# Bohr sets in sumsets I: Compact groups

Anh N. Le  and  Thái Hoàng Lê  Department of Mathematics
Ohio State University
Columbus, OH 43210, USA<EMAIL_ADDRESS>Department of Mathematics
University of Mississippi
University, MS 38677, USA<EMAIL_ADDRESS>
###### Abstract.

Let $G$ be a compact abelian group and $\phi_{1},\phi_{2},\phi_{3}$ be
continuous endomorphisms on $G$. Under certain natural assumptions on the
$\phi_{i}$’s, we prove the existence of Bohr sets in the sumset
$\phi_{1}(A)+\phi_{2}(A)+\phi_{3}(A)$, where $A$ is either a set of positive
Haar measure, or comes from a finite partition of $G$. The first result
generalizes theorems of Bogolyubov and Bergelson-Ruzsa. As a variant of the
second result, we show that for any partition
$\mathbb{Z}=\bigcup_{i=1}^{r}A_{i}$, there exists an $i$ such that
$A_{i}-A_{i}+sA_{i}$ contains a Bohr set for any
$s\in\mathbb{Z}\setminus\\{0\\}$. The latter is a step toward an open question
of Katznelson and Ruzsa.

###### 2020 Mathematics Subject Classification:

Primary: 05B10; Secondary: 11B13, 37A45

##  1\. Introduction and statements of results

Let $G$ be an abelian topological group. For a finite set $\Lambda$ of
characters (i.e. continuous homomorphisms from $G$ to
$S^{1}:=\\{z\in\mathbb{C}:|z|=1\\}$) and $\eta>0$, the set

$B(\Lambda;\eta):=\\{x\in G:|\gamma(x)-1|<\eta\textup{ for any
}\gamma\in\Lambda\\}$

is called a Bohr set or a _Bohr neighborhood of $0$_. We refer to $\eta$ as
the radius and $|\Lambda|$ as the rank (or dimension) of the Bohr set. The set
$B(\Lambda;\eta)$ is also called a Bohr-$(|\Lambda|,\eta)$ set.

If $A,B\subset G$, the sumset and difference set of $A$ and $B$ are $A\pm
B:=\\{a\pm b:a\in A,b\in B\\}$. If $c\in\mathbb{Z}$, we define $cA:=\\{ca:a\in
A\\}$. The study of Bohr sets in sumsets started with the following important
theorem of Bogolyubov [7]111This is reminiscent of Steinhaus’ theorem, which
says that if $A\subset\mathbb{R}$ has positive Lebesgue measure, then $A-A$
contains an open interval around 0..

######  Theorem 1.1 (Bogolyubov [7]).

If $A\subset\mathbb{Z}$ has positive upper Banach density, i.e.

$d^{*}(A):=\lim_{N\rightarrow\infty}\sup_{M\in\mathbb{Z}}\frac{|A\cap[M+1,M+N]|}{N}>0,$

then $A+A-A-A$ contains a Bohr set whose rank and radius depend only on
$d^{*}(A)$.

While it originated from the study of almost periodic functions, Bogolyubov’s
theorem is now a standard tool in additive combinatorics. It was used in
Ruzsa’s proof of Freiman’s theorem [26] and in Gowers’ proof of Szemerédi’s
theorem [14]. See [5, 15] for a recent variant of Bogolyubov’s theorem and its
applications.

The more copies of $A$ are involved, the more structured the sumset is. This
reflects the fact that more convolutions result in smoother functions. Thus, a
natural question is: What is the smallest number of copies of $A$ that will
guarantee the existence of a Bohr set? In $\mathbb{Z}$, it is known that $A-A$
does not necessarily contain a Bohr set, which is a result of Kriz [22]. On
the other hand, Følner [9] proved that there is a Bohr set $B$ such that
$(A-A)\setminus B$ has density 0.

Regarding three copies of $A$, Bergelson and Ruzsa [3] proved the following:

######  Theorem 1.2 (Bergelson-Ruzsa [3]).

Let $r,s,t$ be non-zero integers satisfying $r+s+t=0$. If $A\subset\mathbb{Z}$
has positive upper Banach density, then $rA+sA+tA$ contains a Bohr set whose
rank and radius depend only on $r,s,t$ and $d^{*}(A)$.

The condition $r+s+t=0$ is easily seen to be necessary, by taking
$A=M\mathbb{Z}+1$ for some $M>|r|+|s|+|t|$, since any Bohr set must
necessarily contain 0. In particular, one cannot expect $A+A-A$ to contain a
Bohr set. When $(r,s,t)=(1,1,-2)$, Bergelson-Ruzsa’s theorem generalizes
Bogolyubov’s, since $A+A-2A\subset A+A-A-A$.

###  1.1. Partition results in $\mathbb{Z}$

While the problem of finding Bohr sets in sumsets of sets having positive
density has attracted much attention, the analogous question concerning
partitions of $\mathbb{Z}$ was little studied until recently. Regarding the
latter, there is a well-known problem in additive combinatorics and dynamical
systems, which was popularized by Ruzsa [27, Chapter 5] and Katznelson [21].

######  Question 1.3.

If $\mathbb{Z}=\bigcup_{i=1}^{r}A_{i}$, must there exist
$i\in\\{1,2,\ldots,r\\}$ such that $A_{i}-A_{i}$ contains a Bohr set?

In terms of dynamical systems, Question 1.3 asks if any set of recurrence for
minimal topological systems is also a set of recurrence for minimal isometries
(also known as a set of Bohr recurrence). See [13] for a detailed account of
the history of 1.3, as well as its many equivalent formulations.222In [13],
what we call “Bohr set” is referred to as a “Bohr neighborhood of $0$.” Their
Bohr sets are our Bohr sets translated by any element. While Question 1.3
remains open at the moment and only some partial results were obtained [13,
19], we do have a positive answer when three copies of $A_{i}$ are involved.

######  Theorem 1.4.

Let $\mathbb{Z}=\bigcup_{i=1}^{r}A_{i}$ be a partition.

  1. (a)

For any $s_{1},s_{2}\in\mathbb{Z}\setminus\\{0\\}$, there exists
$i\in\\{1,2,\ldots,r\\}$ such that the set $s_{1}A_{i}-s_{1}A_{i}+s_{2}A_{i}$
contains a Bohr set whose rank and radius depend only on $r$ and
$s_{1},s_{2}$.

  2. (b)

There exists $i\in\\{1,2,\ldots,r\\}$ such that for any
$s\in\mathbb{Z}\setminus\\{0\\}$, the set $A_{i}-A_{i}+sA_{i}$ contains a Bohr
set.

Theorem 1.4 highlights the difference between partition and density since, as
we mentioned earlier, there is a set $A\subseteq\mathbb{Z}$ of positive
density such that $A-A+A$ does not contain a Bohr set.

The expression $s_{1}A_{i}-s_{1}A_{i}+s_{2}A_{i}$ is related to Rado’s
condition on partition regularity [24]. Recall that an equation
$s_{1}x_{1}+s_{2}x_{2}+\cdots+s_{\ell}x_{\ell}=0$ with coefficients in
$\mathbb{Z}\setminus\\{0\\}$ is partition regular if under any finite
partition (or coloring) of $\mathbb{Z}\setminus\\{0\\}$, there exists a
monochromatic solution $(x_{1},x_{2},\ldots,x_{\ell})$. Rado’s theorem says
that the equation $s_{1}x_{1}+s_{2}x_{2}+\cdots+s_{\ell}x_{\ell}=0$ is
partition regular if and only if $\\{s_{1},\ldots,s_{\ell}\\}$ satisfies the
following condition: There exists a nonempty set $J\subset\\{1,\ldots,\ell\\}$
such that $\sum_{i\in J}s_{i}=0$. Using the facts that
$(s_{1}+\ldots+s_{\ell})A\subseteq s_{1}A+\ldots+s_{\ell}A$, and a Bohr set
must contain $0$, Theorem 1.4(a) implies that for $\ell\geq 3$ and
$s_{1},\ldots,s_{\ell}\in\mathbb{Z}\setminus\\{0\\}$, the following are
equivalent:

  1. (1)

For any partition $\mathbb{Z}=\bigcup_{i=1}^{r}A_{i}$, there exists
$i\in\\{1,\ldots,r\\}$ such that $s_{1}A_{i}+\ldots+s_{\ell}A_{i}$ contains a
Bohr set.

  2. (2)

$\\{s_{1},\ldots,s_{\ell}\\}$ satisfies Rado’s condition above.

A novelty of Theorem 1.4(b) is that it guarantees a single set $A_{i}$ that
works for every coefficient $s$ (on the other hand, we do lose control on the
rank and radius of the Bohr set). When $s$ is very large, the set $sA_{i}$ is
small and so its contribution to the sum diminishes. While there is no
consensus on what the answer to 1.3 should be, Theorem 1.4(b) provides
evidence that the answer to 1.3 is either positive or very delicate.

In [13, Table 1, p. 8], Glasscock-Koutsogiannis-Richter summarized results on
Bohr sets in sumsets, pertaining to both density and partition. Our Theorem
1.4 fills in the blank on the Syndeticity333A subset $A$ of a group $G$ is
called syndetic if $G$ can be covered by finitely many translated of $A$.
column and $rA+sA+tA$ row of their table.

###  1.2. Results in compact groups

Bogolyubov’s theorem has been generalized to other groups as well (in more
general groups, the upper Banach density $d^{*}$ can be defined in terms of
Følner sequences or invariant means). Følner [9, 10] extended Bogolyubov’s
theorem to all abelian groups. Answering a question of Hegyvári-Ruzsa [20],
Björlund-Griesmer [6] proved that in any countable discrete abelian group $G$,
for any $A\subset G$ with $d^{*}(A)>0$, for “many” $a\in A$, the set $A+A-A-a$
contains a Bohr set whose rank and radius depend only on $d^{*}(A)$. Very
recently, Griesmer [18] generalized Theorem 1.2 to all countable discrete
abelian groups, though his proof does not give effective bounds for the rank
and radius of the Bohr set in question.

Bergelson-Ruzsa and Bogolyubov first proved their theorems in the cyclic group
$\mathbb{Z}_{N}$, and the statements in $\mathbb{Z}$ follow from a compactness
argument. Likewise, in Björlund-Griesmer [6] and Griesmer [18], certain
compact groups (namely Bohr compactifications and Kronecker factors) play a
prominent role. In view of this “compact first” strategy, the main goal of
this paper is, in fact, to study the existence of Bohr sets in sumsets of
compact groups. Under this investigation, Theorem 1.4 arises as an application
of our general method. In a subsequent paper, we will study the existence of
Bohr sets in arbitrary discrete groups by transferring our results from
compact groups.

Another feature of our work is the consideration of continuous homomorphisms
$\phi:G\rightarrow G$ and the image $\phi(A)$ rather than just dilations $cA$.
This point of view leads to a wider range of applications, for example, linear
maps on vector spaces and multiplication by an element in a ring (see Theorems
1.7 and 1.8 below). This new perspective was also adopted in recent work of
Ackelsberg-Bergelson-Best [1] on Khintchine-type recurrence for actions of an
abelian group (Theorem 1.10 below).

Our main result on Bohr sets in sumsets arising from partitions is as follows.

######  Theorem 1.5.

Let $G$ be a compact abelian group with normalized Haar measure $\mu$ and let
$\phi_{1},\phi_{2}:G\rightarrow G$ be continuous homomorphisms satisfying

  1. (a)

$\phi_{1},\phi_{2}$ are commuting, and

  2. (b)

$\phi_{1}(G),\phi_{2}(G)$ have finite indices in $G$.

Let $G=\bigcup_{i=1}^{r}A_{i}$ be a partition of $G$ into measurable sets.
Then for some $1\leq i\leq r$,

$\phi_{1}(A_{i})-\phi_{1}(A_{i})+\phi_{2}(A_{i})$

contains a Bohr-$(k,\eta)$ set, where $k$ and $\eta$ depend only on $r$,
$[G:\phi_{1}(G)]$ and $[G:\phi_{2}(G)]$.

######  Remark 1.

  * •

If $\mu(A)>0$, then $A+A-A$ is not guaranteed to contain a Bohr set. For a
counterexample, take $G=\mathbb{Z}_{2N}$ for some large $N$ and
$A=\\{a\in\mathbb{Z}_{2N}:a\textup{ is odd}\\}$. In particular, the analogous
version of Theorem 1.5 for sets of positive measure fails.

  * •

We do not know if the commuting condition can be removed entirely, though it
can be slightly relaxed (see Theorem 3.5). For example, commutativity is not
required when $\phi_{1}$ or $\phi_{2}$ is an automorphism (see Remark 5).

  * •

The finite index condition on $\phi_{1}(G)$ cannot be removed, by taking for
example $\phi_{1}=0$ and $\phi_{2}(x)=x$. On the other hand, we do not know
whether the finite index condition on $\phi_{2}(G)$ can be removed. If we let
$\phi_{2}=0$, then the situation amounts to 1.3 itself.

We now turn our attention to density results. In compact abelian groups, the
Haar measure plays the role of the upper Banach density.

######  Theorem 1.6.

Let $G$ be a compact abelian group with normalized Haar measure $\mu$ and
$\phi_{1},\phi_{2},\phi_{3}:G\rightarrow G$ be continuous homomorphisms
satisfying

  1. (a)

$\phi_{1}+\phi_{2}+\phi_{3}=0$,

  2. (b)

$\phi_{1},\phi_{2},\phi_{3}$ are commuting,

  3. (c)

$\phi_{1}(G),\phi_{2}(G),\phi_{3}(G)$ have finite indices in $G$.

Let $A\subseteq G$ be a measurable subset with $\mu(A)=\delta>0$. Then

$\phi_{1}(A)+\phi_{2}(A)+\phi_{3}(A)$

contains a Bohr-$(k,\eta)$ set, where $k$ and $\eta$ depend only on $\delta$
and the indices $[G:\phi_{i}(G)]$ $(1\leq i\leq 3)$.

######  Remark 2.

  * •

The condition $\phi_{1}+\phi_{2}+\phi_{3}=0$ cannot be removed. For a
counterexample, take $G=\mathbb{Z}_{N}$ for some large $N$ and
$A=\\{1,\cdots,\lfloor N/10\rfloor\\}$. Then $A+A+A$ does not contain 0, and
hence not a Bohr set.

  * •

We do not know if the condition on commutativity can be removed entirely,
though it can be weakened (see Theorem 4.4). For example, commutativity is not
required when one of $\phi_{1},\phi_{2}$ and $\phi_{3}$ is an automorphism
(see Remark 6).

  * •

The finite index condition cannot be removed. Indeed, we can take
$G=\mathbb{F}_{2}^{n}$ for some large $n$,
$\phi_{1}(x)=\phi_{2}(x)=x,\phi_{3}(x)=0$. In this setting, Bohr sets are
simply vector subspaces. A construction of Green [17, Theorem 9.4] gives a set
$A$ of size $\geq|G|/4$ such that any subspace contained in $A-A$ must have
codimension $\geq\sqrt{n}$.

As an application, Theorem 1.6 can be used to obtain an effective version of
the aforementioned result of Griesmer [18]. We plan to pursue this idea in a
subsequent paper.

###  1.3. Number-theoretic consequences

As mentioned earlier, the fact that we accommodate homomorphisms in Theorem
1.5 and Theorem 1.6 enables us to generalize Theorem 1.2 and Theorem 1.4 to
number fields and function fields.

In the following, for a subset $A$ of a ring $R$ and $c\in R$, we write

$cA=\\{ca:a\in A\\}$ (1)

and

$A/c=\\{b\in A:bc\in A\\}.$ (2)

The next theorem is true for any number field, but we only state for
$\mathbb{Z}[i]$ for simplicity.

######  Theorem 1.7.

Let $s_{1},s_{2},s_{3}\in\mathbb{Z}[i]\setminus\\{0\\}$ such that
$s_{1}+s_{2}+s_{3}=0$.

  1. (a)

If a set $A\subseteq\mathbb{Z}[i]$ has positive upper density, i.e.

$\overline{d}(A):=\limsup_{N\to\infty}\frac{|A\cap[-N,N]^{d}|}{(2N+1)^{d}}=\delta>0,$

then $s_{1}A+s_{2}A+s_{3}A$ contains a $(k,\eta)$-Bohr set in $\mathbb{Z}[i]$,
where $k$ and $\eta$ depend only on $s_{1},s_{2},s_{3}$ and $\delta$.

  2. (b)

If $\mathbb{Z}[i]=\bigcup_{j=1}^{r}A_{j}$, then for some
$j\in\\{1,2,\ldots,r\\}$, $s_{1}A_{j}-s_{1}A_{j}+s_{2}A_{j}$ contains a
$(k,\eta)$-Bohr set in $\mathbb{Z}[i]$, where $k$ and $\eta$ depend only on
$s_{1},s_{2}$ and $r$.

  3. (c)

If $\mathbb{Z}[i]=\bigcup_{j=1}^{r}A_{j}$, then there exists
$j\in\\{1,2,\ldots,r\\}$ such that $A_{j}-A_{j}+sA_{j}$ contains a Bohr set
for any $s\in\mathbb{Z}[i]\setminus\\{0\\}$.

Here, as a group, we identify $\mathbb{Z}[i]$ with $\mathbb{Z}^{2}$.

Our next result deals with the ring $\mathbb{F}_{q}[t]$ of polynomials over a
finite field ${\mathbb{F}_{q}}$.

######  Theorem 1.8.

Let $s_{1},s_{2},s_{3}\in{\mathbb{F}_{q}}[t]\setminus\\{0\\}$ such that
$s_{1}+s_{2}+s_{3}=0$.

  1. (a)

If a set $A\subseteq{\mathbb{F}_{q}}[t]$ has positive upper density, i.e.

$\overline{d}(A):=\limsup_{N\to\infty}\frac{|\\{x\in{\mathbb{F}_{q}}[t]:\deg
x<N\\}|}{q^{N}}=\delta>0,$

then $s_{1}A+s_{2}A+s_{3}A$ contains a ${\mathbb{F}_{q}}$-vector subspace of
finite codimension in $\mathbb{F}_{q}[t]$, where the codimension depends only
on $s_{1},s_{2},s_{3}$ and $\delta$.

  2. (b)

If ${\mathbb{F}_{q}}[t]=\bigcup_{i=1}^{r}A_{i}$, then for some
$i\in\\{1,\ldots,r\\}$, $s_{1}A_{i}-s_{1}A_{i}+s_{2}A_{i}$ contains a
${\mathbb{F}_{q}}$-vector subspace of finite codimension of
$\mathbb{F}_{q}[t]$, where the codimension depends only on $s_{1},s_{2}$ and
$\delta$.

  3. (c)

If $\mathbb{F}_{q}[t]=\bigcup_{i=1}^{r}A_{i}$, then there exists
$i\in\\{1,2,\ldots,r\\}$ such that $A_{i}-A_{i}+sA_{i}$ contains an
${\mathbb{F}_{q}}$-vector subspace of finite codimension of
$\mathbb{F}_{q}[t]$ for any $s\in{\mathbb{F}_{q}}[t]\setminus\\{0\\}$.

We remark that the special case
$s_{1},s_{2},s_{3}\in{\mathbb{F}_{q}}\setminus\\{0\\}$ of Theorem 1.8(a) is
essentially Corollary 1.4 in [18].

###  1.4. Counting linear patterns

Similarly to the proofs of Bogolyubov [7] and Bergelson-Ruzsa [3]’s theorems,
we deduce Theorem 1.6 from a lower bound (of correct order of magnitude) for
the number of certain linear patterns in $G$. This is straightforward in
Bogolyubov’s case, but less so in Bergelson and Ruzsa’s. Bergelson and Ruzsa
had to count the number of generalized Roth patterns $\\{x,x+ry,x+sy\\}$
(where $r,s\in\mathbb{Z}$) and they deduced this from Szemerédi’s theorem [30]
and Varnavides’ argument [32]. For us, we need to count the number of patterns
$\\{x,x+\phi(y),x+\psi(y)\\}$ (where $\phi$ and $\psi$ are homomorphisms).
This is accomplished by generalizing a Fourier-analytic argument of Bourgain
[8]. Bourgain’s argument, in essence an arithmetic regularity lemma, allows us
to obtain the following Khintchine-type result.

######  Theorem 1.9 (Khintchine-Roth theorem in compact abelian groups).

Let $G$ be a compact abelian group with probability Haar measure $\mu$ and
$\phi,\psi:G\to G$ be continuous homomorphisms such that
$[G:\phi(G)],[G:\psi(G)]$ and $[G:(\phi-\psi)(G)]$ are finite. Let
$f:G\to[0,1]$ be a measurable function with $\int_{G}f\,d\mu=\delta>0$.

Then for any $\epsilon>0$, there exists a constant $c_{1}>0$ that depends only
on $\delta,\epsilon$ and the indices above such that the set

$B=\left\\{y\in
G:\int_{G}f(x)f(x+\phi(y))f(x+\psi(y))\,d\mu(x)>\delta^{3}-\epsilon\right\\}$

has measure at least $c_{1}$. Consequently,

$\iint_{G^{2}}f(x)f(x+\phi(y))f(x+\psi(y))\,d\mu(x)d\mu(y)\geq c_{2}$ (3)

for some positive constant $c_{2}$ depending only on $\delta$ and the indices
above.

Theorem 1.9 was proved independently by Berger-Sah-Sawhney-Tidor [4], under
the hypothesis that $\phi,\psi$ and $\phi-\psi$ are automorphisms, using a
very similar argument. Our execution is slightly different from theirs, in
that we follow Bergelson-Host-McCutcheon-Parreau [2]’s elaboration of
Bourgain’s argument, while they follow Tao [31]’s.

Theorem 1.9 is markedly similar to the following result of Ackelsberg,
Bergelson and Best:

######  Theorem 1.10 ([1, Theorem 1.10]).

Let $G$ be a countable discrete abelian group, and $\phi,\psi:G\to G$ be
homomorphisms such that $[G:\phi(G)],[G:\psi(G)]$ and $[G:(\phi-\psi)(G)]$ are
finite. For any ergodic system $(X,\mathcal{B},\mu,(T_{g})_{g\in G})$, any
$\epsilon>0$, and any $A\in\mathcal{B}$, the set

$B=\left\\{g\in G:\mu(A\cap T^{-1}_{\phi(g)}A\cap
T^{-1}_{\psi(g)}A)>\mu(A)^{3}-\epsilon\right\\}$

is syndetic in $G$.

As discussed in [1, Section 10], the finite index condition in Theorem 1.10 is
necessary. The following result of Fox-Sah-Sawhney-Stoner-Zhao [11], improving
on an earlier result of Mandache [23], shows that the finite index condition
is also necessary in Theorem 1.9.

######  Example 1.

Let $\ell<4$ be arbitrary and $\delta>0$ be sufficiently small in terms of
$l$. Let $G=\mathbb{F}_{2}^{n}\times\mathbb{F}_{2}^{n}$ where $n$ is
sufficiently large, $\phi(u,v)=(u,0),\psi(u,v)=(0,u)$. Then the left hand side
of (3) counts the number of “corners” $\\{(a,b),(a+u,b),(a,b+u)\\}$ in
$\mathbb{F}_{2}^{n}\times\mathbb{F}_{2}^{n}$. [11, Corollary 1.3] states that
there exists a set $A\subset G$ of size $\geq\delta|G|$ such that for any
$u\in\mathbb{F}_{2}^{n}\setminus\\{0\\}$, we have

$\\#\\{(a,b)\in G:(a,b),(a+u,b),(a,b+u)\in A\\}<\delta^{\ell}|G|.$

Hence, the set $B$ in Theorem 1.9 has to be $\\{0\\}\times\mathbb{F}_{2}^{n}$.
But the measure of this set in $G$ goes to $0$ as $n$ goes to infinity.

Regarding Theorem 1.5, we deduce it from the following result, which counts
the number of monochromatic configurations under finite partitions of $G$.

######  Theorem 1.11.

Let $G$ be a compact abelian group with probability Haar measure $\mu$ and let
$\psi,\phi_{1},\ldots,\phi_{k}:G\to G$ be continuous homomorphisms satisfying:

  1. (a)

$\psi,\phi_{1},\ldots,\phi_{k}$ are commuting, and

  2. (b)

$\psi(G),\phi_{1}(G),\ldots,\phi_{k}(G)$ have finite indices in $G$.

Suppose $G=\bigcup_{i=1}^{r}A_{i}$ is a partition of $G$ into measurable sets.
Then

$\sum_{i=1}^{r}\iint_{G^{2}}1_{A_{i}}(\psi(y))1_{A_{i}}(x)1_{A_{i}}(x+\phi_{1}(y))\cdots
1_{A_{i}}(x+\phi_{k}(y))\,d\mu(x)d\mu(y)\geq c_{3}$ (4)

for some positive constant $c_{3}$ depending only on $r,k$ and the indices
above.

######  Remark 3.

  * •

By taking $\psi=0$, we see that the condition $[G:\psi(G)]$ is finite cannot
be removed. However, we do not know whether the condition
$[G:\phi_{i}(G)]<\infty$ is necessary or not.

  * •

Our proof relies heavily on the commuting condition and we do not know if it
can be removed.

When $\psi$ and $\phi$ are dilations, the configuration
$\\{\psi(y),x+\phi_{1}(y),\ldots,x+\phi_{k}(y)\\}$ becomes the Brauer
configuration $\\{y,x,x+y,\ldots,x+ky\\}$. Results on counting such
monochromatic configurations have been established by Serra-Vena [29, Theorem
1.3] for finite abelian groups of bounded torsion. Thus, besides the fact that
it allows for more general homomorphisms, Theorem 1.11 has the advantage of
being uniform over all groups. On the other hand, our finite index condition
is certainly related, and in a sense, dual to Serra-Vena’s bounded exponent
condition [29].

We remark that despite the apparent similarity between (3) and (4), their
proofs are very different. The proof of Theorem 1.11 is “Fourier-free” and its
main ingredient is the Hales-Jewett theorem. Thus, our approach in proving
this theorem is also genuinely different from Serra-Vena’s, which relies on a
removal lemma for groups.

On the quantitative side, our bounds leave much to be desired. Since the proof
of Theorem 1.9 relies on the regularity lemma (Proposition 4.2), in Theorem
1.6, the dependence of $k$ and $\eta$ on $\delta$ and $[G:\phi_{i}(G)]$ is of
tower type. Likewise, since the proof of Theorem 1.11 uses the Hales-Jewett
theorem, the bounds for $k$ and $\eta$ in Theorem 1.5 are even worse. It is an
interesting problem to obtain good bounds for Theorems 1.6 and 1.5, even in
special classes of groups such as $\mathbb{F}_{p}^{n}$. Indeed, Sanders [28,
Theorem A.1] obtained a near optimal bound for Bogolyubov’s theorem in
$\mathbb{F}_{p}^{n}$.

Outline of the paper. In Section 2, we set up notation and collect some basic
facts about Bohr sets, kernels and homomorphisms in compact abelian groups.
Section 3 is devoted to proving results involving partitions, especially,
Theorems 1.5 and 1.11. Theorems 1.6, 1.9 and related density results will be
proved in Section 4. Section 5 contains proofs of results in $\mathbb{Z}$,
number fields and function fields, i.e. Theorems 1.4, 1.7 and 1.8. Lastly, we
present some related open questions in Section 6.

Acknowledgement. We thank Vitaly Bergelson and John Griesmer for many helpful
conversations on sets of recurrence, Bohr sets and related topics. The second
author was partially supported by National Science Foundation Grant
DMS-1702296.

##  2\. Preliminaries

In this section, we gather some background on Bohr sets, kernels and
homomorphisms in compact abelian groups. Most of the results are well-known or
resemble known theorems. We include proofs for the results that we cannot
pinpoint precisely in the literature.

###  2.1. Notation

We write $[N]$ for the set $\\{1,\ldots,N\\}$. If $A$ and $B$ are two
quantities, we write $A=O(B)$ or $A\ll B$ if there is a constant $C$ such that
$|A|\leq CB$. We write $e(x)$ for $e^{2\pi ix}$.

Throughout this paper, $G$ is a Hausdorff compact abelian group with
probability Haar measure $\mu$ and $\Gamma$ is the dual of $G$, written
additively. The relevance of homomorphisms is that if $\gamma\in\Gamma$ and
$\phi:G\rightarrow G$ is a continuous homomorphism, then $\gamma\circ\phi$ is
also an element of $\Gamma$.

If $f:G\rightarrow\mathbb{C}$ is a function, for $t\in G$ we define the
function $f_{t}(x)=f(x+t)$. For $f\in L^{1}(G)$, the Fourier transform of $f$
is the function

$\widehat{f}(\gamma)=\int_{G}f(x)\overline{\gamma(x)}\,d\mu(x)\qquad\textup{
for }\gamma\in\Gamma.$

For $f,g\in L^{2}(G)$, we then have Parseval’s formula

$\int_{G}f(x)\overline{g(x)}\,d\mu(x)=\sum_{\gamma\in\Gamma}\widehat{f}(\gamma)\overline{\widehat{g}(\gamma)}$

and Plancherel’s formula

$\int_{G}|f(x)|^{2}\,d\mu(x)=\sum_{\gamma\in\Gamma}\left|\widehat{f}(\gamma)\right|^{2}.$

###  2.2. Bohr sets

For $\Lambda,\Lambda_{1},\Lambda_{2}\subseteq\Gamma$ and
$\eta_{1},\eta_{2}>0$, it follows from the definition of Bohr sets that

$B(\Lambda_{1};\eta_{1})\cap B(\Lambda_{2};\eta_{2})\supset
B(\Lambda_{1}\cup\Lambda_{2};\min(\eta_{1},\eta_{2}))$

and

$B(\Lambda;\eta_{1})+B(\Lambda;\eta_{2})\subset B(\Lambda;\eta_{1}+\eta_{2}).$

######  Lemma 2.1.

Suppose $f_{1},\ldots,f_{k}\in L^{\infty}(G)$, $\|f_{i}\|_{\infty}\leq 1$ for
all $i=1,\ldots,k$. Let $\phi_{1},\ldots,\phi_{k}$ be homomorphisms
$G\rightarrow G$. Then for any $\eta>0$, the set

$B=\\{t\in
G:\|\widehat{f_{i}}-\widehat{f_{i,\phi_{i}(t)}}\|_{\infty}<\eta\textup{ for
}i=1,\ldots,k\\}$

contains a Bohr set $B(\Lambda;\eta)$ where
$|\Lambda|\leq\frac{4k}{\eta^{2}}$.

###### Proof.

Note that if
$\|\widehat{f_{i}}-\widehat{f_{i,\phi_{i}(t)}}\|_{\infty}\geq\eta$, then for
some $\gamma\in\Gamma$,

$|\widehat{f_{i}}(\gamma)-\widehat{f_{i,\phi_{i}(t)}}(\gamma)|=|1-\gamma(\phi_{i}(t))||\widehat{f_{i}}(\gamma)|\geq\eta.$

This implies that $|1-\gamma(\phi_{i}(t))|\geq\eta$ and
$\gamma\in\Lambda_{i}:=\\{\lambda\in\Gamma:|\widehat{f_{i}}(\lambda)|>\eta/2\\}$.

We have thus shown that

$B(\bigcup_{i=1}^{k}\Lambda_{i}\circ\phi_{i};\eta)\subset B,$

where
$\Lambda_{i}\circ\phi_{i}:=\\{\gamma\circ\phi_{i}:\gamma\in\Lambda_{i}\\}\subset\Gamma$.
By Plancherel’s formula,

$\left(\frac{\eta}{2}\right)^{2}|\Lambda_{i}|\leq\sum_{\lambda\in\Lambda_{i}}\left|\widehat{f_{i}}(\lambda)\right|^{2}\leq
1.$

Therefore, $|\Lambda_{i}|\leq\frac{4}{\eta^{2}}$ and
$|\bigcup_{i=1}^{k}\Lambda_{i}\circ\phi_{i}|\leq\frac{4k}{\eta^{2}}$. ∎

Next lemma is needed in 5.

######  Lemma 2.2.

Let $H$ be a locally compact abelian group, $K$ be a closed subgroup of finite
index $m$. Then $K$ is a Bohr-$(m,|e(1/m)-1|)$ set in $H$.

###### Proof.

Let $\chi_{1},\ldots,\chi_{m}$ be all characters on $H/K$. For any $x\in H/K$
and $1\leq i\leq m$, we have $|\chi_{i}(x)|^{m}=1$, so either $\chi_{i}(x)=1$
or $|\chi_{i}(x)-1|\geq|e(1/m)-1|$. If $\chi_{i}(x)=1$ for all $i$ then $x=0$.
Hence

$\\{0\\}=B(\chi_{1},\ldots,\chi_{m};|e(1/m)-1|).$

The characters $\chi_{i}$ lift to characters $\tilde{\chi_{i}}$ on $H$ by
$\tilde{\chi_{i}}(h)=\chi_{i}(h+K)$. Therefore,

$H=B(\tilde{\chi_{1}},\ldots,\tilde{\chi_{m}};|e(1/m)-1|),$

as desired. ∎

We will also need Bogolyubov’s theorem for compact abelian groups.

######  Lemma 2.3 (Bogolyubov for compact abelian groups, see [26, Lemma
2.1]).

Let $G$ be a compact abelian group with Haar measure $\mu$ and let $A\subseteq
G$ of positive measure. Then $A-A+A-A$ contains a Bohr-$(k,\eta)$ set where
$k,\eta$ depends only on $\mu(A)$.

###  2.3. Kernels

A kernel on $G$ is a non-negative continuous function that satisfies
$\int_{G}K\,d\mu=1$. Specifically, we will utilize the kernels whose Fourier
transforms are also non-negative and supported on given Bohr sets. For a
kernel $K$, we write $\lVert\widehat{K}\rVert_{1}$ to denote
$\sum_{\gamma\in\Gamma}|\widehat{K}(\gamma)|$.

######  Lemma 2.4 (cf. [2, Lemma 4.3]).

Given a finite set $\Lambda\subset\Gamma$ and $\eta\in(0,1/2]$, there exists a
kernel $K$ satisfying the following:

  1. (1)

$K\geq 0,\widehat{K}\geq 0$ and $\int_{G}K\,d\mu=\lVert K\rVert_{1}=1$,

  2. (2)

$\lVert\widehat{K}\rVert_{1}=\lVert K\rVert_{\infty}\leq
1/(C_{0}\eta)^{|\Lambda|}$ for some absolute constant $C_{0}>0$, and

  3. (3)

$K$ vanishes outside the Bohr set $B(\Lambda;\eta)$.

Consequently,

$\mu(B(\Lambda;\eta))\geq(C_{0}\eta)^{|\Lambda|}.$ (5)

We remark that the bound (5) can also be obtained from an elementary covering
argument (see [31]).

###### Proof.

First, for each $\lambda\in\Lambda$, there exists a kernel
$K_{\lambda}:G\to[0,\infty)$ satisfying the following properties:

  1. (1)

$\lVert K_{\lambda}\rVert_{1}=1$,

  2. (2)

$\widehat{K}_{\lambda}\geq 0$,

  3. (3)

$K_{\lambda}$ is supported on $B(\\{\lambda\\};\eta)=\\{x\in
G:|\lambda(x)-1|<\eta\\}$,

  4. (4)

$\|K_{\lambda}\|_{\infty}=K_{\lambda}(0)\leq 1/(C_{0}\eta)$ for some absolute
constant $C_{0}$.

Indeed, let $B=B(\\{\lambda\\};\frac{\eta}{2})$ and let
$K_{\lambda}=\frac{1_{B}}{\mu(B)}*\frac{1_{B}}{\mu(B)}$. Clearly the first and
second properties are satisfied. Additionally, $K_{\lambda}$ is supported on
$B(\\{\lambda\\};\frac{\eta}{2})+B(\\{\lambda\\};\frac{\eta}{2})\subset
B(\\{\lambda\\};\eta)$.

Concerning the last property, we have for every $x\in G$,

$K_{\lambda}(x)=\sum_{\gamma\in\Gamma}\widehat{K}_{\lambda}(\gamma)\gamma(x)$

and so

$\left|K_{\lambda}(x)\right|\leq\sum_{\gamma\in\Gamma}\widehat{K}_{\lambda}(\gamma)=K_{\lambda}(0).$

Therefore, $\lVert
K_{\lambda}\rVert_{\infty}=K_{\lambda}(0)=\frac{1}{\mu(B)}$. Since $\lambda$
is continuous, its image $\lambda(G)$ is a closed subgroup of
$S^{1}=\\{z\in\mathbb{C}:|z|=1\\}$, and so it is either $S^{1}$ or
$\\{z\in\mathbb{C}:z^{q}=1\\}$ for some $q\in\mathbb{N}$. Since $\lambda$ is a
homomorphism, it is measure-preserving (see Lemma 2.7 below). Hence $\mu(B)$
is equal to the normalized Haar measure of the set

$\left\\{z\in S^{1}:|z-1|<\frac{\eta}{2}\right\\}$

in the group $\lambda(G)$. In either case, where $\lambda(G)=S^{1}$ or
$\\{z\in\mathbb{C}:|z|^{q}=1\\}$, we find that $\mu(B)\geq C_{0}\eta$ for some
absolute constant $C_{0}$. Therefore, $\lVert K_{\lambda}\rVert_{\infty}\leq
1/(C_{0}\eta)$.

We now define

$\widetilde{K}=\prod_{\lambda\in\Lambda}K_{\lambda}.$

It follows that $\widetilde{K}\geq 0$ and $\widetilde{K}$ is supported on
$B(\Lambda;\eta)$. Repeatedly using the fact that
$\widehat{fg}(\gamma)=\sum_{\lambda\in\Gamma}\widehat{f}(\lambda)\widehat{g}(\gamma-\lambda)$
for all $f,g\in L^{\infty}(G)$, we have $\widehat{\widetilde{K}}\geq 0$.
Likewise, since $\widehat{K}_{\lambda}(0)=\lVert K_{\lambda}\rVert_{1}=1$, we
have $\lVert\widetilde{K}\rVert_{1}=\widehat{\widetilde{K}}(0)\geq 1$. Upon
defining

$K=\widetilde{K}/\lVert\widetilde{K}\rVert_{1}$

we obtain the desired kernel. ∎

###  2.4. Homomorphisms

We will often make use of the following facts about homomorphisms
$G\rightarrow G$.

######  Lemma 2.5.

Let $\phi:G\rightarrow G$ be a continuous homomorphism such that
$[G:\phi(G)]=m$ is finite. Then for any $\gamma\in\Gamma$, there are at most
$m$ elements $\chi\in\Gamma$ such that $\gamma=\chi\circ\phi$.

###### Proof.

It is easy to see that for each $\gamma\in\Gamma$, the set
$S_{\gamma}:=\\{\chi\in\Gamma:\gamma=\chi\circ\phi\\}$ is either empty, or a
coset of the group $S_{0}$. On the other hand, $S_{0}$ is the annihilator of
the group $\phi(G)$, so by [25, Theorem 2.1.2], it is isomorphic to
$G/\phi(G)$, and hence has cardinality $m$. ∎

######  Lemma 2.6.

Let $\phi,\psi:G\rightarrow G$ be homomorphisms such that $[G:\phi(G)]=m$ and
$[G:\psi(G)]=\ell$ are finite. Then $[G:\phi(\psi(G))]\leq m\ell$ is finite.

###### Proof.

We have $[G:\phi(\psi(G))]=[G:\phi(G)][\phi(G):\phi(\psi(G))]$. It suffices to
show that $[\phi(G):\phi(\psi(G))]\leq\ell$.

Let $x_{1}+\psi(G),\ldots,x_{\ell}+\psi(G)$ be all cosets of $\psi(G)$ in $G$.
Then $\phi(x_{1})+\phi(\psi(G)),\ldots,\phi(x_{\ell})+\phi(\psi(G))$ are all
cosets of $\phi(\psi(G))$ in $\psi(G)$ (these are not necessarily distinct, so
the actual number of cosets may be less than $\ell$), proving the desired
claim. ∎

######  Lemma 2.7.

Let $G,H$ be compact abelian groups and $\mu,\nu$ be the normalized Haar
measures of $G$ and $H$, respectively. Suppose $\phi:G\to H$ is a continuous
surjective homomorphism. Then $\phi_{*}\mu=\nu$ (i.e.
$\nu(B)=\mu(\phi^{-1}(B))$ for any Borel set $B\subset H$).

###### Proof.

Let $\nu_{0}=\phi_{*}\mu$. By the uniqueness of the normalized Haar measure,
it suffices to show that $\nu_{0}$ is a translation-invariant probability
measure on $H$. First, $\nu_{0}$ is a probability measure because
$\nu_{0}(H)=\mu(\phi^{-1}(H))=\mu(G)=1$. Now let $B\subset H$ be a Borel set
and $h_{0}\in H$ be arbitrary. Since $\phi$ is surjective, there exists
$g_{0}\in G$ such that $\phi(g_{0})=h_{0}$. For any $g\in\phi^{-1}(B+h_{0})$,
we have

$\phi(g-g_{0})=\phi(g)-\phi(g_{0})\in B+h_{0}-h_{0}=B.$

Therefore, $\phi^{-1}(B+h_{0})\subseteq\phi^{-1}(B)+g_{0}.$ On the other hand,

$\phi(\phi^{-1}(B)+g_{0})\subseteq B+h_{0}$

and so $\phi^{-1}(B+h_{0})=\phi^{-1}(B)+g_{0}$. Since $\mu$ is translation-
invariant on $G$, it follows that

$\nu_{0}(B+h_{0})=\mu(\phi^{-1}(B+h_{0}))=\mu(\phi^{-1}(B)+g_{0})=\mu(\phi^{-1}(B))=\nu_{0}(B).$

Thus $\nu_{0}$ is translation-invariant on $H$ and so $\nu_{0}=\nu$. ∎

######  Lemma 2.8.

Let $\phi:G\rightarrow G$ be a continuous homomorphism such that
$[G:\phi(G)]=m$ is finite. Then for any measurable set $A\subset G$, we have

$\mu(A)\leq m\mu(\phi(A))$ (6)

and

$\mu(\phi^{-1}(A))\leq m\mu(A).$ (7)

Consequently, if $f\in L^{1}(G)$ is nonnegative, then

$\int_{G}f(x)\,d\mu(x)\geq\frac{1}{m}\int_{G}f(\phi(x))\,d\mu(x).$

###### Proof.

First, since $\phi$ is continuous and $G$ is compact, $\phi(G)$ is a compact
subgroup of $G$. Since $G$ is Hausdorff, $\phi(G)$ is closed. In other words,
$\phi(G)$ is a closed subgroup of $G$.

For each Borel set $B\subset\phi(G)$, $\lambda(B)=m\mu(B)$ defines a
probability measure on $\phi(G)$. Since this measure is translation invariant,
it is equal to the normalized Haar measure on $\phi(G)$. By Lemma 2.7,
$\lambda=\phi_{*}\mu$. This means that for any Borel set $B\subset\phi(G)$, we
have $\mu(\phi^{-1}(B))=m\mu(B)$.

Let $A$ be any Borel set in $G$. Since $A\subset\phi^{-1}(\phi(A))$, we have
hence $\mu(A)\leq\mu(\phi^{-1}(\phi(A))=m\mu(\phi(A))$, and the first
assertion is proved. Applying the first assertion to the set $\phi^{-1}(A)$,
we get the second assertion.

The third assertion follows from the second one, and the fact that $f$ can be
approximated by functions of the form $\sum_{i=1}^{n}c_{i}1_{A_{i}}$ for Borel
sets $A_{i}$ and $c_{i}\geq 0$. ∎

The next lemmas deal with images and preimages of Bohr sets under
homomorphisms.

######  Lemma 2.9.

Let $B\subset G$ be a Bohr-$(k,\eta)$ set and $\phi:G\rightarrow G$ be a
continuous homomorphism. Then $\phi^{-1}(B)$ is also a Bohr-$(k,\eta)$ set.

###### Proof.

If $B=\\{x\in G:|\gamma_{i}(x)-1|<\eta\textup{ for }i=1,\ldots,k\\}$ is a
Bohr-$(k,\eta)$ set, then $\phi^{-1}(B)=\\{x\in
G:|\gamma_{i}\circ\phi(x)-1|<\eta\textup{ for }i=1,\ldots,k\\}$ is also a
Bohr-$(k,\eta)$-set. ∎

The next lemma is more surprising.

######  Lemma 2.10 (cf. Griesmer [18, Lemma 1.7]).

Let $B\subset G$ be a Bohr-$(k,\eta)$ set and $\phi:G\rightarrow G$ be a
continuous homomorphism such that $[G:\phi(G)]=m<\infty$. Then $\phi(B)$
contains a Bohr-$(k^{\prime},\eta^{\prime})$ set, where
$k^{\prime},\eta^{\prime}$ depend on $k,\eta$ and $m$.

###### Proof.

Suppose $B=\\{x\in G:|\gamma_{i}(x)-1|<\eta\text{ for }1\leq i\leq k\\}$ where
$\gamma_{i}\in\Gamma$. Then

$A=\\{x\in G:|\gamma_{i}(x)-1|<\eta/4\text{ for }1\leq i\leq k\\}$

satisfies $A-A+A-A\subseteq B$. The bound (5) implies that
$\mu(A)\geq(C_{0}\eta/4)^{k}$ for some absolute constant $C_{0}>0$.

In view of Lemma 2.8,
$\mu(\phi(A))\geq\mu(A)/m\geq\frac{(C_{0}\eta)^{k}}{4^{k}m}$. Therefore, by
Lemma 2.3, the set $\phi(B)\supseteq\phi(A)-\phi(A)+\phi(A)-\phi(A)$ is a
Bohr-$(k^{\prime},\eta^{\prime})$ set where $k^{\prime},\eta^{\prime}$ depend
only on $\mu(\phi(A))$, which is bounded below by
$\frac{(C_{0}\eta)^{k}}{4^{k}m}$. ∎

###  2.5. Counting lemmas

######  Lemma 2.11 (cf. [8, Lemma 2]).

Let $\phi,\psi:G\rightarrow G$ be continuous homomorphisms such that
$\phi(G),\psi(G)$ have finite indices in $G$. Then for $f_{1},f_{2},f_{3}\in
L^{\infty}(G)$ and $K\in L^{1}(G)$ such that $\widehat{K}\in L^{1}(\Gamma)$,
we have

$\left|\iint_{G^{2}}f_{1}(x)f_{2}(x+\phi(y))f_{3}(x+\psi(y))K(y)\,d\mu(x)d\mu(y)\right|\ll\|\widehat{f_{1}}\|_{\infty}\|f_{2}\|_{2}\|f_{3}\|_{2}\|\widehat{K}\|_{1}$
(8)

where the implied constant depends only on the indices of $\phi(G)$ and
$\psi(G)$ in $G$.

###### Proof.

Since linear combinations of characters are dense in $L^{2}(G)$, without loss
of generality, we can assume $f_{1},f_{2},f_{3}$ and $K$ are equal to their
Fourier series. For $x\in G$, write
$g(x)=\int_{G}f_{2}(x+\phi(y))f_{3}(x+\psi(y))K(y)\,d\mu(y)$.

By Plancherel’s formula,

$\left|\int_{G}f_{1}(x)g(x)\,d\mu(x)\right|=\left|\sum_{\gamma\in\Gamma}\widehat{f_{1}}(\gamma)\widehat{g}(\overline{\gamma})\right|\leq\|\widehat{f_{1}}\|_{\infty}\cdot\|\widehat{g}\|_{1}.$

Thus

$\displaystyle g(x)$ $\displaystyle=$
$\displaystyle\int_{G}f_{2}(x+\phi(y))f_{3}(x+\psi(y))K(y)\,d\mu(y)$
$\displaystyle=$
$\displaystyle\int_{G}\left(\sum_{\gamma_{2},\gamma_{3},\gamma_{0}\in\Gamma}\widehat{f_{2}}(\gamma_{2})\gamma_{2}(x+\phi(y))\widehat{f_{3}}(\gamma_{3})\gamma_{3}(x+\psi(y))\widehat{K}(\gamma_{0})\gamma_{0}(y)\right)\,d\mu(y)$
$\displaystyle=$
$\displaystyle\int_{G}\left(\sum_{\gamma_{2},\gamma_{3},\gamma_{0}\in\Gamma}\widehat{f_{2}}(\gamma_{2})\widehat{f_{3}}(\gamma_{3})\widehat{K}(\gamma_{0})(\gamma_{2}+\gamma_{3})(x)(\gamma_{2}\circ\phi+\gamma_{3}\circ\psi+\gamma_{0})(y)\right)\,d\mu(y)$
$\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}\gamma_{2},\gamma_{3},\gamma_{0}\in\Gamma,\\\
\gamma_{2}\circ\phi+\gamma_{3}\circ\psi+\gamma_{0}=0\end{subarray}}\widehat{f_{2}}(\gamma_{2})\widehat{f_{3}}(\gamma_{3})\widehat{K}(\gamma_{0})(\gamma_{2}+\gamma_{3})(x)$

Consequently,

$\widehat{g}(\gamma)=\sum_{\begin{subarray}{c}\gamma_{2},\gamma_{3},\gamma_{0}\in\Gamma,\\\
\gamma_{2}\circ\phi+\gamma_{3}\circ\psi+\gamma_{0}=0,\\\
\gamma_{2}+\gamma_{3}=\gamma\end{subarray}}\widehat{f_{2}}(\gamma_{2})\widehat{f_{3}}(\gamma_{3})\widehat{K}(\gamma_{0})$

and

$\lVert\widehat{g}\rVert_{1}\leq\sum_{\begin{subarray}{c}\gamma_{2},\gamma_{3},\gamma_{0}\in\Gamma,\\\
\gamma_{2}\circ\phi+\gamma_{3}\circ\psi+\gamma_{0}=0\end{subarray}}|\widehat{f_{2}}(\gamma_{2})|\cdot|\widehat{f_{3}}(\gamma_{3})|\cdot|\widehat{K}(\gamma_{0})|.$

Therefore, it suffices to show that for each $\gamma_{0}\in\Gamma$, we have

$\sum_{\begin{subarray}{c}\gamma_{2},\gamma_{3}\in\Gamma,\\\
\gamma_{2}\circ\phi+\gamma_{3}\circ\psi+\gamma_{0}=0\end{subarray}}|\widehat{f_{2}}(\gamma_{2})|\cdot|\widehat{f_{3}}(\gamma_{3})|\ll\|f_{2}\|_{2}\cdot\|f_{3}\|_{2}.$

By the Cauchy-Schwarz inequality and the Plancherel’s formula, the left hand
side is at most

$\displaystyle\|f_{2}\|_{2}\cdot\left(\sum_{\gamma_{2}}\left(\sum_{\gamma_{3}\circ\psi=-\gamma_{0}-\gamma_{2}\circ\phi}|\widehat{f_{3}}(\gamma_{3})|\right)^{2}\right)^{1/2}$
(9) $\displaystyle\ll$
$\displaystyle\|f_{2}\|_{2}\cdot\left(\sum_{\gamma_{2}}\sum_{\gamma_{3}\circ\psi=-\gamma_{0}-\gamma_{2}\circ\phi}|\widehat{f_{3}}(\gamma_{3})|^{2}\right)^{1/2}$
$\displaystyle\ll$
$\displaystyle\|f_{2}\|_{2}\cdot\left(\sum_{\gamma_{3}}|\widehat{f_{3}}(\gamma_{3})|^{2}\right)^{1/2}$
$\displaystyle=$ $\displaystyle\|f_{2}\|_{2}\cdot\|f_{3}\|_{2}.$

In (9), we use the fact that for each $\xi\in\Gamma$, there are at most
$[G:\psi(G)]$ values of $\gamma_{3}$ such that $\gamma_{3}\circ\psi=\xi$.
Likewise, in (2.5), we use the fact that for each $\xi\in\Gamma$, there are at
most $[G:\phi(G)]$ values of $\gamma_{2}$ such that $\gamma_{2}\circ\phi=\xi$.
Both of these facts follow from Lemma 2.5. ∎

######  Remark 4.

Lemma 2.11 is not true without the assumption on finite indices. As a
counterexample, we let $\phi(x)=x,\psi(x)=2x$ and $G=\mathbb{F}_{2}^{k}$ for
some large $k$. Let $n=|G|=2^{k}$. For each $i=1,\ldots,n$, define

  * •

$\widehat{f_{1}}(\gamma_{i})=\widehat{f_{3}}(\gamma_{i})=1$, so
$f_{1}(x)=f_{3}(x)=n\cdot 1_{x=0}$.

  * •

$\widehat{f_{2}}(\gamma_{i})=a_{i}$, where $a_{i}\geq 0$.

  * •

$\widehat{K}(\gamma_{i})=b_{i}$, where $b_{i}\geq 0$.

Then (8) says that

$n(a_{1}b_{1}+\cdots+a_{n}b_{n})^{2}\ll(a_{1}^{2}+\cdots+a_{n}^{2})(b_{1}+\cdots+b_{n})^{2}.$

This is false by taking $a_{1}=b_{1}=1$ and $a_{i}=b_{i}=0$ for $i\neq 1$.

While the previous lemma involves the configuration $x,x+\phi(y),x+\psi(y)$,
the next one is concerned with $x,x+\phi(y)$ and $\psi(y)$. Its proof is
almost identical and so we only highlight the differences.

######  Lemma 2.12.

Let $\phi,\psi:G\rightarrow G$ be continuous homomorphism such that
$\phi(G),\psi(G)$ have finite indices in $G$. Then for $f_{1},f_{2},f_{3}\in
L^{\infty}(G)$, we have

$\left|\iint_{G^{2}}f_{1}(x)f_{2}(x+\phi(y))f_{3}(\psi(y))\,d\mu(x)d\mu(y)\right|\ll\|\widehat{f_{1}}\|_{\infty}\|f_{2}\|_{2}\|f_{3}\|_{2}$
(11)

where the implicit constant depends only on the indices of $\phi(G)$ and
$\psi(G)$ in $G$.

###### Proof.

Similar to the proof of Lemma 2.11, without loss of generality, we can assume
$f_{1},f_{2},f_{3}$ are equal to their Fourier series. For $x\in G$, write
$g(x)=\int_{G}f_{2}(x+\phi(y))f_{3}(\psi(y))\,d\mu(y)$ and then by
Plancherel’s formula,

$\left|\int_{G}f_{1}(x)g(x)\,d\mu(x)\right|=\left|\sum_{\gamma\in\Gamma}\widehat{f_{1}}(\gamma)\widehat{g}(\overline{\gamma})\right|\leq\|\widehat{f_{1}}\|_{\infty}\cdot\|\widehat{g}\|_{1}.$

Moreover, we also have

$\displaystyle g(x)$ $\displaystyle=$
$\displaystyle\int_{G}\left(\sum_{\gamma_{2},\gamma_{3}\in\Gamma}\widehat{f_{2}}(\gamma_{2})\gamma_{2}(x+\phi(y))\widehat{f_{3}}(\gamma_{3})\gamma_{3}(\psi(y))\right)\,d\mu(y)$
$\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}\gamma_{2},\gamma_{3}\in\Gamma,\\\
\gamma_{2}\circ\phi+\gamma_{3}\circ\psi=0\end{subarray}}\widehat{f_{2}}(\gamma_{2})\widehat{f_{3}}(\gamma_{3})\gamma_{2}(x).$

As a consequence,

$\widehat{g}(\gamma)=\sum_{\begin{subarray}{c}\gamma_{2},\gamma_{3}\in\Gamma,\\\
\gamma_{2}\circ\phi+\gamma_{3}\circ\psi=0,\\\
\gamma_{2}=\gamma\end{subarray}}\widehat{f_{2}}(\gamma_{2})\widehat{f_{3}}(\gamma_{3})=\widehat{f_{2}}(\gamma)\sum_{\begin{subarray}{c}\gamma_{3}\in\Gamma,\\\
\gamma\circ\phi+\gamma_{3}\circ\psi=0\end{subarray}}\widehat{f_{3}}(\gamma_{3})$

and so

$\|\widehat{g}\|_{1}\leq\sum_{\begin{subarray}{c}\gamma_{2},\gamma_{3}\in\Gamma,\\\
\gamma_{2}\circ\phi+\gamma_{3}\circ\psi=0\end{subarray}}\left|\widehat{f_{2}}(\gamma_{2})\right|\left|\widehat{f_{3}}(\gamma_{3})\right|.$

On the other hand, we have

$\displaystyle\sum_{\begin{subarray}{c}\gamma_{2},\gamma_{3}\in\Gamma,\\\
\gamma_{2}\circ\phi+\gamma_{3}\circ\psi=0\end{subarray}}\left|\widehat{f_{2}}(\gamma_{2})\right|\left|\widehat{f_{3}}(\gamma_{3})\right|$
$\displaystyle=$
$\displaystyle\sum_{\gamma_{2}\in\Gamma}\left(\left|\widehat{f_{2}}(\gamma_{2})\right|\sum_{\gamma_{3}\in\Gamma,\atop{\gamma_{3}\circ\psi=-\gamma_{2}\circ\phi}}\left|\widehat{f_{3}}(\gamma_{3})\right|\right)$
(12) $\displaystyle\leq$
$\displaystyle\left(\sum_{\gamma_{2}\in\Gamma}\left|\widehat{f_{2}}(\gamma_{2})\right|^{2}\right)^{1/2}\left(\sum_{\gamma_{2}\in\Gamma}\left(\sum_{\gamma_{3}\in\Gamma,\atop{\gamma_{3}\circ\psi=-\gamma_{2}\circ\phi}}\left|\widehat{f_{3}}(\gamma_{3})\right|\right)^{2}\right)^{1/2}$
$\displaystyle\ll$ $\displaystyle\lVert
f_{2}\rVert_{2}\left(\sum_{\gamma_{2}\in\Gamma}\sum_{\gamma_{3}\in\Gamma,\atop{\gamma_{3}\circ\psi=-\gamma_{2}\circ\phi}}\left|\widehat{f_{3}}(\gamma_{3})\right|^{2}\right)^{1/2}$
$\displaystyle\ll$ $\displaystyle\lVert
f_{2}\rVert_{2}\left(\sum_{\gamma_{3}\in\Gamma}\left|\widehat{f_{3}}(\gamma_{3})\right|^{2}\right)^{1/2}$
$\displaystyle=$ $\displaystyle\lVert f_{2}\rVert_{2}\lVert f_{3}\rVert_{2}.$

In (12), we use the fact that for each $\xi\in\Gamma$, there are
$\leq[G:\psi(G)]$ values of $\gamma_{3}$ such that $\gamma_{3}\circ\psi=\xi$
while (2.5) follows from the fact that there are $\leq[G:\phi(G)]$ values of
$\gamma_{2}$ such that $\gamma_{2}\circ\phi=\xi$. ∎

##  3\. Bohr sets and partitions

###  3.1. Monochromatic configurations

We make some preparations before the proof of Theorem 1.11. In this section,
we only need $G$ to be a commutative semigroup with neutral element. Fix $k+1$
commuting (semigroup) homomorphisms
$\psi,\phi_{1},\ldots,\phi_{k}:G\rightarrow G$. We write

$\Phi_{m}=\\{\psi^{i_{0}}\circ\phi_{1}^{i_{1}}\circ\cdots\circ\phi_{1}^{i_{k}}:0\leq
i_{0},i_{1},\ldots,i_{k}\leq m\\}\cup\\{0\\}$

(where $\phi^{i}$ is the $i$-th composition of $\phi$).

For formal variables $x_{1},\ldots,x_{n}$, we write

$S_{m}(x_{1},\ldots,x_{n})=\left\\{\sum_{i=1}^{n}\xi_{i}(x_{i}):\xi_{i}\in\Phi_{m}\right\\}$

and we refer to $S_{m}(x_{1},\ldots,x_{n})$ as the $S_{m,n}$-set with
generators $x_{1},\ldots,x_{n}$.

For an element $x=\sum_{i=1}^{n}\xi_{i}(x_{i})\in S_{m}(x_{1},\ldots,x_{n})$,
by the support of $x$ we mean the set $\\{i\in[n]:\xi_{i}\neq 0\\}$. The goal
of this section is to prove the following:

######  Theorem 3.1.

For any $r>0$, there exist $n$ and $m$ such that under any $r$-coloring of
$S_{m}(x_{1},\ldots,x_{n})$, there is a monochromatic configuration

$\\{\psi(y),x,x+\phi_{1}(y),\ldots,x+\phi_{k}(y)\\},$

where $x,y$ have nonempty and disjoint supports.

The fact that the supports of $x$ and $y$ are nonempty and disjoint will be
crucial in our applications (Theorem 1.11 and Proposition 5.1). Theorem 3.1
follows from Proposition 3.3 below whose proof requires the multidimensional
Hales-Jewett theorem (for a reference, see [16, Theorem 7, p.40]). We recall
the theorem here for reader’s convenience.

The set $[t]^{N}=\\{(x_{1},\ldots,x_{N}):x_{i}\in[t]\\}$ is called a cube of
dimension $N$ over $t$ elements. Let $[N]=A_{0}\cup A_{1}\ldots A_{m}$ be any
disjoint partition of $[N]$, where $A_{i}\neq\varnothing$ for $i\neq 0$
($A_{0}$ may be empty), and $f:A_{0}\rightarrow[t]$ be any map. Define a map
$g:[t]^{m}\rightarrow[t]^{N}$ by assigning to each
$(y_{1},\ldots,y_{m})\in[t]^{m}$ the element $(x_{1},\ldots,x_{N})\in[t]^{N}$,
where

$x_{i}=\begin{cases}f(i),\quad&\textup{if}\ i\in A_{0}\\\
y_{j},\quad&\textup{if}\ i\in A_{j}\text{ for }j\in[m].\end{cases}$ (14)

A combinatorial space of dimension $m$ is the image of $g$ for some choice of
$A_{0},A_{1},\ldots,A_{m}$ and $f$. We can now state:

######  Theorem 3.2 (Multidimensional Hales-Jewett).

For any $r,t,m$, there exists a number $N=HJ(t,m;r)$ such that whenever
$[t]^{N}$ is $r$-colored, there must be a monochromatic combinatorial space of
dimension $m$.

Using this, we can prove the following proposition:

######  Proposition 3.3.

For any $r>0$ and $\ell>0$, there exist $n=n(k,\ell,r)$ and $m=m(k,\ell,r)$
such that under any $r$-coloring of $S_{m}(x_{1},\ldots,x_{n})$, there are
elements $y_{1},\ldots,y_{\ell}\in S_{m}(x_{1},\ldots,x_{n})$ with nonempty
and disjoint supports, such that for each $i\in[\ell]$, the elements

$\psi(y_{i})+\sum_{1\leq j\leq i-1}\xi_{j}(y_{j})\qquad\textup{ where
}\xi_{j}\in\\{0,\psi,\phi_{1},\ldots,\phi_{k}\\}$

have the same color (i.e. their color depends only on $i$).

###### Proof.

The number of colors $r$ will be fixed throughout. We will proceed by
induction on $\ell$. When $\ell=1$ the statement is obvious. Suppose the
statement is true for $\ell$, we will prove it is true for $\ell+1$.

Write $n^{\prime}=n(k,\ell,r),m^{\prime}=m(k,\ell,r)$. We define
$m=m(k,\ell,r):=|\Phi_{m^{\prime}+1}|+1$,
$N:=HJ(|\Phi_{m^{\prime}+1}|,n^{\prime};r)$ and $n=n(k,\ell+1,r):=1+N$.

Consider an arbitrary $r$-coloring of $S_{m}(x_{1},\ldots,x_{n})$. An
$r$-coloring of $S_{m}(x_{1},\ldots,x_{n})$ induces an $r$-coloring of
$\Phi_{m^{\prime}+1}^{N}$ by assigning to
$(a_{1},\ldots,a_{N})\in(\Phi_{m^{\prime}+1})^{N}$ the color of

$\psi(x_{n})+\sum_{i=1}^{N}\phi\circ a_{i}(x_{i}).$

Since $N=HJ(|\Phi_{m^{\prime}+1}|,n^{\prime};r)$, there is a disjoint
partition

$[N]=A_{0}\cup A_{1}\cdots\cup A_{n^{\prime}},\quad
A_{i}\neq\varnothing\,\forall i\neq 0$

and functions $f_{i}\in\Phi_{m^{\prime}+1}$ for $i\in A_{0}$ such that when
$\xi_{1},\xi_{2},\ldots,\xi_{n^{\prime}}$ range over $\Phi_{m^{\prime}+1}$,
all the elements

$\psi(x_{n})+\sum_{i=1}^{N}\phi\circ a_{i}(x_{i}),$

with

$a_{i}=\begin{cases}f_{i},\quad&\textup{if}\ i\in A_{0}\\\
\zeta_{j},\quad&\textup{if}\ i\in A_{j}\text{ for
}j\in[n^{\prime}]\end{cases}$

have the same color.

Write $z_{j}=\sum_{i\in A_{j}}\phi(x_{i})$ for $1\leq j\leq n^{\prime}$, and
$z_{n^{\prime}+1}=x_{n+1}+\sum_{i\in A_{0}}f_{i}(x_{i})$. Then all the $z_{j}$
have nonempty and disjoint supports, and all elements of the form

$\psi(z_{n^{\prime}+1})+\sum_{j=1}^{n^{\prime}}\zeta_{j}(z_{j}),\qquad\zeta_{j}\in\Phi_{m^{\prime}+1},$

have the same color.

By the inductive hypothesis, there exists a sequence $y_{1},\ldots,y_{\ell}\in
S_{m^{\prime}}(z_{1},\ldots,z_{n}^{\prime})$ having nonempty and disjoint
supports such that for each $i=1,\ldots,\ell$, the elements

$\psi(y_{i})+\sum_{1\leq j\leq i-1}\xi_{j}(y_{j})\qquad\textup{ where
}\xi_{j}\in\\{0,\psi,\phi_{1},\ldots,\phi_{k}\\}$

have the same color. We now set $y_{\ell+1}=z_{n^{\prime}+1}$. Clearly the
elements

$\phi(y_{\ell+1})+\sum_{1\leq j\leq\ell}\xi_{j}(y_{j})\qquad\textup{ where
}\xi_{j}\in\\{0,\psi,\phi_{1},\ldots,\phi_{k}\\}$

are of the form

$\phi(z_{n^{\prime}+1})+\sum_{j=1}^{n^{\prime}}\zeta_{j}(z_{j}),\qquad\zeta_{j}\in\Phi_{m^{\prime}+1},$

and so they have the same color. Thus Proposition 3.3 is proved. ∎

###### Proof of Theorem 3.1.

Applying Proposition 3.3 with $\ell=r+1$, we can find a sequence
$y_{1},\ldots,y_{r+1}$ satisfying the conclusion of that proposition. Let
$c(i)$ be the color of

$\psi(y_{i})+\sum_{1\leq j\leq i-1}\xi_{j}(y_{j})\qquad\textup{ where
}\xi_{j}\in\\{0,\psi,\phi_{1},\ldots,\phi_{k}\\}.$

Then there exist $1\leq u<v\leq r+1$ such that $c(u)=c(v)$. Hence the elements

$\psi(y_{u}),\psi(y_{v}),\psi(y_{v})+\phi_{1}(y_{u}),\ldots,\psi(y_{v})+\phi_{k}(y_{u}),$

have the same color, and we are done (with $x=\psi(y_{v}),y=y_{u}$). ∎

###  3.2. Proofs of Theorem 1.11 and Theorem 1.5

Using Theorem 3.1 we can now prove Theorem 1.11, which we recall for
convenience:

######  Theorem.

Suppose $\phi,\psi_{1},\ldots,\psi_{k}:G\to G$ are continuous homomorphisms
satisfying:

  1. (1)

$\psi(G),\phi_{1}(G),\ldots,\phi_{k}(G)$ have finite indices in $G$, and

  2. (2)

$\psi,\phi_{1},\ldots,\phi_{k}$ are commuting.

Suppose $G=\bigcup_{i=1}^{r}A_{i}$ is a partition of $G$ into measurable sets.
Then

$\sum_{i=1}^{r}\iint_{G^{2}}1_{A_{i}}(\psi(t))1_{A_{i}}(x)1_{A_{i}}(x+\phi_{1}(t))\cdots
1_{A_{i}}(x+\phi_{k}(t))\,d\mu(x)d\mu(t)\geq c_{3}$

for some positive constant $c_{3}$ depending only on $r,k$ and the
aforementioned indices.

###### Proof.

Consider the set $S_{m}(x_{1},\ldots,x_{n})$ given by Theorem 3.1, where we
now let $x_{1},\ldots,x_{n}$ vary over $G$. Note that for any
$\psi\in\Phi_{m}$, we have $[G:\psi(G)]<\infty$ by Lemma 2.6. Let $R$ be the
set of all pairs $(z,t)$ where $z,t\in S_{m}(x_{1},\ldots,x_{n})$ have
nonempty and disjoint supports.

Suppose $G=\bigcup_{i=1}^{r}A_{i}$. For $i\in[r]$, we define

$T_{i}:=\int_{G^{2}}1_{A_{i}}(\psi(y))1_{A_{i}}(x)1_{A_{i}}(x+\phi_{1}(y))\cdots
1_{A_{i}}(x+\phi_{k}(y))\,d\mu(x)d\mu(y).$

Let $(z,t)\in R$ be arbitrary, and suppose

$z=\sum_{u\in U}\zeta_{u}(x_{u})\qquad\textup{and}\qquad t=\sum_{v\in
V}\xi_{v}(x_{v})$

where $U,V\subset[n]$ are nonempty and disjoint and
$\zeta_{u},\xi_{v}\in\Phi_{m}\setminus\\{0\\}$. We have

$\displaystyle\int_{G^{m}}1_{A_{i}}(\psi(t))1_{A_{i}}(z)1_{A_{i}}(z+\phi_{1}(t))\cdots
1_{A_{i}}(z+\phi_{k}(t))\,d\mu(x_{1})\cdots d\mu(x_{m})$ $\displaystyle=$
$\displaystyle\int_{G^{m}}1_{A_{i}}\left(\sum_{v\in
V}\psi(\xi_{v}(x_{v}))\right)1_{A_{i}}\left(\sum_{u\in
U}\zeta_{u}(x_{u})\right)1_{A_{i}}\left(\sum_{u\in
U}\zeta_{u}(x_{u})+\phi_{1}\left(\sum_{v\in
V}\xi_{v}(x_{v})\right)\right)\cdots$ $\displaystyle\qquad
1_{A_{i}}\left(\sum_{u\in U}\zeta_{u}(x_{u})+\phi_{k}\left(\sum_{v\in
V}\xi_{v}(x_{v})\right)\right)\,d\mu(x_{1})\cdots d\mu(x_{m})$
$\displaystyle\ll$ $\displaystyle\int_{G^{m}}1_{A_{i}}\left(\sum_{v\in
V}\psi(x_{v})\right)1_{A_{i}}\left(\sum_{u\in
U}x_{u}\right)1_{A_{i}}\left(\sum_{u\in U}x_{u}+\phi_{1}\left(\sum_{v\in
V}x_{v}\right)\right)\cdots$ $\displaystyle\qquad 1_{A_{i}}\left(\sum_{u\in
U}x_{u}+\phi_{k}\left(\sum_{v\in V}x_{v}\right)\right)\,d\mu(x_{1})\cdots
d\mu(x_{m})$ $\displaystyle=$
$\displaystyle\int_{G^{2}}1_{A_{i}}(\psi(y))1_{A_{i}}(x+\phi_{1}(y))\cdots
1_{A_{i}}(x+\phi_{k}(y))\,d\mu(x)\,d\mu(y)$ $\displaystyle=$ $\displaystyle
T_{i},$

by $|U|+|V|$ applications of Lemma 2.8.

Now Theorem 3.1 implies that

$\displaystyle 1$ $\displaystyle\leq$
$\displaystyle\int_{G^{m}}\sum_{i=1}^{r}\sum_{(z,t)\in
R}1_{A_{i}}(\psi(t))1_{A_{i}}(z)1_{A_{i}}(z+\phi_{1}(t))\cdots
1_{A_{i}}(z+\phi_{k}(t))\,d\mu(x_{1})\cdots d\mu(x_{m})$ $\displaystyle\leq$
$\displaystyle\sum_{i=1}^{r}\int_{G^{m}}\sum_{(z,t)\in
R}1_{A_{i}}(\psi(t))1_{A_{i}}(z)1_{A_{i}}(z+\phi_{1}(t))\cdots
1_{A_{i}}(z+\phi_{k}(t))\,d\mu(x_{1})\cdots d\mu(x_{m})$ $\displaystyle\ll$
$\displaystyle\sum_{i=1}^{r}T_{i},$

thus finishing the proof. ∎

To prove Theorem 1.5, we will need the following proposition. With an eye to
potential applications, we state and prove a slightly stronger version than
what is needed.

######  Proposition 3.4.

Let $\phi,\psi:G\to G$ be commuting continuous homomorphisms with images
having finite indices. Suppose $f_{1},\ldots,f_{r}:G\to[0,1]$ are measurable
functions such that $\sum_{i=1}^{r}f_{i}\geq 1$ pointwise. For $w\in G$,
define

$R_{i}(w)=\iint f_{i}(\psi(y))f_{i}(x+w)f_{i}(x+\phi(y))\ d\mu(x)d\mu(y).$

Then there are $c,k,\eta>0$ depending only on $r$ and the indices above such
that for some $i\in[r]$, the set $\\{w\in G:R_{i}(w)>c\\}$ contains a
Bohr-$(k,\eta)$ set.

###### Proof.

For $i\in[r]$, let $A_{i}=\\{x\in G:f_{i}(x)\geq 1/r\\}$. Since
$\sum_{i=1}^{r}f_{i}\geq 1$ pointwise, $G=\bigcup_{i=1}^{r}A_{i}$. In light of
Theorem 1.11, there exists a constant $c$ depending only on $r$ and the
indices and an $i\in[r]$ such that

$\iint_{G^{2}}1_{A_{i}}(x)1_{A_{i}}(x+\phi(y))1_{A_{i}}(\psi(y))\
d\mu(x)d\mu(y)>c.$

It then follows that

$R_{i}(0)\geq\frac{c}{r^{3}}.$

On the other hand, by Lemma 2.12, for every $w\in G$,

$|R_{i}(w)-R_{i}(0)|\ll\lVert\hat{f}-\widehat{f_{w}}\rVert_{\infty},$

where the implicit constant depends only on the indices of
$\phi(G),\psi(G),(\phi-\psi)(G)$ in $G$. Hence, there exists a constant
$c^{\prime}$ such that $R_{i}(w)\geq\frac{c}{2r^{3}}$ if

$\lVert\widehat{f_{i}}-\widehat{f_{i,w}}\rVert_{\infty}<c^{\prime}.$

By Lemma 2.1, the set of such $w$ contains a Bohr-$(k,\eta)$ set, where $k$
and $\eta$ depend only on $c^{\prime}$. ∎

Theorem 1.5 is now a special case of the next theorem with $\psi_{1}=\phi_{2}$
and $\psi_{2}=\phi_{1}$.

######  Theorem 3.5.

Let $G=\bigcup_{i=1}^{r}A_{i}$ be a partition into measurable sets. Let
$\phi_{1},\phi_{2},\psi_{1},\psi_{2}:G\rightarrow G$ be continuous
homomorphisms satisfying the followings:

  1. (1)

$\phi_{2}\circ\psi_{2}=\phi_{1}\circ\psi_{1}$,

  2. (2)

$\psi_{1}\circ\psi_{2}=\psi_{2}\circ\psi_{1}$,

  3. (3)

$\phi_{1}(G),\psi_{1}(G),\psi_{2}(G)$ have finite indices in $G$.

Then for some $1\leq i\leq r$, the set
$\phi_{1}(A_{i})-\phi_{1}(A_{i})+\phi_{2}(A_{i})$ contains a Bohr-$(k,\eta)$
set, where $k$ and $\eta$ depend only on $r$ and the indices of
$\phi_{1}(G),\psi_{1}(G),\psi_{2}(G)$ in $G$.

###### Proof.

Suppose $G=\bigcup_{i=1}^{r}A_{i}$. We apply Proposition 3.4 with
$f_{i}=1_{A_{i}}$ and $(\psi_{1},\psi_{2})$ in place of $(\psi,\phi)$. Then
for some $i$, the set $\\{w\in G:R_{i}(w)>c\\}$ contains a Bohr-$(k,\eta)$ set
$B$. This means that for $w\in B$, there exist $x,y\in G$ such that

$x+w,\psi_{2}(y),x+\psi_{1}(y)\in A_{i}.$

Since

$\phi_{1}(x+w)+\phi_{2}(\psi_{2}(y))-\phi_{1}(x+\psi_{1}(y))=\phi_{1}(w),$

we conclude that
$\phi_{1}(B)\subset\phi_{1}(A_{i})+\phi_{2}(A_{i})-\phi_{1}(A_{i})$. Our
theorem now follows from Lemma 2.10. ∎

######  Remark 5.

If $\phi_{1}$ is an automorphism and $[G:\phi_{2}]<\infty$ then the hypothesis
of Theorem 3.5 is also satisfied. Indeed, we let
$\psi_{1}=\phi_{1}^{-1}\circ\phi_{2}$ and $\psi_{2}=\textup{Id}$. Then the
first two conditions of Theorem 3.5 are satisfied. As for the third condition,
we have $\psi_{1}(G)=\phi_{1}^{-1}\circ\phi_{2}(G),$ which has finite index in
$G$ by Lemma 2.6. Similarly, we see that if $\phi_{2}$ is an automorphism and
$[G:\phi_{1}]<\infty$, then the conditions of Theorem 3.5 is also satisfied.

##  4\. Bohr sets and sets of positive measure

###  4.1. A regularity lemma

The goal of this section is to prove Proposition 4.2. As mentioned in the
introduction, this argument has its genesis in Bourgain [8]. Bourgain’s ideas
were elaborated by Tao [31], who proved Roth’s theorem in compact abelian
groups that are 2-divisible; and by Bergelson-Host-McCutcheon-Parreau [2,
Theorem 4.1], who proved Roth’s theorem for dilations on the torus
$\mathbb{R}/\mathbb{Z}$. We streamline and generalize Bergelson-Host-
McCutcheon-Parreau’s argument to deal with homomorphisms on arbitrary compact
abelian groups. This generalization requires non-trivial modifications;
especially, we will make use of Lemma 2.1 and Lemma 2.4.

######  Lemma 4.1 (cf. [2, Lemma 4.2]).

Let $\phi,\psi:G\rightarrow G$ be continuous homomorphisms such that
$\phi(G),\psi(G)$ and $(\phi-\psi)(G)$ have finite indices in $G$. For $f\in
L^{\infty}(G)$, define

$J(f)=\iint_{G^{2}}f(x)f(x+\phi(y))f(x+\psi(y))\,d\mu(x)d\mu(y).$

Then for any measurable functions $f,g:G\to[0,1]$,

$|J(f)-J(g)|\ll\|\widehat{f}-\widehat{g}\|_{\infty},$

where the implicit constant depends only on the aforementioned indices.

###### Proof.

We have

$\displaystyle J(f)-J(g)$ $\displaystyle=$
$\displaystyle\iint_{G^{2}}(f-g)(x)\cdot f(x+\phi(y))\cdot
f(x+\psi(y))\,d\mu(x)d\mu(y)$
$\displaystyle+\iint_{G^{2}}g(x)\cdot(f-g)(x+\phi(y))\cdot
f(x+\psi(y))\,d\mu(x)d\mu(y)$ $\displaystyle+\iint_{G^{2}}g(x)\cdot
g(x+\phi(y))\cdot(f-g)(x+\psi(y))\,d\mu(x)d\mu(y).$

The lemma now follows from Lemma 2.11 and the assumptions on $\phi$ and
$\psi$. ∎

######  Proposition 4.2 (Regularity Lemma).

Let $f:G\to[0,1]$ be a measurable function with $\int_{G}f\,d\mu=\delta>0$.
Let $\phi,\psi:G\to G$ be continuous homomorphisms such that $\phi(G),\psi(G)$
and $(\phi-\psi)(G)$ have finite indices in $G$. Then for every $\epsilon>0$,
there exist a constant $C$ that depends only on $\delta,\epsilon$ and the
indices above, a kernel $K:G\to\mathbb{R}_{\geq 0}$, and a decomposition
$f=f_{st}+f_{er}+f_{un}$ such that

  1. (1)

$\lVert K\rVert_{\infty}<C$,

  2. (2)

$\lVert f_{st}\rVert_{\infty}\leq 1$, $\lVert f_{er}\rVert_{\infty}\leq 2$ and
$\lVert f_{un}\rVert_{\infty}\leq 2$,

  3. (3)

$J^{\prime}(f_{st}):=\displaystyle\iint_{G^{2}}f_{st}(x)f_{st}(x+\phi(t))f_{st}(x+\psi(t))K(t)\,d\mu(x)d\mu(t)>\delta^{3}-\epsilon$,

  4. (4)

$\lVert f_{er}\rVert_{2}<\epsilon$,

  5. (5)

$\lVert\widehat{f}_{un}\rVert_{\infty}\lVert\widehat{K}\rVert_{1}<\epsilon$.

###### Proof.

For $t\in G$, let

$d(t):=\max\left(\|\widehat{f}-\widehat{f_{t}}\|_{\infty},\|\widehat{f}-\widehat{f_{\phi(t)}}\|_{\infty},\|\widehat{f}-\widehat{f_{\psi(t)}}\|_{\infty}\right).$

Fixing $\epsilon>0$, we define sequences $\eta_{n}\in(0,1/2]$,
$\kappa_{n}\in(0,\infty)$ and finite sets $\Lambda_{n}\subseteq\Gamma$
recursively as follows:

First set $\eta_{0}=1/2$. For $n\geq 0$, Lemma 2.1 implies that there exists a
set $\Lambda_{n}\in\Gamma$ with $|\Lambda_{n}|\leq 12/\eta_{n}^{2}$ such that
$D(\eta_{n}):=\\{t\in G:d(t)\leq\eta_{n}\\}$ contains a Bohr set
$B(\Lambda_{n};\eta_{n})$. For $\eta\in(0,1/2]$, define
$\nu(\eta)=(C_{0}\eta)^{12/\eta^{2}}$ where $C_{0}$ is the constant found in
Lemma 2.4; in particular,
$\nu(\eta_{n})=(C_{0}\eta_{n})^{12/\eta_{n}^{2}}\leq(C_{0}\eta_{n})^{|\Lambda_{n}|}$.
Put

$\kappa_{n}=\nu(\eta_{n})^{-1/2}\text{ and
}\eta_{n+1}=\min\left\\{\eta_{n},\frac{\epsilon^{2}}{4\kappa_{n}^{2}},\epsilon\nu\left(\frac{\epsilon}{2\kappa_{n}}\right)\right\\}.$

In view of Lemma 2.4, for $n\geq 0$, there is a kernel $K_{n}:G\to[0,\infty)$
such that

$\widehat{K}_{n}\geq 0,\|\widehat{K_{n}}\|_{1}=\lVert K_{n}\rVert_{\infty}\leq
1/\nu(\eta_{n})$

and $K_{n}$ is supported on $B(\Lambda_{n};\eta_{n})\subseteq D(\eta_{n})$. We
define

$f_{n}=f*K_{n}.$

Claim 1:

$\|\widehat{f}-\widehat{f_{n}}\|_{\infty}=\sup_{\gamma\in\Gamma}\left|\widehat{f}(\gamma)(1-\widehat{K_{n}}(\gamma))\right|\leq\eta_{n}.$

Indeed, by construction, $K_{n}$ is supported on $D(\eta_{n})$; and every
$t\in D(\eta_{n})$ satisfies
$\left|\widehat{f}(\gamma)(1-\gamma(t))\right|\leq\eta_{n}$ for all
$\gamma\in\Gamma$. Therefore, for all $\gamma\in\Gamma$,

$\displaystyle\left|\widehat{f}(\gamma)\right|\left|1-\widehat{K_{n}}(\gamma)\right|$
$\displaystyle\leq$
$\displaystyle\left|\widehat{f}(\gamma)\right|\int_{G}K_{n}(x)\left|1-\overline{\gamma(x)}\right|\,d\mu(x)$
$\displaystyle=$
$\displaystyle\left|\widehat{f}(\gamma)\right|\int_{D(\eta_{n})}K_{n}(x)\left|1-\overline{\gamma(x)}\right|\,d\mu(x)$
$\displaystyle=$
$\displaystyle\int_{D(\eta_{n})}K_{n}(x)\left|\widehat{f}(\gamma)\right|\left|1-\overline{\gamma(x)}\right|\,d\mu(x)$
$\displaystyle\leq$
$\displaystyle\eta_{n}\int_{D(\eta_{n})}K_{n}(x)\,d\mu(x)\leq\eta_{n}.$

Claim 2:

$\lVert f_{n+1}-f_{n}\rVert_{2}^{2}\leq\lVert f_{n+1}\rVert_{2}^{2}-\lVert
f_{n}\rVert_{2}^{2}+2\eta_{n+1}\kappa_{n}^{2}.$

Indeed, we have

$\displaystyle\lVert f_{n+1}-f_{n}\rVert_{2}^{2}$ $\displaystyle=$
$\displaystyle\lVert\widehat{f_{n+1}}-\widehat{f_{n}}\rVert_{2}^{2}$
$\displaystyle=$
$\displaystyle\lVert\widehat{f_{n+1}}\rVert_{2}^{2}-\lVert\widehat{f_{n}}\rVert_{2}^{2}-\sum_{\gamma\in\Gamma}\left(-\widehat{f_{n+1}}(\gamma)\overline{\widehat{f_{n}}(\gamma)}+\overline{\widehat{f_{n+1}}(\gamma)}\widehat{f_{n}}(\gamma)\right)$
$\displaystyle=$
$\displaystyle\lVert\widehat{f_{n+1}}\rVert_{2}^{2}-\lVert\widehat{f_{n}}\rVert_{2}^{2}+2\sum_{\gamma\in\Gamma}\left|\widehat{f}(\gamma)\right|_{2}^{2}\widehat{K_{n}}(\gamma)\left(\widehat{K_{n}}(\gamma)-\widehat{K_{n+1}}(\gamma)\right)$
$\displaystyle\leq$
$\displaystyle\lVert\widehat{f_{n+1}}\rVert_{2}^{2}-\lVert\widehat{f_{n}}\rVert_{2}^{2}+2\sum_{\gamma\in\Gamma}\left|\widehat{f}(\gamma)\right|^{2}\widehat{K_{n}}(\gamma)\left(1-\widehat{K_{n+1}}(\gamma)\right)$
$\displaystyle\leq$
$\displaystyle\lVert\widehat{f_{n+1}}\rVert_{2}^{2}-\lVert\widehat{f_{n}}\rVert_{2}^{2}+2\sup_{\gamma\in\Gamma}|\widehat{f}(\gamma)|\left(1-\widehat{K_{n+1}}(\gamma)\right)\cdot\|\widehat{K_{n}}\|_{1}$
$\displaystyle\leq$
$\displaystyle\lVert\widehat{f_{n+1}}\rVert_{2}^{2}-\lVert\widehat{f_{n}}\rVert_{2}^{2}+2\eta_{n+1}\kappa_{n}^{2}$

and the claim is proved.

Since $\eta_{n+1}\leq\epsilon^{2}/(4\kappa_{n}^{2})$, we have

$\lVert f_{n+1}-f_{n}\rVert_{2}^{2}\leq\lVert f_{n+1}\rVert_{2}^{2}-\lVert
f_{n}\rVert_{2}^{2}+\epsilon^{2}/2.$

Let $M$ be the smallest integer such that $M\geq 2/\epsilon^{2}$. Then

$\sum_{n=0}^{M-1}\lVert f_{n+1}-f_{n}\rVert_{2}^{2}\leq\lVert
f_{M}\rVert_{2}^{2}-\lVert f_{0}\rVert_{2}^{2}+M\epsilon^{2}/2\leq
1+M\epsilon^{2}/2\leq M\epsilon^{2}.$

Therefore there exists $0\leq n\leq M-1$ such that

$\lVert f_{n+1}-f_{n}\rVert_{2}\leq\epsilon.$

From now on, we fix this $n$. Next consider the expression

$I_{n}(t)=\int_{G}f_{n}(x)f_{n}(x+\phi(t))f_{n}(x+\psi(t))\,d\mu(x)\;\;\text{
for }t\in G.$

We have

$|I_{n}(0)-I_{n}(t)|\leq\|f_{n}-(f_{n})_{\phi(t)}\|_{1}+\|f_{n}-(f_{n})_{\psi(t)}\|_{1}.$

Note that

$\displaystyle\|f_{n}-(f_{n})_{\phi(t)}\|^{2}_{1}$ $\displaystyle\leq$
$\displaystyle\|f_{n}-(f_{n})_{\phi(t)}\|_{2}^{2}=\|(f-f_{\phi(t)})*K_{n}\|_{2}^{2}$
$\displaystyle=$
$\displaystyle\sum_{\gamma\in\Gamma}\left|\widehat{K_{n}}(\gamma)\right|^{2}\left|\widehat{f}(\gamma)\right|^{2}\left|1-\gamma(\phi(t))\right|^{2}$
$\displaystyle\leq$
$\displaystyle\|\widehat{K_{n}}\|_{1}d(t)^{2}\leq\kappa_{n}^{2}d(t)^{2}.$

The same estimate holds for $\|f_{n}-(f_{n})_{\psi(t)}\|_{1}^{2}$. Hence
$|I_{n}(0)-I_{n}(t)|\leq 2\kappa_{n}d(t)$ for any $t\in G$.

Since $I_{n}(0)=\lVert f_{n}\rVert_{1}^{3}=\lVert
f\rVert_{1}^{3}\geq\delta^{3}$, it follows that

$I_{n}(t)\geq\delta^{3}-2\kappa_{n}d(t)\text{ for all }t\in G.$

Note that $d(t)\leq\epsilon/(2\kappa_{n})$ for $t$ in the set
$D(\epsilon/(2\kappa_{n}))$ and so $I_{n}(t)\geq\delta^{3}-\epsilon$ in this
set.

Let $\eta=\epsilon/(2\kappa_{n})$. In view of Lemma 2.4, there exists a kernel
$K$ supported on $D(\eta)$ such that $\lVert K\rVert_{\infty}\leq
1/\nu(\eta)$. We then have

$J^{\prime}(f_{n}):=\int_{G}I_{n}(t)K(t)\,d\mu(t)\geq(\delta^{3}-\epsilon)\int_{B(\epsilon/(2\kappa_{n}))}K(t)\,d\mu(t)\geq\delta^{3}-\epsilon.$

Letting $f_{st}=f_{n}$, $f_{er}=f_{n+1}-f_{n}$ and $f_{un}=f-f_{n+1}$, we
obtain

  1. (1)

$\lVert K\rVert_{\infty}\leq 1/\nu(\epsilon/(2\kappa_{n}))\leq
1/\nu(\epsilon/(2\kappa_{M}))$,

  2. (2)

$\lVert f_{er}\rVert_{2}=\lVert f_{n+1}-f_{n}\rVert_{2}<\epsilon$,

  3. (3)

$\lVert\hat{f}_{un}\rVert_{\infty}\lVert
K\rVert_{\infty}=\lVert\hat{f}-\hat{f}_{n+1}\rVert_{\infty}\lVert
K\rVert_{\infty}<\eta_{n+1}/\nu(\eta)<\epsilon$ because
$\eta_{n+1}\leq\epsilon\nu(\epsilon/(2\kappa_{n}))=\epsilon\nu(\eta).$

  4. (4)

$J^{\prime}(f_{st})=J^{\prime}(f_{n})\geq\delta^{3}-\epsilon$.

Our proof finishes. ∎

###  4.2. Proof of density results

The goal of this section is to prove Theorem 1.6 and Theorem 1.9. First we
recall Theorem 1.9 for the reader’s convenience.

######  Theorem (Khintchine-Roth theorem for compact abelian groups).

Let $f:G\to[0,1]$ be a mensurable function with $\int_{G}f\,d\mu>\delta$. Let
$\phi,\psi:G\to G$ be continuous homomorphisms such that
$[G:\phi(G)],[G:\psi(G)]$ and $[G:(\phi-\psi)G]$ are finite. Then for every
$\epsilon>0$, there exists a constant $c_{1}$ that depends only on
$\delta,\epsilon$ and the indices above such that the set

$B=\left\\{t\in
G:\int_{G}f(x)f(x+\phi(t))f(x+\psi(t))\,d\mu(x)d\mu(t)>\delta^{3}-\epsilon\right\\}$

has measure greater than $c_{1}$. As a consequence, there exists a constant
$c_{2}$ that depends only on $\delta$ and indices of $\phi(G),\psi(G)$ such
that

$J(f):=\iint_{G^{2}}f(x)f(x+\phi(t))f(x+\psi(t))d\mu(x)d\mu(t)>c_{2}.$

###### Proof.

Fix $\epsilon>0$ and let constant $C$, kernel $K$ and the decomposition
$f=f_{st}+f_{er}+f_{un}$ be as found in Proposition 4.2. Define

$J^{\prime}(f):=\iint_{G^{2}}f(x)f(x+\phi(t))f(x+\psi(t))K(t)\ d\mu(x)d\mu(t)$

and

$J^{\prime}(f_{st}):=\iint_{G^{2}}f_{st}(x)f_{st}(x+\phi(t))f_{st}(x+\psi(t))K(t)\
d\mu(x)d\mu(t).$

Applying the decomposition $f=f_{st}+f_{er}+f_{un}$ and expanding
$J^{\prime}(f)$, we see that the difference $J^{\prime}(f)-J^{\prime}(f_{st})$
will have $26$ terms. The terms that contain $f_{er}$ can be bounded by
$4\epsilon$ since for $f_{1},f_{2},f_{3}\in L^{\infty}(G)$,

$\iint_{G^{2}}f_{1}(x)f_{2}(x+\phi(t))f_{3}(x+\psi(t))K(t)\,d\mu(x)d\mu(t)\leq\max_{i}\lVert
f_{i}\rVert_{\infty}^{2}\max_{i}\lVert f_{i}\rVert_{1}\lVert K\rVert_{1}.$

On the other hand, in view of Lemma 2.11, the terms containing $f_{un}$ are
bounded by $O(\lVert\hat{f}_{un}\rVert_{\infty}\lVert\widehat{K}\rVert_{1})$
which is $O(\epsilon)$ thanks to the properties of the decomposition.
Therefore,

$J^{\prime}(f)>J^{\prime}(f_{st})-O(\epsilon)>\delta^{3}-c_{0}\epsilon$ (15)

where the constant $c_{0}$ depends only on the indices of $\phi(G),\psi(G)$
and $(\phi-\psi)(G)$ in $G$.

Define

$I_{f}(t)=\int_{G}f(x)f(x+\phi(t))f(x+\psi(t))\,d\mu(x)$

and

$B=\\{t\in G:I_{f}(t)>\delta^{3}-2c_{0}\epsilon\\}.$

We then have

$J^{\prime}(f)=\int_{G}I_{f}(t)K(t)\,d\mu(t)=\int_{B}I_{f}(t)K(t)\,d\mu(t)+\int_{G\setminus
B}I_{f}(t)K(t)\,d\mu(t)\leq\\\
\int_{B}K(t)\,d\mu(t)+(\delta^{3}-2c_{0}\epsilon)\int_{G\setminus
B}K(t)\,d\mu(t)\leq\lVert K\rVert_{\infty}\mu(B)+(\delta^{3}-2c_{0}\epsilon).$
(16)

Combining (15) and (16), we deduce that

$\mu(B)>c_{0}\epsilon/\lVert K\rVert_{\infty}.$

Letting $c_{1}=c_{0}\epsilon/\lVert K\rVert_{\infty}$, we obtain the first
part of the theorem.

Now we have

$J(f)=\int_{G}I_{f}(t)\,d\mu(t)>(\delta^{3}-2c_{0}\epsilon)c_{1}.$

Letting $c_{2}=c_{1}(\delta^{3}-2c_{0}\epsilon)$, we obtain the second part of
the theorem. ∎

In order to prove Theorem 1.6, we need the following proposition. For our
future applications, we will state and prove a slightly more general version
than what is necessary.

######  Proposition 4.3.

Suppose $\phi,\psi:G\rightarrow G$ are continuous homomorphisms such that
$\phi(G),\psi(G),(\phi-\psi)(G)$ have finite indices in $G$. Let
$f:G\rightarrow[0,1]$ such that $\int_{G}f\,d\mu=\delta>0$. For $w\in G$,
define

$R(w)=\iint_{G^{2}}f(x+w)f(x+\phi(y))f(x+\psi(y))\,d\mu(x)d\mu(y)$

Then there are $c,k,\eta>0$ depending only on $\delta$ and the indices above
such that the set $\\{w\in G:R(w)>c\\}$ contains a Bohr-$(k,\eta)$ set.

###### Proof.

By Lemma 2.11, we have

$|R(w)-R(0)|\ll\lVert\hat{f}-\widehat{f_{w}}\rVert_{\infty}$

where implicit constant depends only on the indices of
$\phi(G),\psi(G),(\phi-\psi)(G)$ in $G$. By Theorem 1.9, we know that $R(0)>c$
for some constant $c>0$ depending on these indices and $\delta$. It follows
that there exists a constant $c^{\prime}$ such that $R(w)>c/2$ if

$\lVert\hat{f}-\widehat{f_{w}}\rVert_{\infty}<c^{\prime}.$ (17)

Lemma 2.1 implies that the set of such $w$ contains a Bohr-$(k,\eta)$ set,
where $k$ and $\eta$ depend only on $c^{\prime}$. ∎

We can now formulate and prove our main theorem for sets of positive measure.

######  Theorem 4.4.

Let $\phi_{1},\phi_{2},\phi_{3},\psi_{1},\psi_{2}:G\rightarrow G$ be
continuous homomorphisms satisfying the following

  1. (1)

$\phi_{1}+\phi_{2}+\phi_{3}=0$,

  2. (2)

$\phi_{1}\circ\psi_{1}=\phi_{2}\circ\psi_{2}$,

  3. (3)

$\phi_{3}(G),\psi_{1}(G),\psi_{2}(G),(\psi_{1}+\psi_{2})(G)$ have finite
indices in $G$.

The for any measurable set $A\subset G$, $\mu(A)=\delta>0$, the set
$\phi_{1}(A)+\phi_{2}(A)+\phi_{3}(A)$ contains a Bohr-$(k,\eta)$ set, where
$k$ and $\eta$ depend only on $\delta$ and the indices above.

###### Proof.

Applying Proposition 4.3 for $f=1_{A}$ and $\psi_{1},\psi_{2}$ in place of
$\phi$ and $\psi$, we see that there exists a Bohr-$(k,\eta)$ set $B$ such
that for all $w\in B$, there are $x,y\in G$ such that

$x+w,x+\psi_{1}(y)\textup{ and }x-\psi_{2}(y)\in A.$

Note that

$\phi_{3}(x+w)+\phi_{1}(x+\psi_{1}(y))+\phi_{2}(x-\psi_{2}(y))=\phi_{3}(w)$

and so that $\phi_{1}(A)+\phi_{2}(A)+\phi_{3}(A)\supseteq\phi_{3}(B)$. Our
theorem then follows from Lemma 2.10. ∎

Theorem 1.6 is now a special case of Theorem 4.4 when $\psi_{1}=\phi_{2}$ and
$\psi_{2}=\phi_{1}$.

######  Remark 6.

If $\phi_{1}$ is an automorphism and $[G:\phi_{2}(G)],[G:\phi_{3}(G)]<\infty$
then the hypothesis of Theorem 4.4 is also satisfied. Indeed, we let
$\psi_{1}=\phi_{1}^{-1}\circ\phi_{2}$ and $\psi_{2}=\textup{Id}$. Then the
first two conditions of Theorem 4.4 are satisfied. As for the third condition,
we have $\psi_{1}(G)=\phi_{1}^{-1}\circ\phi_{2}(G)$ and
$(\psi_{1}+\psi_{2})(G)=\phi_{1}^{-1}\circ(\phi_{2}+\phi_{1})(G)$. Both of
these have finite indices in $G$ by Lemma 2.6.

##  5\. Bohr sets in sumsets in number fields and function fields

In this section we prove Theorems 1.4, 1.7 and 1.8 using a strategy similar to
Bergelson and Ruzsa’s proof of Theorem 1.2. To prove Theorem 1.2, one could
embed $A\cap[N]$ naturally in $\mathbb{Z}_{N}$, and invoke the counting result
(for example, Theorem 1.9) in $\mathbb{Z}_{N}$. However, one has to deal with
the “wraparound effect”: A solution to $s_{1}x+s_{2}y+s_{3}z=0$ in
$\mathbb{Z}_{N}$ does not necessarily correspond to a solution in
$\mathbb{Z}$. To overcome this issue, Bergelson and Ruzsa embedded $A\cap[N]$
in $\mathbb{Z}_{N^{\prime}}$ for some $N^{\prime}\gg N$. Then $A\cap[N]$
remains dense in $\mathbb{Z}_{N^{\prime}}$ and a solution in
$\mathbb{Z}_{N^{\prime}}$ found in $A\cap[N]$ is now a solution in
$\mathbb{Z}$.

For partitions, the corresponding counting result would be Theorem 1.11.
However, if this theorem were applied directly, we would have a partition of
the whole group $\mathbb{Z}_{N^{\prime}}$ which again causes the wrap-around
effect. To avoid this problem, we need to modify Theorem 1.11 so that it
allows for partitions of a subset $[-N,N]\subset\mathbb{Z}_{N^{\prime}}$
instead of the whole group.

######  Proposition 5.1.

For any $k,\ell,r>0$, there is a constant $c(k,\ell,r)>0$ such that the
following holds: For sufficiently large $N$, if
$[-N,N]=\bigcup_{i=1}^{r}A_{i}$, then for some $1\leq i\leq r$, we have

$\sum_{|x|,|y|\leq N}1_{A_{i}}(\ell y)1_{A_{i}}(x)1_{A_{i}}(x+y)\cdots
1_{A_{i}}(x+ky)\geq c(k,\ell,r)N^{2}.$

######  Remark 7.

Proposition 5.1 also follows from Frankl-Graham-Rödl [12, Theorem 1], but our
proof shows that it is directly in line with Theorem 1.11. Furthermore, our
proof easily generalizes to other rings such as $\mathbb{Z}[i]$ and
${\mathbb{F}_{q}}[t]$.

###### Proof.

We applying Theorem 3.1 with $\psi(y)=\ell y$ and $\phi_{j}(y)=jy$ for $1\leq
j\leq k$. Then there exist $m$ and $n$ depending only on $r$ and $k$ such that
for any $r$-coloring of $S_{m}(x_{1},\ldots,x_{n})$, there are $x$ and $y$ of
nonempty and disjoint support such that the configuration

$\\{\ell y,x,x+y,\ldots,x+ky\\}$

is monochromatic.

Note that elements of $S_{m}(x_{1},\ldots,x_{n})$ are all linear forms in
$x_{1},\ldots,x_{n}$ with bounded integer coefficients. We now let
$x_{1},\ldots,x_{n}$ vary over $[-cN,cN]$ where $c$ is a small constant. Then
$S_{m}(x_{1},\ldots,x_{n})\subset[-N,N]$. Under the partition
$[-N,N]=\bigcup_{i=1}^{r}A_{i}$, each set $S_{m}(x_{1},\ldots,x_{n})$ contains
a monochromatic configuration $\\{\ell y,x,x+y,\ldots,x+ky\\}$. There are $\gg
N^{n}$ monochromatic configurations arising in this way. However, a
configuration may come from many different sets $S_{m}(x_{1},\ldots,x_{n})$.
We will show that the number of tuples $(x_{1},\ldots,x_{n})$ giving rise to
the same configuration $\\{\ell y,x,x+y,\ldots,x+ky\\}$ is $\ll N^{n-2}$.

Indeed, let $I,J$ be disjoint nonempty subsets of $[n]$ such that $x$ and $y$
are linear combinations with bounded coefficients of $(x_{i})_{i\in I}$ and
$(x_{j})_{j\in J}$, respectively. For fixed $I$ and $J$, the number of choices
for $(x_{i})_{i\in I}$ is $\ll N^{|I|-1}$, since any choice of $(|I|-1)$ of
the $x_{i}$’s gives at most one value for the remaining $x_{i}$. For the same
reason, the number of choices for $(x_{j})_{j\in J}$ is $\ll N^{|J|-1}$. Since
there are finitely many pairs $(I,J)$, we see that the number of
$(x_{1},\ldots,x_{n})$ that give rise to $\\{\ell y,x,x+y,\ldots,x+ky\\}$ is
$\ll N^{n-2}$. Hence the number of monochromatic configurations in $[N]$ is
$\gg N^{2}$, and we are done. ∎

Our next statement is essentially a diagonalization argument.

######  Proposition 5.2.

Let $\mathcal{P}$ denote an arbitrary partition
$\mathbb{Z}=\bigcup_{i=1}^{r}A_{i}$. Then there exists some $1\leq i\leq r$
with the following property: For every $\ell\geq 0$, there is a constant
$c(\ell,\mathcal{P})$ such that

$\sum_{|x|,|y|\leq N}1_{A_{i}}(y)1_{A_{i}}(x)1_{A_{i}}(x+\ell y)\geq
c(\ell,\mathcal{P})N^{2}$

for infinitely many $N\in\mathbb{N}$.

###### Proof.

Invoking Proposition 5.1, for each $k\in\mathbb{N}$, there is $i=f(k)$ such
that for infinitely many $N$, we have

$\sum_{|x|,|y|\leq N}1_{A_{i}}(y)1_{A_{i}}(x)1_{A_{i}}(x+y)\cdots
1_{A_{i}}(x+ky)\geq c(k,1,r)N^{2}.$

Hence there exist an $i\in\\{1,\ldots,r\\}$ and an infinite set $K$ such that
$f(k)=i$ for all $k\in K$.

Let $\ell$ be arbitrary and pick $k\in K,k\geq\ell$. We have, for infinitely
many $N$,

$\sum_{|x|,|y|\leq N}1_{A_{i}}(y)1_{A_{i}}(x)1_{A_{i}}(x+\ell y)\\\
\geq\sum_{|x|,|y|\leq N}1_{A_{i}}(y)1_{A_{i}}(x)1_{A_{i}}(x+y)\cdots
1_{A_{i}}(x+ky)\geq c(k,1,r)N^{2},$

thus proving the desired claim. ∎

######  Remark 8.

In the proof above, we do not have any control on $c(k,1,r)$ since we do not
have control on $k$. As a result, the constant $c(\ell,\mathcal{P})$ above
depends on the partition. It is interesting to see if this dependence is
indeed necessary.

We can now prove Theorem 1.4.

###### Proof of Theorem 1.4(a).

Let $\mathbb{Z}=\bigcup_{i=1}^{r}A_{i}$ be an arbitrary partition and
$s_{1},s_{2}\in\mathbb{Z}\setminus\\{0\\}$. Without loss of generality, we
assume $s_{1},s_{2}>0$. For a set $A\subset\mathbb{Z}$ and $N>0$, we write
$A^{(N)}$ to denote $A\cap[-N,N]$.

By Proposition 5.1, there exist $i\in[r]$ and an infinite set $\mathcal{N}$
such that

$\sum_{|x|,|y|\leq
N}1_{A_{i}^{(N)}}(s_{1}y)1_{A_{i}^{(N)}}(x)1_{A_{i}^{(N)}}(x+s_{2}y)\geq
cN^{2}$ (18)

for any $N\in\mathcal{N}$, where $c>0$ is a constant independent of $N$.

Let $N^{\prime}$ be the smallest odd integer greater than $(2s_{1}+s_{2})N$.
We identify $\mathbb{Z}_{N^{\prime}}$ with
$[-\frac{N^{\prime}-1}{2},\frac{N^{\prime}-1}{2}]$. Then (18) implies that

$\sum_{x,y\in\mathbb{Z}_{N^{\prime}}}1_{A_{i}^{(N)}}(s_{1}y)1_{A_{i}^{(N)}}(x)1_{A_{i}^{(N)}}(x+s_{2}y)\geq
c^{\prime}N^{\prime 2}$ (19)

for some constant $c^{\prime}>0$ independent of $N$. Define

$R(w)=\sum_{x,y\in\mathbb{Z}_{N^{\prime}}}1_{A^{(N)}_{i}}(s_{1}y)1_{A^{(N)}_{i}}(x+w)1_{A^{(N)}_{i}}(x+s_{2}y).$

Then by the same argument as the proof of Proposition 3.4, the set
$\\{w\in\mathbb{Z}_{N^{\prime}}:R(w)>0\\}$ contains a Bohr-$(k,\eta)$ set in
$\mathbb{Z}_{N^{\prime}}$, where $k$ and $\eta$ are independent of $N$. Note
that $R(w)>0$, implies there are $a,a^{\prime},a^{\prime\prime}\in
A^{(N)}_{i}$ and $x,y\in\mathbb{Z}_{N^{\prime}}$ such that

$s_{1}y\equiv a,\quad x+w\equiv a^{\prime},\quad\textup{and}\quad
x+s_{2}y\equiv a^{\prime\prime}\pmod{N^{\prime}}.$

Therefore,

$s_{1}w=s_{1}(x+w)-s_{1}(x+s_{2}y)+s_{2}(s_{1}y)\equiv
s_{1}a^{\prime}-s_{1}a^{\prime\prime}+s_{2}a\pmod{N^{\prime}}.$

If $|w|\leq N$ then this congruence is an equality in $\mathbb{Z}$ due to the
way we choose $N^{\prime}$ and the fact that
$|a|,|a^{\prime}|,|a^{\prime\prime}|\leq N$. We have thus proved that, for
each $N\in\mathcal{N}$, there exist
$x_{1},\ldots,x_{k}\in[-\frac{N^{\prime}-1}{2},\frac{N^{\prime}-1}{2}]$ such
that

$(s_{1}A_{i}-s_{1}A_{i}+s_{2}A_{i})/s_{1}\supset[-N,N]\cap\left\\{w\in\mathbb{Z}:\left|e\left(\frac{x_{j}w}{N^{\prime}}\right)-1\right|<\eta\quad\forall
j=1,\ldots,k\right\\}.$

Here we are using the notation $A/c$ defined in (2).

As $N\in\mathcal{N},N\to\infty$ and by passing to a subsequence if necessary,
the sequence $(\frac{x_{1}}{N^{\prime}},\ldots,\frac{x_{k}}{N^{\prime}})$
converges to a point $(\alpha_{1},\ldots,\alpha_{k})$ in
$(\mathbb{R}/\mathbb{Z})^{k}$. Hence,

$(s_{1}A_{i}-s_{1}A_{i}+s_{2}A_{i})/s_{1}\supset\left\\{w\in\mathbb{Z}:\left|e\left(\alpha_{j}w\right)-1\right|<\frac{\eta}{2}\quad\forall
j=1,\ldots,k\right\\}.$

This implies that

$s_{1}A_{i}-s_{1}A_{i}+s_{2}A_{i}\supset\left\\{n\in\mathbb{Z}:\left|e\left(\frac{\alpha_{j}n}{s_{1}}\right)-1\right|<\frac{\eta}{2}\quad\forall
j=1,\ldots,k\right\\}\cap s_{1}\mathbb{Z}.$

Since $s_{1}\mathbb{Z}$ is a Bohr set and the intersection of two Bohr sets is
a Bohr set, our proof finishes. ∎

###### Proof of Theorem 1.4(b).

We proceed similarly to part (a), using Proposition 5.2 instead of Proposition
5.1. Let $\mathcal{P}$ be an arbitrary partition
$\mathbb{Z}=\bigcup_{i=1}^{r}A_{i}$. Let $i$ be given by Proposition 5.2. Let
$s\in\mathbb{N}$ be arbitrary. Then there is an infinite set
$\mathcal{N}_{s}\subset\mathbb{N}$ such that for any $N\in\mathcal{N}_{s}$, we
have

$\sum_{|x|,|y|\leq
N}1_{A_{i}^{(N)}}(y)1_{A_{i}^{(N)}}(x)1_{A_{i}^{(N)}}(x+sy)\geq
c(s,\mathcal{P})N^{2}.$ (20)

for some constant $c(s,\mathcal{P})>0$ independent of $N$. Note that

$w=(w+x)-(w+sy)+sy.$

The rest is identical to part (a). ∎

###  5.1. Sumsets in $\mathbb{Z}[i]$

Even though the corresponding tori in the cases of $\mathbb{Z}[i]$ and
${\mathbb{F}_{q}}[t]$ are slightly different from $\mathbb{Z}$, the general
approaches are very similar. Therefore, we will be brief and highlight only
the differences.

The following proposition is needed for the proof of Theorem 1.7(b,c). We omit
its proof since it is identical to the ones of Propositions 5.1 and 5.2.

######  Proposition 5.3.

  1. (a)

Let $b,a_{1},\ldots,a_{k}\in\mathbb{Z}[i]$ and $r>0$. There is a constant
$c=c(b,a_{1},\ldots,a_{k},r)$ $>0$ such that the following holds: For $N$
sufficiently large, if $[-N,N]^{2}=\bigcup_{j=1}^{r}A_{j}$, then for some
$1\leq j\leq r$, we have

$\sum_{x,y\in[-N,N]^{2}}1_{A_{j}}(by)1_{A_{j}}(x)1_{A_{j}}(x+a_{1}y)\cdots
1_{A_{i}}(x+a_{k}y)\geq cN^{4}.$

  2. (b)

Let $\mathcal{P}$ denote an arbitrary partition
$\mathbb{Z}[i]=\bigcup_{j=1}^{r}A_{j}$. Then there exists some $1\leq j\leq r$
with the following property: For each $\ell\in\mathbb{Z}[i]$, there is a
constant $c(\ell,\mathcal{P})$ such that

$\sum_{x,y\in[N]^{2}}1_{A_{j}}(y)1_{A_{j}}(x)1_{A_{j}}(x+\ell y)\geq
c(\ell,\mathcal{P})N^{4}.$

for infinitely many $N\in\mathbb{N}$.

###### Proof of Theorem 1.7 (a).

Suppose $A\subset\mathbb{Z}[i]$ has $\overline{d}(A)=\delta>0$. Then for
infinitely many $N$, we have $|A^{(N)}|\geq\delta N^{2}$, where
$A^{(N)}=A\cap[-N,N]^{2}$.

Let $N^{\prime}=2(|s_{1}|+|s_{2}|+|s_{3}|)N+1$. We identify
$[-\frac{N^{\prime}-1}{2},\frac{N^{\prime}-1}{2}]^{2}$ with
$\mathbb{Z}_{N^{\prime}}\times\mathbb{Z}_{N^{\prime}}$. By Theorem 1.6, the
set $s_{1}A+s_{2}A+s_{3}A$ contains a Bohr set in
$\mathbb{Z}_{N^{\prime}}\times\mathbb{Z}_{N^{\prime}}$, which is of the form

$\left\\{(w,v)\in\mathbb{Z}_{N^{\prime}}\times\mathbb{Z}_{N^{\prime}}:\left|e\left(\frac{wx_{j}+vy_{j}}{N^{\prime}}\right)-1\right|<\eta\quad\forall
j=1,\ldots,k\right\\}$

for some
$x_{1},\ldots,x_{k},y_{1},\ldots,y_{k}\in[-\frac{N^{\prime}-1}{2},\frac{N^{\prime}-1}{2}]$,
where $k$ and $\eta$ depend only on $\delta$ and $s_{1},s_{2},s_{3}$.

If $(w,v)$ is in the Bohr set above and $|w|,|v|\leq N$, then there exist
$a,a^{\prime},a^{\prime\prime}\in A^{(N)}$ such that

$(w,v)=s_{1}a+s_{2}a^{\prime}+s_{3}a^{\prime\prime},$

where the equality is in $\mathbb{Z}[i]$ and not just in
$\mathbb{Z}_{N^{\prime}}\times\mathbb{Z}_{N^{\prime}}$. Hence,

$s_{1}A+s_{2}A+s_{3}A\supset[-N,N]^{2}\cap\left\\{(w,v)\in\mathbb{Z}[i]:\left|e\left(\frac{wx_{j}+vy_{j}}{N^{\prime}}\right)-1\right|<\eta\quad\forall
j=1,\ldots,k\right\\}.$

Letting $N$ go to infinity along some subsequence, we have that

$s_{1}A+s_{2}A+s_{3}A\supset\left\\{(w,v)\in\mathbb{Z}[i]:\left|e\left(w\alpha_{j}+v\beta_{j}\right)-1\right|<\frac{\eta}{2}\quad\forall
j=1,\ldots,k\right\\},$

where $(\alpha_{1},\ldots,\alpha_{k},\beta_{1},\ldots,\beta_{k})$ is a limit
point of
$(\frac{x_{1}}{N^{\prime}},\ldots,\frac{x_{k}}{N^{\prime}},\frac{y_{1}}{N^{\prime}},\ldots,\frac{y_{k}}{N^{\prime}})$,
and we are done. ∎

###### Proof of Theorem 1.7(b).

Using Proposition 5.3(a) and arguing similarly to the proof of Theorem 1.4(a),
we see that for some $1\leq j\leq r$, for infinitely many $N$, we have

$(s_{1}A_{j}-s_{1}A_{j}+s_{2}A_{j})/s_{1}\supset[-N,N]^{2}\cap\left\\{(w,v)\in\mathbb{Z}[i]:\left|e\left(\frac{wx_{j}+vy_{j}}{N^{\prime}}\right)-1\right|<\eta\quad\forall
j=1,\ldots,k\right\\}$

Letting $N$ go to infinity, we have

$(s_{1}A_{j}-s_{1}A_{j}+s_{2}A_{j})/s_{1}\supset\left\\{(w,v)\in\mathbb{Z}[i]:\left|e\left(w\alpha_{j}+v\beta_{j}\right)-1\right|<\frac{\eta}{2}\quad\forall
j=1,\ldots,k\right\\},$

where $(\alpha_{1},\ldots,\alpha_{k},\beta_{1},\ldots,\beta_{k})$ is a limit
point of
$(\frac{x_{1}}{N^{\prime}},\ldots,\frac{x_{k}}{N^{\prime}},\frac{y_{1}}{N^{\prime}},\ldots,\frac{y_{k}}{N^{\prime}})$.
Note that

$w\alpha_{j}+v\beta_{j}=\Re((w+iv)(\alpha_{j}-i\beta_{j}))$

and hence,

$s_{1}A-s_{1}A+s_{2}A\supset\left\\{z\in\mathbb{Z}[i]:\left|e\left(\Re\left(z\frac{\alpha_{j}-i\beta_{j}}{s_{1}}\right)\right)-1\right|<\frac{\eta}{2}\quad\forall
j=1,\ldots,k\right\\}\cap s_{1}\mathbb{Z}[i],$

which is a Bohr set by Lemma 2.2. ∎

The proof of Theorem 1.7(c) is similar to part (b), using Proposition 5.3(b)
instead of Proposition 5.3(a).

###  5.2. Sumsets in ${\mathbb{F}_{q}}[t]$

Let $p$ be a prime and $q$ be a power of $p$. First, let us introduce some
standard facts about ${\mathbb{F}_{q}}[t]$. Let
${\mathbb{K}}={\mathbb{F}_{q}}(t)$ be the field of fractions of
${\mathbb{F}_{q}}[t]$. For $f/g\in{\mathbb{K}}$ we define
$|f/g|=q^{\deg(f)-\deg(g)}$ and $|0|=0$. The completion of ${\mathbb{K}}$ with
respect to $|\cdot|$ is
${\mathbb{K}_{\infty}}=\mathbb{F}_{q}((\frac{1}{t}))=\left\\{\sum_{i=-\infty}^{n}a_{i}t^{i}:a_{i}\in{\mathbb{F}_{q}},n\in\mathbb{Z}\right\\}$.
Let
$\mathbb{T}_{q}=\left\\{\sum_{i=-\infty}^{-1}a_{i}t^{i}:a_{i}\in{\mathbb{F}_{q}}\right\\}$.
Then ${\mathbb{F}_{q}}[t],{\mathbb{K}},{\mathbb{K}_{\infty}},\mathbb{T}_{q}$
are the analogs of $\mathbb{Z},\mathbb{Q},\mathbb{R}$ and
$\mathbb{R}/\mathbb{Z}$, respectively.

For $x\in{\mathbb{F}_{q}}$, we write
$e_{q}(x)=e\left(\frac{\textup{Tr}(x)}{p}\right)$, where
$\textup{Tr}:{\mathbb{F}_{q}}\rightarrow{\mathbb{F}_{p}}$ is the trace
map.444That is, $\textup{Tr}(x)$ is the trace of the $\mathbb{F}_{p}$-linear
map $y\mapsto xy$ from ${\mathbb{F}_{q}}$ to ${\mathbb{F}_{q}}$, when
${\mathbb{F}_{q}}$ is viewed as a $\mathbb{F}_{p}$-vector space. In
particular, $\textup{Tr}(x)\in\mathbb{F}_{p}$. It can be checked that
$x\mapsto e_{q}(ax)$ (where $a\in{\mathbb{F}_{q}}$) are all the additive
characters of ${\mathbb{F}_{q}}$.

If $\alpha=\sum_{i=-\infty}^{n}a_{i}t^{i}\in{\mathbb{K}_{\infty}}$, we write
$(\alpha)_{-1}=a_{-1}$ and $E(\alpha)=e_{q}(a_{-1})$. It can be checked that
$f\mapsto E(f\alpha)$, where $\alpha\in\mathbb{T}_{q}$, are all the continuous
characters of ${\mathbb{F}_{q}}[t]$. This also shows that $\mathbb{T}_{q}$ is
the dual of ${\mathbb{F}_{q}}[t]$.

Any Bohr set $B$ in ${\mathbb{F}_{q}}[t]$ is of the form

$B=\left\\{f\in{\mathbb{F}_{q}}[t]:\left|E(f\alpha_{i})-1\right|<\eta\textup{
for }i=1,\ldots,k\right\\},$

where $\alpha_{1},\ldots,\alpha_{k}\in\mathbb{T}_{q}$. If $\eta<|e(1/p)-1|$
then

$B=\left\\{f\in{\mathbb{F}_{q}}[t]:\textup{Tr}((f\alpha_{i})_{-1})=0\textup{
for }i=1,\ldots,k\right\\}.$

This is an ${\mathbb{F}_{p}}$-subspace and not necessarily an
${\mathbb{F}_{q}}$-subspace. However, it contains the
${\mathbb{F}_{q}}$-subspace

$\left\\{f\in{\mathbb{F}_{q}}[t]:(f\alpha_{i})_{-1}=0\textup{ for
}i=1,\ldots,k\right\\}.$

We write $G_{N}=\\{f\in{\mathbb{F}_{q}}[t]:\deg(f)<N\\}$. For a set
$A\subset{\mathbb{F}_{q}}[t]$, we write $A^{(N)}$ for $A\cap G_{N}$. Moreover,
for each $N$, we fix a polynomial $P_{N}\in{\mathbb{F}_{q}}[t]$ of degree $N$.
Then $G_{N}\cong{\mathbb{F}_{q}}[t]/(P_{N})$. While $G_{N}$ is already a
group, we work with ${\mathbb{F}_{q}}[t]/(P_{N})$ since the multiplication
$f\mapsto sf$ is a homomorphism on the latter.

Using the same arguments as in Propositions 5.1 and 5.2, we can prove the
following:

######  Proposition 5.4.

  1. (a)

Let $b,a_{1},\ldots,a_{k}\in{\mathbb{F}_{q}}[t]$ and $r>0$. There is a number
$c=c(q,b,a_{1},\ldots,a_{k},r)>0$ such that the following holds. For $N$
sufficiently large, if $G_{N}=\bigcup_{j=1}^{r}A_{i}$, then for some $1\leq
i\leq r$, we have

$\sum_{x,y\in G_{N}}1_{A_{i}}(by)1_{A_{i}}(x)1_{A_{i}}(x+a_{1}y)\cdots
1_{A_{i}}(x+a_{k}y)\geq cq^{2N}.$

  2. (b)

Let $\mathcal{P}$ denote an arbitrary partition
${\mathbb{F}_{q}}[t]=\bigcup_{i=1}^{r}A_{i}$. Then there exists some $1\leq
i\leq r$ with the following property: For each $\ell\in{\mathbb{F}_{q}}[t]$,
there is a constant $c(\ell,\mathcal{P})$ such that

$\sum_{x,y\in{\mathbb{F}_{q}}[t]}1_{A_{i}}(y)1_{A_{i}}(x)1_{A_{i}}(x+\ell
y)\geq c(\ell,\mathcal{P})q^{2N}$

for infinitely many $N\in\mathbb{N}$.

###### Proof of Theorem 1.8.

We will sketch the proof of Theorem 1.8(b). Parts (a) and (c) can be proved
along the same lines.

Let ${\mathbb{F}_{q}}[t]=\bigcup_{i=1}^{r}A_{i}$ be an arbitrary partition and
$s_{1},s_{2}\in{\mathbb{F}_{q}}[t]\setminus\\{0\\}$. By Proposition 5.4(a), we
know that there exist $1\leq i\leq r$ and an infinite set $\mathcal{N}$ such
that

$\sum_{x,y\in
G_{N}}1_{A_{i}^{(N)}}(s_{1}y)1_{A_{i}^{(N)}}(x)1_{A_{i}^{(N)}}(x+s_{2}y)\gg
q^{2N}$ (21)

for each $N\in\mathcal{N}$.

Let $N^{\prime}=\max(\deg s_{1},\deg s_{2})+N$. We identify $G_{N^{\prime}}$
with ${\mathbb{F}_{q}}[t]/(P_{N^{\prime}})$. Arguing similarly to the proof of
Theorem 1.4(a) and using (21), we find that

$(s_{1}A_{i}-s_{1}A_{i}+s_{2}A_{i})/s_{1}\supset
G_{N}\cap\left\\{w\in{\mathbb{F}_{q}}[t]:(w\frac{x_{i}}{P_{N^{\prime}}})_{-1}=0\quad\forall
j=1\ldots,k\right\\},$

for some $x_{1},\ldots,x_{k}\in G_{N^{\prime}}$.

Letting $N\rightarrow\infty$ and using compactness of $\mathbb{T}_{q}$, we
have

$(s_{1}A_{i}-s_{1}A_{i}+s_{2}A_{i})/s_{1}\supset\left\\{w\in{\mathbb{F}_{q}}[t]:(w\alpha_{i})_{-1}=0\quad\forall
j=1,\ldots,k\right\\}$

for some $\alpha_{1},\ldots,\alpha_{k}\in\mathbb{T}_{q}$. Therefore,

$s_{1}A_{i}-s_{1}A_{i}+s_{2}A_{i}\supset\left\\{f\in{\mathbb{F}_{q}}[t]:(f\frac{\alpha_{i}}{s_{1}})_{-1}=0\quad\forall
j=1,\ldots,k\right\\}\cap s_{1}{\mathbb{F}_{q}}[t],$

which is clearly an ${\mathbb{F}_{q}}$-subspace of bounded codimension. ∎

##  6\. Open questions

Theorem 1.4(b) says that in any partition $\mathbb{Z}=\bigcup_{i=1}^{r}A_{i}$,
there exists an $i\in\\{1,\ldots,r\\}$ such that $A_{i}-A_{i}+sA_{i}$ contains
a Bohr set for every $s\in\mathbb{Z}\setminus\\{0\\}$. Inspired by Katznelson
and Ruzsa’s question, Theorem 1.4(b) naturally gives rise to the following
question.

######  Question 6.1.

Suppose $A\subseteq\mathbb{Z}$ does not contain a Bohr set and
$B\subseteq\mathbb{Z}$ such that $B+sA$ contains a Bohr set for every
$s\in\mathbb{Z}\setminus\\{0\\}$. Must it be true that $B$ contains a Bohr
set?

An positive answer to 6.1 would lead to a resolution of Katznelson-Ruzsa’s
question. However, it is likely that the answer to 6.1 is negative.

As mentioned in the introduction, we do not know whether the commuting
conditions in Theorem 1.5 and Theorem 1.6 can be removed entirely or not. It
is interesting to answer the following.

######  Question 6.2.

Can the commuting conditions in Theorem 1.5 and Theorem 1.6 be removed?

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