ANH N. LE AND THÁI HOÀNG LÊ Abstract. Let G be a compact abelian group and φ1,φ2,φ3 be continuous endomor- phisms on G . Under certain natural assumptions on the φ i 's, we prove the existence of Bohr sets in the sumset φ 1 (A) + φ 2 (A) + φ 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 Z = U e 1 A , there exists an i such that A i — A i + s A i contains a Bohr set for any s ∈ Z \ {0}. The latter is a step toward an open question of Katznelson and Ruzsa.
1. I NTRODUCTION AND STATEMENTS OF RESULTS
Let G be an abelian topological group. For a finite set A of characters (i.e. continuous homomorphisms from G to S 1 := { z ∈ C : | z | = 1} and η > 0, the set B(A; η) := { x ∈ G : | γ( x ) − 1 | < η for any γ ∈ A} is called a Bohr set or a Bohr neighborhood of 0. We refer to η as the radius and |A| as the rank (or dimension) of the Bohr set. The set B(A; η) is also called a Bohr-(|A|, η) set. If A, B ⊂ G, the sumset and difference set of A and B are A±B := {a±b : a ∈ A,b ∈ B}. If c ∈ Z, we define cA := {ca : a ∈ A}. The study of Bohr sets in sumsets started with the following important theorem of Bogolyubov [7] 1.
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 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) \ 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 ⊂ 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 = MZ + 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 ⊂ A + A – A – A – A 1.1. Part it ion results in 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 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 Z = U j = 1 A j , must there exist i ∈ {1,2,...,r} such that A j − A j 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 Question 1.3, as well as its many equivalent formulations. 2  While Question 1.3 remains open at the moment and only some partial results were obtained [ 1 3, 1 9 ], we do have a positive answer when three copies of A i are involved. Theorem 1.4. Let Z = Uj=1 Aj be a partition. (a) For any $1, $2 € Z\{0}, there exists i ∈ {1, 2, . . ,r} such that the set $1 Ai − $1 Ai + $2 Ai contains a  Bohr set whose rank and radius depend only on r and s1,52. (b) There exists i ∈ {1,2,...,r} such that for any s ∈ Z \ {0}, the set Ai − Ai + s Ai contains a Bohr set. Theorem 1.4 highlights the difference between partition and density since, as we men- tioned earlier, there is a set A ⊂ 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 x 1 + s 2x 2 + · · + s ℓ x ℓ = 0 with coefficients in Z \ {0} is partition regular if under any finite partition (or coloring) of Z \ {0}, there exists a monochromatic solution ( x 1 , x 2 , . . , x ℓ ).  Rado's theorem says that the equation s 1 x 1 + s 2 x 2 + . .  + s e x e = 0 is partition regular if and only if { s 1, . , s e } s atisfies the following condition: There exists a nonempty set J ⊂ {1, . . , l} such that ∑ i ∈ J 8 i = 0. Using the facts that (s 1 + . . + s ℓ ) A ⊆ s 1 A + . . + s ℓ A, and a Bohr set must contain 0, Theorem 1.4 (a) implies that for ℓ ≥ 3 and s 1 , . . , s ℓ ∈ Z \ {0}, the following are equivalent: (1) For any partition Z = U i =1 A i , there exists i ∈ {1,...,r} such that s 1 A i + ... + s ℓ A i contains  a Bohr set. (2) { s 1, . . , s e } s atisfies 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 Question 1.3 should be, Theorem 1.4(b) provides evidence that the answer to Question 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 Syndeticity 3 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 ⊂ G with d*(A) > 0, for "many" α ∈ A, the set A + A − A − α 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 Z N , and the statements in 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 asubsequent 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 φ : G → G and the image φ(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 1.1 0 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 μ and let φ 1 , φ 2 : G → G be continuous homomorphisms satisfying (a) φ1,φ2 are commuting, and (b) φ1(G),φ2(G) have finite indices in G Let G = U i = 1 A i  be a partition of G into measurable sets. Then for some 1 ≤ i ≤ r, φι(A i) − φι(A i) + φι(A i) contains a  Bohr-(k,η) set, where k and η depend only on r, [G : φ 1 (G)] and [G : φ 2 (G)]. Remark 1. ● If μ(A) > 0, then A + A – A is not guaranteed to contain a Bohr set. For a counterex- ample, take G = Z 2 N for some large N and A = { a ∈ Z 2 N : a 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 φ 1 or φ2 is an automorphism (see Remark 5). ● The finite index condition on φ 1 ( G ) cannot be removed, by taking for example φ 1 = 0 and φ 2 ( x ) = x . On the other hand, we do not know whether the finite index condition on φ2(G) can be removed. If we let φ2 = 0, then the situation amounts to Question 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 μ and Ø1,Ø2,Ø3 : G → G be continuous homomorphisms satisfying (α) φ1 + φ2 + φ3 = 0, (b) φ1,Φ2,Φ3 are commuting, (c) φ1(G),φ2(G),φ3(G) have finite indices in G Let A ⊆ G be a measurable subset with μ(A) = δ > 0. Then φ 1 (A) + φ 2 (A) + φ 3 (A) contains a Bohr-(k,η) set, where k and η depend only on δ and the indices [G : φ i (G)] (1 ≤ i ≤ 3) Remark 2. ● The condition φ 1 + φ 2 + φ 3 = 0 cannot be removed. For a counterexample, take G = Z N for some large N and A = {1, . . , [ N /10]}. 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 φ 1 ,φ 2 and φ 3 is an automorphism (see Remark 6 ). ● The finite index condition cannot be removed. Indeed, we can take G = F 2 for some large n, φ 1 x) = x, φ 2 (x) = − x,φ 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 ≥ |G|/4 such that any subspace contained in A – A must have codimension ≥ √n. As an application, Theorem 1.6 can be used to obtain an effective version of the afore- mentioned 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 accom- modate homomorphisms in Theorem 1.5 and Theorem 1.6 enables us to generalize Theo- rem 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 ∈ R, we write
and
The next theorem is true for any number field, but we only state for Z[i] for simplicity. Theorem 1.7. Let s1, s2, s3 ∈ Z[i] \ {0} such that s1 + s2 + s3 = 0. (a) If a set A C Z[i] has positive upper density, i.e.
then s 1 A + s 2 A + s A contains a (k,η)-Bohr set in Z [ i ], where k and η depend only on  $1, $2, $3  a nd  δ . (b) If Z[i] = U j=1 Aj, then for some j ∈ {1,2,. . ,r}, s1Aj − s1Aj + s2Aj contains a (k, η - Bohr set in Z [ i ], where k and η depend only on s 1, s 2 and r . (c)  If Z[i] = U j =1 Aj, then there exists j ∈ {1,2,. . ,r} such that Aj − Aj + sAj contains a  Bohr set for any s ∈ Z[i] \ {0} Here, as a group, we identify Z[i] with Z2 Our next result deals with the ring Fq[t] of polynomials over a finite field Fq. Theorem 1.8. Let s1, s2, s3 ∈ Fq[t] \ {0} such that s1 + s2 + s3 = 0. (a) If a set A ⊆ Fq[t] has positive upper density, i e
then S1A + S2A + S3A contains a Fq-vector subspace of f inite codimension in Fq[t], where the codimension depends only on 81,82,83 and δ (b) If Fq[t] = U j_1 A i, then for some i ∈ {1,...,r}, s 1 A i − s 1 A i + s 2 A i contains a Fq- vector subspace of f finite codimension of Fq[t], where the codimension depends only on S1, S2 and δ . (c)  If Fq[t] = U i =1 A i , then there exists i ∈ {1,2,...,r} such that A i − A i + sA i contains an Fq-vector subspace of f nite codimension of Fq[t] for any s ∈ Fq[t] \ {0}. We remark that the special case s1,s2,s3 € Fq \ {0} of Theorem 1.8(a) is essentially Corollary 1.4 in [ 8 ]. 55 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 ∈ 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 + φ(y),x + ψ(y)} (where φ and ψ 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 a belian group with proba bility Haar measure μ and φ,ψ : G → G be continuous homomor- phisms such that [G : φ(G)], [G : ψ(G)] and [G : (φ − ψ)(G)] are finite. Let f : G → [0,1] be a  measurable function with ʃς f dμ = δ > 0. Then for any ε > 0, there exists a constant c 1 > 0 that depends only on δ,ε and the
for some positive constant c2 depending only on δ and the indices above. Theorem 1.9 was proved independently by Berger-Sah-Sawhney-Tidor [4], under the hypothesis that φ,ψ and φ — ψ 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 φ,ψ : G → G  be homomorphisms such that [G : φ(G)], [G : ψ(G)] and [G : (φ – ψ)(G)] are
is synd etic 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 ℓ < 4 be arbitrary and δ > 0 be sufficiently small in terms of l. Let G = F 2 × F 2 where n is sufficiently large, φ(u,v) = (u,0),ψ(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 F? × F? 6 [11, Corollary 1.3] states that there exists a set A ⊂ G of size ≥ δ | G | such that for any u ∈ F  \ {0}, we have #{(a,b) ∈ G : (a,b), (a + u,b), (a,b + u) ∈ A} < δ ℓ |G|. Hence, the set B in Theorem 1.9 has to be {0} × F 2 . 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 μ and let ψ,φ1,. . ,φk : G → G be continuous homomorphisms satisfying: (a) ψ,φι,. . ,φκ are commuting, and (b) ψ(G),φ1(G),...,φφk(G) have finite indices in G Suppose G = U i=1 Ai is a partition of G into measurable sets. Then
for some positive constant c3 depending only on r,k and the indices above. Remark 3. ● By taking ψ = 0, we see that the condition [ G : ψ( G )] is finite cannot be removed. However, we do not know whether the condition [G : φ i (G)] < ∞ is necessary or not. ● Our proof relies heavily on the commuting condition and we do not know if it can be
When ψ and φ are dilations, the configuration {ψ(y),x +φ1(y),...,x +φφk(y)} becomes the  Brauer configuration {y,x,x+y,...,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 homo- morphisms, 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 The- orem 1.9 relies on the regularity lemma (Proposition 4.2), in Theorem 1.6, the dependence of k and η on δ and [ G : φ i ( G )] is of tower type. Likewise, since the proof of Theorem 1.11 uses the Hales-Jewett theorem, the bounds for k and η 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 F n . Indeed, Sanders [28, Theorem A.1] obtained a near optimal bound for Bogolyubov's theorem in Fn 77 Out line 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 de- voted 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 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 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,...,N }. If A and B are two quantities, we write A = O(B) or A ≪ B if there is a constant C such that | A | ≤ CB. We write e(x) for e2πix Throughout this paper, G is a Hausdorff compact abelian group with probability Haar measure μ and Γ is the dual of G , written additively. The relevance of homomorphisms is that if γ ∈ Γ and φ : G → G is a continuous homomorphism, then γ φ is also an element of  Γ If f : G → C is a function, for t ∈ G we define the function ft(x) = f(x + t). For f ∈ L 1 (G), the Fourier transform of f is the function
For f,g ∈ L 2 (G), we then have Parseval's formula
and Plancherel's formula 2.2. Bohr sets. For A,A 1,A ⊆ Γ and η 1,m > 0, it follows from the definition of Bohr sets that
and
Lemma  2.1.  Suppose f 1, . . , f k ∈ L ∞ (G), ∥ f i ∥ ∞ ≤ 1 for all i = 1, . . ,k  . ,k  . ,φ k be homomorphisms G → G. Then for any η > 0, the set
80 Proof. Note that if ‖ f i , φ i ( t ) || ∞ ≥ η, then for some γ ∈ Γ,
Next lemma is needed in 5. Lemma 2.2. Let H be a locally compact abelian group, K be a closed subgroup of finite ind ex m . Then K is a Bohr-(m, le(1/m) — 1 ) set in H Proof. Let X1, . . , Xm be all characters on H/K . For any x ∈ H/K and 1 ≤ i ≤ m, we have | x i ( x )| m  = 1, so either χ i ( x )  = 1 or | x i ( x ) − 1 | ≥ |e(1/m) − 1 . If χ i ( x ) = 1 for all i then x = 0. Hence {0} = B(x1, . . , Xm; |e(1/m) – 1|). The characters χ i lift to characters χ i on H by χ i (h) = χ i (h + K). Therefore,
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 μ and let A ⊆ G of positive measure. Then A − A + A − A contains a Bohr-(k,n) set where k,n) depends only on µ(A). 2.3. Kernels. A kernel on G is a non-negative continuous function that satisfies f c K dμ = 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 ‖ K ‖ 1 to denote Σ γεг | K (γ) . Lemma 2.4 (cf. [2, Lemma 4.3]). Given a finite set Λ ⊂ Γ and η ∈ (0,1/2], there exists a kernel K satisfying the following: (1) K ≥ 0, K ≥ 0 and f ς K dμ = ||K ||1 = 1, (2) || K || 1 = || K ||∞ ≤ 1/(C o η)|^| for some absolute constant C > 0, and (3) K vanishes outside the Bohr set B(A; η) Co ns equ ently,
We remark that the bound (5) can also be obtained from an elementary covering argu- ment (see [31]). 9 Proof. First, for each λ ∈ Λ, there exists a kernel K λ : G → [0,00) satisfying the following properties: (1) ||K λ ||1 = 1, (2) K λ ≥ 0, (3) K λ is supported on B ( λ } ; η ) = { x ∈ G : |λ( x ) − 1| < η }, (4) ‖ K λ ‖ ∞ ＝ K λ (0) ≤ 1/(C o η)  for some absolute constant C 0. Indeed, let B = B({λ}; f ) and let K λ = μ{B] * μ{B].  Clearly the first and second properties are satisfied.  Additionally, K x is s supported on B({\}; f ) + B({\}; f ) ⊂ B({\}; f ) Concerning the last property, we have for every x E G,
and so
Therefore, ‖ K λ ‖ ∞ ＝ K λ (0) ＝ π ( B )  Since λ is continuous, its image λ( G ) is a closed subgroup of S 1 = { z ∈ C : | z | = 1}, and so it is either S 1 or { z ∈ C : z 9 = 1} for some q ∈ N. Since λ is a homomorphism, it is measure-preserving (see Lemma 2.7 below). Hence μ(B) is equal to the normalized Haar measure of the set in the group λ(G).  In either case, where λ(G) = S 1 or {z ∈ C : |z|9 = 1}, we find that μ(B) ≥ Coη for some absolute constant C 0. Therefore, ‖ K λ ‖ ∞ ≤ 1/(C oη) We now define
It follows that K ≥ 0 and K is supported on B(A;n). Repeatedly using the fact that Likewise, since ƒɡ(γ) = Σλεг ƒ(λ)ğ(γ − λ) for all f,g ∈ L ∞(G), we have K ≥ 0. K λ (0) = || K λ || 1 = 1, we have || K || 1 = K (0) ≥ 1. Upon defining we obtain the desired kernel.
2.4. Homomorphisms. We will often make use of the following facts about homomor- phisms G → G Lemma 2.5. Let φ : G → G be a continuous homomorphism such that [G : φ(G)] = m is finite. Then for any γ ∈ Γ, there are at most m elements χ ∈ Γ such that γ = χ φ Proof. It is easy to see that for each γ ∈ Γ, the set S γ := {χ ∈ Γ : γ = χ ◦ φ} is either empty, or a coset of the group S . On the other hand, S 0 is the annihilator of the group φ( G ), so by [25, Theorem 2.1.2], it is isomorphic to G/φ( G ), and hence has cardinality □ m. Lemma 2.6. Let φ,ψ : G → G be homomorphisms such that [ G : φ(G)] = m and [G : ψ(G)] = ℓ are finite. Then [G : φ(ψ(G))] ≤ mℓ is finite. Proof. We have [G : φ(φ(G))] = [G : φ(G)][φ(G) : φ(φ(G))].  It suffices to show that [φ(G) : φ(ψ(G))] ≤ ℓ Let x 1 + ψ ( G ), . . . , x ℓ + ψ ( G ) be all cosets of ψ ( G ) in G. Then φ ( x 1 )+φ ( ψ ( G )), . . . , φ ( x ℓ )+ φ(ψ(G)) are all cosets of φ(ψ(G)) in ψ(G) (these are not necessarily distinct, so the actual □ number of cosets may be less than ℓ ), proving the desired claim. Lemma 2.7. Let G,H be compact abelian groups and μ,V be the normalized Haar measures of G and H , respectively. Suppose φ : G → H is a continuous surjective homomorphism. Then φ ∗ μ = ν (i.e. ν (B) = μ (φ − 1 (B)) for any Borel set B ⊂ H). Proof. Let ν 0 = φ ∗ μ . By the uniqueness of the normalized Haar measure, it suffices to show that ν 0 is a translation-invariant probability measure on H . First, ν 0 is a probability measure because ν 0 (H) = μ(φ− 1 (H)) = μ(G) = 1. Now let B ⊂ H be a Borel set and h 0 ∈ H be arbitrary. Since φ is surjective, there exists g 0 ∈ G such that φ(g 0 ) = h 0 . For any g ∈ φ−1 (B + ho), we have
and so φ− 1 ( B + h 0 ) = φ− 1 ( B ) + go. Since μ is translation-invariant on G , it follows that
Thus ν 0 is translation-invariant on H and so ν 0 = ν .
Lemma 2.8. Let φ : G → G be a continuous homomorphism such that [G : φ(G)] = m is finite. Then for any measurable set A ⊂ G, we have
and
Consequently, if f ∈ L 1 (G) is nonnegative, then
Proof. First, since φ is continuous and G is compact, φ( G ) is a compact subgroup of G. Since G is Hausdorff, φ( G ) is closed. In other words, φ( G ) is a closed subgroup of G. For each Borel set B ⊂ φ(G), λ(B) = mμ(B) defines a probability measure on φ(G). Since this measure is translation invariant, it is equal to the normalized Haar measure on φ(G). By Lemma 2.7, λ = φ ∗ μ . This means that for any Borel set B ⊂ φ(G), we have μ(φ−1(B)) = mμ(B). Let A be any Borel set in G. Since A ⊂ φ− 1 (φ(A)), we have hence μ(A) ≤ μ(φ− 1 (φ(A)) = mμ(φ(A)), and the first assertion is proved. Applying the first assertion to the set φ− 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 ∑ n =1 c i 1 A t for Borel sets A i and c i ≥ 0. The next lemmas deal with images and preimages of Bohr sets under homomorphisms. Lemma 2.9. Let B ⊂ G be a Bohr-(k,η) set and φ : G → G be a continuous homomor- phism. Then φ-1(B) is also a Bohr-(k,η) set. Proof. If B = {x ∈ G : | γ i (x) − 1| < η for i = 1, . . , k } is a Bohr -(k,η) set, then φ − 1 (B) = { x ∈ G : | γ i o φ ( x ) − 1 | < η for i = 1, . . . , k } is also a Bohr-( k , η )-set. ● The next lemma is more surprising. Lemma 2.10 (cf.  Griesmer [ 18 , Lemma 1.7 ]). Let B ⊂ G be a Bohr-(k,η) set and φ : G → G  be a continuous homomorphism such that [G : φ(G)] = m < ∞. Then φ(B) contains a  Bohr-(k',n') set, where k',n' depend on k,n and m. Proof. Suppose B = {x ∈ G : | γ i (x) − 1 | < η for 1 ≤ i ≤ k } where γ i ∈ Γ . Then A = { x ∈ G : | γ i ( x ) − 1 | < η / 4 for 1 ≤ i ≤ k } satisfies A – A + A – A ⊆ B.  The bound ( 5 ) implies that μ ( A ) ≥ (Con/4) k for some absolute constant C 0 > 0. Therefore, by Lemma 2.3, the In view of Lemma 2.8, μ(φ(Α)) ≥ μ(Α)/m ≥ [Con]* set φ(B)ニφ(A) − φ(A) + φ(A) − φ(A) is a Bohr-(k',η') set where k',η' depend only on μ(φ(Α)), which is bounded below by (Con) k口
2.5. Counting lemmas.
Lemma 2.11 (cf. [8, Lemma 2]). Let φ,ψ : G → G be continuous homomorphisms such that φ(G),ψ(G) have finite indices in G.  Then for fi,f2,f3 ∈ L ∞(G) and K ∈ L 1 (G) such that K ∈ L 1 (Γ), we have
where the implied constant depends only on the indices of φ(G) and ψ(G) in G. Proof. Since linear combinations of characters are dense in L 2 ( G ), without loss of gener- ality, we can assume f1,f2,f3 and K are equal to their Fourier series. For x ∈ G, write g(x) = f G f 2(x + φ(y))f 3(x + ψ(y))K (y) dμ(y) By Plancherel's formula,
Consequently
and
Therefore, it suffices to show that for each 70 ∈ Γ , we have
By the Cauchy-Schwarz inequality and the Plancherel's formula, the left hand side is at most
In (9), we use the fact that for each ξ ∈ Γ , there are at most [ G : ψ( G )] values of γ 3 such that γ3 οψ = ξ. Likewise, in ( 10 ), we use the fact that for each ξ ∈ Γ, there are at most [G : φ(G)] values of γ 2 such that γ 2 Φ = ξ. 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 coun- terexample, we let φ(x) = x,ψ(x) = 2x and G = F ζ for some large k. Let n = |G| = 2k. For each i = 1, . . , n, define
● f 2 ( γ i ) = a i , where a i ≥ 0. ●  K(Yi) = bi, where bi ≥ 0. Then (8) says that
This is false by taking α 1 = b 1 = 1 and α i = b i = 0 for i̸= 1. While the previous lemma involves the configuration x, x+φ(y), x+ψ(y), the next one is concerned with x,x + φ(y) and ψ(y). Its proof is almost identical and so we only highlight the differences. Lemma 2.12.  Let φ,ψ : G → G  be continuous homomorphism such that φ(G),ψ(G) have finite indices in G. Then for f1, f2, f3 ∈ L ∞ (G), we have
where the implicit constant depends only on the indices of φ(G) and ψ(G) in G. Proof. Similar to the proof of Lemma 2.11, without loss of generality, we can assume f1, f2, f3 are equal to their Fourier series.  For x ∈ G, write g(x) = fG f2(x+φ(y))f3(ψ(y)) dμ(y) and then by Plancherel's formula,
Moreover, we also have
As a consequence,
and so On the other hand, we have
In (12), we use the fact that for each ξ ∈ Γ, there are ≤ [ G : ψ( G )] values of γ 3 such that γ3 ◦ ψ = ξ while (13) follows from the fact that there are ≤ [ G : φ(G)] values of γ2 such ■ that γ2 οφ = ξ
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 ψ,φ 1 , . . . ,φ k : G → G.  We write
(where φ ² is the i-th composition of φ ). For formal variables x 1 , . . . , x n , we write
and we refer to Sm(x1,...,xn) as the Sm,n-set with generators x1,...,xn For an element x = ∑ i=1 ξ i ( x i ) ∈ S m ( x 1 , . . , x n ), by the support of x we mean the set { i ∈ [n] : ξ i̸= 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 Sm(x1,...,xn), there is a monochromatic configuration
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 15 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 , . . , x N ) : x i ∈ [t]} is called a cube of dimension N over t elements. Let [N] = A 0 U A 1 . . . A m be any disjoint partition of [N], where A i ≠ ∅ for i ≠ 0 (A 0 may be empty), and f : A 0 → [ t ] be any map. Define a map g : [ t ] m → [ t ] N by assigning to each ( y 1 , . . , y m ) ∈ [ t ] m the element ( x 1 , . . , x N ) ∈ [ t ] N , where
A combinatorial space of dimension m is the image of g for some choice of A 0 , A 1 , . . . , 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 ℓ > 0, there exist n = n(k,ℓ,r) and m = m(k,ℓ,r) such that under any r-coloring of Sm(x1,...,xn), there are elements y1,...,yl ∈ Sm(x1,...,xn) with nonempty and disjoint supports, such that for each i ∈ [ ℓ ], the elements
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 l. When l = 1 the statement is obvious. Suppose the statement is true for l, we will prove it is true for ℓ + 1. Write n' = n(k,ℓ,r),m' = m(k,ℓ,r). We define m = m(k,ℓ,r) := |Φm'+1| + 1, N := HJ(|Φm′+1|,n′;r) and n = n(k,l + 1,r) := 1 + N. Consider an arbitrary r-coloring of S m (x 1 , . . , x n )  An r-coloring of S m (x 1 , . . . , x n ) induces an r-coloring of Φ N ′+1 by assigning to (α1,...,α N ) ∈ (Φ m ′+1) N the color of
Since N = HJ(|Φm'+1|,n';r), there is a disjoint partition
and functions f i ∈ Φ m ′+1 for i ∈ A 0 such that when ξ1,ξ2, . . . ,ξπ ′ range over Φ m ′+1, all the elements
have the same color. Write zj = ∑ i€Aj Φ( xj ) for 1 ≤ j ≤ n ′ , and z n ′ +1 = xn+1 + ∑ i€A 0 fj ( xj ). Then all the zj have nonempty and disjoint supports, and all elements of the form
have the same color By the inductive hypothesis, there exists a sequence y1,...,yl E Sm'(z1,...,zn) having nonempty and disjoint supports such that for each i = 1,..., ℓ, the elements
have the same color. We now set ye+1 = zn'+1. Clearly the elements
Proof of Theorem 3.1. Applying Proposition 3.3 with ℓ = r + 1, we can find a sequence Y1, . . . , Yr+1 satisfying the conclusion of that proposition. Let c(i) be the color of
ψ(Yu), ψ(Yu), ψ(Yu) + φ1(Yu), . . . , ψ(Yv) + φκ(Yu), have the same color, and we are done (with x = ψ( yv ), y = yv ).
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 φ,ψ 1 , . . ,ψ k : G → G are continuous homomorphisms satisfying: (1) ψ(G),φ1(G),...,φk(G) have finite indices in G, and (2) ψ,φι,...,φκ are commuting Suppose G = Uj_1 Ai is a partition of G into measurable sets. Then
for some positive constant C3 depending only on r,k and the aforementioned indices. 17 with Proof. Consider the set Sm ( x 1 , . . , x n ) given by Theorem 3.1, where we now let x 1 , . . , x n vary over G. Note that for any ψ ∈ Φm, we have [G : ψ(G)] < ∞ by Lemma 2.6. Let R be the set of all pairs ( z, t ) where z, t ∈ S m ( x 1 , . . . , x n ) have nonempty and disjoint supports. Suppose G = U i =1 A i  For i ∈ [r], we define
by |U| + |V| applications of Lemma 2.8. Now Theorem 3.1 implies that
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 φ,ψ : G → G be commuting continuous homomorphisms with im- ages having finite indices. Suppose f 1 , . . , f r : G → [0,1] are measurable functions such 18 that ∑ i=1 fi ≥ 1 pointwise. For w ∈ G, define
Then there are c,k,¶ > 0 depending only on r and the indices above such that for some i E [r], the set { w E G : R i ( w ) > C} contains a Bohr-(k,n) set. Proof. For i ∈ [r], let Ai = {x ∈ G : fi(x) ≥ 1/r}. Since ∑ i=1 fi ≥ 1 pointwise, G = U i =1 A i . In light of Theorem 1.11, there exists a constant c depending only on r and the indices and an i ∈ [ r ] such that
It then follows that
On the other hand, by Lemma 2.12, for every w ∈ G, |R i (w) – R i (0)| ≪ ‖ f – f w ‖∞ , where the implicit constant depends only on the indices of φ ( G ),ψ ( G ),ψ ( G ) (φ –ψ )( G ) in G. Hence, there exists a constant c ′ such that R i (w) ≥              if
By Lemma 2.1, the set of such w contains a Bohr-( k, η ) set, where k and η depend only on c ■ Theorem 1.5 is now a special case of the next theorem with ψ 1 = φ 2 and ψ 2 = φ 1. Theorem 3.5. Let G = Uj=1 Aj be a partition into measurable sets. Let φ 1 , φ 2 , ψ 1 , ψ 2 : G → G be continuous homomorphisms satisfying the followings: (1) φ2 ○ ψ2 = φ1 ○ ψ1, (2) ψ1 ◦ψ2 = ψ2 ◦ψ1 , (3) φ1(G),ψ1(G),ψ2(G) have finite indices in G Then for some 1 ≤ i ≤ r, the set φ 1 (A i ) − φ 1 (A i ) + φ 2 (A i ) contains a Bohr-(k,η) set, where k and η depend only on r and the indices of φ 1 ( G ),ψ 1 ( G ),ψ 2 ( G ) in G Proof. Suppose G = U i =1 A i  We apply Proposition 3.4 with f i = 1 A i and ( ψ 1 ,ψ 2 ) in place of ( ψ, φ). Then for some i, the set { w ∈ G : R i ( w ) > c } contains a Bohr -( k, η ) set B. This means that for w ∈ B, there exist x,y ∈ G such that
Since φ 1 (x + w) + φ 2 (ψ 2 y)) − φ 1 (x + ψ 1 (y)) = φ 1 (w), we conclude that φ 1 ( B ) ⊂ φ 1 ( A i ) + φ 2 ( A i ) − φ 1 ( A i ) − φ 1 ( A i ) − Our theorem now follows from □ Lemma 2.10 19 Remark 5. If φ 1 is an automorphism and [ G : φ 2 ] < ∞ then the hypothesis of Theorem 3.5 is also satisfied. Indeed, we let ψ 1 = φ 1 0 φ 2 and ψ 2 = Id. Then the first two conditions of Theorem 3.5 are satisfied. As for the third condition, we have ψ 1 ( G ) = φ − 1 0 φ 2 ( G ), which has finite index in G by Lemma 2.6. Similarly, we see that if Φ 2 is an automorphism and [G : φ 1 ] < ∞, 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 men- tioned 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 R/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 φ,ψ : G → G be continuous homomorphisms such that φ(G),ψ(G) and (φ − ψ)(G) have finite indices in G. For f ∈ L ∞(G), define
where the implicit constant depends only on the aforementioned indices.
Proposition 4.2 (Regularity Lemma). Let f : G → [0,1] be a measurable function with ſς f dμ = δ > 0. Let φ,ψ : G → G be continuous homomorphisms such that φ(G),ψ(G) and (φ−ψ)(G) have finite indices in G . Then for every ∈ > 0, there exist a constant C that depends only on δ,€ and the indices above, a kernel K : G → R≥0, and a decomposition
(5) || fun || ∞ || K || 1 < ε Proof. For t ∈ G, let
Fixing ϵ > 0, we define sequences ηn ∈ (0,1/2), κ n ∈ (0,∞) and finite sets Λ n ⊆ Γ recursively as follows: First set η 0 = 1/2. For n ≥ 0, Lemma 2.1 implies that there exists a set Λ n ∈ Γ with |Λ n | ≤ 12/η 2 such that D( η n ) := { t ∈ G : d( t ) ≤ η n } contains a Bohr set B( Λ n ; η n ). For η ∈ (0,1/2], define ν(η) = (Coη) 12/π 2 where Co is the constant found in Lemma 2.4; in particular, ν(ηη) = (Coηη) 12/ηη ≤ (Coηη) 1 ⁄ ∧ l  Put
In view of Lemma 2.4, for n ≥ 0, there is a kernel K n : G → [0, ∞ ] such that
and K n is supported on B ( Λ n ; η n ) ⊆ D ( η n ) . We define
Claim 1:
Indeed, by construction, K n is supported on D ( η n ); and every t ∈ D ( η n ) satisfies | f ( γ )(1 − γ ( t )) | ≤ ηn for all γ ∈ Γ. Therefore, for all γ ∈ Γ,
Claim 2: Indeed, we have
and the claim is proved. Since ηn+1 ≤ ϵ 2 /(4 κ 2 ), we have
Let M be the smallest integer such that M ≥ 2/ε 2 . Then
From now on, we fix this n. Next consider the expression
We have
Note that
The same estimate holds for ‖ f n − ( f n ) ψ ( t ‖ 2  Hence ‖ I n ( 0 ) − I n ( t ) ‖ 2 κ n d ( t ) for any t E G. Since In(0) = ||ƒn||³ = ||ƒ||³ ≥ δ³, it follows that
Note that d(t) ≤ ϵ/(2кn) for t in the set D(ϵ/(2кn)) and so In(t) ≥ δ 3 — ϵ in this set. 22 Let η = ε/(2 κ n ). In view of Lemma 2.4, there exists a kernel K supported on D( η ) such that ‖K‖∞ ≤ 1/ν(η). We then have
(1) ||K||∞ ≤ 1/ν(ε/(2κπ)) ≤ 1/ν(ε/(2κπ)),
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 → [0,1] be a mensurable function with fG f dµ > δ . Let φ,ψ : G → G be continuous homomorphisms such that [G : φ(G)], [G : ψ(G)] and [G : (φ − ψ)G] are finite. Then for every ∈ > 0, there exists a constant c1 that depends only on δ,€ and the indices a bove such that the set
has measure greater than C1. As a consequence, there exists a constant C2 that depends only on δ and indices of φ(G),ψ(G) such that
Proof. Fix ε > 0 and let constant C , kernel K and the decomposition f = fst + fer + fun be as found in Proposition 4.2. Define
and
difference J ′ ( f ) − J ′ ( f st ) will have 26 terms. The terms that contain f er can be bounded by 4c since for f1,f2,f3 E L ∞ (G),
On the other hand, in view of Lemma 2.11, the terms containing fun are bounded by O(|| Îun||∞|| Î|| 1 ) which is O( ε ) thanks to the properties of the decomposition. Therefore,
where the constant co depends only on the indices of φ( G ),ψ( G ) and (φ − ψ)( G ) in G. 23 Define
and
We then have
Letting c1 = coe/llKlloo, we obtain the first part of the theorem.
In order to prove Theorem 1.6, we need the following proposition. For our future appli- cations, we will state and prove a slightly more general version than what is necessary. Proposition 4.3. Suppose φ,ψ : G → G are continuous homomorphisms such that φ(G),ψ(G),ψ − ψ)(G) have finite indices in G. Let f : G → [0,1] such that SG f dμ = δ > 0. For w ∈ G, define
{w ∈ G : R(w) > c} contains a Bohr-(k,η) set. Proof. By Lemma 2.11, we have
where implicit constant depends only on the indices of φ(G),ψ(G),ψ(G) (φ − ψ)(G) in G. By Theorem 1.9, we know that R(0) > c for some constant c > 0 depending on these indices and δ. It follows that there exists a constant c ′ such that R(w) > c/2 if
Lemma 2.1 implies that the set of such w contains a Bohr-(k,η) set, where k and η depend only on c' We can now formulate and prove our main theorem for sets of positive measure. Theorem 4.4. Let φ1,φ2,φ3,ψ1,ψ2 : G → G be continuous homomorphisms satisfying the following (1) φ1 + φ2 + φ3 = 0, (2) φ1 ◦ψ1 = φ2 ◦ψ2, (3) φ3(G),ψ1(G),ψ2(G),ψ1 +ψ2)(G) have finite indices in G The for any measurable set A ⊂ G, μ(A) = δ > 0, the set φ 1 (A) + φ 2 (A) + φ 3 (A) contains a Bohr-(k,η) set, where k and η depend only on δ and the indices above. Proof. Applying Proposition 4.3 for f = 1 A and ψ 1 ,ψ 2 in place of φ and ψ , we see that there exists a Bohr-(k,η) set B such that for all w ∈ B, there are x,y ∈ G such that x + w, x + ψ1(y) and x − ψ2(y) ∈ A. Note that φ3(x + w) + φ1(x + ψ1(y)) + φ2(x − ψ2(y)) = φ3(w) and so that φ 1 (A) + φ 2 (A) + φ 3 (A)ニφ 3 (B). Our theorem then follows from Lemma 2.10. □ Theorem 1.6 is now a special case of Theorem 4.4 when ψ 1 = φ 2 and ψ 2 = φ 1. Remark 6. If φ 1 is an automorphism and [ G : φ 2 ( G )], [ G : φ 3 ( G )] < ∞ then the hypothesis of Theorem 4.4 is also satisfied. Indeed, we let ψ 1 = φ − 1 o φ 2 and ψ 2 = Id. Then the first two conditions of Theorem 4.4 are satisfied. As for the third condition, we have ψ1(G) = φ −1 ◦ φ2(G) and (ψ1 + ψ2)(G) = φ −1 ◦ (φ2 + φ1)(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 ∩ [ N ] n at- urally in Z N , and invoke the counting result (for example, Theorem 1.9) in 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 Z N does not necessarily correspond to a solution in Z . To overcome this issue, Bergelson and Ruzsa embedded A ∩ [ N ] in Z N ′ for some N ′ ≫ N . Then A ∩ [ N ] remains dense in Z N ′ and a solution in Z N found in A ∩ [ N ] is now a solution in 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 Z N , 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] ⊂ ZN, instead of the whole group. Proposition 5.1. For any k,ℓ,r > 0, there is a constant c(k,ℓ,r) > 0 such that the following holds:  For sufficiently large N, if [-N, N] = Uj=1 Aj, then for some 1 < i < r, we have
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 Z[i] and Fq[t] 25 Proof.  We applying Theorem 3.1 with ψ(y) = ℓy and φ j (y) = jy for 1 ≤ j ≤ k. Then there exist m and n depending only on r and k such that for any r-coloring of Sm(x1,...,xn), there are x and y of nonempty and disjoint support such that the configuration
is monochromatic. Note that elements of Sm(x 1 , . . . , x n ) are all linear forms in x 1 , . . . , x n with bounded integer coefficients. We now let x 1 , . . . , x n vary over [−cN,cN] where c is a small con- stant.  Then S m (x 1 , . . , x n ) ⊂ [− N , N ].  Under the partition [− N , N ] = U i = 1 A i , each set S m ( x 1 , . . . , x n ) contains a monochromatic configuration { ℓy, x, x + y, . . . , x + ky}. There are ≫ N n monochromatic configurations arising in this way. However, a configuration may come from many different sets Sm (x 1 , . . , x n ). We will show that the number of tuples ( x 1 , . . . , x n ) giving rise to the same configuration { ℓy, x, x + y, . . . , x + ky} is ≪ N n − 2 Indeed, let I, J be disjoint nonempty subsets of [ n ] such that x and y are linear combi- nations with bounded coefficients of ( x i ) i ∈ I and ( x j ) j ∈ J , respectively. For fixed I and J , the number of choices for ( x i ) i ∈ I is ≪ 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 ∈ J is ≪ N | J − 1 . Since there are finitely many pairs ( I , J ) , we see that the number of ( x 1 , . . . , x n ) that give rise to { ly, x, x + y, . . . , x + ky} is ≪ N n-2  Hence the number of monochromatic configurations in [ N ] is ≫ N 2 , and we are done.口Our next statement is essentially a diagonalization argument. Proposition 5.2. Let P denote an arbitrary partition Z = Uj=1 Aj.  Then there exists some 1 ≤ i ≤ r with the following property: For every l ≥ 0, there is a constant c(l,P) such that
for infinitely many N ∈ N Proof. Invoking Proposition 5.1, for each k ∈ N, there is i = f(k) such that for infinitely many N, we have
Hence there exist an i ∈ {1, . . , r } and an infinite set K such that f(k) = i for all k ∈ K . Let ℓ be arbitrary and pick k ∈ K,k ≥ ℓ. We have, for infinitely many N,
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( ℓ , 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 Z = U i =1 A i be an arbitrary partition and s 1 , s 2 ∈ Z \ {0}. Without loss of generality, we assume s1,s2 > 0. For a set A ⊂ Z and N > 0, we write A(N) to denote AN [—N,N] By Proposition 5.1, there exist i ∈ [r] and an infinite set N such that
for any N ∈ N , where c > 0 is a constant independent of N . Let N ′ be the smallest odd integer greater than (2s1 + s2 + 1) N . We identify Z N ′ with [- N',-1, N',-1]. Then ( 18 ) implies that
for some constant c > 0 independent of N . Define
Then by the same argument as the proof of Proposition 3.4, the set {w ∈ Z Nי: R(w) > 0} contains a Bohr-( k, η ) set in Z N r , where k and η are independent of N . Note that R( w ) > 0, implies there are a, a', a" ∈ A { N } and x, y ∈ Z N , such that
Therefore s1w = s1(x + w) − s1(x + s2y) + s2(s1y) ≡ s1α ′ − s1α ′ + s2α   (mod N ′ ). If | w | ≤ N then this congruence is an equality in Z thanks to the way we choose N ′ and the fact that | a | , | a ′ | , | a ′′ | ≤ N . We have thus proved that, for each N ∈ N , there exist x1, . . . , xk ∈ [ − N ′ 2 − 1 , N ′ 2 − 1 ] such that
Here we are using the notation A/c defined in ( 2 ). As N ∈ N,N → ∞ and by passing to a subsequence if necessary, the sequence (予, . . . ,予) converges to a point ( α 1 , . . , α k ) in ( R / Z ) k . Hence,
This implies that
Since s 1 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 P be an arbitrary partition Z = Uj-1 Aj. Let i be given by Proposition 5.2. Let s ∈ N be arbitrary. Then there is an infinite set N s ⊂ N such that for any N E N 8 , we have
for some constant c(s,P) > 0 independent of N. Note that
The rest is identical to part (a).
5.1. Sumsets in Z[i]. Even though the corresponding tori in the cases of Z[i] and Fq[t are slightly different from 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.
(a) Let b,α1,...,αk ∈ Z [i] and r > 0. There is a constant c = c(b,α1,...,αk,r) > 0 such that the following holds: For N sufficiently large, if [—N,N]2 = Uj=1 Aj, then for
(b) Let P denote an arbitrary partition Z[i] = Uj=1 Aj. Then there exists some 1 ≤ j ≤ r with the following property: For each ℓ ∈ Z [i], there is a constant c(ℓ,P) such that
for infinitely many N ∈ N Proof of Theorem 1.7 (a). Suppose A ⊂ Z [i] has d (A) = δ > 0. Then for infinitely many N, we have | A(N) | ≥ δN 2, where A(N) = A ∩ [−N,N] 2. Let N' = 2(|$1| + |$2| + |$3|)N + 1.  We identify [ − N' = 1, N' = 1]2 with ZN' × ZN'  By Theorem 1.6, the set s 1 A + s 2 A + s 3 A contains a Bohr set in Z N ′ × Z N ′ , which is of the form
for some x 1 , . 81, 82, 83 If (w,v) is in the Bohr set above and | w | , | v | ≤ N, then there exist a, α ′ , α ′′ ∈ A(N) such that
where the equality is in Z [ i ] and not just in Z N , × Z N , Hence, wx + vy s1A+s2A+s3Aつ[−N,N]2n (w,v) ∈ Z[i] : [e N ′ Letting N go to infinity along some subsequence, we have that
∀j = 1, . . . , k } s 1 A + s 2 A + s 3 A ⊃ 〈 (w,v) ∈ Z [i] : |e (wα j + v β j ) − 1 | < 2 |グwhere (α 1 , . . . ,α k , β 1 , . . . , β k ) is a limit point of (み, . . ,み,み, . . ,み), 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 ≤ j ≤ r, for infinitely many N, we have ( 81 A j − 81 A j + 82 A j )/ 81つ[− N, N ]²∩ ( w, v ) ∈ Z [ i ] : [e Letting N go to infinity, we have ( 81 Aj − s 1 Aj + s 2 Aj )/ s 1つ{ ( w, v ) ∈ Z [i] : |e ( wαj + v βj ) − 1 | < 1 q where (α 1 , . . . ,α κ , β 1 , . . . ,β k ) is a limit point of (予, . . ,予,予, . . . ,予)  Note that wα j + vβ j = R((w + iv)(α j − iβ j )) and hence,
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 F g [ t ]. Let p be a prime and q be a power of p. First, let us introduce some standard facts about F g [ ℓ ]. Let K = F g ( t ) be the field of fractions of F g [ t ]. For f/g ∈ K we define | f/g | = qdeg(f)-deg(g) and |0| = 0. The completion of K with respect to | · | is K ∞ = F q (( f )) = { ∑ i = −∞ α i t i : α i ∈ F q , n ∈ Z } . Let T q = { ∑ i = −∞ α i t i : α i ∈ F q Then Fq[t], K, K ∞ , Tq are the analogs of Z, Q, R and R/Z, respectively. For x ∈ F g, we write e g (x) = e ( π (x) , where Tr : F g → F p is the trace map. 4 It can be checked that x → e q (ax) (where a ∈ Fq) are all the additive characters of Fq. If α = γ n i=−∞ α i t i ∈ K∞, we write (α)−1 = α−1 and E(α) = e q (α−1). It can be checked that f ↔ E(fα), where α ∈ T g , are all the continuous characters of F g [t]. This also shows that T q is the dual of F q [t] Any Bohr set B in Fq[t] is of the form B = {f ∈ Fq[t] : |E(fai) − 1| < η for i = 1, . . . , k} , where α₁, . . . , αk ∈ T g  If η < |e(1/p) − 1| then B = {f ∈ Fg[t] : Tr((fai)-1) = 0 for i = 1, . . . , k} . This is an F p -subspace and not necessarily an F q -subspace. However, it contains the F q - subspace
We write GN = {f E Fq[f] : deg(f) < N}. For a set A ⊂ Fq[f], we write A(N) for A ∩ GN. Moreover, for each N , we fix a polynomial P N ∈ F q [ t ] of degree N . Then G N ∈ F q [ t ]/( P N ). While GN is already a group, we work with Fq[t]/(PN) since the multiplication f → 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. (a) Let b,α1,...,αk ∈ F q [t] and r > 0. There is a number c = c(q,b,α1,...,αk,r) > 0. such that the following holds. For N sufficiently large, if GN = Uj=1 Ai, then for
(b) Let P denote an arbitrary partition Fq[t] = Uj-1 Aj. Then there exists some 1 < i < r with the following property: For each ℓ ∈ Fq[t], there is a constant c(ℓ,P) such that
for infinitely many N ∈ 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 Fq[t] = Uj=1 Aj be an arbitrary partition and s1, s2 ∈ Fq[t] \ {0}. By Proposition 5.4(a), we know that there exist 1 ≤ i ≤ r and an infinite set N such that
for each N E N . Let N ′ = max(deg s 1 , deg s 2 ) + N . We identify G N ′ with Fq [ ℓ] /( P N ′ ). Arguing similarly to the proof of Theorem 1.4(a) and using (21), we find that
for some x1,...,xk E GיאLetting N → ∞ and using compactness of T g , we have ( s 1 A i − s 1 A i + s 2 A i )/ s 1つ{ w ∈ F q [t] : ( w α i )− 1 = 0   ∀ j = 1, . . . , k } for some α1,...,ακ ∈ Tq. Therefore,
which is clearly an Fq-subspace of bounded codimension. 30 ■
6. OPEN QUESTIONS
Theorem 1.4(b) says that in any partition Z = U i =1 A i , there exists an i ∈ {1, . . . ,r} such that A i − A i + sA i contains a Bohr set for every s ∈ Z \ {0}. Inspired by Katznelson and Ruzsa's question, Theorem 1.4 (b) naturally gives rise to the following question. Question 6.1. Suppose A ⊆ Z does not contain a Bohr set and B ⊆ Z such that B + sA contains a Bohr set for every s ∈ Z \ {0}. Must it be true that B contains a Bohr set? An positive answer to Question 6.1 would lead to a resolution of Katznelson-Ruzsa's question. However, it is likely that the answer to Question 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 re moved?