# Additive averages of multiplicative correlation sequences and applications

Sebastián Donoso  ,  Anh N. Le  ,  Joel Moreira  and  Wenbo Sun  Departamento
de Ingeniería Matemática and Centro de Modelamiento Matemático, CNRS IRL 2807,
Universidad de Chile, Beauchef 851, Santiago, Chile<EMAIL_ADDRESS>Department of Mathematics
Ohio State University
231 W. 18th Ave., Columbus, OH 43210<EMAIL_ADDRESS>Mathematics Institute
University of Warwick
Coventry, UK<EMAIL_ADDRESS>Department of Mathematics
Virginia Polytechnic Institute and State University
225 Stanger Street, Blacksburg VA, 24061-1026, USA<EMAIL_ADDRESS>
###### Abstract.

We study sets of recurrence, in both measurable and topological settings, for
actions of $(\mathbb{N},\times)$ and $(\mathbb{Q}^{>0},\times)$. In
particular, we show that autocorrelation sequences of positive functions
arising from multiplicative systems have positive additive averages. We also
give criteria for when sets of the form
$\\{(an+b)^{\ell}/(cn+d)^{\ell}:n\in\mathbb{N}\\}$ are sets of multiplicative
recurrence, and consequently we recover two recent results in number theory
regarding completely multiplicative functions and the Omega function.

###### 2010 Mathematics Subject Classification:

Primary: 37B20 ; Secondary: 37A45, 11N37

The first author was supported by ANID/Fondecyt/1200897 and Centro de
Modelamiento Matemático (CMM), ACE210010 and FB210005, BASAL funds for centers
of excellence from ANID-Chile.

##  1\. Introduction and results

An equation of several variables is _partition regular_ if for any finite
coloring of $\mathbb{N}$, there exists a solution to the equation with all
variables having the same color. An old question of Erdős and Graham [10] asks
whether the equation $x^{2}+y^{2}=z^{2}$ is partition regular. A partial
answer was obtained in 2016, when the conjecture was confirmed in the case of
two colors [15]. In general, the answer to Erdős and Graham’s question is
still unknown even if we just require two of the variables to have the same
color.

######  Conjecture 1.1.

For any finite coloring of $\mathbb{N}$, there exist $x,y$ of the same color
such that $x^{2}+y^{2}$ (or $x^{2}-y^{2}$) is a perfect square.

Frantzikinakis and Host proved in [11] that for any finite coloring of
$\mathbb{N}$, there are $x,y$ of the same color such that $16x^{2}+9y^{2}$ is
a perfect square. Expanding on these ideas, the fourth author established an
analogue of 1.1 where $\mathbb{N}$ is replaced with the ring of integers of a
larger number field (see [22, 23]). However the methods used there do not seem
to apply to $\mathbb{N}$.

The above results were proved by first recasting the combinatorial problems
into questions in ergodic theory, in particular about sets of return times in
multiplicative measure preserving systems. For a semigroup $G$, a _topological
$G$-system_ is a pair $(X,T)$ where $X$ is a compact metric space and $T$ is a
$G$-action on $X$, i.e. for $g\in G$, $T_{g}\colon X\to X$ is a homeomorphism
and for $g,h\in G$, $T_{g}\circ T_{h}=T_{gh}$. Let $\mathcal{B}$ be the Borel
$\sigma$-algebra and $\mu$ be a probability measure on $X$ such that
$\mu(T_{g}^{-1}A)=\mu(A)$ for all $A\in\mathcal{B}$. Then the tuple
$(X,\mathcal{B},\mu,T)$ is called a _measure preserving $G$-system_. By a
_multiplicative measure preserving system_ , we mean a measure preserving
$(\mathbb{N},\times)$-system with $\times$ being the multiplication on
$\mathbb{N}$.

For $x,y\in\mathbb{N}$, $x^{2}+y^{2}$ is a perfect square if and only if
$x=k(m^{2}-n^{2})$ and $y=k(2mn)$ for $k,m,n\in\mathbb{N}$ and $m>n$.
Therefore, to answer the case $x^{2}+y^{2}$ of 1.1 using the approach in [11],
it suffices to solve the following conjecture:

######  Conjecture 1.2.

Let $(X,\mathcal{B},\mu,T)$ be a measure preserving
$(\mathbb{N},\times)$-system and $A\in\mathcal{B}$ with $\mu(A)>0$. There
exist $m,n\in\mathbb{N}$ with $m>n$ such that

$\mu(T_{m^{2}-n^{2}}^{-1}A\cap T_{2mn}^{-1}A)>0.$

The idea of using ergodic theoretical methods to solve combinatorial problems
traces back to Furstenberg’s proof of Szemerédi’s Theorem [12] in 1977. The
novelty in Frantzikinakis and Host [11]’s approach is the usage of additive
averages to study multiplicative measure preserving systems. This idea has the
potential to address other unsolved problems in ergodic Ramsey theory
regarding partition regularity of polynomial equations and a primary goal of
this paper is to begin a systematic study of this new tool and to obtain
potential applications.

###  1.1. Additive averages of multiplicative recurrence sequences

Throughout this paper, for a finite set $E$ and function $f\colon
E\to\mathbb{C}$, we use $\mathop{\mathbb{E}}_{n\in E}f(n)$ to denote the
average $\frac{1}{|E|}\sum_{n\in E}f(n)$. By the von Neumann ergodic theorem,
for a multiplicative measure preserving system $(X,\mathcal{B},\mu,T)$, a set
$A\in\mathcal{B}$ with $\mu(A)>0$ and a multiplicative Følner sequence
$(\Phi_{N})_{N\in\mathbb{N}}$ in $\mathbb{N}$ (see Section 2 for definitions),
we have

(1) $\lim_{N\to\infty}\mathop{\mathbb{E}}_{n\in\Phi_{N}}\mu(A\cap
T_{n}^{-1}A)>0.$

On the other hand, it is not clear whether (1) is still true if
$(\Phi_{N})_{N\in\mathbb{N}}$ is replaced with the additive Følner sequence
$([N])_{N\in\mathbb{N}}$. (Here $[N]$ denotes the set $\\{1,\dots,N\\}$.)

Our first result confirms that this is indeed the case. In fact, we obtain a
more general theorem regarding multiple ergodic averages along subsemigroups
of $(\mathbb{N},\times)$.

######  Theorem 1.3.

Let $(X,\mathcal{B},\mu,T)$ be a measure preserving
$(\mathbb{N},\times)$-system and $A\in\mathcal{B}$ with $\mu(A)>0$. Let $G$ be
a subsemigroup of $(\mathbb{N},\times)$. Then for every $\ell\in\mathbb{N}$,

$\limsup_{N\to\infty}\mathop{\mathbb{E}}_{n\in G\cap[N]}\mu(A\cap
T_{n}^{-1}A\cap\ldots\cap T_{n^{\ell}}^{-1}A)>0.$

Theorem 1.3 follows from the slightly more general Theorem 5.4 below. Besides
$\mathbb{N}$ itself, examples of multiplicative semigroups of $\mathbb{N}$
include $\\{m^{2}+Dn^{2}:m,n\in\mathbb{N}\\}$ for some $D\in\mathbb{N}$,
$\\{a^{n}b^{m}:m,n\in\mathbb{N}\\}$ and
$\\{m^{a}n^{b}q^{c}:m,n,q\in\mathbb{N}\\}$ for $a,b,c\in\mathbb{N}$. We remark
that there are two natural ways to take averages along these semigroups. One
way is to enumerate its elements in increasing order as done in Theorem 1.3.
Alternatively, one can exploit the fact that these semigroups admit
parametrizations and take averages on the parameters. Our next result states
that, also with this scheme, the averages of multiple recurrence sequences are
positive. To formulate our result we need the notion of _parametrized
multiplicative function_ , which describes certain parametrizations of
multiplicative semigroups. We postpone the formal definition to Section 5.4,
but here are some illustrative examples of commutative parametrized
multiplicative functions:

  * •

$f\colon\mathbb{N}^{2}\to\mathbb{N}$ defined by
$f(n_{1},n_{2})=n_{1}^{2}+Dn_{2}^{2}$, for any $D\in\mathbb{N}$,

  * •

$f\colon\mathbb{N}^{2}\to\mathbb{N}$ defined by
$f(n_{1},n_{2})=|n_{1}^{2}+n_{1}n_{2}-n_{2}^{2}|$,

  * •

$f\colon\mathbb{N}^{3}\to\mathbb{N}$ defined by
$f(n_{1},n_{2},n_{3})=n_{1}^{a_{1}}n_{2}^{a_{2}}n_{3}^{a_{3}}$ for any
$a_{1},a_{2},a_{3}\in\mathbb{N}$.

######  Theorem 1.4.

Let $(X,\mathcal{B},\mu,T)$ be a measure preserving
$(\mathbb{N},\times)$-system and $A\in\mathcal{B}$ with $\mu(A)>0$. Let
$f\colon\mathbb{N}^{k}\to\mathbb{N}$ be a commutative parametrized
multiplicative function and $\ell\in\mathbb{N}$. Then

$\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]^{k}}\mu\Big{(}A\cap
T_{f(n)}^{-1}A\cap T_{f(n)^{2}}^{-1}A\cap\cdots\cap
T_{f(n)^{\ell}}^{-1}A\Big{)}>0.$

Theorem 1.4 follows from the more general Theorem 5.16 in Section 5. As an
illustrative special case of Theorem 1.4 we obtain

######  Proposition 1.5.

Let $(X,\mathcal{B},\mu,T)$ be a measure preserving
$(\mathbb{N},\times)$-system and let $A\in\mathcal{B}$ with $\mu(A)>0$. Then

$\liminf_{N\to\infty}\mathop{\mathbb{E}}_{m,n\in[N]}\mu(A\cap
T_{m^{2}+n^{2}}^{-1}A)>0.$

###  1.2. Topological recurrence and applications to number theory

A subset $R$ of the set of positive rational numbers $\mathbb{Q}^{>0}$ is
called a _set of (measurable) multiplicative recurrence_ if for every
$(\mathbb{Q}^{>0},\times)$-system $(X,\mathcal{B},\mu,T)$ and
$A\in\mathcal{B}$ with $\mu(A)>0$, there exists $r\in R$ for which $\mu(A\cap
T_{r}^{-1}A)>0$. By Furstenberg’s correspondence principle, $R$ is a set of
multiplicative recurrence if and only if for any $E\subset\mathbb{N}$ of
positive upper multiplicative density (see Section 2 for definitions), there
exist $x,y\in E$ such that $x/y\in R$. If 1.2 is true, then it follows that
$\\{(m^{2}-n^{2})/(2mn):m,n\in\mathbb{N},m>n\\}$ is a set of multiplicative
recurrence.

There is a weaker notion that suffices for our purpose of studying partition
regularity called sets of topological multiplicative recurrence. A set
$R\subset\mathbb{Q}^{>0}$ is called a _set of topological multiplicative
recurrence_ if for every minimal topological $(\mathbb{Q}^{>0},\times)$-system
$(X,T)$ and $U\subset X$ open and non-empty, there exists $r\in R$ such that
$U\cap T_{r}^{-1}U\neq\emptyset$. It turns out that $R$ is a set of
topological multiplicative recurrence if and only if for any finite coloring
of $\mathbb{N}$, there exist $x,y$ of the same color such that $x/y\in R$ (see
Proposition 2.13 below).

In Section 2 we show that if
$\phi\colon(\mathbb{N},+)\to(\mathbb{Q}^{>0},\times)$ is a homomorphism and
$S\subset\mathbb{N}$ is a set of topological (or measurable) additive
recurrence, then $\phi(S)$ is a set of topological (or measurable)
multiplicative recurrence. Using this fact, and drawing on the many known
examples of sets of additive recurrence (cf. [12, 16, 20]), one can exhibit
examples of sets of multiplicative recurrence: $\\{2^{P(n)}:n\in\mathbb{N}\\}$
where $P\in\mathbb{Z}[x]$ satisfies $P(0)=0$ and $\\{2^{p-1}:p\text{
prime}\\}$. However, all such examples are confined to a sparse semigroup of
the form $\\{a^{n}:n\in\mathbb{N}\\}$, and there are hardly any other known
examples of sets of multiplicative recurrence. Our next result produces a set
of multiplicative recurrence which does not belong to the aforementioned
class.

######  Theorem 1.6.

Let $a,c,\ell\in\mathbb{N}$, $b,d\in\mathbb{Z}$ and let
$R:=\left\\{\left(\frac{an+b}{cn+d}\right)^{\ell}:n\in\mathbb{N}\right\\}$.

  1. (1)

If $a\neq c$, then $R$ is not a set of topological multiplicative recurrence.

  2. (2)

If $a=c$, $b\neq d$ and there exists a prime $p$ such that $p|a$ but $p\nmid
bd$, then $R$ is not a set of topological multiplicative recurrence.

  3. (3)

If $a=c$ and either $a|b$ or $a|d$, then $R$ is a set of topological
multiplicative recurrence. 111In the special case $\ell=a=c=1$ and $b=d=0$,
this result was also observed by Bergelson in [2, Section 5].

Theorem 1.6 follows from combining Propositions 3.2, 3.3 and 3.6.

A function $f\colon\mathbb{N}\to\mathbb{C}$ is called _completely
multiplicative_ if $f(mn)=f(m)f(n)$ for $m,n\in\mathbb{N}$. Klurman and
Mangerel [17] proved that for any completely multiplicative function $f$ with
$|f(n)|=1$ for $n\in\mathbb{N}$, one has
$\liminf_{n\to\infty}|f(n+1)-f(n)|=0$. They remarked that their proof can be
modified to treat the case in which the shift $1$ is replaced by any fixed
$k\in\mathbb{N}$. As an application of Theorem 1.6, we give a strengthening of
their result by allowing both arbitrary shifts and dilations.

######  Corollary 1.7.

Let $f:\mathbb{N}\to\mathbb{C}$ be a completely multiplicative function with
$|f(n)|=1$ for $n\in\mathbb{N}$. Then for all $a,k\in\mathbb{N}$,

$\liminf_{n\to\infty}|f(an+k)-f(an)|=0.$

The proof of Corollary 1.7 is given in Section 3, after the proof of
Proposition 3.6. In the previous corollary, by setting $f=\xi^{\Omega(n)}$,
where $\xi$ is a $q$-th root of $1$ and, for $n\in\mathbb{N}$, $\Omega(n)$ is
the number of prime factors of $n$ counting with multiplicity, one immediately
derives the following enhancement of [17, Corollary 1.3].

######  Corollary 1.8.

For any $a,q,k\in\mathbb{N}$, there exists infinitely many $n\in\mathbb{N}$
such that $\Omega(an+k)\equiv\Omega(an)\bmod q$.

Using (an appropriate version of) the Poincaré recurrence theorem it can be
shown that every subsemigroup of $(\mathbb{Q}^{>0},\times)$ is a set of
multiplicative recurrence. Therefore, to understand which algebraic sets are
sets of multiplicative recurrence, one needs to know which contain a
subsemigroup. Our next theorem characterizes all polynomials of integer
coefficients whose images contain subsemigroups of $(\mathbb{Q}^{>0},\times)$.

######  Theorem 1.9.

For $P\in\mathbb{Z}[x]$, the set $\\{P(n):n\in\mathbb{N}\\}$ contains an
infinite subsemigroup of $(\mathbb{Q}^{>0},\times)$ if and only if
$P(x)=(ax+b)^{d}$ for some $a,d\in\mathbb{N}$, $b\in\mathbb{Z}$ with
$a|b(b-1)$.

Theorem 1.9 follows from the more general Theorem 4.3.

Regarding linear polynomials, we show that containing subsemigroups is not
only a sufficient but also a necessary condition for their images to be sets
of multiplicative recurrence.

######  Theorem 1.10.

For every $a,b\in\mathbb{N}$, $\\{an+b:n\in\mathbb{N}\\}$ is a set of
multiplicative recurrence if and only if $a|b(b-1)$.

Theorem 1.10 is implied by Theorem 3.8 in Section 3.

Lastly, as previously mentioned, it is unknown whether the sets arising from
Pythagorean triples are sets of multiplicative recurrence. However, we show
that these sets do not contain subsemigroups of $(\mathbb{Q}^{>0},\times)$.

######  Proposition 1.11.

Neither of the sets

$\left\\{\frac{m^{2}+n^{2}}{2mn}:m,n\in\mathbb{N}\right\\}\ \text{ and }\
\left\\{\frac{2mn}{m^{2}-n^{2}}:m,n\in\mathbb{N},m>n\right\\}$

contain an infinite subsemigroup of $(\mathbb{Q}^{>0},\times)$.

Proposition 1.11 is proved in Section 4, after Lemma 4.4.

### Organization of the paper

In Section 2 we give the background regarding sets of multiplicative
recurrence in both topological and measurable settings. Section 3 is devoted
to studying the topological recurrence properties of certain algebraic sets
and providing applications in number theory. In particular, Theorem 1.6,
Corollary 1.7, Corollary 1.8 and Theorem 1.10 are proved in this section. In
Section 4, we analyze polynomial configurations and investigate when the image
of $\mathbb{N}$ under a polynomial map contains a semigroup of
$(\mathbb{N},\times)$. Section 5 contains the ergodic theoretical point of
view of this article where Theorem 1.3 and Theorem 1.4 will be proved.
Finally, in Section 6, we study the problem of finding Pythagorean pairs and
triples in dense subsets of finite fields.

### Acknowledgements

We thank Florian Richter for help with the proof of Lemma 3.5 and thank Vitaly
Bergelson for helpful conversations. We also thank the referee for suggestions
that improve the organization and readability of the paper.

##  2\. Background

In this section we collect basic facts about sets of recurrence for the
semigroup $(\mathbb{N},\times)$ of natural numbers and the semigroup
$(\mathbb{Q}^{>0},\times)$ of positive rational numbers, both under
multiplication. Most of the content is well known and presented here for
completeness.

### Notations

In this article, $\mathbb{N}=\\{1,2.\ldots\\}$ denotes the set of positive
natural numbers, and $\mathbb{Q}^{>0}$ denotes the set of positive rational
numbers. For a positive natural number $N$, we let $[N]$ denote the set
$\\{1,\ldots,N\\}$.

###  2.1. Multiplicatively invariant density and Furstenberg’s correspondence

A _Følner sequence_ in a countable semigroup $G$ is a sequence
$\Phi=(\Phi_{N})_{N\in\mathbb{N}}$ of finite subsets of $G$ which are
asymptotically invariant in the sense that for any $g\in G$,

$\lim_{N\to\infty}\frac{|g\Phi_{N}\cap\Phi_{N}|}{|\Phi_{N}|}=1,$

where $g\Phi_{N}=\\{gx:x\in\Phi_{N}\\}$. Given a Følner sequence $\Phi$ on a
semigroup $G$, the _upper density_ of a set $A\subset G$ with respect to
$\Phi$ is the quantity

$\bar{d}_{\Phi}(A):=\limsup_{N\to\infty}\frac{|A\cap\Phi_{N}|}{|\Phi_{N}|}.$

For a subset $A\subset G$ of a semigroup $G$ and an element $g\in G$, we write
$S/g:=\\{x\in G:xg\in A\\}$. It is easy to check from the definition that
$\bar{d}_{\Phi}(A/g)=\bar{d}_{\Phi}(A)$ for any $g\in G$.

The _upper Banach density_ of $A$ is $d^{*}(A):=\sup_{\Phi}\bar{d}_{\Phi}(A)$,
where the supremum is taken over all Følner sequences $\Phi$ on $G$. This
supremum is always achieved, so for each set $A$ there is a Følner sequence
$\Phi$ such that $d^{*}(A)=\bar{d}_{\Phi}(A)$.

A discussion of Følner sequences and invariant densities in the case where
$G=(\mathbb{N},\times)$ is provided in [4]. As a concrete example one can take
the sequence $\Phi_{N}:=\\{x\in\mathbb{N}:x|N!\\}$, but the exact choice will
not be important in our discussion. Note that any Følner sequence for
$(\mathbb{N},\times)$ is also a Følner sequence for
$(\mathbb{Q}^{>0},\times)$.

Given a semigroup $G$, a _measure preserving $G$-system_ is a measure
preserving system $(X,{\mathcal{B}},\mu,T)$ where $(X,{\mathcal{B}},\mu)$ is a
probability space and $T=(T_{g})_{g\in G}$ is an action of $G$ on $X$ by
measure preserving transformations $T_{g}:X\to X$ which satisfy
$T_{gh}=T_{g}\circ T_{h}$ for every $g,h\in G$.

The Furstenberg correspondence principle connects sets with positive upper
density and measure preserving systems. Here is the version we will use. Given
a subset $E$ of a semigroup $G$ and $n\in G$ we let $E/n$ denote $\\{m\in
G:mn\in E\\}$.

######  Theorem 2.1 (Furstenberg’s correspondence principle for semigroups,
cf. [5, Theorem 5.8 and Remark 5.9]).

Let $G$ be a countable commutative semigroup, let $E\subset\mathbb{N}$ and let
$\Phi$ be a Følner sequence in $G$. Then there exists a $G$-system
$(X,{\mathcal{B}},\mu,T)$ and a set $A\in{\mathcal{B}}$ with
$\mu(A)=\bar{d}_{\Phi}(E)$ and, for every $n_{1},\dots,n_{k}\in G$,

$\bar{d}_{\Phi}\big{(}E\cap(E/n_{1})\cap(E/n_{2})\cap\cdots\cap(E/n_{k})\big{)}\geq\mu\big{(}A\cap
T_{n_{1}}^{-1}A\cap\cdots\cap T_{n_{k}}^{-1}A\big{)}.$

###  2.2. Syndetic sets

A subset $S\subset G$ is called _(right) syndetic_ if there exists a finite
set $F\subset G$ such that $S/F:=\bigcup_{x\in F}S/x=G$. The following
proposition is well known and we include its proof for completeness.

######  Proposition 2.2.

A set $S$ is syndetic if and only if it has positive upper density with
respect to any Følner sequence.

###### Proof.

If $S$ is syndetic and $\Phi$ is a Følner sequence, then take $F$ finite so
that $S/F=G$. We have $1=\bar{d}_{\Phi}(G)\leq\sum_{x\in
F}\bar{d}_{\Phi}(S/x)=\bar{d}_{\Phi}(S)\cdot|F|$, so $\bar{d}_{\Phi}(S)>0$.

Conversely, if $S$ is not syndetic and $\Phi$ is a Følner sequence, for every
$N\in\mathbb{N}$ we have that $S/\Phi_{N}\neq G$. Take $x_{N}\in
G\setminus(S/\Phi_{N})$. Then $\Psi_{N}:=x_{N}\Phi_{N}$ is disjoint from $S$.
On the other hand, the sequence $\Psi=(\Psi_{N})_{N\in\mathbb{N}}$ is a Følner
sequence, and so $\bar{d}_{\Psi}(S)=0$. ∎

In order to prove Theorem 1.3, we will need the following theorem of
Furstenberg and Katznelson [14] (see also [4, Page 9]). Recall that a
commutative semigroup is _cancelative_ if and only if it is a subsemigroup of
a group.

######  Theorem 2.3 (Furstenberg-Katznelson [14]).

Let $G$ be a cancelative commutative semigroup and let $T_{1},\dots,T_{\ell}$
be commuting measure preserving actions of $G$ on the probability space
$(X,\mathcal{B},\mu)$. Then for every $A\in\mathcal{B}$ with $\mu(A)>0$, there
exist $\delta>0$ and a syndetic set $S\subset G$ such that for every $g\in S$,

$\mu(A\cap T_{1,g}^{-1}A\cap\cdots\cap T_{\ell,g}^{-1}A)>\delta.$

When $\ell=1$, Theorem 2.3 becomes significantly simpler and is a version of
Khintchine’s recurrence theorem. It turns out that, in this case, it is
possible to extend the scope of Theorem 2.3 by dropping the assumption that
$G$ is cancelative and commutative. This result is probably well known; we
provide a short proof which is almost verbatim taken from the proof of [2,
Theorem 5.1].

######  Theorem 2.4 (Khintchine’s theorem).

Let $G$ be a countable semigroup, let $(T_{g})_{g\in G}$ be an action of $G$
by measure preserving transformations on a probability space
$(X,{\mathcal{B}},\mu)$ and let $A\in\mathcal{B}$ with $\mu(A)>0$. Then for
every $\epsilon>0$ the set

$R:=\big{\\{}g\in G:\mu(A\cap T^{-1}_{g}A)>\mu^{2}(A)-\epsilon\big{\\}}$

is syndetic.

###### Proof.

Suppose, for the sake of a contradiction that $R$ is not syndetic. Choose
$g_{1}\in G\setminus R$ arbitrary. Then, inductively, for each $m\geq 1$ let
$F_{m}=\big{\\{}g_{m}g_{m-1}\cdots g_{n}:1\leq n\leq m\big{\\}}$ and choose
$g_{m+1}\in G\setminus RF_{m}^{-1}$. This is possible if $R$ is not syndetic.
From the construction we have $F_{m}=g_{m}F_{m-1}\cup\\{g_{m}\\}$, and hence
it follows that $F_{m}\cap R\subset\\{g_{m}\\}$ for every $m\in\mathbb{N}$.

Let $N\in\mathbb{N}$ be large, to be determined later depending only on
$\mu(A)$ and $\epsilon$ and let $f=\frac{1}{N}\sum_{n=1}^{N}1_{A}\circ
T_{g_{n}g_{n-1}\cdots g_{1}}$. Clearly $\int_{X}f\,\mathrm{d}\mu=\mu(A)$. By
the Cauchy-Schwarz inequality,

$\displaystyle\mu(A)^{2}$ $\displaystyle\leq$
$\displaystyle\int_{X}f^{2}\,\mathrm{d}\mu=\frac{1}{N^{2}}\sum_{n,m=1}^{N}\mu\big{(}T_{g_{n}g_{n-1}\cdots
g_{1}}^{-1}A\cap T_{g_{m}g_{m-1}\cdots g_{1}}^{-1}A\big{)}$ $\displaystyle=$
$\displaystyle\frac{1}{N^{2}}\left(\sum_{n=1}^{N}\mu(A)+2\sum_{m=2}^{N}\sum_{n=1}^{m-1}\mu\big{(}T_{g_{n}g_{n-1}\cdots
g_{1}}^{-1}A\cap T_{g_{m}g_{m-1}\cdots g_{1}}^{-1}A\big{)}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{N^{2}}\left(N\mu(A)+2\sum_{m=2}^{N}\sum_{n=1}^{m-1}\mu\big{(}A\cap
T_{g_{m}g_{m-1}\cdots g_{n+1}g_{n}}^{-1}A\big{)}\right)$ $\displaystyle=$
$\displaystyle\frac{1}{N}\mu(A)+\frac{2}{N^{2}}\left(\sum_{m=2}^{N}\sum_{g\in
F_{m}}\mu\big{(}A\cap T_{g}^{-1}A\big{)}\right)$ $\displaystyle\leq$
$\displaystyle\frac{1}{N}\mu(A)+\frac{2}{N^{2}}\binom{N}{2}\big{(}\mu^{2}(A)-\epsilon\big{)}$

which is a contradiction if $N$ is large enough. ∎

###  2.3. Sets of multiplicative recurrence

Sets of recurrence for $(\mathbb{Z},+)$-systems, and their connection to
combinatorics were introduced by Furstenberg in his book [13]. Sets of
recurrence for arbitrary countable semigroups were defined and studied in [7],
and the specific case when the semigroup is $(\mathbb{N},\times)$ in [4].

######  Definition 2.5 (Cf. [7, Definition 1.1]).

Let $G$ be a semigroup and let $R\subset G$. We say that $R$ is a _set of
recurrence_ , or a _set of measurable recurrence_ if for any measure
preserving action $T=(T_{g})_{g\in G}$ of $G$ on a probability space $(X,\mu)$
and any $A\subset X$ with $\mu(A)>0$ there exists $r\in R$ such that
$\mu(A\cap T_{r}^{-1}A)>0$.

Examples of sets of recurrence include $G$ itself and any subsemigroup of $G$.
In particular, note that if $G$ admits an identity $e$, then the singleton
$\\{e\\}$ is a set of recurrence. More generally, the usual proof of
Poincaré’s recurrence theorem can be adapted to show that any $\Delta$-set
(i.e., a set of the form $\\{g_{i}g_{i+1}\cdots g_{j}:i\leq j\\}$ for some
sequence $(g_{i})_{i\in\mathbb{N}}$ in $G$) is a set of recurrence.

######  Proposition 2.6.

Let $G,H$ be semigroups and let $\phi:G\to H$ be a homomorphism. If $R\subset
G$ is a set of recurrence, then $\phi(R)$ is a set of recurrence in $H$.

###### Proof.

Composing any measure preserving action of $H$ with $\phi$ yields a measure
preserving action of $G$ on the same space, and the conclusion follows. ∎

We are mainly interested in the semigroup $(\mathbb{N},\times)$. In view of
the following lemma, we will also consider the group
$(\mathbb{Q}^{>0},\times)$. We denote by $d_{\times}^{*}(A)$ the Banach upper
density of $A$ with respect to the multiplicative structure in either
$(\mathbb{N},\times)$ or $(\mathbb{Q}^{>0},\times)$.

The following proposition is well known among experts but we were unable to
find a complete proof in the literature.

######  Proposition 2.7.

Let $R\subset\mathbb{Q}^{>0}$. Then the followings are equivalent:

  1. (1)

$R$ is a set of recurrence, i.e., for any measure preserving action $T$ of
$(\mathbb{Q}^{>0},\times)$ on a probability space $(X,\mu)$ and any $A\subset
X$ with $\mu(A)>0$ there exists $r\in R$ such that $\mu(A\cap T_{r}^{-1}A)>0$.

  2. (2)

For any measure preserving action $T$ of $(\mathbb{N},\times)$ on a
probability space $(X,\mu)$ and any $A\subset X$ with $\mu(A)>0$ there exists
$a,b\in\mathbb{N}$ with $a/b\in R$ such that $\mu(T_{a}^{-1}A\cap
T_{b}^{-1}A)>0$.

  3. (3)

For any $A\subset\mathbb{Q}^{>0}$ with $d_{\times}^{*}(A)>0$ there exists
$r\in R$ such that $A\cap A/r\neq\emptyset$.

  4. (4)

For any $A\subset\mathbb{Q}^{>0}$ with $d_{\times}^{*}(A)>0$ there exists
$r\in R$ such that $d_{\times}^{*}(A\cap A/r)>0$.

  5. (5)

For any $A\subset\mathbb{N}$ with $d_{\times}^{*}(A)>0$ there exist
$a,b\in\mathbb{N}$ with $a/b\in R$ such that $A/a\cap A/b\neq\emptyset$.

  6. (6)

For any $A\subset\mathbb{N}$ with $d_{\times}^{*}(A)>0$ there exist
$a,b\in\mathbb{N}$ with $a/b\in R$ such that $d_{\times}^{*}(A/a\cap A/b)>0$.

###### Proof.

The implications (6)$\Rightarrow$(5) and (4)$\Rightarrow$(3) are trivial. The
implications (1)$\Rightarrow$(4) and (2)$\Rightarrow$(6) follow directly from
the Furstenberg correspondence principle, Theorem 2.1.

The implications (3)$\Rightarrow$(1) and (5)$\Rightarrow$(2) have very similar
proofs, based on an observation of Bergelson, so here we only give the proof
of (5)$\Rightarrow$(2). Suppose we are given a measure preserving action $T$
of $(\mathbb{N},\times)$ on a probability space $(X,\mu)$ and a set $A\subset
X$ with $\mu(A)>0$. Applying [4, Theorem 3.14] we find a set
$P\subset\mathbb{N}$ with $d_{\times}^{*}(P)>0$ and such that for any $a,b\in
P$, $\mu(T_{a}^{-1}A\cap T_{b}^{-1}A)>0$. Using (5) we can then find $a,b\in
P$ with $a/b\in R$, establishing (2).

The implication (2)$\Rightarrow$(1) follows from the fact that any measure
preserving action of $(\mathbb{Q}^{>0},\times)$ induces a natural measure
preserving action of the subsemigroup $(\mathbb{N},\times)$. Finally, the
implication (1)$\Rightarrow$(2) follows from a routine inverse extension
argument, which we provide for completeness.

Let $(T_{n})_{n\in\mathbb{N}}$ be a measure preserving action of
$(\mathbb{N},\times)$ on $(X,\mu)$. The invertible extension of the measure
preserving system $(X,\mu,(T_{n})_{n\in\mathbb{N}})$ is the system
$(Y,\nu,(S_{u})_{u\in\mathbb{Q}^{>0}})$ where

$Y=\left\\{(x_{u})_{u\in\mathbb{Q}^{>0}}\in X^{\mathbb{Q}^{>0}}:(\forall
n\in\mathbb{N})(\forall u\in\mathbb{Q}^{>0})x_{un}=T_{n}x_{u}\right\\},$

$S_{v}(x_{u})_{u\in\mathbb{Q}^{>0}}=(x_{uv})_{u\in\mathbb{Q}^{>0}}\subset
X^{\mathbb{Q}^{>0}}$ and for any cylinder set $C=\prod_{u\in
F}A_{u}\times\prod_{u\in\mathbb{Q}^{>0}\setminus F}X$ where
$F\subset\mathbb{Q}^{>0}$ is finite and $A_{u}\subset X$ are arbitrary
measurable sets we have $\nu(C)=\mu(\bigcap_{u\in F}T_{un}^{-1}A_{u})$, where
$n\in\mathbb{N}$ is any common denominator of all elements of $F$. It is
routine to check that $(Y,\nu,(S_{u})_{u\in\mathbb{Q}^{>0}})$ is indeed a
measure preserving system, and that the projection defined by
$(x_{u})_{u\in\mathbb{Q}^{>0}}\mapsto x_{1}$ is a factor map between the
$(\mathbb{N},\times)$ systems $(Y,\nu,(S_{u})_{u\in\mathbb{N}})$ and
$(X,\mu,(T_{n})_{n\in\mathbb{N}})$. On the other hand, the fact that (1) holds
for the system $(Y,\nu,(S_{u})_{u\in\mathbb{Q}^{>0}})$ implies that (2) holds
for the system $(Y,\nu,(S_{u})_{u\in\mathbb{N}})$. But the property (2) passes
through to factors, so it also holds for the original system
$(X,\mu,(T_{n})_{n\in\mathbb{N}})$. ∎

A weaker notion of recurrence is _topological recurrence_ , which we now
define. Given a semigroup $G$, a _topological $G$-system_ is a pair $(X,T)$
where $X$ is a compact metric space and $T=(T_{g})_{g\in G}$ is an action of
$G$ on $X$ via continuous functions. A topological $G$-system is _minimal_ if
every orbit $\operatorname{orb}(x,T):=\\{T_{g}x:g\in G\\}$ is dense in $X$.

######  Definition 2.8.

Let $G$ be a semigroup and let $R\subset G$. We say that $R$ is a _set of
topological recurrence_ if for any minimal $G$-system and any non-empty open
set $U\subset X$ there exists $r\in R$ such that $U\cap
T_{r}^{-1}U\neq\emptyset$.

One can not drop the minimality condition entirely in this definition, as
otherwise no set $R$ would satisfy it. However, it is possible to slightly
relax the minimality condition and still have an equivalent definition. Recall
that given a topological $G$-system $(X,T)$, a point $x\in X$ is called a
_minimal point_ if its orbit closure $\overline{\operatorname{orb}(x,T)}$ is
minimal.

######  Proposition 2.9.

Let $G$ be a semigroup and let $R\subset G$. Then $R$ is a set of topological
recurrence if and only if for any topological $G$-system $(X,T)$ with a dense
set of minimal points, and any non-empty open set $U\subset X$ there exists
$r\in R$ such that $U\cap T_{r}^{-1}U\neq\emptyset$.

###### Proof.

Let $x\in U$ be a minimal point, let $Y=\overline{\operatorname{orb}(x,T)}$
and let $V:=Y\cap U\neq\emptyset$. Then $(Y,T)$ is a minimal system and $V$ a
non-empty open subset of $Y$, so there exists $r\in R$ such that $V\cap
T_{r}^{-1}V\neq\emptyset$, which implies that $U\cap
T_{r}^{-1}U\neq\emptyset$. ∎

If $G$ is amenable (in particular if $G$ is commutative) then in view of
Bogolyubov-Krilov’s theorem [8], every set of recurrence is a set of
topological recurrence. We remark in passing that this is not true in general
for non-amenable semigroups, and in fact Bergelson conjectures that it is
never true for non-amenable groups (cf. [6] or [3, Conjecture after Exercise
20]).

One can also ask whether every set of topological recurrence is a set of
measurable recurrence. When $G=(\mathbb{N},+)$, a negative answer was given by
Kriz [18]. We show below that we can use Kriz’s example to answer the question
in the negative for the semigroup $(\mathbb{N},\times)$.

######  Proposition 2.10.

Let $G,H$ be commutative semigroups and let $\phi:G\to H$ be a homomorphism.
If $R\subset G$ is a set of topological recurrence, then $\phi(R)$ is a set of
topological recurrence in $H$.

###### Proof.

Let $T$ be a minimal continuous action of $H$ on the compact metric space $X$
and let $U\subset X$ be open and non-empty. Let $S$ be the $G$-action on $X$
defined by $S_{g}=T_{\phi(g)}$, let $x\in U$, let
$Y_{0}:=\overline{\\{S_{g}x:g\in G\\}}$ and let $Y\subset Y_{0}$ be a minimal
$S$-subsystem. Since $T$ is minimal, there exists $h\in H$ such that
$V:=T_{h}^{-1}U\cap Y\neq\emptyset$.

The set $V$ is a non-empty open subset of $Y$, so by the hypothesis there
exists $g\in R$ such that $S_{g}^{-1}V\cap V\neq\emptyset$. This implies that
$S_{g}^{-1}T_{h}^{-1}U\cap T_{h}^{-1}U\neq\emptyset$, and hence
$T_{\phi(g)}^{-1}U\cap U\neq\emptyset$ as well. ∎

######  Corollary 2.11.

There exists $R\subset\mathbb{N}$ which is a set of topological recurrence in
$(\mathbb{N},\times)$ but not a set of measurable recurrence in
$(\mathbb{N},\times)$.

###### Proof.

Let $R\subset\mathbb{N}$ be a set of additive topological recurrence which is
not a set of additive recurrence. Let $\phi:\mathbb{N}\to\mathbb{N}$ be the
map $\phi(n)=2^{n}$. Observe that $\phi:(\mathbb{N},+)\to(\mathbb{N},\times)$
is a homomorphism, so in view of Proposition 2.10 the set $\phi(R)$ is a set
of multiplicative topological recurrence.

Let $(X,\mu,T)$ be an additive measure preserving system, and $A\subset X$
satisfy $\mu(A)>0$ but $\mu(A\cap T^{-n}A)=0$ for all $n\in R$. Consider the
multiplicative measure preserving action $S$ on $(X,\mu)$ given by
$S_{n}=T^{a(n)}$, where $a(n)$ is the unique integer such that
$n=2^{a(n)}b(n)$, with $b(n)$ odd. Since $a(\phi(n))=n$ for all
$n\in\mathbb{N}$, it follows that $\mu(S_{m}^{-1}A\cap A)=0$ for all
$m\in\phi(R)$, and hence that $\phi(R)$ is not a set of multiplicative
recurrence. ∎

######  Remark 2.12.

We note that in general, shifts and dilations of sets of measurable
multiplicative recurrence are not sets of measurable or even topological
multiplicative recurrence. For example, $\\{4^{n}:n\in\mathbb{N}\\}$ is a set
of multiplicative recurrence since it is a semigroup. On the other hand,
$\\{4^{n}+3:n\in\mathbb{N}\\}$ is not since it is contained in $4\mathbb{N}+3$
and it is shown in Theorem 1.10 that the latter set is not a set of
topological multiplicative recurrence. For dilations, the set $\\{3\cdot
4^{n}:n\in\mathbb{N}\\}$ is not a set of multiplicative recurrence as seen in
the following coloring: For $n\in\mathbb{N}$, write $n=3^{k}q$ where
$k\in\mathbb{N}\cup\\{0\\}$ and $q\in\mathbb{N}$ with $3\nmid q$. Then color
$c(n)=0$ if $k$ is even and $c(n)=1$ if $k$ is odd.

######  Proposition 2.13.

Let $R\subset\mathbb{Q}^{>0}\setminus\\{1\\}$. Then the followings are
equivalent:

  1. ((1))

$R$ is a set of topological recurrence, i.e. for any minimal action $T$ of
$(\mathbb{Q}^{>0},\times)$ on a compact metric space $X$ and any non-empty
open set $U\subset X$ there exists $r\in R$ such that $U\cap
T_{r}^{-1}U\neq\emptyset$.

  2. ((2))

For any topological $(\mathbb{Q}^{>0},\times)$-system $(X,T)$ with a dense set
of minimal points and any non-empty open set $U\subset X$ there exist
infinitely many $r\in R$ such that $U\cap T_{r}^{-1}U\neq\emptyset$.

  3. ((3))

For any minimal action $T$ of $(\mathbb{N}^{>0},\times)$ on a compact metric
space $X$ and any non-empty open set $U\subset X$ there exist infinitely many
pairs $(a,b)\in\mathbb{N}^{2}$ with $a/b\in R$ such that $T_{a}^{-1}U\cap
T_{b}^{-1}U\neq\emptyset$.

  4. ((4))

For any finite partition $\mathbb{Q}^{>0}=C_{1}\cup\cdots\cup C_{s}$ there
exists $i\in\\{1,\dots,s\\}$, $r\in R$ such that $C_{i}\cap
C_{i}/r\neq\emptyset$.

  5. ((5))

For any finite partition $\mathbb{N}=C_{1}\cup\cdots\cup C_{s}$ there exist
$i\in\\{1,\dots,s\\}$, $a,b\in\mathbb{N}$ with $a/b\in R$ such that $C_{i}\cap
C_{i}/r\neq\emptyset$.

Moreover, any of the conditions from Proposition 2.7 implies any of the
conditions from this lemma.

###### Proof.

Similar to Proposition 2.7, the proof Proposition 2.13 is routine, with the
exception of the implication (1)$\Rightarrow$(3), which we prove here. Given a
minimal multiplicative topological system $(X,(T_{n})_{n\in\mathbb{N}})$, let
$(Y,(S_{u})_{u\in\mathbb{Q}^{>0}})$ be its invertible extension, defined so
that

$Y=\left\\{(x_{u})_{u\in\mathbb{Q}^{>0}}\in X^{\mathbb{Q}^{>0}}:(\forall
n\in\mathbb{N})(\forall u\in\mathbb{Q}^{>0})x_{un}=T_{n}x_{u}\right\\}\subset
X^{\mathbb{Q}^{>0}}$

and $S_{v}(x_{u})_{u\in\mathbb{Q}^{>0}}=(x_{uv})_{u\in\mathbb{Q}^{>0}}$. Given
a cylinder set $C=\prod_{u\in F}A_{u}\times\prod_{u\in\mathbb{Q}^{>0}\setminus
F}X\subset X^{\mathbb{Q}^{>0}}$, where $F\subset\mathbb{Q}^{>0}$ is finite and
$A_{u}\subset X$ are open sets, we have $C\cap Y\neq\emptyset$ if and only if
$\bigcap_{u\in F}T_{un}^{-1}A_{u}\neq\emptyset$, where $n\in\mathbb{N}$ is any
common denominator of all elements of $F$. For any such $C$ and any $x\in Y$,
since $(X,T)$ is minimal the orbit of $x_{n^{-1}}$ is dense in $X$ under $T$
and hence for some $m\in\mathbb{N}$ we have $T_{m}x_{n^{-1}}\in\bigcap_{u\in
F}T_{un}^{-1}A_{u}$ so that $T_{nmu}x_{n^{-1}}\in A_{u}$ for all $u\in F$.
Since $x\in Y$ and $nmu\in\mathbb{N}$, we have that
$T_{nmu}x_{n^{-1}}=x_{mu}$, and hence $(S_{m}x)_{u}=x_{mu}\in A_{u}$ for all
$u\in F$. We conclude that $(Y,(S_{m})_{m\in\mathbb{N}})$ is also a minimal
$(\mathbb{N},\times)$ system. Since the projection defined by
$(x_{u})_{u\in\mathbb{Q}^{>0}}\mapsto x_{1}$ is a factor map between
$(Y,(S_{m})_{m\in\mathbb{N}})$ and $(X,T)$, the property (3) passes through to
factors, and the property (1) for $(Y,(S_{u})_{u\in\mathbb{Q}^{>0}})$ implies
property (3) for $(Y,(S_{m})_{m\in\mathbb{N}})$, this finishes the proof. ∎

We remark in passing that items (4) and (6) of Proposition 2.7 also have an
analogue for topological recurrence, using the notion of piecewise syndetic
set.

######  Corollary 2.14.

Let $\ell\in\mathbb{N}$. A set $R\subset\mathbb{Q}^{>0}$ is a set of
topological multiplicative recurrence if and only if
$R^{\prime}=\\{r^{\ell}:r\in R\\}$ is a set of topological multiplicative
recurrence.

###### Proof.

Suppose $R\subset\mathbb{Q}^{>0}$ is a set of topological multiplicative
recurrence. In view of Proposition 2.10, we have $R^{\prime}$ is a set of
topological multiplicative recurrence.

Now let $R$ be such that $R^{\prime}=\\{r^{\ell}:r\in R\\}$ is a set of
topological multiplicative recurrence. Letting $\chi$ be a finite coloring of
$\mathbb{N}$, we claim that there are $x,y\in\mathbb{N}$ with
$\chi(x)=\chi(y)$ and $x/y\in R$. Consider the coloring $\tilde{\chi}$ defined
as follows: for any $x\in\mathbb{N}$, write $x=y_{x}^{\ell}z_{x}$ where
$y_{x},z_{x}\in\mathbb{N}$ and $z_{x}$ is not divisible by any $\ell$-th
power. (Note that $y_{x}$ and $z_{x}$ are uniquely determined by $x$.) Then
for such $x$, define $\tilde{\chi}(x)=\chi(y_{x})$. Since $R^{\prime}$ is a
set of topological multiplicative recurrence, there exist $a,b\in\mathbb{N}$
such that $\tilde{\chi}(a)=\tilde{\chi}(b)$ and $a/b=r^{\ell}$ for some $r\in
R$. By writing $a=y_{a}^{\ell}z_{a}$ and $b=y_{b}^{\ell}z_{b}$, we have that
$r^{\ell}=a/b=(y_{a}/y_{b})^{\ell}\cdot z_{a}/z_{b}$. Since $z_{a}$ and
$z_{b}$ are not divisible by any $\ell$-th power, it follows that
$z_{a}=z_{b}$ and $y_{a}/y_{b}=r$. Moreover,
$\chi(y_{a})=\tilde{\chi}(a)=\tilde{\chi}(b)=\chi(y_{b})$. Hence,
$\chi(y_{a})=\chi(y_{b})$ and $y_{a}/y_{b}\in R$; so our proof finishes. ∎

##  3\. Sets of multiplicative recurrence and applications

###  3.1. Rational functions

The purpose of this section is to prove Theorem 1.6. We start with the
following necessary criteria for a set to be a set of topological
multiplicative recurrence.

######  Lemma 3.1.

For any $\epsilon,M\in\mathbb{R}$ satisfying $0<\epsilon<1<1+\epsilon<M$, the
set $S:=\left((\epsilon,1-\epsilon)\cup(1+\epsilon,M)\right)\cap\mathbb{Q}$ is
not a set of topological multiplicative recurrence. In other words, if $S$
does not have $0,1$ or $\infty$ as an accumulation point, then $S$ is not a
set of topological multiplicative recurrence.

###### Proof.

Let $\epsilon$ and $M$ be as in the hypothesis of the lemma. Choose
$c_{1},c_{2},a\in\mathbb{R}$ such that

$c_{1}<1-\epsilon<1+\epsilon<c_{2}$

and

$a>\max\\{c_{2}/\epsilon,M/c_{1}\\}.$

Then it follows that

$\bigcup_{k\in\mathbb{Z}}(c_{1}a^{k},c_{2}a^{k})\subset(0,\epsilon)\cup(1-\epsilon,1+\epsilon)\cup(M,\infty).$

Let $S^{1}=\\{z\in\mathbb{C}:|z|=1\\}$ and consider topological
$(\mathbb{Q}^{>0},\times)$-system $(S^{1},T)$ defined as
$T_{q}(e(x))=e(\log_{a}q)e(x)$ for every $q\in\mathbb{Q}^{>0}$ and $x\in[0,1)$
(here $e(x):=e^{2\pi ix}$). Let
$\epsilon_{1}=-\log_{a}c_{1},\epsilon_{2}=\log_{a}c_{2}$ and let
$B=\\{e(x):x\in(1-\epsilon_{1},1+\epsilon_{2})\\}$ be a neighborhood of $e(0)$
in $S^{1}$. We have $T_{q}e(0)\in B$ if any only if
$\log_{a}q\in\bigcup_{k\in\mathbb{Z}}(k-\epsilon_{1},k+\epsilon_{2})$, or
equivalently
$q\in\bigcup_{k\in\mathbb{Z}}(a^{-\epsilon_{1}}a^{k},a^{\epsilon_{2}}a^{k})$.
It follows that

$\\{q\in\mathbb{Q}^{>0}:T_{q}e(0)\in
B\\}\subset\bigcup_{k\in\mathbb{Z}}(a^{-\epsilon_{1}}a^{k},a^{\epsilon_{2}}a^{k})\subset(0,\epsilon)\cup(1-\epsilon,1+\epsilon)\cup(M,\infty).$

Since every set of topological multiplicative recurrence must intersect
$\\{q\in\mathbb{Q}^{>0}:T_{q}e(0)\in B\\}$ we conclude that $R$ must intersect
$(0,\epsilon)\cup(1-\epsilon,1+\epsilon)\cup(M,\infty)$. Therefore the set $S$
defined in Lemma 3.1 is not a set of topological multiplicative recurrence. ∎

We now prove parts (1) and (2) of Theorem 1.6.

######  Proposition 3.2 (Part (1) of Theorem 1.6).

Let $a,c,\ell\in\mathbb{N}$, $b,d\in\mathbb{Z}$ with $a\neq c$. Then
$S=\\{(an+b)^{\ell}/(cn+d)^{\ell}:n\in\mathbb{N}\\}$ is not a set of
topological multiplicative recurrence.

###### Proof.

First note that $(an+b)^{\ell}/(cn+d)^{\ell}\to(a/c)^{\ell}$ as $n\to\infty$.
Therefore, for any $\epsilon>0$, the set
$S\setminus((a/c)^{\ell}-\epsilon,(a/c)^{\ell}+\epsilon)$ is finite. Moreover,
since $a\neq c$, for sufficiently small $\epsilon$, we have
$((a/c)^{\ell}-\epsilon,(a/c)^{\ell}+\epsilon)$ does not contain $0,1$. Hence,
$S$ does not have $0,1$ or $\infty$ as an accumulation point, and by Lemma
3.1, the set $S$ is not a set of topological multiplicative recurrence. ∎

######  Proposition 3.3 (Part (2) of Theorem 1.6).

Let $a,\ell\in\mathbb{N}$ and $b,d\in\mathbb{Z}$. Suppose that $b\neq d$ and
there exists a prime $p$ for which $p|a$ but $p\nmid bd$. Then
$S=\\{(an+b)^{\ell}/(an+d)^{\ell}:n\in\mathbb{N}\\}$ is not a set of
topological multiplicative recurrence.

###### Proof.

Since $S$ is a subset of
$S^{\prime}=\\{(pn+b)^{\ell}/(pn+d)^{\ell}:n\in\mathbb{N}\\}$, it suffices to
prove that $S^{\prime}$ is not a set of topological multiplicative recurrence.
In view of Corollary 2.14, we need to prove that
$S^{\prime\prime}=\\{(pn+b)/(pn+d):n\in\mathbb{N}\\}$ is not a set of
topological multiplicative recurrence.

Because $b\neq d$, there exists $u\in\mathbb{N}$ such that $p^{u}$ does not
divide $b-d$. Let $\chi$ be the finite coloring of $\mathbb{N}$ defined
recursively as

$\chi(n)=\begin{cases}n\bmod p^{u}&\text{ if }p\nmid n\\\ \chi(n/p)&\text{ if
}p|n.\end{cases}$

We shall show that $S^{\prime\prime}$ is not a set of recurrence for the
coloring $\chi$. By contradiction, assume there exist monochromatic
$x,y\in\mathbb{N}$ such that

(2) $\frac{x}{y}=\frac{pn+b}{pn+d}.$

Write $x=p^{k_{x}}x_{0}$ and $y=p^{k_{y}}y_{0}$ where
$k_{x},k_{y}\in\mathbb{N}\cup\\{0\\}$ and $p\nmid x_{0}y_{0}$. By the
definition of the coloring $\chi$, we have that $x_{0}\equiv y_{0}\bmod
p^{u}$.

From (2), we have $x(pn+d)=y(pn+b)$. It follows that
$p^{k_{x}}x_{0}(pn+d)=p^{k_{y}}y_{0}(pn+b)$. Since $pn+b$ and $pn+d$ are
coprime to $p$, we deduce that $k_{x}=k_{y}$ and $x_{0}(pn+d)=y_{0}(pn+b)$. As
$x_{0}\equiv y_{0}\bmod p^{u}$, it must be that $pn+d\equiv pn+b\bmod p^{u}$,
or equivalently, $b\equiv d\bmod p^{u}$. This contradicts our assumption that
$p^{u}$ does not divide $b-d$ and our proof finishes. ∎

For the proof of part (3) of Theorem 1.6, we need a recurrence result in
topological dynamical systems:

######  Lemma 3.4.

Let $(X,T)$ be a topological $(\mathbb{N},+)$-system where $X$ is a compact
metric space with the metric $d$. Suppose that there exists $x\in X$ such that
$\\{T^{n}x:n\in\mathbb{N}\\}$ is dense in $X$. Then for every $y\in
X,\epsilon>0$ and $N\in\mathbb{N}$, there exists $m_{0}\in\mathbb{N}$ such
that $d(T^{i+m_{0}}x,T^{i}y)<\epsilon$ for all $i\in[N]$.

###### Proof.

For $i\in[N]$, the map $T^{i}:X\to X$ is continuous, hence uniformly
continuous. Thus, there exists $\delta>0$ such that for $w,z\in X$,
$d(w,z)<\delta$ implies $d(T^{i}w,T^{i}z)<\epsilon$ for all $i\in[N]$. Since
$\\{T^{n}x:n\in\mathbb{N}\\}$ is dense in $X$, for any $y\in Y$ we can find
$m_{0}\in\mathbb{N}$ such that $d(T^{m_{0}}x,y)<\delta$. Then the choice of
$\delta$ implies that $d(T^{i+m_{0}}x,T^{i}y)<\epsilon$ for all $i\in[N]$. ∎

Next we prove the following special case of part (3) of Theorem 1.6.

######  Lemma 3.5.

For any $a\in\mathbb{N}$, the sets $\\{(an+1)/(an):n\in\mathbb{N}\\}$ and
$\\{(an-1)/(an):n\in\mathbb{N}\\}$ are sets of topological multiplicative
recurrence.

###### Proof.

We only prove the first set is a set of topological multiplicative recurrence
as the proof for the second set is the same. Fix $r\in\mathbb{N}$ with $r\geq
1$. Let $\mathbb{N}_{0}$ denote $\mathbb{N}\cup\\{0\\}$ and let
$X_{0}=[r]^{\mathbb{N}_{0}}=\\{(w(n))_{n\in\mathbb{N}_{0}}:w(n)\in[r]\\}$.
Define a metric $d$ on $X_{0}$ as the following: For
$w=(w(n))_{n\in\mathbb{N}_{0}}$ and $z=(z(n))_{n\in\mathbb{N}_{0}}$,

$d(w,z)\colon=\sum_{n\in\mathbb{N}_{0}}\frac{|w(n)-z(n)|}{2^{n}}.$

Under this metric, $X_{0}$ becomes a compact metric space and $d(w,z)<1$
implies $w(0)=z(0)$. Let $T:X_{0}\to X_{0}$ be the left shift which is defined
as $T(w(n))_{n\in\mathbb{N}_{0}}=(w(n+1))_{n\in\mathbb{N}_{0}}$. Then $T$ is a
continuous map.

Let $c:\mathbb{N}\to[r]$ be a finite coloring of $\mathbb{N}$. We need to find
$x,y\in\mathbb{N}$ of the same color and $n\in\mathbb{N}$ such that
$y/x=(an+1)/an$. This is equivalent to finding $\\{x,x+k\\}$ of the same color
with $ak|x$ (indeed, writing $x=nak$ and $y=x+k$ we have $y/x=(an+1)/an$). By
regarding $c$ as an element of $X_{0}$, we define $X$ to be the closure of
$\\{T^{n}c:n\in\mathbb{N}_{0}\\}$ in $X_{0}$. It follows that $(X,T)$ is a
topological $(\mathbb{N}_{0},+)$-system. Let $(Y,T)$ be a minimal subsystem of
$(X,T)$. 222The existence of minimal subsystems follows from an application of
Zorn’s lemma (cf. [1, Theorem 1.4]). For each color $j\in[r]$, let $U_{j}$ be
the cylinder $U_{j}=\\{x\in X_{0}:x(0)=j\\}$. By minimality, the return times
of points in $Y\cap U_{j}$ back to $Y\cap U_{j}$ are uniformly bounded, i.e.
there exists $N_{j}\in\mathbb{N}$ such that for all $y\in Y\cap U_{j}$, there
exists $n=n(y)\in[N_{j}]$ satisfying $T^{n}y\in U_{j}\cap Y$ (for example, see
[1, Chapter 1]). Setting $N_{0}=\max\\{N_{j}:j\in[r]\\}$, we have that for all
$y\in Y$ and $n\in\mathbb{N}$, there exists $1\leq n^{\prime}\leq N_{0}$ for
which $y(n)=y(n+n^{\prime})$.

Fix $y_{0}\in Y$. By Lemma 3.4, there exists $m_{0}\in\mathbb{N}$ such that
$d(T^{i+m_{0}}c,T^{i}y_{0})<1$ for all $i\in[aN_{0}!]$ (the number $aN_{0}!$
is chosen here because later we will use the fact that $an^{\prime}|aN_{0}!$
for all $1\leq n^{\prime}\leq N_{0}$). In particular, we have
$c(m_{0}+i)=y_{0}(i)$ for $1\leq i\leq aN_{0}!$. Let $i_{0}\in[aN_{0}!]$
satisfy $m_{0}+i_{0}\equiv 0\bmod aN_{0}!.$ From previous paragraph, there
exists $1\leq n^{\prime}\leq N_{0}$ such that
$y_{0}(i_{0})=y_{0}(i_{0}+n^{\prime})$. Hence it follows that
$c(m_{0}+i_{0})=c(m_{0}+i_{0}+n^{\prime})$. By writing $x=m_{0}+i_{0}$ and
$k=n^{\prime}$, we have $c(x)=c(x+k)$ with $ak|x$, finishing our proof. ∎

We are now ready to prove part (3) of Theorem 1.6.

######  Proposition 3.6 (Part (3) of Theorem 1.6).

For $a,b,d,\ell\in\mathbb{Z}$ with $a,\ell>0$ and $a|b$ or $a|d$,

$S=\left\\{\left(\frac{an+b}{an+d}\right)^{\ell}:n\in\mathbb{N}\right\\}$

is a set of topological multiplicative recurrence.

###### Proof.

In view of Corollary 2.14, it suffices to show that

$S^{\prime}=\left\\{\frac{an+b}{an+d}:n\in\mathbb{N}\right\\}$

is a set of topological multiplicative recurrence. Without loss of generality,
assume $a|d$ and $d=ad_{1}$. If $b=d$, then $S^{\prime}=\\{1\\}$ which is
trivially a set of multiplicative recurrence. If $b>d$, then we have

$S^{\prime}=\left\\{\frac{a(n+d_{1})+b-ad_{1}}{a(n+d_{1})}:n\in\mathbb{N}\right\\}$

which contains the set

$\left\\{\frac{a(n+d_{1})+b-ad_{1}}{a(n+d_{1})}:n+d_{1}\equiv
0\bmod(b-ad_{1})\right\\}=\left\\{\frac{am+1}{am}:m\in\mathbb{N}\right\\}.$

By Lemma 3.5, the last set is a set of topological multiplicative recurrence,
and hence so is $S^{\prime}$.

On the other hand, if $b<d$, then $S^{\prime}$ contains the set

$\left\\{\frac{a(n+d_{1})+b-ad_{1}}{a(n+d_{1})}:n+d_{1}\equiv
0\bmod(ad_{1}-b)\right\\}=\left\\{\frac{am-1}{am}:m\in\mathbb{N}\right\\}$

which is a again a set of topological multiplicative recurrence. ∎

Next we prove Corollary 1.7.

###### Proof of Corollary 1.7.

Let $f:\mathbb{N}\to\mathbb{C}$ be a completely multiplicative function with
$|f(n)|=1$ for $n\in\mathbb{N}$ and let $a,k\in\mathbb{N}$. We need to prove
that,

$\liminf_{n\to\infty}|f(an+k)-f(an)|=0.$

Let $\epsilon>0$. Consider the topological $(\mathbb{N},\times)$-system
$(S^{1},T)$ where $S^{1}=\\{z\in\mathbb{C}:|z|=1\\}$ and $T_{n}e^{2\pi
ix}=f(n)e^{2\pi ix}$ for all $x\in[0,1)$ and $n\in\mathbb{N}$. Let
$A=\\{e^{2\pi ix}\colon x\in(-\epsilon/2,\epsilon/2)\\}\subset S^{1}$. Then by
Proposition 3.6 and Proposition 2.13, there exist infinitely many
$n\in\mathbb{N}$ such that $T_{(an+k)m}A\cap T_{anm}A\neq\emptyset$ for some
$m\in\mathbb{N}$ (which may depend on $n$). This implies $f((an+k)m)A\cap
f(anm)A\neq\emptyset$, or equivalently, $f(m)f(an+k)A\cap
f(m)f(an)A\neq\emptyset$. Since multiplication by $f(m)$ is an isometry, we
have

$f(an+k)A\cap f(an)A\neq\emptyset.$

It follows that $|f(an+k)-f(an)|\leq\epsilon$ for infinitely many $n$. Since
$\epsilon$ is arbitrary, we have $\liminf_{n\to\infty}|f(an+k)-f(an)|=0$. ∎

######  Remark 3.7.

Theorem 1.6 implies that for any finite coloring of $\mathbb{N}$ and every
$k\in\mathbb{N}$, there exist $a,n\in\mathbb{N}$ such that $an|x$ and the set

$\left\\{x,x\frac{an+k}{an}\right\\}$

is monochromatic. It is natural to ask if, more generally, we can find
$a,n\in\mathbb{N}$ for which $an|x$ and the set

$\left\\{x,x\frac{an+1}{an},x\frac{an+2}{an},\dots,x\frac{an+k}{an}\right\\}$

is monochromatic It turns out that the answer is negative, even in the case
$a=1$ and $k=2$, as shown by the following example.

Let $c:\mathbb{N}\to\\{1,2\\}$ be the $3$-Rado coloring: For $m\in\mathbb{N}$,
write $m=3^{k}q$ where $k\in\mathbb{N}\cup\\{0\\}$ and $3\nmid q$. Then color
$c(m)=1$ if $q\equiv 1\mod 3$ and $c(m)=2$ if $q\equiv 2\mod 3$. Note that in
every $3$ consecutive integers, there is a number congruent to $1\bmod 3$ and
another congruent to $2\bmod 3$, therefore there are no $3$ consecutive
numbers with the same color. On the other hand, observe that $c(x)=c(y)$ if
and only if $c(xz)=c(yz)$. Indeed, this is obvious if $z$ is a power of $3$;
and if $z$ is co-prime to $3$ then $c(xz)\equiv c(x)z\bmod 3$. If the set

$\left\\{\frac{x}{n}n,\frac{x}{n}(n+1),\frac{x}{n}(n+2)\right\\}$

were monochromatic for some $x,n\in\mathbb{N}$ with $n|x$, so would be the set
$\\{n,n+1,n+2\\}$, a contradiction.

###  3.2. Linear polynomials

Let $(X,T)$ be an $(\mathbb{N},\times)$-system. If $X$ is a finite set, we say
that $(X,(T_{n})_{n\in\mathbb{N}})$ is a _finite
$(\mathbb{N},\times)$-system._ The following theorem classifies which sets of
the form $a\mathbb{N}+b$ that are sets of multiplicative recurrence, and hence
implies Theorem 1.10.

######  Theorem 3.8.

Let $a,b\in\mathbb{N}$ and $S=\\{an+b:n\in\mathbb{N}\\}$. The followings are
equivalent:

  1. (i)

$a|b(b-1)$.

  2. (ii)

$S$ is a multiplicative semigroup.

  3. (iii)

$S$ contains an infinite multiplicative semigroup.

  4. (iv)

$S$ is a set of measurable recurrence for $(\mathbb{N},\times)$-systems.

  5. (v)

$S$ is a set of topological recurrence for finite
$(\mathbb{N},\times)$-systems.

###### Proof.

(i) $\Rightarrow$ (ii): If $a|b(b-1)$, then for any $ax+b,ay+b\in S$, we have

$(ax+b)(ay+b)=a\left(axy+bx+by+\frac{b(b-1)}{a}\right)+b\in S.$

(ii) $\Rightarrow$ (iii): Obvious.

(iii) $\Rightarrow$ (iv): This follows from the Poincaré Recurrence Theorem.

(iv) $\Rightarrow$ (v): Obvious.

(v) $\Rightarrow$ (i): We prove the contrapositive. Let $a,b$ be such that
$a\nmid b(b-1)$. It suffices to construct a finite
$(\mathbb{N},\times)$-system $(X,(T_{n})_{n\in\mathbb{N}})$ such that whenever
$n\in S$, the permutation $T_{n}$ fixes no element. Let $a^{\prime}=a/(a,b)$.
Observe that $b\not\equiv 1\bmod a^{\prime}$, for otherwise $b(b-1)$ would be
a multiple of $a=(a,b)a^{\prime}$. We need to consider separate cases:

Case 1: $(a^{\prime},b)=1$. In this case every $n\in S$ satisfies
$(n,a^{\prime})=1$ and $n\equiv b\not\equiv 1\bmod a^{\prime}$. Let
$X=\big{\\{}x\in\\{0,1,\dots,a^{\prime}-1\\}:(x,a^{\prime})=1\big{\\}}$. For
$n\in\mathbb{N}$, let $T_{n}\colon X\to X$ be defined as follows. Let $A$ be
the set of all $q\in\mathbb{N}$ such that the all prime divisors of $q$ also
divide $a^{\prime}$. Any natural number $n$ can be written uniquely as
$n=uv=u(n)v(n)$ where $u\in A,v\in\mathbb{N}$ and $(v,a^{\prime})=1$. For
$x\in X$, define $T_{n}x=xv(n)\bmod a^{\prime}$. Notice that $v(nm)=v(n)v(m)$,
so $(T_{n})_{n\in\mathbb{N}}$ is indeed an action of $(\mathbb{N},\times)$ on
$X$. Since every $n\in S$ satisfies $(n,a^{\prime})=1$ and $n\not\equiv 1\bmod
a^{\prime}$, $v(n)=n\not\equiv 1\bmod a^{\prime}$. Therefore, for $n\in S$, we
have $T_{n}$ does not fix any point in $X$.

Case 2: $(a^{\prime},b)>1$. Let $p$ be a prime which divides $(a^{\prime},b)$
and let $k\in\mathbb{N}$ be such that $p^{k-1}|b$ but $p^{k}\nmid b$. Let
$X=\mathbb{Z}_{k}$ and let $T_{n}\colon X\to X$, $n\in\mathbb{N}$ be defined
as follows. For $n\in\mathbb{N}$, let $u(n)$ and $v(n)$ be the unique non-
negative integers satisfying $n=p^{u(n)}v(n)$ and $p\nmid v(n)$. For $x\in X$,
define $T_{n}x:=x+u(n)\bmod k$. Since $u(nm)=u(n)+u(m)$,
$(T_{n})_{n\in\mathbb{N}}$ indeed defines an action of $(\mathbb{N},\times)$
on $X$. Because $(a^{\prime},b/(a,b))=1$ and $p|a^{\prime}$, it follows that
$p\nmid b/(a,b)$. On the other hand, $p^{k-1}|b$. Therefore, $p^{k-1}|(a,b)$,
and hence $p^{k}|a=(a,b)a^{\prime}$. Thus for any $n\in S$, we have
$u(n)=k-1$, which implies that $T_{n}x=x-1\bmod k$ for every $x\in X$. In
particular, $T_{n}$ does not fix any element of $X$. ∎

A polynomial $P\in\mathbb{Z}[x]$ is called _divisible_ if for any
$q\in\mathbb{N}$, there exists $n\in\mathbb{N}$ such that $q|P(n)$. As proven
in [16], the set $\\{P(n):n\in\mathbb{N}\\}$ is a set of additive topological
(and measurable) recurrence if and only if $P$ is divisible.

######  Corollary 3.9.

Let $P\in\mathbb{Z}[x]$. If $S=\\{P(n):n\in\mathbb{N}\\}$ is a set of
topological multiplicative recurrence, then
$R=\\{P(n)(P(n)-1):n\in\mathbb{N}\\}$ is a set of measurable additive
recurrence, i.e. the polynomial $P(P-1)$ is divisible.

###### Proof.

Assume $P(P-1)$ is not divisible. Then there exists $a\in\mathbb{N}$ such that
$a\nmid P(n)(P(n)-1)$ for all $n\in\mathbb{N}$. Hence a

(3) $\\{P(n):n\in\mathbb{N}\\}\subset\bigcup_{\begin{subarray}{c}0\leq b\leq
a-1\\\ a\nmid b(b-1)\end{subarray}}a\mathbb{N}+b.$

In view of Theorem 3.8, for each $b\in\\{0,\dots,a-1\\}$ such that $a\nmid
b(b-1)$, the set $a\mathbb{N}+b$ is not a set of multiplicative recurrence.
Invoking [7, Theorem 2.7 (b)] it follows that the union in the right hand side
of (3) can not be a set of recurrence. ∎

######  Example 3.10.

As an application of Corollary 3.9, the set $\\{n^{2}+3:n\in\mathbb{N}\\}$ is
not a set of topological multiplicative recurrence. This is because the
polynomial $Q(n)=n^{4}+5n^{2}+6$ is not divisible, as $5\nmid Q(n)$ for any
$n\in\mathbb{N}$.

Any polynomial $P\in\mathbb{Z}[x]$ with $P(0)=0$ is divisible, so in that case
the set $S=\\{P(n):n\in\mathbb{N}\\}$ is a set of measurable additive
recurrence. However, this set is not a set of multiplicative recurrence in
general as the following example shows:

######  Example 3.11.

Let $p$ be a prime. The set $S=\\{pn^{2}:n\in\mathbb{N}\\}$ is not a set of
topological multiplicative recurrence. Indeed, let
$E=\\{p^{2k}\ell:k\in\mathbb{N}\cup\\{0\\},\ell\in\mathbb{N},p\nmid\ell\\}$
and
$F=\\{p^{2k+1}\ell:k\in\mathbb{N}\cup\\{0\\},\ell\in\mathbb{N},p\nmid\ell\\}$.
Then $E\cup F=\mathbb{N}$. For every $x,y\in E$ or $x,y\in F$, $x/y$ has the
form $p^{2k}\ell_{1}/\ell_{2}$ for some $k\in\mathbb{Z}$ and
$\ell_{1},\ell_{2}\in\mathbb{N}$ with $p\nmid\ell_{1}\ell_{2}$. It follows
that $x/y\not\in S$, hence $S$ is not a set of multiplicative recurrence.

Example 3.11 can also be used to show that the converse to Corollary 3.9 is
false in general. Indeed, take $P(n)=pn^{2}$. Then
$P(n)(P(n)-1)=pn^{2}(pn^{2}-1)$ is divisible, but $\\{P(n):n\in\mathbb{N}\\}$
is not set of topological multiplicative recurrence.

##  4\. Algebraic sets that contain multiplicative semigroups

As mentioned in Section 2, every subsemigroup of $(\mathbb{N},\times)$ is a
set of multiplicative recurrence. With the ultimate goal of understanding
which algebraically defined sets are sets of multiplicative recurrence, it is
natural to also study sets which are (or contain) a semigroup. In this section
we investigate this question for sets that are the image of a polynomial or
the image of a rational function.

###  4.1. Polynomial sets that contain multiplicative semigroups

In this section we prove Theorem 1.9. We first need some lemmas.

######  Lemma 4.1.

Let $d\geq 2$ and
$P(x)=a_{d}x^{d}+a_{d-2}x^{d-2}+\ldots+a_{1}x+a_{0}\in\mathbb{Q}[x]$ such that
at least one of $a_{0},a_{1},\ldots,a_{d-2}$ is non-zero and $a_{d}>0$. Then
$S=\\{P(n):n\in\mathbb{N}\\}$ does not contain any infinite subsemigroup of
$(\mathbb{Q}^{>0},\times)$.

###### Proof.

First we show that, for any $D\in\mathbb{N}\setminus\\{1\\}$, there are only
finitely many $n\in\mathbb{N}$ such that $D^{d}P(n)=P(Dn)$. Indeed, otherwise,
the polynomial $D^{d}P(x)-P(Dx)$ would have infinitely many roots, and hence
$D^{d}P(x)=P(Dx)$ for all $x\in\mathbb{R}$. It would follow that
$P(D^{k})=P(1)(D^{k})^{d}$ for all $k\in\mathbb{N}$. This implies that the
equation $P(x)=P(1)x^{d}$ has infinitely many solutions. So $P(x)=P(1)x^{d}$
for all $x\in\mathbb{R}$. This is contradicts our assumption that at least one
of $a_{0},a_{1},\ldots,a_{d-2}$ is nonzero.

Next we assume, for the sake of a contradiction, that there exists
$D\in\mathbb{N}\setminus\\{1\\}$ such that $\\{D^{n}:n\in\mathbb{N}\\}\subset
S$. Then there exist infinitely many $m>n\in\mathbb{N}$ such that

(4) $D^{d}P(n)=P(m).$

Using the first paragraph of the proof, there are infinitely many solutions to
(4) with $m\neq Dn$. Next, observe that (4) can be written as

(5) $a_{d}(m^{d}-(Dn)^{d})=\sum_{j=0}^{d-2}a_{j}(D^{d}n^{j}-m^{j}).$

Note that for all $x,y\in\mathbb{N}$ with $x\neq y$,
$|x^{d}-y^{d}|>(\max\\{x,y\\})^{d-1}$. Hence if $m\neq Dn$ for some
$m,n\in\mathbb{N}$, then $|m^{d}-(Dn)^{d}|>(\max\\{m,Dn\\})^{d-1}$. On the
other hand,

$\left|\sum_{j=0}^{d-2}a_{j}(D^{d}n^{j}-m^{j})\right|\leq 2(d-1)\max_{0\leq
j\leq d-2}\\{|a_{j}|\\}(\max\\{m,Dn\\})^{d-2}.$

Therefore, from (5) we derive that

$a_{d}\max\\{m,Dn\\}^{d-1}<2(d-1)\max_{0\leq j\leq
d-2}\\{|a_{j}|\\}\max\\{m,Dn\\}^{d-2}$

for infinitely many $m,n\in\mathbb{N}$. This is impossible and hence it yields
the desired contradiction. ∎

######  Lemma 4.2.

Let $P\in\mathbb{Z}[x]$. Then there exist $a,b\in\mathbb{Z}$ and
$Q\in\mathbb{Q}[x]$ such that $P(x)=Q(ax+b)$ and writing
$Q(x)=a_{d}x^{d}+a_{d-1}x^{d-1}+\cdots+a_{1}x+a_{0}$ we have $a_{d-1}=0$. If
the leading coefficient of $P$ is positive, then $a>0$.

###### Proof.

Let $P(x)=b_{d}x^{d}+\cdots+b_{0}$ and factor it over $\mathbb{C}$ as
$b_{d}\prod_{i=1}^{d}(x-\alpha_{i})$, where
$\alpha_{1},\dots\alpha_{d}\in\mathbb{C}$ are the roots of $P$ counted with
multiplicity. It follows that $\sum_{i=1}^{d}\alpha_{i}=-b_{d-1}/b_{d}$. Let
$a=db_{d}$, $b=b_{d-1}$ and define $Q(y):=P((y-b)/a)$. Observe that $Q(y)=0$
if and only if $y=a\alpha_{i}+b$ for some $1\leq i\leq d$. Therefore the sum
of the roots of $Q$ is $0$, which implies that $a_{d-1}=0$. ∎

We are ready to prove Theorem 1.9. In fact we prove the following more general
result.

######  Theorem 4.3.

Let $P\in\mathbb{Z}[x]$ and $S=\\{P(n):n\in\mathbb{N}\\}$. The following are
equivalent:

  1. (i)

$S$ is an infinite subsemigroup of $(\mathbb{Q}^{>0},\times)$.

  2. (ii)

$S$ contains an infinite subsemigroup of $(\mathbb{Q}^{>0},\times)$.

  3. (iii)

$P(x)=(ax+b)^{d}$ for some $a,d\in\mathbb{N}$, $b\in\mathbb{Z}$ with
$a|b(b-1)$.

###### Proof.

(iii) $\Rightarrow$ (i): Assume $Q(x)=(ax+b)^{d}$ with $a|b(b-1)$. Then for
any $x,y\in\mathbb{N}$,

$Q(x)Q(y)=(ax+b)^{d}(ay+b)^{d}=\left(a\left(axy+bx+by+\frac{b(b-1)}{a}\right)+b\right)^{d}\in
S.$

Hence $S$ is a multiplicative semigroup.

(i) $\Rightarrow$ (ii): Trivial.

(ii) $\Rightarrow$ (iii): Assume that $S=\\{P(n):n\in\mathbb{N}\\}$ contains
an infinite multiplicative semigroup of $\mathbb{Q}^{>0}$. Then the leading
coefficient of $P$ is positive. Write $P(x)=Q(ax+b)$ as in Lemma 4.2, noting
that $a>0$. Then we have
$\\{P(n):n\in\mathbb{N}\\}=\\{Q(an+b):n\in\mathbb{N}\\}\subset\\{Q(n):n\in\mathbb{N}\\}$.
It then follows that $\\{Q(n):n\in\mathbb{N}\\}$ contains a semigroup. By
Lemma 4.1, $Q(x)=a_{d}x^{d}$ for some $d\in\mathbb{N}$ and
$a_{d}\in\mathbb{Q}$. Since $\\{Q(n):n\in\mathbb{N}\\}$ contains a semigroup,
there exist $x,y,z\in\mathbb{N}$ such that $Q(x)Q(y)=Q(z)$. In other words,
$a_{d}^{2}(xy)^{d}=a_{d}z^{d}$, or $a_{d}=(z/(xy))^{d}$. Let
$c=z/(xy)\in\mathbb{Q}^{>0}$, we get $a_{d}=c^{d}$. Hence
$P(x)=(c(ax+b))^{d}=(cax+cb)^{d}$.

By abuse of notation, assume that $P(x)=(ax+b)^{d}$ for some
$a,b\in\mathbb{Q}$. Since $P\in\mathbb{Z}[x]$, we have $a,b\in\mathbb{Z}$.
Because $S$ contains an infinite multiplicative semigroup, there exists
$D\in\mathbb{N}\setminus\\{1\\}$ such that for all $k\in\mathbb{N}$, there
exists $x\in\mathbb{N}$ and $(ax+b)^{d}=D^{kd}$. In other words, $ax+b=D^{k}$.
It follows that $D^{k}\equiv b\mod a$ for all $k\in\mathbb{N}$. Hence we have
simultaneously that $D^{2k}\equiv b^{2}$ and $D^{2k}\equiv b\mod a$. This
implies that $b^{2}\equiv b\mod a$, or $a|b(b-1)$, which finishes the proof. ∎

###  4.2. Rational functions that do not contain a semigroup

We prove Proposition 1.11 in this section. In addition, we show that many
other rational functions of interest do not contain infinite multiplicative
semigroups.

######  Lemma 4.4.

Let $\ell\geq 2$, $q_{1},q_{2},\ldots,q_{\ell}\in\mathbb{R}\backslash\\{0\\}$
be non-zero and $|q_{i}|\neq|q_{j}|$ for all $i\neq j$. Let
$a_{1},\ldots,a_{\ell}\in\mathbb{R}$ be such that
$a_{1}q_{1}^{k}+\ldots+a_{\ell}q_{\ell}^{k}=0$ for infinitely many
$k\in\mathbb{N}$. Then $a_{1}=a_{2}=\ldots=a_{\ell}=0$.

###### Proof.

By contradiction, assume some $a_{i}$ are non-zero. By discarding those
$a_{i}$ that are zero, without loss of generality, we may assume that none of
the $a_{i}$ are zero.

Assume $|q_{1}|=\max\\{|q_{i}|:1\leq i\leq\ell\\}>0$. Then we have

$|a_{1}q_{1}^{k}|=|a_{2}q_{2}^{k}+\ldots+a_{\ell}q_{\ell}^{k}|.$

It follows that

$|a_{1}|=|a_{2}(q_{2}/q_{1})^{k}+\ldots+a_{\ell}(q_{\ell}/q_{q})^{k}|$

for infinitely many $k\in\mathbb{N}$. But this is impossible since the right
hand side approaches zero as $k\to\infty$. ∎

We are ready to prove Proposition 1.11.

###### Proof of Proposition 1.11.

We only give a proof that the set
$S=\\{(m^{2}+n^{2})/(2mn):m,n\in\mathbb{N}\\}$ does not contain an infinite
subsemigroup of $(\mathbb{Q}^{>0},\times)$ as the proof for the other set is
the same. By contradiction, assume that there exist $p,q\in\mathbb{Z},p\neq
q,(p,q)=1$ such that $\\{(p/q)^{k}:k\in\mathbb{N}\\}\subset S$. By replacing
$(p/q)$ by $(p/q)^{2}$, we can assume $p,q>0$. Then it follows that for every
$k\in\mathbb{N}$, there exist $m,n\in\mathbb{N}$ such that

(6) $\frac{p^{k}}{q^{k}}=\frac{m^{2}+n^{2}}{2mn}.$

We can dividing both $m$ and $n$ by $gcd(m,n)$ and (6) still holds. Therefore,
without loss of generality, we may assume $(m,n)=1$. Then

$2p^{k}mn=q^{k}(m^{2}+n^{2}).$

Since $(p,q)=1$, it follows that $q^{k}|2mn$. On the other hand, since
$(m,n)=1$, $(mn,m^{2}+n^{2})=1$. Thus $mn|q^{k}$. It follows that $q^{k}=mn$
or $q^{k}=2mn$.

If $q^{k}=mn$, then $2p^{k}=m^{2}+n^{2}$. Since $(m,n)=1$, $m=q_{1}^{k}$ and
$n=q_{2}^{k}$ for some $q_{1}q_{2}=q$. Hence

(7) $2p^{k}=(q_{1}^{2})^{k}+(q_{2}^{2})^{k}.$

If $q^{k}=2mn$, then $p^{k}=m^{2}+n^{2}$. Similarly to above,
$m=(q_{1})^{k}/2$ and $n=q_{2}^{k}$ for some $q_{1}q_{2}=q$. We then have
$p^{k}=q_{1}^{2k}/4+q_{2}^{2k}$, or equivalently,

(8) $4p^{k}=(q_{1}^{2})^{k}+4(q_{2}^{2})^{k}.$

Since the set $\\{(q_{1},q_{2})\in\mathbb{N}^{2}:q_{1}q_{2}=q\\}$ is finite,
there must be some fixed $q_{1},q_{2}$ such that either (7) or (8) is true for
infinitely many $k\in\mathbb{N}$. But, in view of Lemma 4.4, this is
impossible, and hence we obtained the desired contradiction. ∎

Let $\ell_{1},\ell_{2},\ell_{3},\ell_{4}\in\mathbb{Z}$ be pairwise distinct
and let

(9)
$S=\left\\{\frac{(m+\ell_{1}n)(m+\ell_{2}n)}{(m+\ell_{3}n)(m+\ell_{4}n)}:m,n\in\mathbb{N}\right\\}.$

It is shown in [11] that the set $S$ in (9) is a set of measurable
multiplicative recurrence. In the following proposition, we show that $S$ does
not contain any infinite subsemigroup of $(\mathbb{Q}^{>0},\times)$.

######  Proposition 4.5.

Let $\ell_{1},\ell_{2},\ell_{3},\ell_{4}\in\mathbb{Z}$ be pairwise distinct.
Then the set

$S=\left\\{\frac{(m+\ell_{1}n)(m+\ell_{2}n)}{(m+\ell_{3}n)(m+\ell_{4}n)}:m,n\in\mathbb{N}\right\\}$

does not contain any infinite subsemigroup of $(\mathbb{Q}^{>0},\times)$.

###### Proof.

By contradiction, assume there exist $p\in\mathbb{Z},q\in\mathbb{N}$ with
$(p,q)=1$ such that for all $k\in\mathbb{N}$, there exist $m,n\in\mathbb{N}$
such that

$\frac{(m+\ell_{1}n)(m+\ell_{2}n)}{(m+\ell_{3}n)(m+\ell_{4}n)}=\frac{p^{k}}{q^{k}}.$

Dividing $m,n$ by $gcd(m,n)$, we can assume $(m,n)=1$. We then have

$q^{k}(m+\ell_{1}n)(m+\ell_{2}n)=p^{k}(m+\ell_{3}n)(m+\ell_{4}n).$

It follows that $q^{k}|p^{k}(m+\ell_{3}n)(m+\ell_{4}n)$. But since $(p,q)=1$,
it implies that $q^{k}|(m+\ell_{3}n)(m+\ell_{4}n)$. Similarly,
$p^{k}|(m+\ell_{1}n)(m+\ell_{2}n)$. Let

$t=\frac{(m+\ell_{1}n)(m+\ell_{2}n)}{p^{k}}=\frac{(m+\ell_{3}n)(m+\ell_{4}n)}{q^{k}}\in\mathbb{Z}.$

Therefore $t|gcd((m+\ell_{1}n)(m+\ell_{2}n),(m+\ell_{3}n)(m+\ell_{4}n))$.

For $i,j\in\\{1,2,3,4\\}$, let $d=gcd(m+\ell_{i}n,m+\ell_{j}n)$. Then
$d|(\ell_{j}-\ell_{i})n$ and
$d|((\ell_{j}-\ell_{i})m+(\ell_{j}-\ell_{i})\ell_{i}n)$. Therefore,
$d|(\ell_{j}-\ell_{i})m$. Since $(m,n)=1$, it follows that
$d|(\ell_{j}-\ell_{i})$. We deduce that
$t|(\ell_{3}-\ell_{1})(\ell_{3}-\ell_{2})(\ell_{4}-\ell_{1})(\ell_{4}-\ell
2)$. We then have

$(m+\ell_{1}n)(m+\ell_{2}n)=tp^{k}$

and

$(m+\ell_{3}n)(m+\ell_{4}n)=tq^{k}.$

Since $(m+\ell_{1}n,m+\ell_{2}n)|(\ell_{2}-\ell_{1})$, we deduce that there
exists some $k_{0}=k_{0}(\ell_{1},\ell_{2},\ell_{3},\ell_{4})$ such that

(10) $m+\ell_{1}n=u_{1}p_{1}^{k-k_{0}}\mbox{ and
}m+\ell_{2}n=u_{2}p_{2}^{k-k_{0}}$

where $(p_{1},p_{2})=1$, $p_{1}p_{2}=p$ and $u_{1}u_{2}=tp^{k_{0}}$.
Similarly, there exists $k_{1}=k_{1}(\ell_{1},\ell_{2},\ell_{3},\ell_{4})$
such that

(11) $m+\ell_{3}n=v_{1}q_{1}^{k-k_{1}}\mbox{ and
}m+\ell_{2}n=v_{2}q_{2}^{k-k_{1}}$

where $(q_{1},q_{2})=1$, $q_{1}q_{2}=q$ and $v_{1}v_{2}=tq^{k_{1}}$.

Since there are only finitely many choices of $t,u_{i},v_{i},p_{i},q_{i}$ for
$i=1,2$, it follows that there exist fixed $u_{i},v_{i},p_{i},q_{i}$ for
$i=1,2$ for which there are infinitely many $k\in\mathbb{N}$ such that (10)
and (11) simultaneously have solutions $(m,n)\in\mathbb{N}^{2}$. Solving these
equations, we get

$n=\frac{u_{2}p_{2}^{k-k_{0}}-u_{1}p_{1}^{k-k_{0}}}{\ell_{2}-\ell_{1}}=\frac{v_{2}q_{2}^{k-k_{1}}-v_{1}q_{1}^{k-k_{1}}}{\ell_{3}-\ell_{4}}.$

In particular,

$\frac{u_{2}}{(\ell_{2}-\ell_{1})p_{2}^{k_{0}}}\times
p_{2}^{k}-\frac{u_{1}}{(\ell_{2}-\ell_{1})p_{1}^{k_{0}}}\times
p_{1}^{k}-\frac{v_{2}}{(\ell_{3}-\ell_{4})q_{2}^{k_{1}}}\times
q_{2}^{k}+\frac{v_{1}}{(\ell_{3}-\ell_{4})q_{1}^{k_{1}}}\times q_{1}^{k}=0$

for infinitely many $k\in\mathbb{N}$. Since $p_{1},p_{2},q_{1},q_{2}$ are
pairwise coprime, we can apply Lemma 4.4 to show that $u_{i}=v_{i}=0$ for
$i=1,2$. It follows from (10) that $m=n=0$. This contradicts our assumption
that $m,n\in\mathbb{N}$ finishing the proof. ∎

##  5\. Additive averages of multiplicative recurrence sequences

###  5.1. Multiplicative subsemigroups of $\mathbb{N}$

Recall from Section 2 that a subset $S\subset G$ is syndetic if there exists a
finite set $F\subset G$ such that $\bigcup_{x\in F}S/x=G$. As Proposition 2.2
shows, every syndetic set $F\subset G$ has positive density with respect to
any Følner sequence in $G$. The next result shows that in the case when
$(G,\times)$ is a subsemigroup of $(\mathbb{N},\times)$, then a
(multiplicative) syndetic subset of $G$ has positive “upper density” also with
respect to the “additive” sequence $(G\cap[N])_{N\in\mathbb{N}}$.

######  Lemma 5.1.

Let $(G,\times)$ be a subsemigroup of $(\mathbb{N},\times)$ and $S$ be a
syndetic subset of $G$. Then

$\limsup_{N\to\infty}\mathop{\mathbb{E}}_{n\in G\cap[N]}1_{S}(n)>0.$

Moreover, if $G$ has positive lower additive density in $\mathbb{N}$, then

$\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n\in G\cap[N]}1_{S}(n)>0.$

###### Proof.

We first claim that for any non-empty set $E\subset\mathbb{N}$ and $c>0$, we
have

(12) $\limsup_{N\to\infty}\frac{|E\cap[N]|}{|E\cap[cN]|}>0.$

In fact, by contradiction, assume

$\limsup_{N\to\infty}\frac{|E\cap[N]|}{|E\cap[cN]|}=0.$

Then there exists $N_{0}$ such that for all $N\geq N_{0}$, we have

$\frac{|E\cap[N]|}{|E\cap[cN]|}<\frac{1}{2c}.$

Or equivalently,

$|E\cap[cN]|>2c|E\cap[N]|.$

Inductively, we have for all $n\in\mathbb{N}$,

(13) $|E\cap[c^{n}N_{0}]|>(2c)^{n}|E\cap[N_{0}]|.$

But since $|E\cap[c^{n}N_{0}]|\leq c^{n}N_{0}$, (13) cannot hold for $n$ very
large. This is a contradiction. This finishes the proof of the claim.

We now return to the proof of Lemma 5.1. Since $S$ is syndetic in $G$, there
exists a finite subset $F\subset G$ such that $S/F=G$. It follows that

$\sum_{m\in F}1_{S}(mn)\geq 1\,\,\,\mbox{ for all }n\in G.$

Therefore

(14) $\sum_{n\in[N]\cap G}\sum_{m\in F}1_{S}(mn)\geq|G\cap[N]|.$

Let $M=\max\\{m:m\in F\\}$. Since each $n\leq MN$ is repeated at most $|F|$
times in the sum appearing in the left hand side of (14), we deduce that

$\sum_{n\in G\cap[MN]}1_{S}(n)\geq\frac{|G\cap[N]|}{|F|}.$

Therefore

(15) $\mathop{\mathbb{E}}_{n\in
G\cap[MN]}1_{S}(n)\geq\frac{|G\cap[N]|}{|F||G\cap[MN]|}$

In view of the claim,

(16) $\limsup_{N\to\infty}\frac{|G\cap[N]|}{|G\cap[MN]|}>0.$

(16) and (15) imply $\overline{d}_{G}(S)>0$.

Assume $G$ has positive lower additive density. In this case, we have

$\frac{|G\cap[N]|}{|G\cap[MN]|}=\frac{|G\cap[N]|}{N}\cdot\frac{N}{|G\cap[MN]|}\geq\frac{\delta}{M}\mbox{
for sufficiently large }N.$

Hence, for sufficiently large $N$,

(17) $\mathop{\mathbb{E}}_{n\in G\cap[MN]}1_{S}(n)\geq\frac{\delta}{M|F|}.$

We now have

(18) $\mathop{\mathbb{E}}_{n\in
G\cap[N]}1_{S}(n)=\frac{1}{|G\cap[N]|}\sum_{n\in G\cap[N]}1_{S}(n)\geq\\\
\frac{|G\cap[M\lfloor N/M\rfloor|}{|G\cap[N]|}\cdot\frac{1}{|G\cap[M\lfloor
N/M\rfloor]|}\sum_{n\in G\cap[M\lfloor
N/M\rfloor]}1_{S}(n)=\frac{|G\cap[M\lfloor
N/M\rfloor|}{|G\cap[N]|}\mathop{\mathbb{E}}_{n\in G[M\lfloor
N/M\rfloor]}1_{S}(n).$

Moreover,

(19) $\frac{|G\cap[M\lfloor N/M\rfloor|}{|G\cap[N]|}\geq\frac{|G\cap[M\lfloor
N/M\rfloor|}{M\lfloor N/M\rfloor}\cdot\frac{M\lfloor
N/M\rfloor}{N}>\frac{|G\cap[M\lfloor N/M\rfloor|}{M\lfloor
N/M\rfloor}\cdot\frac{M(N/M-1)}{N}>\frac{\delta}{2}$

for sufficiently large $N$. Combining (17), (18) and (19), for sufficiently
large $N$,

$\mathop{\mathbb{E}}_{n\in G\cap[N]}1_{S}(n)>\frac{\delta^{2}}{2M|F|}$

This implies $\underline{d}_{G}(S)>0$. This finishes our proof. ∎

######  Remark 5.2.

Lemma 5.1 implies that a multiplicatively syndetic set in $\mathbb{N}$ has
positive upper density. On the contrary, it is not true that an additively
syndetic set has positive upper multiplicative density. For example, the set
$2\mathbb{N}+1$ is additively syndetic but has zero multiplicative density
with respect to any multiplicative Følner sequence.

It is worth noting that in (12), one can not replace $\limsup$ with $\liminf$.
In fact, we have:

######  Proposition 5.3.

There exists a subsemigroup $G$ of $(\mathbb{N},\times)$ such that

$\liminf_{N\to\infty}\frac{|G\cap[N]|}{|G\cap[2N]|}=0.$

###### Proof.

We construct $G$ by putting primes into $G$ in an inductive way as follows.
Let $1\leq N_{1}<N_{2}<\ldots$ be a sequence of integers with the exact values
of $N_{k}$ to be chosen later. Put all primes in the intervals
$[N_{1},2N_{1}]\cup[N_{2},2N_{2}]\cup\ldots$ into $G$. Other elements of $G$
are generated from those primes. Then for $k\in\mathbb{N}$, the set
$G\cap[N_{k+1}]$ is subset of the set $\\{p_{1}^{\ell_{1}}\cdots
p_{m}^{\ell_{m}}:p_{1},\ldots,p_{m}\in\mathbb{P}\cap[N_{k}],m\in\mathbb{N}\\}$
where $\mathbb{P}$ denotes the set of primes. Since $|\mathbb{P}\cap
N_{k}|<N_{k}$, we can see

$|G\cap[N_{k+1}]|<(\log_{2}N_{k+1})^{N_{k}}<(2\log N_{k+1})^{N_{k}}.$

On the other hand, using the Prime Number Theorem, we have

$\big{|}G\cap[N_{k+1},2N_{k+1}]\big{|}\geq\big{|}\mathbb{P}\cap[N_{k+1},2N_{k+1}]\big{|}=\frac{N_{k+1}}{\log
N_{k+1}}\big{(}1+o_{k\to\infty}(1)\big{)}$

Hence

(20) $\frac{|G\cap[N_{k+1}]|}{|G\cap[N_{k+1},2N_{k+1}]|}\leq\frac{(2\log
N_{k+1})^{N_{k}+1}}{N_{k+1}}\big{(}1+o_{k\to\infty}(1)\big{)}$

If we choose the sequence $(N_{k})$ growing fast enough so that the quantity
on the right hand side of (20) approaches zero as $k\to\infty$, then we have
the conclusion. ∎

###  5.2. Averages along subsemigroups of $(\mathbb{N},\times)$

Theorem 1.3 now follows from the next result.

######  Theorem 5.4.

Let $(X,\mathcal{B},\mu,T)$ be a measure preserving
$(\mathbb{N},\times)$-system and $A\in\mathcal{B}$ with $\mu(A)>0$. Let
$(G,\times)$ be a subsemigroup of $(\mathbb{N},\times)$. Then for every
$\ell\in\mathbb{N}$,

$\limsup_{N\to\infty}\mathop{\mathbb{E}}_{n\in G\cap[N]}\mu(A\cap
T_{n}^{-1}A\cap\ldots\cap T_{n^{\ell}}^{-1}A)>0.$

If $\liminf_{N\to\infty}\frac{|G\cap[N]|}{N}>0$ (i.e. $G$ has positive lower
density), then we have the stronger bound

$\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n\in G\cap[N]}\mu(A\cap
T_{n}^{-1}A\cap\ldots\cap T_{n^{\ell}}^{-1}A)>0.$

######  Remark 5.5.

Theorem 5.4 is no longer true if we replace the Følner sequence
$([N])_{N\in\mathbb{N}}$ by arbitrary additive Følner sequence on
$\mathbb{N}$. For example, take $(X,\mathcal{B},\mu,T)$ to be the
$(\mathbb{N},\times)$-system where $X=\mathbb{T}$, $\mu$ is the Lebesgue
measure and $T_{n}x\coloneqq x+\log n$, and let $A$ be a small interval on
$X$. Then there exist arbitrary long intervals $[M,N)$ of integers such that
$A\cap T^{-1}_{n}A=\emptyset$ for all $n\in[M,N)$. Therefore we can find a
Følner sequence $\Phi$ consisting of longer and longer such intervals with the
property that

$\lim_{N\to\infty}\mathop{\mathbb{E}}_{n\in\Phi_{N}}\mu(A\cap T_{n}A)=0.$

###### Proof.

In view of Theorem 2.3, there exists $\delta>0$ such that the set

$S=\\{n\in G:\mu(A\cap T_{n}^{-1}A\cap\ldots\cap
T_{n^{\ell}}^{-1}A)>\delta\\}$

is syndetic in $(G,\times)$. We then have

$\mathop{\mathbb{E}}_{n\in G\cap[N]}\mu(A\cap T_{n}^{-1}A\cap\ldots\cap
T_{n^{\ell}}^{-1}A)\geq\delta\mathop{\mathbb{E}}_{n\in G\cap[N]}1_{S}.$

Theorem 5.4 now follows from Lemma 5.1. ∎

We conclude this section by remarking that one can not replace the $\liminf$
and $\limsup$ in Theorem 5.4 with limits. This is because the limits may not
exist as the following proposition shows.

######  Proposition 5.6.

There exists a measure preserving system $(\mathbb{N},\times)$-system
$(X,\mathcal{B},\mu,T)$ and a set $A\in\mathcal{B}$ such that

(21) $\lim_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]}\mu(A\cap T_{n}^{-1}A)$

does not exist.

###### Proof.

Let $\chi\colon\mathbb{N}\to S^{1}$ be the completely multiplicative function
$\chi(n)=n^{i}=e^{i\log n}$ for $n\in\mathbb{N}$. It is well known that

(22) $\lim_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]}\chi(n)$

does not exist. Consider the $(\mathbb{N},\times)$-system
$(X=S^{1},\mu,(T_{n})_{n\in\mathbb{N}})$ where $\mu$ is the normalized
Lebesgue measure on $X$ and $T_{n}(x):=\chi(n)x$ for $x\in X$ and
$n\in\mathbb{N}$. We show that this system satisfies our condition. By
contradiction, assume for every $A\subset X$ measurable, the limit in (21)
exists.

Let $A$ be an arbitrary measurable subset of $X$. Let $\mathcal{H}$ be the
smallest closed subspace of $L^{2}(X)$ that contains $T_{n}1_{A}$ for all
$n\in\mathbb{N}$ and constant functions. By our assumption, for every
$g\in\mathcal{H}$,

$\lim_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]}\int_{X}1_{A}(T_{n}x)\overline{g}(x)\,d\mu(x)$

exists. Let

$\mathcal{F}=\left\\{f\in
L^{2}(X):\lim_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]}\int_{X}1_{A}(T_{n}x)\cdot\overline{f}(x)\,d\mu(x)\mbox{
exists }\right\\}.$

It follows that $\mathcal{H}\subset\mathcal{F}$. On the other hand, for all
$g\in\mathcal{H}^{\perp}$, we have

$\int_{X}1_{A}(T_{n}x)\cdot\overline{g}(x)\,\mu(x)=0.$

Hence

$\lim_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]}\int_{X}1_{A}(T_{n}x)\cdot\overline{g}(x)\,d\mu(x)$

exists and is equal to $0$. In particular,
$\mathcal{H}^{\perp}\subset\mathcal{F}$. Therefore $\mathcal{F}=L^{2}(X)$. In
conclusion, for every $A\in\mathcal{B}$ and every $f\in L^{2}(X)$, the limit

$\lim_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]}\int_{X}1_{A}(T_{n}x)\cdot\overline{f}(x)\,\mu(x)$

exists. By approximating $f$ with simple functions, we get for all $f\in
L^{2}(X)$, the limit

$\lim_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]}\int_{X}f(T_{n}x)\cdot\overline{f}(x)\,\mu(x)$

exists. However, this is not true as the following example shows. Let $f\colon
X\to\mathbb{C}$ be defined by $f(x)=x$ for $x\in S^{1}$. Then
$f(T_{n}x)=\chi(n)x$. We then have

$\int_{X}f(T_{n}x)\overline{f}(x)d\mu=\chi(n).$

Hence the limit

(23) $\lim_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]}\int_{X}f\cdot
T_{n}\overline{f}\,d\mu$

does not exist. We reach a desired contradiction. ∎

######  Remark 5.7.

One may ask whether the limit in (21) might exist after replacing Cesàro
averages with a different averaging scheme (for example logarithmic averages).
However, for any given averaging scheme, there exists a complete
multiplicative function $\chi$ for which the limit in (22) does not exist.
Hence we can carry a similar construction as in proof of Proposition 5.6 to
show that there exists a system and set $A$ for which the limit in (21) with
respect to such an averaging scheme does not exist.

###  5.3. Subordinated semigroups

In this section, we introduce the notion of subordinated semigroups of
$(\mathbb{N}^{k},\ast)$ and some basic properties of these semigroups for
later uses.

######  Definition 5.8 (Subordinated semigroups).

We say that a semigroup $(\mathbb{N}^{k},\ast)$ is subordinated to the
Euclidean norm $\|\cdot\|$ in $\mathbb{N}^{k}$ if there exists a constant
$C>0$ such that

(24) $\|n*m\|\leq C\|n\|\|m\|\text{ for all }n,m\in\mathbb{N}^{k}.$

We remark that (24) is equivalent of saying that
$\sup_{n,m\in\mathbb{N}^{k}}\frac{\|n*m\|}{\|n\|\|m\|}$ is finite. Also note
that, up to modifying the constant $C$, in the definition of subordinated
semigroup we may change the Euclidean norm $\|\cdot\|$ by any other equivalent
norm.

The following lemma follows immediately from the definition.

######  Lemma 5.9.

Let $(\mathbb{N}^{k},\ast)$ be a semigroup subordinated to the Euclidean norm
in $\mathbb{N}^{k}$. Then for every $m\in\mathbb{N}^{k}$, there exists
$K=K(m)\in\mathbb{N}$ such that for all $N\in\mathbb{N}$ and
$n\in\mathbb{N}^{k}$ with $\|n\|\leq N$, we have that $\|m*n\|\leq KN$.

A class of subordinated semigroups of particular interest for us is given by
semigroups $(\mathbb{N}^{k},\ast)$ whose operation is induced by the usual
matrix multiplication in a subsemigroup of $\mathcal{M}_{d\times
d}(\mathbb{Z})$ for some $d\in\mathbb{N}$. Here $\mathcal{M}_{d\times
d}(\mathbb{Z})$ denotes the set of $d\times d$-matrices with integer
coefficients.

######  Definition 5.10.

Let $d,k\in\mathbb{N}$. We say that a semigroup $(\mathbb{N}^{k},\ast)$ is
induced by $\mathcal{M}_{d\times d}(\mathbb{Z})$ (via $\psi$) if there exists
a linear injection $\psi\colon\mathbb{N}^{k}\to\mathcal{M}_{d\times
d}(\mathbb{Z})$ (meaning that $\psi$ is a linear function of $\mathbb{N}^{k}$
in every entry) such that $\psi(\mathbb{N}^{k})$ is a semigroup of
$\mathcal{M}_{d\times d}(\mathbb{Z})$ not contained in
$\\{A\in\mathcal{M}_{d\times d}(\mathbb{Z}):\det(A)=0\\}$ and such that

$n\ast m=\psi^{-1}(\psi(n)\psi(m)).$

######  Lemma 5.11.

A semigroup $(\mathbb{N}^{k},\ast)$ induced by $\mathcal{M}_{d\times
d}(\mathbb{Z})$ is subbordinated to the Euclidean norm in $\mathbb{N}^{k}$.

###### Proof.

We regard $\mathcal{M}_{d\times d}(\mathbb{Z})$ as a subset of
$\mathbb{R}^{d^{2}}$, and let $N\in M_{k,d^{2}}(\mathbb{R})$ and $M\in
M_{d^{2},k}(\mathbb{R})$ be such that $\psi(x)=Nx$ and $\psi^{-1}(y)=My$ for
$x\in\mathbb{N}^{k}$, $y\in\psi(\mathbb{N}^{k})$. For any matrix $A\in
M_{d_{1},d_{2}}(\mathbb{R})$, denote
$\|A\|=\sup_{x\in\mathbb{R}^{d_{1}}\setminus\\{0\\}}\|Ax\|/\|x\|$. We have
that

$\|n\ast
m\|=\|\psi^{-1}(\psi(n)\cdot\psi(m))\|\leq\|M\|\|\psi(n)\cdot\psi(m)\|\leq\|M\|\|\psi(n)\|\|\psi(m)\|\leq\|M\|\|N\|^{2}\|n\|\|m\|.$

∎

The next lemma states that a (multiplicative) syndetic subset in a semigroup
subordinated to the Euclidean norm has positive (additive) lower density; this
phenomenon was already observed in Lemma 5.1.

######  Lemma 5.12.

Let $(\mathbb{N}^{k},\ast)$ be a semigroup subordinated to the Euclidean norm
in $\mathbb{N}^{k}$ and $(Z,\ast)$ be a subsemigroup of
$(\mathbb{N}^{k},\ast)$ with

$\overline{d}_{+}(Z)\coloneqq\limsup_{N\to\infty}\frac{\big{|}[N]^{k}\cap
Z\big{|}}{N^{k}}=1.$

Let $S\subseteq Z$ be a syndetic subset (with respect to the $\ast$
operation). Then,

$\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]^{k}\cap Z}1_{S}(n)>0.$

###### Proof.

Since $S$ is syndetic, there exists a finite set $F\subset Z$ such that
$Z=F^{-1}\ast S$. Then for every $x\in Z$ we have that $1\leq\sum_{n\in
F}1_{S}(n\ast x)$. By Lemma 5.9, there exist $K=\max\\{K(n)\colon n\in F\\}$
and $N_{0}:=\max\\{N_{0}(n)\colon n\in F\\}$ such that for all $N\geq N_{0}$,
$m*n\in f([KN]^{k})$ for all $n\in[N]^{k}$ and $m\in F$. So for all $N\geq
N_{0}$,

$1\leq\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]^{k}\cap Z}\sum_{m\in
F}1_{S}(m*n)=\liminf_{N\to\infty}\sum_{m\in
F}\mathop{\mathbb{E}}_{n\in[N]^{k}\cap
Z}1_{S}(m*n)\leq\liminf_{N\to\infty}\frac{1}{N^{k}}\sum_{n\in[KN]^{k}\cap
Z}1_{S}(n),$

where in the last inequality we used the fact that the map $n\to m*n$ is
injective from $Z$ to $Z$ for all $m\in Z$. Therefore,

(25) $\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n\in[mN]^{k}\cap
Z}1_{S}(n)>K^{-k}>0.$

Now for all $N\geq KN_{0}$, let $M=[N/K]$, then (25) implies that

$\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]^{k}\cap
Z}1_{S}(n)\geq\liminf_{N\to\infty}\Bigl{(}\frac{KM}{N}\Bigr{)}^{k}\mathop{\mathbb{E}}_{n\in[KM]^{k}\cap
Z}1_{S}(n)\\\
>\liminf_{N\to\infty}\Bigl{(}\frac{KM}{N}\Bigr{)}^{k}K^{-k}>\liminf_{N\to\infty}\Bigl{(}\frac{1}{K}-\frac{1}{N}\Bigr{)}^{k}.$

So $\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]^{k}\cap Z}1_{S}(n)\geq
K^{-k}$ and we are done. ∎

###  5.4. Parametrized multiplicative functions

In this section, we introduce a class of functions that parametrize
multiplicative subsemigroups of $\mathbb{Q}^{>0}$.

######  Definition 5.13 (Parametrized multiplicative function).

For $k\in\mathbb{N}$, we say that $f\colon\mathbb{N}^{k}\to\mathbb{Q}^{\geq
0}$ is a _parametrized multiplicative function_ if

$\bar{d}_{+}\big{(}\\{n\in\mathbb{N}^{k}:f(n)=0\\}\big{)}=0$

and there exists an operation
$\ast\colon\mathbb{N}^{k}\times\mathbb{N}^{k}\to\mathbb{N}^{k}$ so that
$(\mathbb{N}^{k},\ast)$ is a semigroup subordinated to the euclidean norm in
$\mathbb{N}^{k}$ and

$f(n*m)=f(n)f(m)$

for all $n,m\in\mathbb{N}^{k}$. We say that $f$ is a _commutative parametrized
multiplicative function_ if one can choose $(\mathbb{N}^{k},\ast)$ to be a
cancelative commutative semigroup.

######  Remark 5.14.

We could disallow functions $f:\mathbb{N}^{k}\to\mathbb{Q}^{\geq 0}$ that take
the value $0$, but this would forbid some of the natural examples we exhibit
below. On the other hand, we are interested in functions that take values in
the multiplicative group $(\mathbb{Q}^{>0},\times)$, so as a compromise we
allow parametrized multiplicative functions to take the value $0$, as long as
only on a set of zero additive density.

This definition leads to a situation where we may have a function
$f:\mathbb{N}^{k}\to\mathbb{R}$ which is only defined on a full density subset
$D$ of $\mathbb{N}^{k}$; in such cases we denote by

$\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]^{k}}f(n):=\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]^{k}}1_{D}(n)f(n).$

We sometimes write $f\colon(\mathbb{N}^{k},\ast)\to\mathbb{Q}^{\geq 0}$ for a
parametrized multiplicative function to stress the semigroup operation for
which $f$ is a multiplicative function.

A typical example of a parametrized multiplicative function arising from a
semigroup $(\mathbb{N}^{k},\ast)$ induced by $\mathcal{M}_{d\times
d}(\mathbb{Z})$ is the one induced by the determinant function. That is
$f\colon(\mathbb{N}^{k},\ast)\to\mathbb{Q}^{\geq 0}$,
$n\mapsto|\det(\psi(n))|$, or more generally $n\mapsto\xi(|\det(\psi(n))|)$
for any multiplicative function
$\xi\colon|\det(\psi(\mathbb{N}^{k}))|\to\mathbb{Q}^{\geq 0}$. Here we point
out that since $\psi(\mathbb{N}^{k})$ is not contained in
$\\{A\in\mathcal{M}_{d\times d}(\mathbb{Z}):\det(A)=0\\}$, the polynomial
$\det(\psi(n))$ is non trivial and therefore
$\bar{d}\big{(}\\{n\in\mathbb{N}^{k}:\det(\psi(n))=0\\}\big{)}=0$. (To see
this, note that for $(n_{1},\ldots,n_{k-1})\in\mathbb{N}^{k-1}$ there are at
most $d$ positive integers $n_{k}\in\mathbb{N}$ such that
$\det(\psi(n_{1},\ldots,n_{k-1},n_{k}))=0$.)

######  Example 5.15.

In the following, we present some examples of parametrized multiplicative
functions.

  1. (1)

Consider the function $f$ given by $f=\det\circ\psi$, where
$\psi\colon\mathbb{N}^{k}\to\mathcal{M}_{2\times 2}(\mathbb{Z})$,
$(n_{1},n_{2})\mapsto\begin{pmatrix}n_{1}&-n_{2}\\\ n_{2}&n_{1}\end{pmatrix}$
is a linear injection. By Lemma 5.11, the semigroup induced by
$\mathcal{M}_{2\times 2}(\mathbb{Z})$ via $\psi$ is subbordinated to the
Euclidean norm in $\mathbb{N}^{2}$. Thus we have that
$f(n_{1},n_{2})=n_{1}^{2}+n_{2}^{2}$ is a parametrized multiplicative
function.

  2. (2)

If $k=a^{2}+b^{2}$ for some $a,b\in\mathbb{N}$, then
$f(n_{1},n_{2})=k(n_{1}^{2}+n_{2}^{2})$ is a parametrized multiplicative
function, since $f=\det\circ\psi$ with
$\psi\colon\mathbb{N}^{2}\to\mathcal{M}_{2\times 2}(\mathbb{Z})$ being the
linear injection $(n_{1},n_{2})\mapsto\begin{pmatrix}a&-b\\\
b&a\end{pmatrix}\begin{pmatrix}n_{1}&-n_{2}\\\ n_{2}&n_{1}\end{pmatrix}$.

  3. (3)

Similar to the previous examples, the linear injection
$\psi:(n_{1},n_{2})\mapsto\begin{pmatrix}n_{1}&-Dn_{2}\\\
n_{2}&n_{1}\end{pmatrix}$ for $D\in\mathbb{Z}$ gives us the paramatrized
multiplicative function
$f(n_{1},n_{2})=|\det\circ\psi(n_{1},n_{2})|=|n_{1}^{2}+Dn_{2}^{2}|$.

  4. (4)

Given two parametrized multiplicative functions,
$f_{1}\colon(\mathbb{N}^{d_{1}},\ast_{1})\to\mathbb{Q}^{\geq 0}$ and
$f_{2}\colon(\mathbb{N}^{d_{2}},\ast_{2})\to\mathbb{Q}^{\geq 0}$, we may
define for $\ell_{1},\ell_{2}\in\mathbb{Z}$ the function
$f\colon(\mathbb{N}^{d_{1}}\times\mathbb{N}^{d_{2}},\ast_{1}\times\ast_{2})\to\mathbb{Q}$,
such that $f(n_{1},n_{2})=f_{1}^{\ell_{1}}(n_{1})f_{2}^{\ell_{2}}(n_{2})$, if
$f_{1}(n_{1})f_{2}(n_{2})\neq 0$ and $f(n_{1},n_{2})=0$ otherwise. Since
$(\mathbb{N}^{d_{i}},\ast_{i})$, $i=1,2$ is subordinated to the Euclidean norm
in $\mathbb{N}^{d_{i}}$, we get that
$(\mathbb{N}^{d_{1}}\times\mathbb{N}^{d_{2}},\ast_{1}\times\ast_{2})$ is
subordinated to the Euclidean norm in
$\mathbb{N}^{d_{1}}\times\mathbb{N}^{d_{2}}$. It is not hard to check that the
upper density of
$\\{(n_{1},n_{2})\in\mathbb{N}^{d_{1}}\times\mathbb{N}^{d_{2}}:f(n_{1},n_{2})=0\\}$
equals 0. This implies that $f$ is a parametrized multiplicative function.
This also shows that $1/f_{1}$ is a parametrized multiplicative function.

In particular, for any $p,q\in\mathbb{Z}$ the function
$f\colon\mathbb{N}^{2}\to\mathbb{Q}^{>0}$, $(m,n)\mapsto m^{p}n^{q}$ is a
parametrized multiplicative function.

  5. (5)

Let $K$ be a finite field extension of $\mathbb{Q}$ and $\mathcal{O}_{K}$ be
the ring of integers of $K$. Let $d=[K:\mathbb{Q}]$ and
$\mathcal{B}=\\{b_{1},\dots,b_{d}\\}$ be an integral basis of
$\mathcal{O}_{K}$, i.e., every $x\in\mathcal{O}_{K}$ can be written as
$x=n_{1}b_{1}+\dots+n_{d}b_{d}$ for some $n_{1},\dots,n_{d}\in\mathbb{Z}$ in a
unique way. Let $N_{K}(x)$ denote the norm of $x$ in $K$. Since
$N_{K}(x)N_{K}(y)=N_{K}(xy)$, the function
$f\colon\mathbb{N}^{d}\to\mathbb{Q}^{>0}$ given by
$f(n_{1},\dots,n_{d})=\big{|}N_{K}(n_{1}b_{1}+\dots+n_{d}b_{d})\big{|}$ is a
parametrized multiplicative function. We remark that in this example, $f$ can
be written as $f=\det\circ\psi$ for some linear injection
$\psi\colon\mathbb{Z}^{d}\to\mathcal{M}_{d\times d}(\mathbb{Q})$.

As a special case, let $\omega=e^{2\pi i/3}$ and $K=\mathbb{Q}(\omega)$. Then
$\\{1,\omega,\omega^{2}\\}$ is an integral basis of $\mathcal{O}_{K}$. For
every $x=a_{0}+a_{1}\omega+a_{2}\omega^{2}$ and
$y=b_{0}+b_{1}\omega+b_{2}\omega^{2}$ with
$a_{0},a_{1},a_{2},b_{0},b_{1},b_{2}\in\mathbb{Q}$, note that

$\begin{split}&\quad
xy=(a_{0}b_{0}+a_{1}b_{2}+a_{2}b_{1})+(a_{0}b_{1}+a_{1}b_{0}+a_{2}b_{2})\omega+(a_{1}b_{1}+a_{0}b_{2}+a_{2}b_{0})\omega^{2}\\\
&=\begin{pmatrix}b_{0}&b_{1}&b_{2}\end{pmatrix}\cdot\begin{pmatrix}a_{0}&a_{1}&a_{2}\\\
a_{2}&a_{0}&a_{1}\\\ a_{1}&a_{2}&a_{0}\end{pmatrix}\cdot\begin{pmatrix}1\\\
\omega\\\ \omega^{2}\end{pmatrix}.\end{split}$

Therefore the function

$\begin{split}&\quad
f(a_{0},a_{1},a_{2}):=|N_{K}(a_{0}+a_{1}\omega+a_{2}\omega^{2})|=\left|\det\begin{pmatrix}a_{0}&a_{1}&a_{2}\\\
a_{2}&a_{0}&a_{1}\\\ a_{1}&a_{2}&a_{0}\end{pmatrix}\right|\\\
&=|(a_{0}+a_{1}+a_{2})(a_{0}^{2}+a_{1}^{2}+a_{2}^{2}-a_{0}a_{1}-a_{0}a_{2}-a_{1}a_{2})|=|a_{0}^{3}+a_{1}^{3}+a_{2}^{3}-3a_{0}a_{1}a_{2}|\end{split}$

is a parametrized multiplicative function.

  6. (6)

Let $p\colon\mathbb{Z}\to\mathbb{Z}$ be the minimal polynomial of
$A\in\mathcal{M}_{n\times n}(\mathbb{Z})$ and $d=\operatorname{deg}(p)$. Let
$\psi\colon\mathbb{N}^{d}\to\mathcal{M}_{n\times n}(\mathbb{Z})$ be such that
$\psi(n_{0},\dots,n_{d-1}):=n_{0}I_{n}+n_{1}A+\dots+n_{d-1}A^{d-1}$ (which is
a linear injection since $d$ is the degree of the minimal polynomial of $A$).
Since $p(A)=0$, $\psi(\mathbb{N}^{d})$ is a semigroup and
$\psi^{-1}\colon\psi(\mathbb{N}^{d})\to\mathbb{N}$ is a well defined map. So
we may denote

$n\ast m:=\psi^{-1}(\psi(n)\psi(m)).$

Also note that $\psi(\mathbb{N}^{d})$ does not consists of only non-invertible
matrices. Then the group $(\mathbb{N}^{d},\ast)$ is induced by
$\mathcal{M}_{n\times n}(\mathbb{Z})$ (via the map $\psi$), and so
$\left|\det\circ\psi\right|\colon\mathbb{N}^{d}\to\mathbb{Z}$ is a
parametrized multiplicative function.

To illustrate this class of examples, consider the Fibonacci matrix
$A=\begin{pmatrix}1&1\\\ 1&0\end{pmatrix}$. Its minimal polynomial is of
degree 2 and then we can consider
$\psi\colon\mathbb{N}^{2}\to\mathcal{M}_{2\times 2}(\mathbb{Z})$, given by

$\psi(n,m)=nI+mA=\begin{pmatrix}n+m&m\\\ m&n\end{pmatrix}$

Hence $|\det\circ\psi(n,m)|=|n^{2}-m^{2}+nm|$ is a parametrized multiplicative
function. More generally, consider a matrix $A=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}$ not being a multiple of the identity. Then its minimal
polynomial is of degree 2 and therefore

$\left|\det(nI+mA)\right|=\left|\det\begin{pmatrix}n+am&bm\\\
cm&n+dm\end{pmatrix}\right|=|n^{2}+(a+d)mn+(ad-bc)m^{2}|$

is a parametrized multiplicative function.

  7. (7)

Consider the quaternion semigroup of matrices
$\\{M_{n_{1},n_{2},n_{3},n_{4}}\colon n_{1},n_{2},n_{3},n_{4}\in\mathbb{N}\\}$
where

$M_{n_{1},n_{2},n_{3},n_{4}}=\begin{pmatrix}n_{1}&-n_{2}&-n_{3}&-n_{4}\\\
n_{2}&n_{1}&-n_{4}&n_{3}\\\ n_{3}&n_{4}&n_{1}&-n_{2}\\\
n_{4}&-n_{3}&n_{2}&n_{1}\end{pmatrix}.$

Via the linear injection $\psi\colon\mathbb{N}^{4}\to\mathcal{M}_{4\times
4}(\mathbb{Z})$ with $(n_{1},n_{2},n_{3},n_{4})\mapsto
M_{n_{1},n_{2},n_{3},n_{4}}$ we obtain the parametrized multiplicative
functions
$f(n_{1},n_{2},n_{3},n_{4})=\det\circ\psi(n_{1},n_{2},n_{3},n_{4})=(n_{1}^{2}+n_{2}^{2}+n_{3}^{2}+n_{4}^{2})^{2}$
and
$g(n_{1},n_{2},n_{3},n_{4})=(\det)^{1/2}(\psi(n_{1},n_{2},n_{3},n_{4}))=n_{1}^{2}+n_{2}^{2}+n_{3}^{2}+n_{4}^{2}$.

We remark that the functions in examples (1)–(6) are all _commutative_
parametrized multiplicative functions, however it is not clear whether the one
in example (7) is, as the quaternion semigroup is non-commutative.

###  5.5. Averages along parametrized multiplicative functions

In this section we prove Theorem 1.4. In fact we will establish the following
more general version which also deals with commutative parametrized functions
that take values in $\mathbb{Q}^{\geq 0}$. Recall the convention adopted in
Remark 5.14.

######  Theorem 5.16.

Let $(X,\mathcal{B},\mu,T)$ be a measure preserving
$(\mathbb{N},\times)$-system and $A\in\mathcal{B}$ with $\mu(A)>0$. Let
$f\colon\mathbb{N}^{k}\to\mathbb{Q}^{\geq 0}$ be a commutative parametrized
multiplicative function and $\ell\in\mathbb{N}$. Then

$\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]^{k}}\mu\Big{(}A\cap
T_{f(n)}^{-1}A\cap T_{f(n)^{2}}^{-1}A\cap\cdots\cap
T_{f(n)^{\ell}}^{-1}A\Big{)}>0.$

###### Proof.

The proof resembles that of Theorem 5.4. Let $(\mathbb{N}^{k},\ast)$ be the
semigroup that gives rise to the parametrized multiplicative function $f$. Let
$Z$ be the set of all $n\in\mathbb{N}^{k}$ with $f(n)>0$. Since $f$ is a
homomorphism, $(Z,\ast)$ is a subsemigroup of $(\mathbb{N}^{k},\ast)$. For
each $n\in Z$, let $R_{n}:=T_{f(n)}$. Denote by $n^{\ast k}$ the iterate
product $n^{\ast k}=n\ast n\cdots\ast n$ ($k$-times). By Theorem 2.3, there
exists $\delta>0$ such that the set

$S=\\{n\in Z:\mu(A\cap R_{n}^{-1}A\cap\ldots\cap
R_{n^{\ast\ell}}^{-1}A)>\delta\\}$

is syndetic in $(Z,\ast)$. Since $\overline{d}_{+}(Z)=1$ by definition, we
have

$\displaystyle\quad\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]^{k}}\mu(A\cap
T_{f(n)}^{-1}A\cap T_{f(n)^{2}}^{-1}A\cap\cdots T_{f(n)^{\ell}}^{-1}A)$
$\displaystyle=\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]^{k}\cap
Z}\mu(A\cap T_{f(n)}^{-1}A\cap T_{f(n)^{2}}^{-1}A\cap\cdots
T_{f(n)^{\ell}}^{-1}A)$
$\displaystyle\geq\delta\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]^{k}\cap
Z}1_{S}(n)>0,$

where the last inequality comes from Lemma 5.12. ∎

As an immediate corollary of Theorem 1.4, we conclude that for all
multiplicative measure preserving system $(X,\mathcal{B},\mu,T)$ and
$A\in\mathcal{B}$ with $\mu(A)>0$, the following averages are positive:

  1. (1)

$\liminf_{N\to\infty}\mathop{\mathbb{E}}_{m,n\in[N]}\mu(T_{m^{a}}^{-1}A\cap
T_{n^{b}}^{-1}A)$ for $a,b\in\mathbb{N}$;333We do not know how to establish
this without using Theorem 1.4 (i.e. using only Theorem 1.3).

  2. (2)

$\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n,m\in[N]}\mu(A\cap
T_{n^{2}+m^{2}}^{-1}A)$ (which implies Proposition 1.5);

  3. (3)

$\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n,m\in[N]}\mu(A\cap
T_{n^{2}+Dm^{2}}^{-1}A)$ for all $D\in\mathbb{N}$;

  4. (4)

$\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n,m\in[N]}\mu(A\cap
T_{k(n^{2}+m^{2})}^{-1}A\cap T_{(k(n^{2}+m^{2}))^{2}}A)$ for all $k$ of the
form $a^{2}+b^{2}$ for some $a,b\in\mathbb{Z}$.

######  Remark 5.17.

We remark that item (4) above fails if $k$ does not have the form
$a^{2}+b^{2}$. Indeed, if this is the case, there exists a prime $p\equiv
3\pmod{4}$ such that $p$ has odd exponent in prime factorization of $k$. For
$n\in\mathbb{N}$ let $f_{p}(n)$ denote the exponent of $p$ in the prime
factorization of $n$. We then have $f_{p}(k(m^{2}+n^{2}))$ is odd for all
$m,n\in\mathbb{N}$. For $n\in\mathbb{N}$, color $n$ with $C(n)\equiv
f_{p}(n)\pmod{2}$. It follows that $\\{k(m^{2}+n^{2}):m,n\in\mathbb{N}\\}$ is
not a set of topological multiplicative recurrence.

###  5.6. Single recurrence and non-commutative semigroups

Next theorem shows that in the special case $\ell=1$ of Theorem 1.4, one can
drop the assumption that $f$ arises from a commutative semigroup.

######  Theorem 5.18.

Let $(X,\mathcal{B},\mu,T)$ be an $(\mathbb{N},\times)$ measure preserving
system and $A\in\mathcal{B}$ with $\mu(A)>0$. Let
$f\colon(\mathbb{N}^{k},\ast)\to\mathbb{Q}^{\geq 0}$ be a parametrized
multiplicative function. Then

$\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]^{k}}\mu(A\cap
T_{f(n)}^{-1}A)>0.$

###### Proof.

Let $(\mathbb{N}^{k},\ast)$ be the semigroup that gives the parametrized
multiplicative function $f$. Let $Z$ be the set of all $n\in\mathbb{N}^{k}$
with $f(n)>0$. Since $f$ is a homomorphism, $(Z,\ast)$ is a subsemigroup of
$(\mathbb{N}^{k},\ast)$. For $n\in Z$, denote $R_{n}:=T_{f(n)}$. By Theorem
2.4, the set

$S:=\big{\\{}n\in Z:\mu(A\cap R^{n}A)>\mu^{2}(A)/2\big{\\}}$

is syndetic in $(Z,\ast)$. Since $\mu(A\cap
R^{n}A)\geq\tfrac{1}{2}\mu(A)^{2}1_{S}(n)$ for all $n\in Z$ and
$\overline{d}_{+}(Z)=1$, it suffices to show that

$\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n\in[N]^{k}\cap Z}1_{S}(n)>0.$

The last inequality follows from Lemma 5.12, so our proof finishes. ∎

######  Remark 5.19.

As an immediate corollary of Theorem 5.18, we conclude that for all measure
preserving $(\mathbb{N},\times)$-system $(X,\mathcal{B},\mu,T)$ and
$A\in\mathcal{B}$ with $\mu(A)>0$,

$\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n_{1},n_{2},n_{3},n_{4}\in[N]}\mu(A\cap
T_{n_{1}^{2}+n_{2}^{2}+n_{3}^{2}+n_{4}^{2}}^{-1}A)>0.$

##  6\. Density regularity of Pythagorean triples in finite fields

###  6.1. Pythagorean pairs in finite fields.

In what follows we explore the question of finding Pythagorean pairs in sets
of positive density in finite fields. We start with a result in number theory,
which follows from [21, Theorem 5A] (see also the proof of Corollary 5B
there).

######  Theorem 6.1 ([21, Theorem 5A]).

Let $k\in\mathbb{N}$, let $F$ be a finite field, let $Q=\\{x^{2}\colon x\in
F\\}$ and let $a_{1},\dots,a_{k}\in F$ be arbitrary and pairwise distinct.
Then

$\frac{\Big{|}(Q-a_{1})\cap\cdots\cap(Q-a_{k})\Big{|}}{|F|}=\frac{1}{2^{k}}+o_{k,|F|\to\infty}(1).$

In next proposition, we show that 1.1 is true when $\mathbb{N}$ is replaced by
a finite field. In fact, we prove a stronger result which involves “density
regularity” rather than “partition regularity”.

######  Proposition 6.2.

For every $\delta>0$ there is $M>0$ such that whenever $\mathbb{F}$ is a
finite field with $|\mathbb{F}|>M$ and $A\subset\mathbb{F}$ having
$|A|>\delta|\mathbb{F}|$, there exist $x,y\in A$ with $x^{2}+y^{2}$ being a
perfect square.

###### Proof.

Since $|A|>\delta|\mathbb{F}|$, the set $B:=\\{x^{2}:x\in A\\}$ has
$|B|>\tfrac{\delta}{2}|\mathbb{F}|$. Our task is to show that there exist
$x,y\in B$ such that $x+y$ is a perfect square when $|\mathbb{F}|$ is
sufficiently large. Denote by $Q$ the set of perfect squares in $\mathbb{F}$;
we will show that there exists $x\in B$ such that $(Q-x)\cap B\neq\emptyset$.
This will follow if we show that the set $Q_{B}:=\bigcup_{x\in B}(Q-x)$
satisfies $|Q_{B}|>(1-\delta/2)|\mathbb{F}|$.

Let $Q^{c}:=\mathbb{F}\setminus Q$ and $Q_{B}^{c}:=\mathbb{F}\setminus Q_{B}$.
Then we have $Q_{B}^{c}=\bigcap_{x\in B}(Q^{c}-x)$ and we want to show that
$|Q_{B}^{c}|<\tfrac{\delta}{2}|\mathbb{F}|$. Let $k\in\mathbb{N}$ be such that
$1/2^{k}<\frac{\delta}{2}$. In view of Theorem 6.1,
$|Q_{B}^{c}|=\frac{1}{2^{k}}+o_{k,|\mathbb{F}|\to\infty}(1)$, so if
$|\mathbb{F}|$ is large enough we conclude that indeed
$|Q_{B}^{c}|<\tfrac{\delta}{2}$, as desired. ∎

######  Remark 6.3.

In contrast to Proposition 6.2, it is not true that every set of positive
additive density of $\mathbb{N}$ contains $x,y$ such that $x^{2}+y^{2}$ is a
perfect square. For example, consider the set $4\mathbb{N}+1$.

###  6.2. Pythagorean triples in finite fields

It is also natural to ask whether an extension of Proposition 6.2 stating that
there exist $x,y,z\in A$ with $x^{2}+y^{2}=z^{2}$ still holds. We show that
this is false and characterize all equations of the form
$ax^{2}+by^{2}+cz^{2}=0$ for which the analogous results are true.

######  Proposition 6.4.

For $a,b,c\in\mathbb{Z}$, the followings are equivalent:

  1. (i)

For every $\delta>0$, there is $M$ such that whenever $\mathbb{F}$ is a finite
field with $|\mathbb{F}|>M$ and $A\subseteq\mathbb{F}$ with
$|A|>\delta|\mathbb{F}|$, there exist $x,y,z\in A$ such that
$ax^{2}+by^{2}+cz^{2}=0$.

  2. (ii)

$a+b+c=0$.

######  Remark 6.5.

We remark that related questions are also raised and addressed in [9, 19].

###### Proof.

Suppose that $a+b+c=0$ and $\mathbb{F}$ is a finite field. Let $\delta>0$ and
$A\subset\mathbb{F}$ with $|A|>\delta|\mathbb{F}|$. Set $B:=\\{x^{2}\colon
x\in A\\}$. Then $|B|\geq|A|/2>\delta/2|\mathbb{F}|$. It suffices to find
$x,t\in\mathbb{F}$ such that $x,x+ct,x-bt\in B$, but this immediately follows
from Szemerédi Theorem for finite groups ([24, Theorem 10.5]).

Now assume that $a+b+c\neq 0$ and let $p>2$ be a prime. Let $m=|a+b+c|>0$ and
$q$ be the smallest prime such that $q>\max\\{m,5\\}$. For $0\leq i\leq m-1$
and $0\leq j\leq q-1$, denote

$B_{i,j}=\Big{\\{}x\in\mathbb{Z}\colon iN/m\leq x<(i+1)p/m,x\equiv
j\pmod{q}\text{ and $x$ is a perfect square}\pmod{p}\Big{\\}}$

and let $A_{i,j}=\\{x\in\mathbb{F}\colon x^{2}\in B_{i,j}\\}$. Then
$\sum_{i=0}^{m-1}\sum_{j=0}^{p-1}|A_{i,j}|=p$.

We say that $(i,j)$ is _good_ if $mj\neq ip\mod q$. Let $U$ be the set of
$(i,j),0\leq i\leq m-1,0\leq j\leq q-1$ which is good, and $V$ be the set of
$(i,j)$ which is not good. Using the estimation $|A_{i,j}|\leq 2|B_{i,j}|\leq
2+2p/mq$ and the fact that $|V|\leq m$, we have that

$\sum_{(i,j)\text{ is good}}|A_{i,j}|\geq p-2m-4p/q.$

By the pigeonhole principle, if $p$ is sufficiently large, then there exists a
good pair $(i,j)$ such that $|A_{i,j}|\geq(1-5/q)p/mq$. It suffices to show
that $ax^{2}+by^{2}+cz^{2}\neq 0$ for all $x,y,z\in A_{i,j}$.

Indeed, for any $x,y,z\in B_{i,j}$, denote $x^{\prime}=mx-ip$, $y^{\prime}=my-
ip$ and $z^{\prime}=mz-ip$. Then $0\leq x^{\prime},y^{\prime},z^{\prime}<p$.
If $ax+by+cz=0\mod p$, then $m(ax+by+cz)=0\mod p$. By the construction of
$x^{\prime},y^{\prime}$ and $z^{\prime}$, we have that
$ax^{\prime}+by^{\prime}+cz^{\prime}=0\mod p$. Assume that $ax+by+cz=kp$ and
$ax^{\prime}+by^{\prime}+cz^{\prime}=rp$ for some $k,r\in\mathbb{Z}$, then
$(mk-r)p=(a+b+c)ip=mip$. In other words, $mk-r=mi$. This implies that $m|r$.
On the other hand, since $-mp<ax^{\prime}+by^{\prime}+cz^{\prime}=rp<mp$, we
must have that $r=0$ and $ax^{\prime}+by^{\prime}+cz^{\prime}=0$. Therefore,

$0=ax^{\prime}+by^{\prime}+cz^{\prime}\equiv(mj-ip)(a+b+c)=\pm m(mj-ip)\mod
q.$

This is impossible since $(i,j)$ is a good pair. This completes the proof.∎

######  Remark 6.6.

An example due to Frantzikinakis shows that there exists a subset of
$\mathbb{N}$ of positive multiplicative density that contains no triple
$x,y,z$ such that $x^{2}+y^{2}=z^{2}$. The example is as follows: let
$(\mathbb{T}\coloneqq\mathbb{R}/\mathbb{Z},\mu,T)$ be the multiplicative
measure preserving system where $\mu$ is the Lebesgue measure and
$T_{n}x=n^{2}x$ for all $x\in\mathbb{T}$. Let $x_{0}$ be a generic point for
$\mu$. Fix $\delta>0$ small. Then the set $E=\\{n\in\mathbb{N}:n^{2}x_{0}\mod
1\in[1/2-\delta,1/2+\delta]\\}$ has multiplicative density $2\delta>0$. For
$x,y,z\in E$, $(x^{2}+y^{2})x_{0}\mod 1\in[0,2\delta]\cup[1-2\delta,1)$ and
$z^{2}x_{0}\in[1/2-\delta,1/2+\delta]$. Hence $(x^{2}+y^{2})x_{0}\neq
z^{2}x_{0}\mod 1$. It follows that $x^{2}+y^{2}\neq z^{2}$.

##  7\. Open questions

In this section, we collect some open questions regarding sets of
multiplicative recurrence that naturally arise from our study. In [11, Problem
6], Frantzikinakis and Host ask whether the set
$\\{(m^{2}+n^{2})/m^{2}:m,n\in\mathbb{N}\\}$ is a set of topological
multiplicative recurrence. In the spirit of this paper, we can ask the average
version of this question, and some other related configurations:

######  Question 7.1.

Let $(X,\mathcal{B},\mu,T)$ be a multiplicative measure preserving system and
$A\in\mathcal{B}$ with $\mu(A)>0$. Is it true that

$\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n,m\in[N]}\mu(T_{n^{2}+m^{2}}^{-1}A\cap
T_{m^{2}}^{-1}A)>0,$

or

$\liminf_{N\to\infty}\mathop{\mathbb{E}}_{n,m\in[N]}\mu(T_{n^{2}+n}^{-1}A\cap
T_{m^{2}}^{-1}A)>0?$

Theorem 1.6 gives criteria for the image of a Möbius transformation to be a
set of topological multiplicative recurrence. However there are cases not
covered by this theorem. For example, we do not know whether
$\\{(6n+3)/(6n+2):n\in\mathbb{N}\\}$ is a set of topological multiplicative
recurrence. Because of this reason, we ask:

######  Question 7.2.

For $a\in\mathbb{N}$, $b,d\in\mathbb{Z}$, is it true that
$S=\\{(an+b)/(an+d):n\in\mathbb{N}\\}$ is a set of topological multiplicative
recurrence if and only if $a|b$ or $a|d$?

For a polynomial $P\in\mathbb{Z}[x]$, Theorem 4.3 shows that
$\\{P(n):n\in\mathbb{N}\\}$ contains an infinite multiplicative semigroup if
and only if $P(x)=(ax+b)^{d}$ for some $a,d\in\mathbb{N}$, $b\in\mathbb{Z}$
with $a|b(b-1)$ and $d\in\mathbb{N}$. Here we can ask:

######  Question 7.3.

For which polynomial $P\in\mathbb{Z}[x]$ is $\\{P(n):n\in\mathbb{N}\\}$ a set
of topological multiplicative recurrence?

Even the answer for the following question is unknown:

######  Question 7.4.

Is $\\{n^{2}+1:n\in\mathbb{N}\\}$ a set of topological multiplicative
recurrence?

It is known that the set of shifted primes $\mathbb{P}-1=\\{p-1:p\text{ is a
prime}\\}$ and $\mathbb{P}+1=\\{p+1:p\text{ is a prime}\\}$ are sets of
additive recurrence. Hence it is of interest to ask:

######  Question 7.5.

Are $\mathbb{P}-1$ and $\mathbb{P}+1$ sets of topological multiplicative
recurrence?

We remark that neither of the sets $\mathbb{P}-1$ and $\mathbb{P}+1$ contain a
multiplicative semigroup, since for any $a\in\mathbb{N}$ we can factor
$a^{2}-1=(a-1)(a+1)$ and $a^{3}+1=(a+1)(a^{2}-a+1)$ and hence $\mathbb{P}+1$
does not contain a perfect square and $\mathbb{P}-1$ does not contain a
perfect cube.

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