big{(}u+(2k+1)\cdot 2^{N}\big{)}=v_{2}(u)$, hence
$\chi_{A_{0}}\big{(}u+(2k+1)\cdot 2^{N}\big{)}=\chi_{A_{0}}(u)$. Also
$\chi_{A_{0}}\big{(}(2k+1)\cdot 2^{N}\big{)}=N\bmod 2=0=\chi_{A_{0}}(0)$.
Hence $\operatorname{Per}_{U}(\chi_{A_{0}})$ contains the coset
$2^{N}+2^{N+1}\mathbb{Z}$ and therefore is generic.

$\mathbb{Z}$$2\mathbb{Z}+1$$0$$2\mathbb{Z}$$4\mathbb{Z}+2$$1$$4\mathbb{Z}$$8\mathbb{Z}+4$$0$$8\mathbb{Z}$$16\mathbb{Z}+8$$1$$16\mathbb{Z}$$\ddots$
Figure 4: A tree that leaves the value at $k=0$ undefined.

Note that both sets considered in the example are “almost tree-founded”,
meaning they are defined in the same way as before by the tree shown in Figure
4, except for the value at $k=0$ which must be assigned manually as either $0$
or $1$. We can try to generalize and ask whether the property of being
strongly generic will be preserved if instead of one point we allow such
arbitrary assignment on some small subset of $G$, e.g. a set of elements
corresponding to a nowhere dense set of infinite branches. We will demonstrate
that the answer is quite simply negative after we formalize the necessary
notions. However, the sets with the described property will be of interest in
the next section.

######  Definition 3.23.

Let $(T,\mathcal{H},d,v)$ be as in Definition 3.13 (ii), except we no longer
assume that each $g\in G$ belongs to some coset corresponding to a leaf of
$T$. Thus the set $G_{L}=\bigcup_{\eta\in T_{L}}d_{\eta}H_{\eta}$ is not
necessarily equal to $G$, but we still define $f_{T}:G_{L}\to T_{L}$ as
before. Assume that the set of infinite branches of $T$ is nowhere dense, i.e.
every $\eta\in T$ extends to a leaf.

  1. (i)

A function $f:G\to\\{0,1\\}$ is _almost founded_ on $(T,\mathcal{H},d,v)$ if
$f\restriction G_{L}=v\circ f_{T}$.

  2. (ii)

A function $f:G\to\\{0,1\\}$ is _almost tree-founded_ if it is almost founded
on some $(T,\mathcal{H},d,v)$ as above.

  3. (iii)

A subset $A\subseteq G$ is _almost tree-founded_ if $\chi_{A}$ is almost tree-
founded.

$G$$2G+(1,0)$$0$$2G$$4G+(2,0)$$1$$4G$$8G+(4,0)$$0$$8G$$16G+(8,0)$$1$$16G$$\ddots$
Figure 5: An almost tree-founded set that need not be strongly generic.

######  Example 3.24.

Assume $A\subseteq G=\mathbb{Z}\times\mathbb{Q}$ is a subset such that
$\chi_{A}$ is almost founded on the tree in Figure 5. Then $A$ is strongly
generic if and only if $\chi_{A}$ is constant on $\\{0\\}\times\mathbb{Q}$.
Indeed: if $\chi_{A}$ is constant on $\\{0\\}\times\mathbb{Q}$, then the
argument is the same as in Example 3.22 with the additional remark that
$\\{0\\}\times\mathbb{Q}\subseteq\operatorname{Per}(\chi_{A})$. On the other
hand, if $\chi_{A}$ is not constant on $\\{0\\}\times\mathbb{Q}$, take
$a,b\in\mathbb{Q}$ such that $(0,a)\in A$ and $(0,b)\notin A$. Then the set
$A\setminus\big{(}(0,a-b)+A\big{)}$ is non-empty and contained in
$\\{0\\}\times\mathbb{Q}$, hence not generic.

The following variation on Example 3.22 will be used later:

######  Example 3.25.

Assume $G$ is an infinite profinite group. Take a sequence

$G=F_{0}\geqslant\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-0.85358pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to17.07164pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{\textbackslash}}}}}}\hss}\hskip
0.0pt}}}}F_{1}\geqslant\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-0.85358pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to17.07164pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{\textbackslash}}}}}}\hss}\hskip
0.0pt}}}}F_{2}\geqslant\scalebox{0.5}[0.5]{\leavevmode\hbox{\raise-0.85358pt\hbox{\set@color\leavevmode\hbox
to0.0pt{\hss\hbox
to17.07164pt{\leavevmode\hbox{\set@color\resizebox{}{}{{\leavevmode\hbox{\set@color{{\textbackslash}}}}}}\hss}\hskip
0.0pt}}}}\ldots$

of clopen subgroups of $G$, thus of finite index, and let
$F_{\infty}:=\bigcap_{n=0}^{\infty}F_{n}$. We define
$v:G\to\mathbb{N}\cup\\{\infty\\}$ so that $v(x)$ is the highest
$n\in\mathbb{N}\cup\\{\infty\\}$ such that $x\in F_{n}$. Similarly as for
valuations, we have that $v(xy)\geqslant\min\\{v(x),v(y)\\}$ and these values
are equal if $v(x)\neq v(y)$.

Consider $A=\bigcup_{n=1}^{\infty}(F_{2n-1}\setminus F_{2n})\subseteq G$, so
that

$\chi_{A}(x)=\begin{cases}0&\text{if }\ x\in F_{\infty},\\\ v(x)\bmod
2&\text{if }\ x\in G\setminus F_{\infty}.\end{cases}$

We will prove that $A$ is strongly generic. By Proposition 3.11, it suffices
to check that $A$ is locally periodic. Fix a finite $U\subseteq G$ and take an
even natural number $N>v(u)$ for all $u\in U\setminus F_{\infty}$. We claim
that $F_{N}\setminus F_{N+1}\subseteq\operatorname{Per}_{U}(\chi_{A})$.
Indeed, fix $t\in F_{N}\setminus F_{N+1}$ and $u\in U$. If $u\in U\setminus
F_{\infty}$, then $v(t)=N>v(u)$, so $v(ut)=v(u)$ and so
$\chi_{A}(ut)=\chi_{A}(u)$. Furthermore, if $u\in F_{\infty}$, then
$\chi_{A}(ut)=N\bmod 2=0=\chi_{A}(u)$. Hence
$\operatorname{Per}_{U}(\chi_{A})$ contains $F_{N}\setminus F_{N+1}$, so it is
generic.

We also prove that $A$ is not periodic. In fact, we claim that
$\operatorname{Per}(\chi_{A})=F_{\infty}$, which clearly has infinite index.
Note that

$\operatorname{Per}(\chi_{A})=\\{t\in G:At^{-1}=A\\}$

and take any $t\in F_{\infty}$. Then $F_{n}t^{-1}=F_{n}$ for any
$n\in\mathbb{N}$, thus

$At^{-1}=\bigcup_{n=1}^{\infty}(F_{2n-1}\setminus
F_{2n})t^{-1}=\bigcup_{n=1}^{\infty}(F_{2n-1}\setminus F_{2n})=A$

and so $t\in\operatorname{Per}(\chi_{A})$. Now take any $t\in G\setminus
F_{\infty}$ and let $n=v(t)\in\mathbb{N}$ so that $t\in F_{n}\setminus
F_{n+1}$. If $n$ is even, then

$\displaystyle A$ $\displaystyle=(F_{1}\setminus
F_{2})\cup\ldots\cup(F_{n-1}\setminus
F_{n})\cup(F_{n+1}\phantom{t^{-1}}\setminus
F_{n+2}\phantom{t^{-1}})\cup\ldots$ $\displaystyle At^{-1}$
$\displaystyle=(F_{1}\setminus F_{2})\cup\ldots\cup(F_{n-1}\setminus
F_{n})\cup(F_{n+1}t^{-1}\setminus F_{n+2}t^{-1})\cup\ldots$

It follows that $A\cap F_{n}\subseteq F_{n+1}$ and $At^{-1}\cap F_{n}\subseteq
F_{n+1}t^{-1}$. In particular, the sets on the left side of the inclusions are
disjoint. But they are also non-empty, hence $A\neq At^{-1}$. If $n$ is odd,
we show in a similar manner that $A\neq At^{-1}$. In either case
$t\notin\operatorname{Per}(\chi_{A})$, as desired.

###  3.3 Uniformly strongly generic sets

In this subsection we investigate a particular strengthening of strong
genericity. Assume $G$ is a group definable in a first-order structure $M$ and
let $M\preccurlyeq N$. Recall that one of the central ideas of the thesis is
to relate the Ellis groups of $S_{\mathrm{ext},G}(M)$ and
$S_{\mathrm{ext},G}(N)$ using image algebras. This could be particularly
effective in the following case: assume some image algebra consists of
definable sets. Then it can be naturally lifted to $N$. If the lifted sets
still generate a generic $G^{N}$-algebra, we can extend it to an image algebra
and possibly conclude some relation between the corresponding Ellis groups.

However, given a strongly generic definable subset $A\subseteq G$, it may not
be the case that $A^{N}\subseteq G^{N}$ is also strongly generic. For
instance, assume $N$ is $\aleph_{0}$-saturated. Then $A^{N}$ is strongly
generic if and only if $A$ satisfies the following property:

For each Boolean term $\tau(X_{1},\ldots,X_{n})$ there is
$m_{\tau}\in\mathbb{N}$ such that for any $g_{1},\ldots,g_{n}\in G$ the set
$\tau(g_{1}A,\ldots,g_{n}A)$ is either empty or $m_{\tau}$-generic, (3)

where a set $B\subseteq G$ is said to be $m$-generic if
$G=h_{1}B\cup\ldots\cup h_{m}B$ for some $h_{1},\ldots,h_{m}\in G$. In fact,
(3) holds if and only if $A^{N}$ is strongly generic in $G^{N}$ for every
$N\succcurlyeq M$. For this reason we devote some attention to investigating
the property.

######  Definition 3.26.

Assume $G$ is an arbitrary group and $\mathcal{A}$ is a $G$-algebra.

  1. (a)

When $m\in\mathbb{N}$, an element $B\in\mathcal{A}$ is said to be _$m$
-generic_ provided that $1=h_{1}B\vee\ldots\vee h_{m}B$ for some
$h_{1},\ldots,h_{m}\in G$.

  2. (b)

An element $A\in\mathcal{A}$ is called _uniformly strongly generic_ ,
abbreviated USG, if it satisfies the property (3).

Although the generality of the definition will be useful at times, we mainly
focus on uniformly strongly generic subsets of $G$, that is, sets $A\subseteq
G$ that are USG as elements of the $G$-algebra $\mathcal{P}(G)$. The next two
remarks are straightforward to prove:

######  Remark 3.27.

An element $A\in\mathcal{A}$ is uniformly strongly generic if and only if it
satisfies (3) restricted to terms $\tau$ of the form

$\tau(X_{1},\ldots,X_{I},Y_{1},\ldots,Y_{J})=X_{1}\wedge\ldots
X_{I}\wedge(Y_{1})^{c}\wedge\ldots\wedge(Y_{J})^{c}.$

######  Remark 3.28.

Assume $\varphi:\mathcal{A}\to\mathcal{B}$ is a homomorphism of $G$-algebras
and $A\in\mathcal{A}$ is USG. Then $\varphi(A)\in\mathcal{B}$ is USG.

Propositions 3.10 and 3.11 and the corresponding notions have a natural
counterpart in the uniform setting.

######  Definition 3.29.

Assume $f\in 2^{G}$.

  1. (a)

$f$ is _uniformly self-replicating_ if it satisfies the condition

$(\forall\underset{\text{finite}}{U\subseteq
G})(\exists\underset{\text{finite}}{V\subseteq G})(\forall g\in G)(\exists
h\in G)\,h\odot(f\restriction U)\subseteq f\restriction Vg,$

where additionally $|V|$ depends only on $|U|$.

  2. (b)

$f$ is _uniformly almost periodic_ in $2^{G}$ if for each finite $U\subseteq
G$ there is a finite $W\subseteq G$ such that $\operatorname{cl}(G\odot
f)\subseteq W^{-1}\odot[f\restriction U]$, and $|W|$ depends only on $|U|$.

  3. (c)

$f$ is _uniformly locally periodic_ if for each finite $U\subseteq G$ there is
$m\in\mathbb{N}$ such that $\operatorname{Per}_{U}(f)$ is $m$-generic, where
$m$ depends only on $|U|$.

######  Remark 3.30.

For any $A\subseteq G$, the following are equivalent:

  1. (i)

$A$ is uniformly strongly generic;

  2. (ii)

$\chi_{A}$ is uniformly self-replicating;

  3. (iii)

$\chi_{A}$ is uniformly almost periodic in $2^{G}$;

  4. (iv)

$\chi_{A}$ is uniformly locally periodic.

###### Proof.

Essentially repeat the reasoning from Propositions 3.10 and 3.11. ∎

If $A\subseteq G$, we have the following implications:

$A\text{ is periodic}\implies A\text{ is USG}\implies A\text{ is strongly
generic}.$

None of these can be reversed without extra assumptions. We later show (see
Corollary 4.15 and Theorem 4.25) that in many groups there are strongly
generic sets that are not USG. Now let us show examples of uniformly strongly
generic sets that are not periodic.

######  Example 3.31.

For any finite group $\mathcal{G}\neq\\{e\\}$ let
$G=\mathcal{G}^{\omega}\rtimes_{\varphi}\operatorname{Sym}(\omega)$, where the
underlying action of the semidirect product is
$\varphi_{\sigma}(s)=s\circ\sigma^{-1}$ for
$\sigma\in\operatorname{Sym}(\omega)$, $s\in\mathcal{G}^{\omega}$. Consider
the subset $A\subseteq G$ defined as

$A=\\{s\in\mathcal{G}^{\omega}:s(0)=e\\}\times\operatorname{Sym}(\omega).$

We will show that $A$ is uniformly strongly generic but not periodic. Given
$n\in\omega$, $g\in\mathcal{G}$, let

$A_{n}^{g}=\\{s\in\mathcal{G}^{\omega}:s(n)=g\\}\times\operatorname{Sym}(\omega),$

so that $A=A_{0}^{e}$. Then for any $\langle s,\sigma\rangle\in G$ we have
that

$\displaystyle\langle s,\sigma\rangle A$ $\displaystyle=\\{\langle
s,\sigma\rangle\langle t,\tau\rangle:t(0)=e\\}$ $\displaystyle=\\{\langle
s\varphi_{\sigma}(t),\sigma\tau\rangle:t(0)=e\\}$ $\displaystyle=\\{\langle
s\varphi_{\sigma}(t),\sigma\tau\rangle:(s\varphi_{\sigma}(t))\big{(}\sigma(0)\big{)}=s(\sigma(0))\\}$
$\displaystyle=A_{\sigma(0)}^{s(\sigma(0))}.$

It follows that the $G$-algebra $\mathcal{A}\leqslant\mathcal{P}(G)$ generated
by $A$ is

$\mathcal{A}=\\{C\times\operatorname{Sym}(\omega):C\subseteq\mathcal{G}^{\omega}\text{
is clopen}\\},$

where the topology on $\mathcal{G}^{\omega}$ is that of a product of finite
discrete groups. In particular, $\mathcal{A}$ is infinite, so $A$ is not
periodic. On the other hand, consider any non-empty set $B\in\mathcal{A}$ of
the form

$B=a_{1}A\cap\ldots\cap a_{I}A\cap b_{1}(G\setminus A)\cap\ldots\cap
b_{J}(G\setminus A),$

where $a_{i},b_{j}\in G$, and write it as

$B=A_{n_{1}}^{g_{1}}\cap\ldots\cap A_{n_{I}}^{g_{I}}\cap(G\setminus
A_{m_{1}}^{h_{1}})\cap\ldots\cap(G\setminus A_{m_{J}}^{h_{J}}).$

Then $B$ is $|\mathcal{G}|^{I}\cdot 2^{J}$-generic. Therefore $A$ is uniformly
strongly generic.

We show in Appendix A that the Ellis group of $S(\mathcal{A})$ is isomorphic
to $\mathcal{G}$.

######  Example 3.32.

Let $G=\mathbb{F}_{x,y}$ denote the free group on generators $x$, $y$. Write
$S=\\{x,y,x^{-1},y^{-1}\\}$ and define $S^{*}$ to be the set of all words over
$S$, i.e. finite sequences of elements of $S$. For any non-empty irreducible
word $w\in S^{*}$ let $T_{w}\subseteq G$ denote the set of elements
represented by irreducible words beginning with $w$. Consider the subset
$A:=T_{x}\subseteq G$. Clearly it is not periodic because

$\\{x^{n}A:n\in\omega\\}=\\{T_{x^{n+1}}:n\in\omega\\}$

is infinite. We will check that $A$ is uniformly strongly generic.

We claim that for any $a\in A$, the set $a^{-1}A$ contains $T_{y^{n}}\cup
T_{y^{-n}}$ for sufficiently large $n\in\omega$. Indeed, let $xuy^{k}$ be the
irreducible word representing $a$, where $k\in\mathbb{Z}$ and $u\in S^{*}$
does not end with $y$ nor $y^{-1}$. Now let $n>|k|$ and take any $g\in
T_{y^{n}}\cup T_{y^{-n}}$. If $g\in T_{y^{n}}$, we can write it as an
irreducible word $y^{n}v$, where $v\in S^{*}$ does not begin with $y^{-1}$.
Then $ag=xuy^{n+k}v$ is an irreducible representation, hence $ag\in A$. If
$g\in T_{y^{-n}}$, we show in a similar way that $ag\in A$. Hence
$T_{y^{n}}\cup T_{y^{-n}}\subseteq a^{-1}A$ whenever $n>|k|$.

Also for any $b\notin A$, the set $b^{-1}(G\setminus A)$ contains
$T_{y^{n}}\cup T_{y^{-n}}$ for sufficiently large $n\in\omega$. To see this,
let $u\in S^{*}$ be an irreducible word representing $b$. Then $u$ does not
begin with $x$, hence $u^{-1}x$ is irreducible and $b^{-1}A=T_{u^{-1}x}$.
Clearly $T_{y^{n}}\cup T_{y^{-n}}$ is disjoint from this set when $n\in\omega$
is sufficiently large.

It suffices to show that any non-empty set $B$ in the $G$-algebra generated by
$A$ is $2$-generic. Take any such $B$ and assume without loss of generality
that $e\in B$ and $B$ is of the form

$B=a_{1}^{-1}A\cap\ldots\cap a_{I}^{-1}A\cap b_{1}^{-1}(G\setminus
A)\cap\ldots\cap b_{J}^{-1}(G\setminus A),$

so that $a_{i}\in A$ and $b_{j}\notin A$. It then follows from the previous
observations that $T_{y^{n}}\cup T_{y^{-n}}\subseteq B$ for some $n\in\omega$.
But the last set is $2$-generic because

$G=T_{y^{n}}\cup y^{2n-1}T_{y^{-n}},$

thus also $B$ is $2$-generic. Therefore $A$ is uniformly strongly generic.

For the sake of the next example we introduce some notation. When a group $G$
acts on a set $X$ and $x\in X$, let

$d_{x}A=\\{g\in G:gx\in A\\}.$

The set $\mathcal{P}(X)$ carries a natural structure of a $G$-algebra such
that the map $d_{x}:\mathcal{P}(X)\to\mathcal{P}(G)$ is a $G$-algebra
homomorphism.

######  Remark 3.33.

Assume $G$ acts transitively on $X$. Assume $A\subseteq X$ is uniformly
strongly generic and the set $\\{gA:g\in G\\}$ is infinite. Then for any $x\in
X$ the set $d_{x}A$ is uniformly strongly generic and not periodic.

###### Proof.

By Remark 3.28, $d_{x}A$ is USG. Furthermore, because of the transitivity of
the action, $d_{x}$ is a monomorphism, so the family $\\{g\cdot d_{x}A:g\in
G\\}$ is infinite. Hence $d_{x}A$ is not periodic. ∎

######  Example 3.34.

Let $G=\text{PSL}(2,\mathbb{R})$ and consider the usual action of $G$ on the
projective line $\mathbf{P}(\mathbb{R})=\mathbb{R}\cup\\{\infty\\}$. Let
$A=[0,1)\subseteq\mathbf{P}(\mathbb{R})$ and

$B=d_{0}A=\\{M\in G:M\cdot 0\in[0,1)\\}.$

We will check that $B$ is uniformly strongly generic and not periodic. Using
Remark 3.33, it suffices to show that $A$ is uniformly strongly generic and
has infinite orbit. Given distinct $y,z\in\mathbf{P}(\mathbb{R})$, let $[y,z)$
denote the usual interval if $y<z$ and $\mathbf{P}(\mathbb{R})\setminus[z,y)$
otherwise, where $\mathbb{R}<\infty$. Note that the action of $G$ on
$\mathbf{P}(\mathbb{R})$ is $2$-transitive, thus

$\\{M\cdot[0,1):M\in G\\}=\\{[y,z):y,z\in\mathbf{P}(\mathbb{R}),y\neq z\\}.$

In particular, $A=[0,1)$ has infinite orbit.

Furthermore, take any non-empty set $C$ of the form

$C=a_{1}A\cap\ldots a_{I}A\cap b_{1}(X\setminus A)\cap\ldots\cap
b_{J}(X\setminus A),$

where $a_{1},\ldots,a_{I},b_{1},\ldots,b_{J}\in G$. Then $C$ contains an
interval of the form $[y,z)$ for some $y<z$. We can find $M\in G$ such that
$M\cdot[y,z)=[z,y)$, so $C$ is $2$-generic. Therefore $A$ is uniformly
strongly generic.

The last example is based on [HPP06, Remark 5.2 (iv)] where $G$ was an example
of a group definable in an o-minimal structure that is not definably amenable.
The lack of definable amenability is not a coincidence, as made clear by the
next theorem due to Newelski.

######  Theorem 3.35.

Assume $A\subseteq G$ is uniformly strongly generic and the formula
$\varphi(x;y)\equiv x\in yA$ does not have the independence property.
Moreover, assume that the structure $(G,\cdot,A)$ is definably amenable. Then
$A$ is periodic.

###### Proof.

Assume for contradiction that $A$ is not periodic and take a translation
invariant Keisler measure $\mu$ on $G$. By Theorem 3.7, $\varphi(x;y)$ is
unstable, so by [She90, Theorem 4.7 (2)], for some $n\in\mathbb{N}$ and
$\eta\in\\{0,1\\}^{n}$ the formula

$\psi(x;y_{1},\ldots,y_{n}):=\bigwedge_{i=1}^{n}\varphi(x;y_{i})^{\eta_{i}}$

has the strict order property. Thus for any $K<\omega$ we can find
$\overline{b_{0}},\ldots,\overline{b_{K}}\subseteq G$ such that

$\psi(G;\overline{b}_{0})\supsetneq\psi(G;\overline{b}_{1})\supsetneq\ldots\supsetneq\psi(G;\overline{b}_{K}).$

Since $A$ is USG, we can find $m\in\mathbb{N}$ such that
$\psi(G;\overline{b})\setminus\psi(G;\overline{c})$ is either empty or
$m$-generic for any $\overline{b},\overline{c}\subseteq G$. It follows that
$\mu\big{(}\psi(G;\overline{b}_{k})\setminus\psi(G;\overline{b}_{k+1})\big{)}\geqslant\frac{1}{m}$
for $k<K$, which is a contradiction when $K>m$. ∎

We have shown three examples of a group with a subset that is uniformly
strongly generic but not periodic. However, given a concrete group $G$, it is
not always easy to verify whether it has a subset with this property. In
particular, we were unable to answer the following question:

######  Question 3.36.

Does the group $(\mathbb{Z},+)$ have a uniformly strongly generic subset which
is not periodic?

Note that any such subset would have to be quite complicated; for instance,
when $A\subseteq\mathbb{Z}$ is such a set, by the last theorem the formula
$\varphi(x;y)\equiv x\in y+A$ must have the independence property. We share
some ideas related to the problem in Appendix B.

##  4 The Ellis group across models

Consider a group $G$ definable in a first-order structure $M$. We wish to
investigate the relation between Ellis groups computed in $M$ and its
extension $M^{*}$ in various scenarios. Although the ultimate goal is to
relate the Ellis groups of the “full” flows $S_{\mathrm{ext},G}(M)$ and
$S_{\mathrm{ext},G}(M^{*})$, the task seems very difficult in general.
Therefore we focus on some variants of the problem which we find more
tractable.

Our approach is based on the following well known observation, which resembles
the idea of localization from stability theory. Consider any $G$-algebra
$\mathcal{A}\leqslant\mathcal{P}(G)$. By Theorem 2.24, $E(S(\mathcal{A}))$ is
isomorphic to $S(\mathcal{A}^{d})$ as a $G$-flow and semigroup, so we may
compute the Ellis group of $S(\mathcal{A})$ directly from
$S(\mathcal{A}^{d})$. Consider any direct system $\langle\mathcal{A}_{i}:i\in
I\rangle$ of $G$-subalgebras $\mathcal{A}_{i}\leqslant\mathcal{A}$ with
inclusions such that $\mathcal{A}=\bigcup_{i\in I}\mathcal{A}_{i}$, e.g. the
system of all finitely generated $G$-subalgebras. Then $\mathcal{A}^{d}$ is
the union and the direct limit of the system $\langle\mathcal{A}_{i}^{d}:i\in
I\rangle$. By Stone duality, the $G$-flow $S(\mathcal{A}^{d})$ is the
projective limit of the inverse system $\langle S(\mathcal{A}_{i}^{d}):i\in
I\rangle$ with restrictions. By Remark 2.23, these restrictions are also
semigroup epimorphisms with respect to $\ast$.

Let $\pi_{j}:S(\mathcal{A}^{d})\to S(\mathcal{A}_{j}^{d})$ denote the
restriction. Choose a minimal ideal
$I\trianglelefteqslant_{m}S(\mathcal{A}^{d})$ with an idempotent $u\in J(I)$.
Then $I_{j}:=\pi_{j}[I]\trianglelefteqslant_{m}S(\mathcal{A}_{j}^{d})$ and
$u_{j}:=\pi_{j}(u)\in J(I_{j})$ and $\pi_{j}[uI]=u_{j}I_{j}$ for each $j\in
I$. From here it is not hard to show that the group $uI$ is isomorphic to the
projective limit of the system $\langle u_{j}I_{j}:j\in I\rangle$.

It follows that the Ellis group of $S(\mathcal{A})$ can be retrieved from the
Ellis groups of $S(\mathcal{A}_{j})$. Thus a good approach to understanding
the “full” Ellis group may be to identify and study some tractable
$G$-subalgebras of the target algebra together with their “partial” Ellis
groups. In this section we focus on a particular algebra
$\mathcal{SBP}\leqslant\mathcal{P}(G)$, introduced below. This algebra is
strictly related to the algebra of all externally definable subsets of $G$ at
least in the case when $G$ is definable in a densely ordered o-minimal
structure, as shown in Corollary 4.34.

######  Definition 4.1.

Let $X$ be a topological space. We say that $A\subseteq X$ has the _strong
Baire property_ (abbreviated SBP) if it satisfies any of the following
equivalent conditions:

  * •

$A=U\Delta M$ for some open $U\subseteq X$ and nowhere dense $M\subseteq X$;

  * •

$\operatorname{bd}(A)$ has empty interior;

  * •

$\operatorname{int}(A)\cup\operatorname{int}(X\setminus A)$ is dense.

Throughout the section $G$ is a topological group (meaning that inversion and
multiplication are continuous) satisfying various assumptions specified
individually in subsections. The family of all subsets of $G$ having the
strong Baire property is a $G$-algebra, which we denote as $\mathcal{SBP}$.
The plan is as follows: given an arbitrary
$\mathcal{A}\leqslant\mathcal{SBP}$, we first describe the Ellis group of
$S(\mathcal{A})$. Next we set up $G$ as a group definable in a model $M$, take
an elementary extension $M\preccurlyeq M^{*}$ and find a natural
$G^{*}$-subalgebra $\mathcal{A}^{*}\leqslant\mathcal{SBP}^{*}$ corresponding
to $\mathcal{A}$. Finally, we apply our description to relate the Ellis groups
of $S(\mathcal{A})$ and $S(\mathcal{A}^{*})$.

###  4.1 Groups with profinite topology

In this subsection we consider an arbitrary group $G$ equipped with the
profinite topology, which makes it a topological group. We first show that
$\mathcal{SBP}$ includes virtually all sets considered in the previous
section, namely:

######  Proposition 4.2.

If $A\subseteq G$ is almost tree-founded, then $A$ has SBP with respect to the
profinite topology.

Although for an almost tree-founded $A\subseteq G$ the set
$\operatorname{int}(A)\cup\operatorname{int}(A^{c})$ clearly contains the
“tree-founded part” of $A$, which is dense in the tree, it does not directly
follow that it is dense in the profinite topology, so the proposition needs a
proper proof. We first state a useful lemma:

######  Lemma 4.3.

Assume $H$ is a group and $H_{0},F_{0}\leqslant H$ are its subgroups. Then
$[H_{0}:H_{0}\cap F_{0}]\leqslant[H:F_{0}]$. If additionally $F_{0}$ has
finite index and some coset of $H_{0}$ is disjoint from some coset of $F_{0}$,
then the inequality is strict.

###### Proof.

Consider the map $\tau:H_{0}/(H_{0}\cap F_{0})\to H/F_{0}$ given by
$\tau\big{(}x(H_{0}\cap F_{0})\big{)}=xF_{0}$. It is injective since if
$x,y\in H_{0}$ and $xF_{0}=yF_{0}$, then $y^{-1}x\in H_{0}\cap F_{0}$ and
$x(H_{0}\cap F_{0})=y(H_{0}\cap F_{0})$. It follows that $|H_{0}/(H_{0}\cap
F_{0})|\leqslant|H/F_{0}|$, which proves the first part. Under the additional
assumption, $\tau$ is not onto: take $y,z\in H$ with $yH_{0}\cap
zF_{0}=\varnothing$, equivalently $H_{0}\cap y^{-1}zF_{0}=\varnothing$. Then
clearly $y^{-1}zF_{0}$ is not in the image of $\tau$. Consequently, the
inequality is strict. ∎

###### Proof of Proposition 4.2..

Take a tree $(T,\mathcal{H},d,v)$ on which $\chi_{A}$ is almost founded.
Assume for contradiction that there is a subgroup $F\leqslant G$ of finite
index with a coset $aF$ disjoint from
$\operatorname{int}A\cup\operatorname{int}A^{c}$. Let $Q$ denote the set of
all $\eta\in T$ satisfying $d_{\eta}H_{\eta}\cap aF\neq\varnothing$. In
particular, $T_{L}\cap Q=\varnothing$.

We claim that there are $\eta\in Q$ and $i,j\in\omega$ such that
$\eta^{\frown}i\notin Q$, $\eta^{\frown}j\in Q$ and

$[H_{\eta}:H_{\eta}\cap F]\text{ is the smallest among all }\eta\in Q.$
($\dagger$)

Indeed, take any $\eta_{0}\in Q$ satisfying $(\dagger)$ and note that by Lemma
4.3, passing to any extension $\eta_{0}\subseteq\eta\in Q$ preserves the
minimality. By assumption, there is a leaf $\theta\in T_{L}$ extending
$\eta_{0}$ and then $\theta\notin Q$. Pick $\eta\in T$ and $i\in\omega$ such
that $\eta_{0}\subseteq\eta\subseteq\eta^{\frown}i\subseteq\theta$ and
$\eta\in Q$, $\eta^{\frown}i\notin Q$. On the other hand, $\eta^{\frown}j\in
Q$ for some $j\in\omega$. Then $\eta,i,j$ satisfy the desired properties.

Since $\eta\in Q$, we can assume that $a\in d_{\eta}H_{\eta}$, so that
$a^{-1}d_{\eta^{\frown}i}\in H_{\eta}$. It follows that the group
$H:=H_{\eta}$, its subgroup
$H_{0}:=H_{\eta^{\frown}i}=H_{\eta^{\frown}j}\leqslant H$ and the subgroup
$F_{0}:=H\cap F\leqslant H$ of finite index satisfy all assumptions of Lemma
4.3, since $a^{-1}d_{\eta^{\frown}i}H_{0}\cap F_{0}=\varnothing$. Therefore
$[H_{\eta^{\frown}j}:H_{\eta^{\frown}j}\cap F]<[H_{\eta}:H_{\eta}\cap F]$,
contradicting $(\dagger)$. ∎

Given $\mathcal{A}\leqslant\mathcal{SBP}$, computing the Ellis group of
$S(\mathcal{A})$ is most convenient when $\mathcal{A}$ is d-closed because of
Theorem 2.24. This situation is not uncommon, since the next proposition shows
that $\mathcal{SBP}^{d}=\mathcal{SBP}$. It follows that when
$\mathcal{A}\leqslant\mathcal{SBP}$ is arbitrary, the d-closure of
$\mathcal{A}$ is still contained in $\mathcal{SBP}$.

######  Proposition 4.4.

$\mathcal{SBP}$ is a d-closed $G$-algebra.

###### Proof.

By Remark 2.20 (iv), it suffices the check the following: if
$A\in\mathcal{SBP}$ and $B\subseteq G$ is in the pointwise closure of
$\\{Ag^{-1}:g\in G\\}$, then $B\in\mathcal{SBP}$. Write
$H\leqslant_{\text{fin}}G$ when $H$ is a subgroup of $G$ of finite index.
Assume for contradiction that $B\notin\mathcal{SBP}$, so there is a coset $Fc$
of a subgroup $F\leqslant_{\text{fin}}G$ in which both $B$ and $B^{c}$ are
dense. Shifting $B$ if necessary, we can assume that $c=e$.

Let $Fc_{1},\ldots,Fc_{n}$ be a list of all right cosets of $F$. Since
$A\in\mathcal{SBP}$, for each $1\leqslant i\leqslant n$ we can find a coset
$H_{i}d_{i}\subseteq Fc_{i}$ of some $H_{i}\leqslant_{\text{fin}}G$ on which
$\chi_{A}$ is constant. By taking the intersection, we can assume that all
$H_{i}$ equal some $H\leqslant_{\text{fin}}G$ and clearly $H\subseteq F$. Let
$Hf_{1},\ldots,Hf_{m}$ denote all right cosets of $H$ in $F$. By assumption,
we can find $a_{j},b_{j}\in Hf_{j}$ such that $a_{j}\in B,b_{j}\notin B$.

Since $B$ is in the pointwise closure of $\\{Ag^{-1}:g\in G\\}$, we can find
$g\in G$ such that $a_{j}\in Ag^{-1},b_{j}\notin Ag^{-1}$ for $j=1,\ldots,m$.
Pick $i\in\\{1,\ldots,n\\}$ so that $g\in Fc_{i}$. Then $d_{i}g^{-1}\in
Fc_{i}g^{-1}=F$, so we can find $j\in\\{1,\ldots,n\\}$ such that
$d_{i}g^{-1}\in Hf_{j}$. It follows that $a_{j}g,b_{j}g\in Hf_{j}g=Hd_{i}$ and
$a_{j}g\in A,b_{j}g\notin A$, contradicting the fact that $\chi_{A}$ is
constant on $Hd_{i}$. ∎

The same proof yields the following generalization of Proposition 4.4:

######  Remark 4.5.

Assume $\mathcal{N}_{0}$ is a family of finite index normal subgroups of $G$
closed under finite intersection. Let $\mathcal{SBP}(\mathcal{N}_{0})$ denote
the $G$-algebra of subsets of $G$ having SBP with respect to the topology
generated by cosets of subgroups from $\mathcal{N}_{0}$. Then
$\mathcal{SBP}(\mathcal{N}_{0})$ is d-closed.

Take any $G$-subalgebra $\mathcal{A}\leqslant\mathcal{SBP}$. We are going to
explicitly compute the Ellis group of $S(\mathcal{A})$. Let

$\mathcal{N}:=\\{N\leqslant G:N\text{ is a normal subgroup of finite
index}\\}.$

The set $\mathcal{N}_{\mathcal{A}}=\mathcal{N}\cap\mathcal{A}^{d}$ with
reverse inclusion is a directed set. Consider the projective system of groups
$(G/N)_{N\in\mathcal{N}_{\mathcal{A}}}$, where for
$N,N^{\prime}\in\mathcal{N}_{\mathcal{A}}$, $N\subseteq N^{\prime}$ we have
the natural quotient map $\pi_{N,N^{\prime}}:G/N\to G/N^{\prime}$.

######  Theorem 4.6.

Assume $\mathcal{A}\leqslant\mathcal{SBP}$. The Ellis group of
$S(\mathcal{A})$ is isomorphic to the projective limit

$\mathcal{G}:=\projlim_{N\in\mathcal{N}_{\mathcal{A}}}G/N.$

Let $\langle\mathcal{N}_{\mathcal{A}}\rangle$ denote the $G$-algebra generated
by $\mathcal{N}_{\mathcal{A}}$. In order to prove the theorem, we will need
two lemmas:

######  Lemma 4.7.

Any $G$-algebra $\mathcal{B}\leqslant\mathcal{P}(G)$ consisting of periodic
sets is d-closed.

###### Proof.

Given any $N\in\mathcal{B}$ and $q\in\beta G$, we need to show that
$d_{q}N\in\mathcal{B}$. Using Corollary 3.6, we can assume that
$N\in\mathcal{N}$. Pick $a\in G$ such that $aN\in q$. We have that
$Ng^{-1}=Na^{-1}$ for $g\in aN$. Thus $Ng^{-1}=Na^{-1}$ holds ultimately as
$\hat{g}\to q$, so by Remark 2.20 (iii),

$d_{q}N=\lim_{\hat{g}\to q}Ng^{-1}=Na^{-1}=a^{-1}N\in\mathcal{B}.\qed$

In particular, $\langle\mathcal{N}_{\mathcal{A}}\rangle$ is d-closed, so
$S(\langle\mathcal{N}_{\mathcal{A}}\rangle)$ equipped with $\ast$ is a
semigroup.

######  Lemma 4.8.

The semigroup $\big{(}S(\langle\mathcal{N}_{\mathcal{A}}\rangle),\ast\big{)}$
is a group isomorphic to $\mathcal{G}$.

###### Proof.

For $N\in\mathcal{N}_{\mathcal{A}}$, define
$\varphi_{N}:S(\langle\mathcal{N}_{\mathcal{A}}\rangle)\to G/N$ so that
$\varphi_{N}(q)$ is the unique coset in $G/N$ that belongs to $q$. Each
$\varphi_{N}$ is a homomorphism by Lemma 2.19 and is clearly continuous, where
$G/N$ is treated as a discrete group. Applying the universal property of
$\mathcal{G}$ to the family
$\langle\varphi_{N}:N\in\mathcal{N}_{\mathcal{A}}\rangle$, we get a continuous
homomorphism
$\varphi:S(\langle\mathcal{N}_{\mathcal{A}}\rangle)\to\mathcal{G}$ satisfying
the usual diagram commutativity. It is easy to see that $\varphi$ is
injective. It is also surjective, because it has dense image and
$S(\langle\mathcal{N}_{\mathcal{A}}\rangle)$ is compact. Hence $\varphi$ is an
isomorphism of semigroups, thus in particular,
$S(\langle\mathcal{N}_{\mathcal{A}}\rangle)$ is a group. ∎

###### Proof of Theorem 4.6..

Without loss of generality we can assume that $\mathcal{A}$ is d-closed, since
in general by Theorem 2.24, we have the semigroup isomorphism
$E(S(\mathcal{A}))\cong S(\mathcal{A}^{d})$ and $E(S(\mathcal{A}^{d}))\cong
S((\mathcal{A}^{d})^{d})=S(\mathcal{A}^{d})$, hence the Ellis group of
$S(\mathcal{A})$ is the same as that of $S(\mathcal{A}^{d})$. Also
$\mathcal{N}_{\mathcal{A}^{d}}=\mathcal{N}_{\mathcal{A}}$.

We first note that for every $A\in\mathcal{A}$:

$A\text{ is nowhere dense}\iff A\text{ is not generic}\iff A\text{ has empty
interior}.$

Indeed, the left-to-right implications are easy and the third one to close the
circle follows from SBP. In particular, the non-generic sets in $\mathcal{A}$
are precisely the nowhere dense sets, so they form a proper $G$-ideal in
$\mathcal{A}$. Therefore we can find $p\in S(\mathcal{A})$ such that every
$A\in p$ is generic. Such $p$ is a generic point of the flow $S(\mathcal{A})$,
so by Fact 2.5, there is a unique minimal subflow
$I\trianglelefteqslant_{m}S(\mathcal{A})$ consisting of all such ultrafilters
$p$.

By Remark 2.23, the restriction $\pi:S(\mathcal{A})\to
S(\langle\mathcal{N}_{\mathcal{A}}\rangle)$ is a semigroup epimorphism.
Therefore by Lemma 4.8 and Corollary 2.9, it remains to show that
$I\cap\pi^{-1}[\\{\hat{e}\\}]\subseteq J(I)$. So take any $u\in I$ such that
$\pi(u)=\hat{e}$, which means that $\mathcal{N}_{\mathcal{A}}\subseteq u$.
Assume for contradiction that $u\ast u\neq u$ and take $A\in\mathcal{A}$ such
that $A\in u$ and $A\notin u\ast u$, i.e. $A\setminus d_{u}A\in u$. Since
$u\in I$, the set $A\setminus d_{u}A$ has non-empty interior, so $aF\subseteq
A\setminus d_{u}A$ for some $a\in G$ and $F\in\mathcal{N}$. Letting
$B:=a^{-1}A\in\mathcal{A}$ we get $F\subseteq B\setminus d_{u}B$.

If $F\in\mathcal{A}$, the rest of the proof is easy: we have that $F\in u$, so
$B\in u$. But $e\in F$ so $e\notin d_{u}B$, which means $B\notin u$, a
contradiction. In general we will replace $B$ with some $C\in\mathcal{A}$ and
$F$ with some $N\in\mathcal{N}_{\mathcal{A}}$ with similar properties, so that
this argument will work.

######  Claim.

There is $C\in\mathcal{A}$ of the form $\tau(f_{1}B,\ldots,f_{m}B)$, where
$f_{1},\ldots,f_{m}\in F$ and $\tau(X_{1},\ldots,X_{m})$ is a nonconstant
positive Boolean term, such that $C$ is a union of cosets of $F$.

_Proof of Claim._ Suppose not and take
$C=\tau(f_{1}B,\ldots,f_{m}B)\in\mathcal{A}$, where $f_{1},\ldots,f_{m}\in F$
and $\tau(X_{1},\ldots,X_{m})$ is as above, such that the number of cosets of
$F$ on which $\chi_{C}$ is constant is maximal. Take a coset $Fb$ on which
$\chi_{C}$ is not constant. Since $C$ has SBP, there is a coset $Hc\subseteq
Fb$ of a subgroup $H\leqslant G$ of finite index on which $\chi_{C}$ is
constant. We can assume that $b=c$ and so $Hb\subseteq Fb$, hence $H\subseteq
F$. Let $g_{1}H,\ldots,g_{n}H$, where $g_{1},\ldots,g_{n}\in F$, be a list of
all left cosets of $H$ in $F$.

Two cases are possible: either $\chi_{C}\restriction Hb\equiv 0$ or
$\chi_{C}\restriction Hb\equiv 1$. In the first case consider the set

$C^{\prime}:=\bigcap_{i=1}^{n}g_{i}C\in\mathcal{A}.$

Any coset of $F$ contained in or disjoint from $C$ will remain so with respect
to $C^{\prime}$. Moreover, $\chi_{C^{\prime}}\restriction Fb\equiv 0$, which
contradicts maximality in the definition of $C$. In the second case we
consider

$C^{\prime}:=\bigcup_{i=1}^{n}g_{i}C\in\mathcal{A}$

and arrive at a contradiction in a similar manner. This proves the claim. ∎

Take $C\in\mathcal{A}$ as in the claim. Since $F\cap d_{u}B=\varnothing$, for
any $f\in F$ we have that $fB\notin u$. Also $F\subseteq B$, so since $C$ is a
nontrivial positive Boolean combination of sets $fB$, where $f\in F$, it
follows that $F\subseteq C$ and $C\notin u$. By definition $C$ is periodic,
hence by Corollary 3.6, it can be expressed as a union of cosets of some
$N\in\mathcal{N}_{\mathcal{A}}$. Since $e\in C$, we have that $N\subseteq C$
and therefore $N\notin u$, contradicting the fact that
$\mathcal{N}_{\mathcal{A}}\subseteq u$. ∎

######  Example 4.9.

Let $G=(\mathbb{Z},+)$ and consider the set $A_{0}\subseteq\mathbb{Z}$ from
Example 3.22, defined by

$\chi_{A_{0}}(k)=\begin{cases}0&\text{if }\ k=0,\\\ v_{2}(k)\bmod 2&\text{if
}\ k\neq 0.\end{cases}$

Let $\mathcal{A}=\langle A_{0}\rangle\leqslant\mathcal{P}(\mathbb{Z})$. We
will show that the Ellis group of $S(\mathcal{A})$ is $\mathbb{Z}_{2}$, the
additive group of the $2$-adic integers.

First note that for any $n\in\mathbb{N}$ we have
$2^{n}\mathbb{Z}\in\mathcal{A}$. Indeed, the set

$B:=A_{0}\cap(4^{n}+A_{0})=\bigcup_{k=0}^{n-1}2^{2k+1}(2\mathbb{Z}+1)\in\mathcal{A}$

is a union of an odd number of cosets of $4^{n}\mathbb{Z}$. Therefore it is
periodic and $\operatorname{Per}(B)=4^{n}\mathbb{Z}$, so
$4^{n}\mathbb{Z}\in\mathcal{A}$ by Corollary 3.5 and therefore
$2^{n}\mathbb{Z}\in\mathcal{A}$ as well.

So
$\\{2^{n}\mathbb{Z}:n\in\mathbb{N}\\}\subseteq\mathcal{N}\cap\mathcal{A}^{d}$.
It remains to show the reverse inclusion and apply Theorem 4.6, because
$\projlim\mathbb{Z}/2^{n}\mathbb{Z}=\mathbb{Z}_{2}$. Assume for contradiction
that there is some subgroup $N\in\mathcal{N}\cap\mathcal{A}^{d}$ of the form
$N=(2^{n}s)\mathbb{Z}$ where $s>1$ is odd. Using Fact 2.22, it is easy to see
that $\mathcal{A}^{d}$ is generated as a Boolean algebra by
$\\{d_{q}(k+A_{0}):q\in S(\beta\mathbb{Z}),k\in\mathbb{Z}\\}$, so $N$ can be
expressed as

$N=\tau(d_{q_{1}}(k_{1}+A_{0}),\ldots,d_{q_{m}}(k_{m}+A_{0})),$

where $k_{i}\in\mathbb{Z}$, $q_{i}\in\beta\mathbb{Z}$ and
$\tau(X_{1},\ldots,X_{m})$ is a Boolean term. On the one hand, for any
$j\in\mathbb{N}$ we have that $A_{0}\mathbin{\Delta}(2^{j}+A_{0})\subseteq
2^{j}\mathbb{Z}$, so

$\displaystyle N\mathbin{\Delta}(2^{j}+N)$
$\displaystyle\subseteq\bigcup_{i=1}^{m}d_{q_{i}}(k_{i}+A_{0})\mathbin{\Delta}\big{(}2^{j}+d_{q_{i}}(k_{i}+A_{0})\big{)}$
$\displaystyle=\bigcup_{i=1}^{m}\big{(}k_{i}+d_{q_{i}}(A_{0})\big{)}\mathbin{\Delta}\big{(}k_{i}+d_{q_{i}}(2^{j}+A_{0})\big{)}$
$\displaystyle=\bigcup_{i=1}^{m}k_{i}+d_{q_{i}}\big{(}A_{0}\mathbin{\Delta}(2^{j}+A_{0})\big{)}$
$\displaystyle\subseteq\bigcup_{i=1}^{m}k_{i}+d_{q_{i}}(2^{j}\mathbb{Z}).$

Since each $d_{q_{i}}(2^{j}\mathbb{Z})$ is a coset of $2^{j}\mathbb{Z}$ (see
Lemma 4.7), the set $N\mathbin{\Delta}(2^{j}+N)$ is contained in a union of
$m$ cosets of $2^{j}\mathbb{Z}$. On the other hand,

$N\mathbin{\Delta}(2^{j}+N)=(2^{n}s)\mathbb{Z}\mathbin{\Delta}(2^{j}+(2^{n}s)\mathbb{Z})\supseteq(2^{n}s)\mathbb{Z}.$

When $j\in\mathbb{N}$ is sufficiently big, it is a contradiction.

######  Example 4.10.

Assume $(G,\cdot)$ is a group definable in a stable structure $M$ and let
$\mathcal{A}=\operatorname{Def}_{G}(M)$ denote the $G$-algebra of subsets of
$G$ definable in $M$. By stability, $\mathcal{A}$ is equal to
$\operatorname{Def}_{\mathrm{ext},G}(M)$, the $G$-algebra of subsets of $G$
externally definable in $M$. It follows that $\mathcal{A}$ is d-closed. It is
known that in this case the Ellis group of $S(\mathcal{A})=S_{G}(M)$ is
isomorphic to $G/G^{0}$ computed in a monster model. We will reprove this fact
using the results from this subsection. Note that it does not follow
immediately from Theorem 4.6 since $\mathcal{A}$ need not be contained in
$\mathcal{SBP}$.

Observe that $\mathcal{B}=\langle\mathcal{N}_{\mathcal{A}}\rangle$ is the
family of all strongly generic sets in $\mathcal{A}$. Indeed, one inclusion
follows from the fact that periodic sets are strongly generic and the other
one from Theorem 3.7 and Corollary 3.6. It follows that $\mathcal{B}$ is a
(unique) maximal generic subalgebra of $\mathcal{A}$, so by Corollary 2.35, it
is an image algebra.

By Corollary 2.37, the Ellis group of $S(\mathcal{A})$ is isomorphic to the
Ellis group of $S(\mathcal{B})$. Since
$\mathcal{N}_{\mathcal{B}}=\mathcal{N}_{\mathcal{A}}$ and
$\mathcal{B}\leqslant\mathcal{SBP}$, we get from Theorem 4.6 that the Ellis
group of $S(\mathcal{B})$ is isomorphic to
$\projlim_{N\in\mathcal{N}_{\mathcal{A}}}G/N$, which is clearly isomorphic to
$G/G^{0}$ computed in a monster model.

The same reasoning proves that more generally:

######  Corollary 4.11.

Assume $\mathcal{A}\leqslant\mathcal{P}(G)$ is d-closed and all strongly
generic sets in $\mathcal{A}$ are periodic. Then the Ellis group of
$S(\mathcal{A})$ is profinite and isomorphic to
$\projlim_{N\in\mathcal{N}_{\mathcal{A}}}G/N$.

Now we consider two scenarios of making $G$ a group definable in a model $M$
and lifting a $G$-algebra to an elementary extension $M^{*}\succcurlyeq M$.

1\. Fix a subfamily $\mathcal{N}_{0}\subseteq\mathcal{N}$ closed under finite
intersection and taking supergroups from $\mathcal{N}$ and denote by
$\mathcal{SBP}(\mathcal{N}_{0})$ the $G$-algebra of subsets of $G$ having SBP
with respect to the topology generated by cosets of subgroups from
$\mathcal{N}_{0}$. Take any $G$-algebra
$\mathcal{N}_{0}\subseteq\mathcal{A}\leqslant\mathcal{SBP}(\mathcal{N}_{0})$.
By Remark 4.5, we have that $\mathcal{SBP}(\mathcal{N}_{0})$ is d-closed, so
$\mathcal{A}^{d}\leqslant\mathcal{SBP}(\mathcal{N}_{0})$.

######  Lemma 4.12.

We have $\mathcal{SBP}(\mathcal{N}_{0})\cap\mathcal{N}=\mathcal{N}_{0}$, so in
particular $\mathcal{A}^{d}\cap\mathcal{N}=\mathcal{N}_{0}$.

###### Proof.

The right-to-left inclusion is obvious. In order to prove the other one, fix
$N\in\mathcal{SBP}(\mathcal{N}_{0})\cap\mathcal{N}$ and let

$\mathcal{C}=\\{C\in G/N:(\exists
A\in\langle\mathcal{N}_{0}\rangle)\,C\subseteq A\subseteq G\setminus N\\}.$

Clearly $\mathcal{C}$ is finite, so we can find
$B\in\langle\mathcal{N}_{0}\rangle$ containing the union of $\mathcal{C}$ and
disjoint from $N$. Since $B$ is clopen and $N$ has SBP, we can find a coset
$K\subseteq G\setminus B$ of a subgroup in $\mathcal{N}_{0}$ contained in $N$
or $G\setminus N$. We claim that the first possibility must hold.

Assume for contradiction that $K\subseteq G\setminus N$. Then $K\cap
C\neq\varnothing$ for some $N\neq C\in G/N$. Let $A$ denote a union of
finitely many translates of $K$ by elements of $N$ covering $C$. Then
$A\in\langle\mathcal{N}_{0}\rangle$ and $C\subseteq A\subseteq G\setminus N$,
hence $C\in\mathcal{C}$, contradicting the fact that $K\subseteq G\setminus
B$.

It follows that $K\subseteq N$. Let $K$ be a coset of $F\in\mathcal{N}_{0}$.
Then $F\leqslant N$ and the conclusion follows because $\mathcal{N}_{0}$ is
closed under taking supergroups. ∎

Consider the structure $M=(G,\cdot,A)_{A\in\mathcal{A}}$ and take an
elementary extension $M\preccurlyeq
M^{*}=(G^{*},\cdot,A^{*})_{A\in\mathcal{A}}$. The family
$\\{A^{*}:A\in\mathcal{A}\\}$ is a $G$-algebra, but not necessarily a
$G^{*}$-algebra, so let $\mathcal{A}^{*}$ denote the $G^{*}$-algebra generated
by that family. Also let
$\mathcal{N}_{0}^{*}=\\{N^{*}:N\in\mathcal{N}_{0}\\}$, which makes sense as
$\mathcal{N}_{0}\subseteq\mathcal{A}$, and

$\mathcal{N}^{*}:=\\{N\leqslant G^{*}:N\text{ is a normal subgroup of finite
index }\\}.$

Clearly $\mathcal{N}_{0}^{*}\subseteq\mathcal{N}^{*}$ is a family closed under
finite intersection and taking supergroups from $\mathcal{N}^{*}$ and
$\mathcal{N}_{0}^{*}\subseteq\mathcal{A}^{*}\leqslant\mathcal{SBP}(\mathcal{N}_{0}^{*})$,
hence as before
$(\mathcal{A}^{*})^{d}\leqslant\mathcal{SBP}(\mathcal{N}_{0}^{*})$ and
$(\mathcal{A}^{*})^{d}\cap\mathcal{N}^{*}=\mathcal{N}_{0}^{*}$. Let
$\mathcal{E}(M)$ and $\mathcal{E}(M^{*})$ denote the Ellis group of
$S(\mathcal{A})$ and $S(\mathcal{A}^{*})$, respectively. By Theorem 4.6, we
have that

$\displaystyle\mathcal{E}(M)\cong\projlim_{N\in\mathcal{N}_{0}}G/N$ and
$\displaystyle\mathcal{E}(M^{*})\cong\projlim_{N^{*}\in\mathcal{N}_{0}^{*}}G^{*}/N^{*}.$

The inverse systems $(G/N)_{N\in\mathcal{N}_{0}}$ and
$(G^{*}/N^{*})_{N^{*}\in\mathcal{N}_{0}^{*}}$ are isomorphic. Therefore
$\mathcal{E}(M)\cong\mathcal{E}(M^{*})$, so here the Ellis group is preserved
between models.

In the next example we assume that the reader is familiar with the relation
$\preccurlyeq^{*}$ between models, defined in the beginning of Section 2 in
[New12b].

######  Example 4.13.

Consider a group $G$ definable in an arbitrary structure $M$. Let
$\operatorname{Def}_{\mathrm{ext},G}(M)$ denote the $G$-algebra of subsets of
$G$ externally definable in $M$ and

$\displaystyle\mathcal{N}_{0}:=\operatorname{Def}_{\mathrm{ext},G}(M)\cap\mathcal{N},$
$\displaystyle\mathcal{A}:=\operatorname{Def}_{\mathrm{ext},G}(M)\cap\mathcal{SBP}(\mathcal{N}_{0}).$

Now let $M\preccurlyeq^{*}M^{*}$. Since
$\mathcal{A}\subseteq\operatorname{Def}_{\mathrm{ext},G}(M)$, for each
$A\in\mathcal{A}$, the structure $M^{*}$ provides an interpretation
$A^{M^{*}}\in\operatorname{Def}_{\mathrm{ext},G}(M^{*})$. Define
$\mathcal{A}^{*}\leqslant\operatorname{Def}_{\mathrm{ext},G}(M^{*})$ as the
$G^{*}$-algebra generated by the family $\\{A^{M^{*}}:A\in\mathcal{A}\\}$. We
also define $\mathcal{N}_{0}^{*}$ and $\mathcal{N}^{*}$ in the same way as in
the last scenario.

Since
$\mathcal{N}_{0}\subseteq\mathcal{A}\leqslant\mathcal{SBP}(\mathcal{N}_{0})$,
the conclusions from the last scenario remain true here by the same reasoning.
In particular, for the two corresponding algebras
$\mathcal{A}\leqslant\operatorname{Def}_{\mathrm{ext},G}(M)$ and
$\mathcal{A}^{*}\leqslant\operatorname{Def}_{\mathrm{ext},G}(M^{*})$ we have
that the Ellis groups of $S(\mathcal{A})$ and $S(\mathcal{A}^{*})$ are
isomorphic.

2\. Consider any structure $M=(G,\cdot,C,N)$, where $(G,\cdot)$ is a group,
$C\subseteq G$ and $\mathcal{N}_{0}:=\\{N(c,G):c\in C\\}$ is a family of
normal subgroups of $G$ of uniformly bounded index. Take an elementary
extension $M\preccurlyeq M^{*}=(G^{*},\cdot,C^{*},N^{*})$ and let
$\mathcal{N}_{0}^{*}:=\\{N^{*}(c,G^{*}):c\in C^{*}\\}$. Define
$\overline{\mathcal{N}_{0}}$ as the closure of $\mathcal{N}_{0}$ under finite
intersection and supergroups from $\mathcal{N}$, and
$\overline{\mathcal{N}_{0}^{*}}$ in the same way. For any $G$-algebra
$\mathcal{A}\leqslant\mathcal{SBP}(\overline{\mathcal{N}_{0}})$ containing
$\mathcal{N}_{0}$ and $G^{*}$-algebra
$\mathcal{A}^{*}\leqslant\mathcal{SBP}(\overline{\mathcal{N}_{0}^{*}})$
containing $\mathcal{N}_{0}^{*}$, denote by $\mathcal{E}(M)$ the Ellis group
of $S(\mathcal{A})$ and by $\mathcal{E}(M^{*})$ the Ellis group of
$S(\mathcal{A}^{*})$. Reasoning as in the previous scenario, we get

$\displaystyle\mathcal{E}(M)\cong\projlim_{N\in\overline{\mathcal{N}_{0}}}G/N$
and
$\displaystyle\mathcal{E}(M^{*})\cong\projlim_{N^{*}\in\overline{\mathcal{N}_{0}^{*}}}G^{*}/N^{*}.$

This time the Ellis groups need not be isomorphic because the set $A^{*}$ may
be essentially bigger than $A$. However, $\mathcal{E}(M)$ is still a