4 Proof of Theorem 2

North-Western European Journal of Mathematics

W N

M

E J

On the sum of dilates in R ^d

Mario Huicochea¹

Received: May 24, 2018/Accepted: November 2, 2020/Online: March 6, 2021

Abstract

LetAbe a nonempty finite subset ofZ^dwhich is not contained in a hyper- plane,q∈Zwith|q|>1 andm∈Zsuch that|q|+ 2d−1≤m≤(|q|+ 2d−1)². In this paper it is shown that

|A+q·A| ≥ m

|q|+ 2d−1

!

|A| −c

wherecdepends only onq, dandm. In particular, takingm= (|q|+ 2d−1)², this results confirms a conjecture of A. Balog and G. Shakan.

Keywords:sumsets, dimension, affine hull.

msc: 11B30, 11B13.

1 Introduction

We denote byZ,NandRthe set of integers, natural numbers and real numbers, respectively; we consider 0<N. In this paper,dwill denote a nonnegative integer.

For any nonempty subsetsAandA⁰ofR^dandq∈R, set A+A⁰:={a+a⁰: a∈A,a⁰∈A⁰}

−A:={−a: a∈A} q·A:={qa: a∈A}.

LetV be aR-vector space. Anaffine subspaceW ofV is a translation of a linear subspaceW⁰ ofV; we write dimW:= dimW⁰. The minimal affine subspace con- tainingAis known as theaffine hulland we will denote it by affA; set dimA:=

dim affA. The canonical basis ofR^d will be denoted by{e₁,e₂, . . . ,e_d}. SetZ^d :=

nP_d

j=1n_je_j: n₁, n₂, . . . , n_d∈Z

oand0:=P_d

i=10e_i.

The study of the sum of dilates has had new and interesting results in the last few years, see Balog and Shakan 2014, Balog and Shakan 2015, Cilleruelo, Hamidoune, and O. 2009, Cilleruelo, M., and Vinuesa 2010, Du, Cao, and Sun 2015,

1CONACyT/UAZ

Fiz Pontiveros 2013, Hamidoune and Rué 2011, Ljujic 2013, Plagne 2011, Shakan 2016. Most of these results deal with the sum of dilations in one-dimensional spaces.

However, A. Balog and G. Shakan proved the following high-dimensional result.

Theorem 1 – LetAbe a nonempty finite subset ofZ^d withdimA=d >1andq∈Z such that|q|>1. Then

|A+q·A| ≥(|q|+d+ 1)|A| −c (1)

wherecis a constant which depends only onqandd.

Proof. See Balog and Shakan 2015, Thm. 3.

Balog and Shakan conjectured that the coefficient|q|+d+ 1 in (1) could be improved to|q|+ 2d−1, see Balog and Shakan 2015, Conj. 1. Furthermore, they gave an example which shows that|q|+ 2d−1 is the best possible. This conjecture is the main motivation of this paper. To state the main result, write for alli, j∈Z,

ci,j:= 4^j4i.

Theorem 2 – LetAbe a nonempty finite subset ofZ^d,q∈Zwith|q|>1andm∈Zsuch that|q|+ 2 dimA−1≤m≤(|q|+ 2 dimA−1)². Then

|A+q·A| ≥ m

|q|+ 2 dimA−1|A| −c_dimA,m.

The particular case dimA=dandm= (|q|+ 2 dimA−1)²confirms the conjecture of Balog and Shakan.

We sketch the content of this paper. In Section 2 we state auxiliary results that will be needed later. In Section 3 we shall study the partitionsA=A₁]A₂such that

(A1+q·A)∩(A2+q·A) =∅. (2)

Also, in Section 3, we study the partitionsA=A₁]A₂which satisfy the stronger condition: for alli, j, i⁰, j⁰∈ {1,2}with (i, j),(i⁰, j⁰), we have that

(A_i+q·A_j)∩(A_i⁰+q·A_j⁰) =∅.

The most important results of Section 3 will be Lemma 7 and Lemma 8, and they will be fundamental tools in the proof of Theorem 2. The proof of Theorem 2 will be done by induction on dimAandm. We will take a partitionA=A1]A2with A1 andA2nonempty satisfying (2), and we will proceed using the hypothesis of induction depending on whether

2. Preliminaries

i) dimA₁<dimAor dimA₂<dimA.

ii) dimA1= dimA2= dimA.

Case ii) follows from the ideas that Balog and Shakan used in Balog and Shakan 2014, Balog and Shakan 2015 and Shakan 2016. Case i) is the one where more work needs to be done. The conclusion of the proof of Theorem 2 is done in Section 4.

2 Preliminaries

In this section we will state some auxiliary results that will be needed in the proof of Theorem 2. We start with a fundamental result.

Theorem 3 – LetA₁andA₂be nonempty finite subsets ofR^d. Then

|A₁+A₂| ≥ |A₁|+|A₂| −1.

Proof. See Grynkiewicz 2013, Thm. 3.1.

For any affine subspaceV ofR^d, we denote byR^d/V the set of equivalence classes with respect to the relationa∼bif{a}+V ={b}+V; we considerR^d/V with its usualR−vector space structure, and we denote byπ_V :R^d→R^d/V the canonical projection. For any subsetAofR^d/V, we have that the dimension of the affine hull ofAis at most|A| −1. This implies easily the following fact.

Remark 1 – LetA1andA2be nonempty finite subsets ofR^d. Then

|π_affA1(A₁+A₂)| ≥1 + dimπ_affA1(aff(A1+A₂))

= 1 + dim aff(A1+A₂)−dim affA1

= 1 + dim(A₁+A2)−dimA₁. We shall need two consequences of Theorem 3.

Corollary 1 – LetA1andA2be nonempty finite subsets ofR^d. Then

|A₁+A₂| ≥ |π_affA1(A₂)|(|A₁| −1) +|A₂| ≥ |π_affA1(A₂)||A₁|.

Proof. Since|πaffA₁(A₂)| ≤ |A2|, it suffices to prove the left-hand side inequality. Set π:=π_affA1andπ(A₂) ={z₁, z₂, . . . , z_n}. For eachi∈ {1,2, . . . , n}, writeB_i:=π⁻¹(z_i)∩A₂. On the one hand, for eachi, j∈ {1,2, . . . , n}withi,j, we have thatB_i andB_j are contained in distinct translations of affA1; thereforeA₁+B_iandA₁+B_jare contained in distinct translations of affA1and hence

|A1+A2|= Xn

i=1

|A1+B_i|. (3)

On the other hand, for eachi∈ {1,2, . . . , n}, Theorem 3 leads to

|A₁+B_i| ≥ |A₁|+|B_i| −1. (4)

Finally

|A1+A2|=

n

X

i=1

|A1+Bi|

by (3)

≥ Xn

i=1

(|A₁| −1) +|B_i|

by (4)

= Xn

i=1

(|A₁| −1) + Xn

i=1

|B_i|

=|π(A2)|(|A1| −1) +|A2|.

Corollary 2 – Let A₁ andA₂ be nonempty finite subsets ofR^d. Setd₁ := dimA₁, d₂:= dimA₂andd₃:= dim(A₁+A₂). Then

|A₁+A₂| ≥(1 +d₃−d₁)|A₁|+ (1 +d₃−d₂)|A₂| −(1 +d₃−d₁)(1 +d₃−d₂).

Proof. The proof is by induction ond₃. Ifd₃= 0, then|A₁+A₂|=|A₁|=|A₂|= 1 and d₁=d₂= 0 so the statement is trivial. We assume from now on thatd₃>0 and that the statement holds for all 0≤d⁰< d₃. Ifd₃=d₁andd₃=d₂, then the statement follows from Theorem 3. It remains to complete the induction whend₁ < d₃ or d₂< d₃; without loss of generality assume thatd₁< d₃. Also, translating if necessary, we assume thatA₁andA₂contain the origin; hence

dim(affA1∩affA2) + dim(affA1+ affA2) = dim affA1+ dim affA2. (5) Setπ :=πaffA₁. Write π(A₂) ={z1, z2, . . . , zn} andBi := π⁻¹(z_i)∩A2 for each i ∈ {1,2, . . . , n}. Takei∈ {1,2, . . . , n}, and note thatB_i is contained in a translation of affA1. Moreover, since B_i ⊆A₂, it is also contained in affA2, and thereforeB_i is contained in a translation of affA1∩affA2. Insomuch as aff(A1+A₂) = affA1+ affA2, we conclude by (5) that

dimB_i≤dim(affA1∩affA2) =d₁+d₂−d₃. (6) SinceBiis contained in a translation of affA1, the affine hull ofA1+Biis a translation of affA1; in particular dimA1+Bi =d1 < d3. Thus we can apply the induction hypothesis on the pair (A₁, B_i), and we obtain that

|A1+Bi| ≥ |A1|+ (1 +d1−dimBi)|Bi| −(1 +d1−dimBi)

(Cont. next page)

2. Preliminaries

≥ |A₁|+ (1 +d₃−d₂)|B_i| −(1 +d₃−d₂).

by (6)

(7) Setπ_i:=π_affBi. Recall that dim(A₁+B_i) =d₁. Hence Remark 1 applied to the pair (B_i, A1) leads to

|π_i(B_i+A₁)| ≥1 +d₁−dimB_i. (8)

Corollary 1 applied to (Bi, A1) leads to

|B_i+A₁| ≥ |π_i(A₁)||B_i|, (9)

and thus

|A₁+B_i| ≥ |π_i(A₁)||B_i|

by (9)

=|π_i(B_i+A₁)||B_i|

≥(1 +d₁−dimB_i)|B_i|

by (8)

≥(1 +d3−d2)|Bi|.

by (6)

(10)

>From Remark 1 applied to the pair (A₁, A₂), n=|π(A₁+A₂)| ≥1 +d₃−d₁.

On the one hand, (7) yields

1+d₃−d₁

X

i=1

|A1+Bi| ≥

(1 +d₃−d₁)|A₁|+ (1 +d₃−d₂)











1+d₃−d₁

X

i=1

|B_i|











−(1 +d₃−d₁)(1 +d₃−d₂). (11) On the other hand, (10) leads to

Xn

i=2+d₃−d₁

|A1+Bi| ≥(1 +d3−d2) Xn

i=2+d₃−d₁

|Bi|. (12)

For eachi, j∈ {1,2, . . . , n}withi,j,A1+Bi andA1+Bj are contained in distinct translations of affA1; in particular they are disjoint and therefore

|A1+A2|= Xn

i=1

|A1+B_i|. (13)

Finally

|A₁+A₂|= Xn

i=1

|A₁+B_i|

by (13)

≥(1 +d₃−d₁)|A₁|+ (1 +d₃−d₂)|A₂|−

(1 +d₃−d₁)(1 +d₃−d₂),

by (11),(12)

and this completes the induction.

For any nonempty subsetsA₁andA₂ofR^d, set

δ(A1, A2) :=











1 if dimA1= dim(A1∪A2) or dimA2= dim(A1∪A2);

0 otherwise.

Lemma 1 – LetA1andA2be nonempty subsets ofR^dsuch thatdim(A1∪A2) =dand q∈Z. Then

dim(A1+q·A2)≥d−1 +δ(A1, A2)

Proof. SinceA1+q·A2 contains a copy ofA1, if dimA1 =d, then d= dimA1 ≤ dim(A1+q·A2). In the same way, if dimA2=d, then dimq·A2=d and thereby d= dimq·A2≤dim(A1+q·A2).

Now we prove that dim(A1+q·A2)≥d−1 for arbitraryA1andA2. Setπ:=πaffA₁. Hence

dimπ(A1+q·A2) = dim(A1+q·A2)−dimA1

dimπ(A₁∪A₂) = dim(A₁∪A₂)−dimA₁. (14) Thus, sinceπis linear, we have that inR^d/affA1

dimπ(A1+q·A2) = dimπ(q·A2) = dimπ(A2). (15) Also, insomuch as|π(A₁)|= 1,

dimπ(A₁∪A₂) = dim(π(A₁)∪π(A₂))≤(dimπ(A₂)) + 1. (16) Then

dim(A₁+q·A₂) = dimπ(A₁+q·A₂) + dimA₁

by (14)

= dimπ(A2) + dimA1

by (15)

(Cont. next page)

2. Preliminaries

≥(dimπ(A₁∪A₂))−1 + dimA₁

by (16)

= (d−dimA1)−1 + dimA1

by (14)

=d−1.

For any affine subspaceV ofR^d, we say thatV isdefined overZif there are linear equations overZsuch thatVis its solution space. Notice that if there isA⊆Z^dsuch that affA=V, thenV is defined overZ. We will need some results of the geometry of numbers. We start with an easy consequence of Cassels 1997, Ch. 1 Cor. 3.

Lemma 2 – LetV be a linear subspace ofR^ddefined overZ. Then there is{f₁,f2, . . . ,fd} a basis ofZ^d such that{f₁,f₂, . . . ,f_dimV}is a basis ofZ^d∩V.

For any nonempty subsetAofZ^d, we denote byhAithe subgroup ofZ^dgenerated byA. We say thatAisZ^d-reducedifhA−Ai=Z^d; if no confusion is possible withd, we simply say thatAisreduced. Note that ifAisZ^d-reduced, then dimA=d. For arbitrary subsetsAofZ^dwith dimA=d,hA−Aiis a sublattice ofZ^d. Thus we get the following fact.

Remark 2 – LetAbe a nonempty finite subset ofZ^d. If dimA=d, there is a bijective affine mapφ:R^d→R^dsuch thatφ(A) is reduced.

Given an ordered basis B={f₁,f2, . . . ,f_d} ofZ^d and an ordered subset B⁰ = {f_i1,fi₂, . . . ,fi_k}ofB, define

π_B,B⁰:R^d−→R^k, π_B,B⁰







 Xd

i=1

z_if_i









= Xk

j=1

z_ije_j.

Note thatπ_B,B⁰(Z^d) =Z^k. Moreover, we have the following trivial fact.

Remark 3 – LetBbe an ordered basis ofZ^d andB⁰be a nonempty ordered subset ofBwithk:=|B⁰|. IfAisZ^d-reduced, thenπ_B,B⁰(A) isZ^k-reduced.

Before we conclude this section, we recall two results of Balog and Shakan. LetAbe a nonempty subset ofZ^d,q∈Zandπ:Z^d→Z^d/q·Z^dthe canonical projection. We say thatAisq-domainifπ(A) =Z^d/q·Z^d.

Lemma 3 – LetAbe a nonempty subset ofZ^d andq∈Zwith|q|>1. IfAisq-domain, then

|A+q·A| ≥

d+|q|^d

|A| −d(d+ 1) 2 |q|^d.

Proof. See Balog and Shakan 2015, Lemma 1.

Lemma 4 – LetAbe a nonempty subset ofZ^d,q∈Zwith|q|>1andπ:Z^d→Z^d/q·Z^d be the canonical projection. Setπ(A) ={z₁, z₂, . . . , z_n}and takeb₁,b₂, . . .b_n∈Asuch that π(b_i) =z_i for eachi∈ {1,2, . . . , n}. WriteB_i:=π⁻¹(z_i)∩AandB⁰_i :=q⁻¹·(B_i− {b_i})for eachi∈ {1,2, . . . , n}. For alli∈ {1,2, . . . , n}, ifB⁰_i is notq-domain, then

|B_i+q·A| ≥ |B_i+q·B_i|+ min

1≤j≤n

|B_j|.

Proof. See Balog and Shakan 2015, Lemma 4.

3 q-weak partitions and q-partitions

LetA₁andA₂be nonempty subsets ofZ^dandq∈Z. We say that (A₁, A₂) is aq-weak pairif

(A1+q·(A1∪A2))∩(A2+q·(A1∪A2)) =∅.

We say that (A₁, A₂) is aq-pairif, for alli, j, i⁰, j⁰∈ {1,2}with (i, j),(i⁰, j⁰), we have that

(A_i+q·Aj)∩(A_i⁰+q·Aj⁰) =∅.

Hence we have that if (A₁, A₂) is aq-pair, then it is aq-weak pair.

Remark 4 – LetB1, B2, . . . , Bnbe nonempty subsets ofZ^d,{1,2. . . , n}=I1]I2a parti- tion withI1andI2nonempty, andq∈Z.

i) Assume that for alli, j∈ {1,2, . . . , n}withi,j,







 Bi+q·

n

[

i=1

Bi









∩







 Bj+q·

n

[

i=1

Bi









=∅.

ThenS

i∈I₁B_i,S

i∈I₂B_i

is aq-weak pair.

ii) Assume that for alli, j, i⁰, j⁰∈ {1,2, . . . , n}with (i, j),(i⁰, j⁰), B_i+q·B_j

∩

B_i⁰+q·B_j⁰

=∅.

ThenS

i∈I₁B_i,S

i∈I₂B_i

is aq-pair.

LetAbe a subset ofZ^d,A=A1]A2a partition andq∈Z. We say thatA=A1]A2is aq-weak partitionif (A1, A2) is aq-weak pair. We say thatA=A1]A2is aq-partition if (A1, A2) is aq-pair.

3.q-weak partitions andq-partitions

Remark 5 – LetBbe an ordered basis ofZ^d,B⁰be a nonempty ordered subset of BandAa subset ofZ^d.

i) Ifπ_B,B⁰(A) =A1]A2is aq-weak partition, then A=

π_B,B^{−1 0}(A₁)∩A ]

π⁻_B,B^{1 0}(A₂)∩A is aq-weak partition.

ii) Ifπ_B,B⁰(A) =A₁]A₂is aq-partition, then A=

π_B,B^{−1 0}(A₁)∩A ]

π⁻_B,B^{1 0}(A₂)∩A is aq-partition.

The next two results study the existence ofq-weak partitions andq-partitions ofA whenAisZ^d-reduced and|A|is small.

Lemma 5 – LetAbe a reduced subset ofZ^d andq∈Zwith|q|>1. If|A| ≤2d, then there is aq-partitionA=A₁]A₂.

Proof. The proof will be done by induction ond. First assume thatd= 1. Since AisZ-reduced,hA−Ai=Z, and therefore|A|>1. Insomuch as|A| ≤2d = 2, we conclude that|A|= 2; writeA={a₁,a₂}. We have thatA={a₁} ] {a₂}is aq-partition and the basis of induction is proven. From now on we assume thatd >1 and that the claim holds for all 1≤d⁰< d. Letπ:Z^d→Z^d/q·Z^dbe the canonical projection and writeπ(A) ={z1, z2, . . . , zn}. SetBi:=π⁻¹(zi)∩Afor eachi∈ {1,2, . . . , n}. Inasmuch as hA−Ai=Z^d, we have thatπ(A)−π(A) generatesZ^d/q·Z^dand thereforen≥2 (since

|q|>1). We deal with two cases.

• Assume that for alli, j, i⁰, j⁰∈ {1,2, . . . , n}with (i, j),(i⁰, j⁰), B_i+q·B_j

∩

B_i⁰+q·B_j⁰

=∅.

Then Remark 4 ii) implies thatA=B₁](A\B₁) is aq-partition.

• Assume that there arei, j, i⁰, j⁰∈ {1,2, . . . , n}with (i, j),(i⁰, j⁰) such that B_i+q·B_j

∩

B_i⁰+q·B_j⁰

,∅. (17)

Sinceπ(Bi+q·Bj) ={zi}andπ(Bi⁰+q·Bj⁰) ={zi⁰}, we get from (17) thati=i⁰. Leta,a⁰∈Bi,b∈Bjandb⁰∈Bj⁰ be such that

a+qb=a⁰+qb⁰. (18)

Insomuch as (i, j),(i⁰, j⁰) andi=i⁰, we notice thata,a⁰ (otherwiseb=b⁰ and therebyj =j⁰). LetV be the linear subspace ofR^d generated bya−a⁰.

Sincea−a⁰∈Z^d\ {0}, note thatV is a 1-dimensional subspace ofR^ddefined overZ. From Lemma 2, there is an ordered basisB:={f₁,f₂, . . . ,f_d}ofZ^dsuch that{f_d}is a basis ofZ^d∩V. Set the ordered setB⁰:={f₁,f₂, . . . ,f_d−1}. On the one hand,π_B,B⁰(A) isZ^d−1-reduced by Remark 3. On the other hand, since a,a⁰, we also have thatb,b⁰. However, (18) implies thata−a⁰,b−b⁰∈V, and hence

π_B,B⁰(a) =π_B,B⁰(a⁰) and π_B,B⁰(b) =π_B,B⁰(b⁰). (19) Sincea,a⁰,b,b⁰, and|A| ≤2d, we get that|π_B,B⁰(A)| ≤2d−2 from (19).

Thenπ_B,B⁰(A)⊆Z^d−1 satisfies the hypothesis of induction, and therefore there is aq-partitionπ_B,B⁰(A) =A⁰₁]A⁰₂. SetA₁:=π⁻_B,B^{1 0}(A⁰₁)∩AandA₂:=

π⁻_B,B^{1 0}(A⁰₂)∩A. Remark 5 ii) implies thatA=A1]A2is aq-partition, and this

concludes the induction.

Lemma 6 – LetAbe a reduced subset ofZ^dandq∈Zwith|q|>1. If|A| ≤2d+ 1, then there is aq-weak partitionA=A₁]A₂such that|A₁|= 1.

Proof. Translating if necessary, we assume that0∈A. Letπ:Z^d→Z^d/q·Z^d be the canonical projection. Since

hAi=hA− {0}i=hA−Ai=Z^d,

we have thatπ(A) generates the groupZ^d/q·Z^d. Insomuch as|q|>1, the group Z^d/q·Z^d has rankd. Therefore any subset which generatesZ^d/q·Z^d has at leastd nonneutral elements; hence, since0∈A, we obtain that

|π(A)| ≥d+ 1. (20)

Writeπ(A) ={z1, z2, . . . , zn}andBi:=π⁻¹(Bi)∩Afor eachi∈ {1,2, . . . , n}. Without loss of generality assume that

|B1| ≤ |B2| ≤. . .≤ |Bn|. (21)

On the one hand, for alli, j∈ {1,2, . . . , n}withi,j, (B_i+q·A)∩

B_j+q·A

=∅.

Thus Remark 4 i) implies thatA=B₁](A\B₁) is aq-weak partition. On the other hand, insomuch as|A| ≤2d+ 1, we get from (20) and (21) that|B₁|= 1.

LetAbe a nonempty subset ofR^d, andA=A₁]A₂ a partition withA₁and A2 nonempty. We say thatA=A1]A2 islow-dimensionalif dimA1 <dimAor dimA2 < dimA. We say that A= A1]A2 is flat if there is i ∈ {1,2} such that dimAi<dimAand|πaffA_i(A)| ≤2(dimA−dimAi). The previous two lemmas will have useful applications as we shall see in the next two results.

3.q-weak partitions andq-partitions

Lemma 7 – LetAbe a reduced subset ofZ^dandq∈Zwith|q|>1. Assume that there is a flat partition ofA. Hence there is aq-partitionA=A₁]A₂such that, ifd₁:= dimA₁ andd₂:= dimA₂, then

|A+q·A| ≥|A1+q·A1|+|A2+q·A2|+ 2δ(A1, A2)|A|

+ 2(d−d₁)|A₁|+ 2(d−d₂)|A₂| −2(1 +d−d₁)(1 +d−d₂).

Proof. LetA=A⁰₁]A⁰₂ be a partition with dimA⁰₁<dimAand|πaffA⁰₁(A)| ≤2(d− dimA⁰₁). SetV := affA⁰₁ andk:=d−dimV; translating if necessary, assume that 0∈A⁰₁soV is a linear subspace ofR^d defined overZ. From Lemma 2, there is an ordered basisB:={f₁,f2, . . . ,f_d}ofZ^dsuch that{f_k+1,f_k+2, . . . ,fd}is an ordered basis ofV∩Z^d. SetB⁰:={f₁,f2, . . . ,f_k}andπ:=π_B,B⁰. The function

φ:R^k→R^d/V , φ







 Xk

i=1

aie_i









=πV







 Xk

i=1

aif_i









is an isomorphism andφ◦π=πV. Thus

|π(A)|=|π_V(A)| ≤2k. (22)

Note thatπ(A) isZ^k-reduced by Remark 3. Thus, from (22), the assumptions of Lemma 5 are satisfied byπ(A), and therefore there is aq-partitionπ(A) =A⁰⁰₁]A⁰⁰₂. TakingA₁:=π⁻¹(A⁰⁰₁)∩AandA₂:=π⁻¹(A⁰⁰₂)∩A, Remark 5 ii) implies thatA=A₁]A₂ is aq-partition. Setδ:=δ(A₁, A₂). The definition ofq-partition yields that the sets Ai+q·Ajfori, j∈ {1,2}are pairwise disjoint. Thus

|A+q·A|=|A₁+q·A₁|+|A₂+q·A₂|+|A₁+q·A₂|+|A₂+q·A₁|. (23) Now Lemma 1 applied to (A₁, A₂) and (A₂, A₁) leads to

dim(A1+q·A2)≥d−1 +δ

dim(A₂+q·A₁)≥d−1 +δ. (24)

We apply Corollary 2 to (A1, q·A2) and (A2, q·A1), and we get by (24) that

|A₁+q·A₂| ≥(δ+d−d₁)|A₁|+ (δ+d−d₂)|A₂| −(δ+d−d₁)(δ+d−d₂)

|A2+q·A1| ≥(δ+d−d1)|A1|+ (δ+d−d2)|A2| −(δ+d−d1)(δ+d−d2). (25) Thus, from (23) and (25), we get that

|A+q·A| ≥|A₁+q·A₁|+|A₂+q·A₂|+ 2δ|A|

+ 2(d−d1)|A1|+ 2(d−d2)|A2| −2(1 +d−d1)(1 +d−d2).

Lemma 8 – LetAbe a reduced subset ofZ^dandq∈Zwith|q|>1. Assume that there is a low-dimensionalq-weak partition. Also assume that all low-dimensionalq-weak partitions ofAare not flat. Hence there is aq-weak partitionA=A₁]A₂such that, if d₁:= dimA₁andd₂:= dimA₂, then

|A+q·A| ≥ |A1+q·A|+|A2+q·A2|+ 2(d−d2)(|A2| −1)

|A+q·A| ≥ |A₂+q·A|+|A₁+q·A₁|+ 2(d−d₁)(|A₁| −1) and

|A+q·A| ≥|A₁+q·A₁|+|A₂+q·A₂|+ min{|A₁|,|A₂|}

+ 2(d−d1)|A1|+ 2(d−d2)|A2| −2 max{(d−d1),(d−d2)}.

Proof. For any low-dimensionalq-weak partitionA=A⁰₁]A⁰₂, we have by assump- tion that it is not flat; thus, since|π_R^d(A)|= 1 = 2(d−dimR^d) + 1, we get that that for allq-weak partitionA=A⁰₁]A⁰₂,

|π_affA⁰

i(A)| ≥2(d−dimA⁰_i) + 1 ∀i∈ {1,2}. (26)

>From the family of low-dimensional partitions ofA(which is not empty by as- sumption), takeA=A⁰₁]A⁰₂with min{dimA⁰₁,dimA⁰₂}minimal ( in particular min{dimA⁰₁,dimA⁰₂}< d). The proof of the statement is divided into two cases.

• Assume that|π_affA⁰

1(A)| ≥2(d−dimA⁰₁) + 2 or|π_affA⁰

2(A)| ≥2(d−dimA⁰₂) + 2;

without loss of generality suppose that

|π_affA⁰₁(A)| ≥2(d−dimA⁰₁) + 2. (27) TakeA1 :=A⁰₁andA2:=A⁰₂, and writeπ1:=πaffA₁ andπ2 :=πaffA₂. Since A=A1]A2isq-weak,A1+q·AandA2+q·Aare disjoint. Thus

|A+q·A|=|A₁+q·A|+|A₂+q·A|. (28)

Now, fori∈ {1,2}, we have thatA_i+q·(A∩affAi) andA_i+q·(A\affAi) are disjoint so

|A1+q·A| ≥ |A1+q·A1|+|A1+q·(A\affA1)|

|A₂+q·A| ≥ |A₂+q·A₂|+|A₂+q·(A\affA2)|. (29) On the one hand,

|A₁+q·(A\affA1)| ≥ |π₁(q·(A\affA1))||A₁|

by Cor. 1

=|π₁(A\affA1)||A₁|

≥(|π1(A)| −1)|A1|

(Cont. next page)

3.q-weak partitions andq-partitions

≥(2(d−d₁) + 1)|A₁|.

by (27)

(30) On the other hand,

|A2+q·(A\affA2)| ≥ |π2(q·(A\affA2))||A2|

by Cor. 1

=|π2(A\affA2)||A2|

≥(|π₂(A)| −1)|A₂|

≥2(d−d2)|A2|.

by (26)

(31)

>From (28)-(31), note that

|A+q·A| ≥ |A1+q·A|+|A2+q·A2|+ 2(d−d2)|A2|

|A+q·A| ≥ |A₂+q·A|+|A₁+q·A₁|+ (2(d−d₁) + 1)|A₁|, and

|A+q·A| ≥|A₁+q·A₁|+|A₂+q·A₂|+|A₁| + 2(d−d₁)|A₁|+ 2(d−d₂)|A₂|, and this concludes this case.

• Assume that

|π_affA⁰₁(A)|= 2(d−dimA⁰₁) + 1

|πaffA⁰₂(A)|= 2(d−dimA⁰₂) + 1. (32) Without loss of generality assume that dimA⁰₁ = min{dimA⁰₁,dimA⁰₂}. Set V := affA⁰₁andk:=d−dimV; translating if necessary, we assume that0∈A⁰₁ soV is a linear subspace ofR^d defined overZ. From Lemma 2, there is an ordered basisB:={f₁,f₂, . . . ,f_d}ofZ^dsuch that{f_k+1,f_k+2, . . . ,f_d}is an ordered basis ofV∩Z^d. SetB⁰:={f₁,f2, . . . ,f_k}andπ:=π_B,B⁰. The function

φ:R^k→R^d/V , φ







 Xk

i=1

a_ie_i









=π_V







 Xk

i=1

a_if_i









is an isomorphism andφ◦π=πV. Thus

|π(A)|=|π_V(A)|= 2k+ 1. (33)

Note thatπ(A) isZ^k-reduced by Remark 3. Thus, from (33), the assumptions of Lemma 6 is satisfied byπ(A), and therefore there is aq-weak partition

π(A) =A⁰⁰₁]A⁰⁰₂with|A⁰⁰₁|= 1. TakingA₁:=π⁻¹(A⁰⁰₁)∩AandA₂:=π⁻¹(A⁰⁰₂)∩A, Remark 5 i) implies thatA=A₁]A₂is aq-weak partition. Furthermore, since

|A⁰⁰₁|= 1,

|πV(A1)|=|π(A1)|= 1

|π_V(A₂)|=|π(A₂)|=|π(A)| −1. (34)

Note thatA=A1]A2is low-dimensional sinceA1is contained in a translation ofV by (34). From the minimality of dimA⁰₁= min{dimA⁰₁,dimA⁰₂}, we have that dimA1≥dimA⁰₁; however, sinceA1is contained in a translation ofV = affA⁰₁ by (34), we conclude that dimA₁ = dimA⁰₁, and therefore affA1 is a translation ofV. Writeπ₁:=π_affA1 andπ₂:=π_affA2; thusπ_V =π₁. Inasmuch as affA1is a translation ofV, we get from (34) thatA₂does not intersect affA1; therefore

A2=A\affA1. (35)

By (33) and (34),

|π1(A2)|=|π(A)| −1 = 2(d−dimA1). (36) SinceA=A₁]A₂isq-weak, it follows thatA₁+q·AandA₂+q·Aare disjoint.

Thus

|A+q·A|=|A1+q·A|+|A2+q·A|. (37)

Now, fori∈ {1,2}, we have thatA_i+q·(A∩affAi) andA_i+q·(A\affAi) are disjoint so

|A1+q·A| ≥ |A1+q·A1|+|A1+q·(A\affA1)|

|A₂+q·A| ≥ |A₂+q·A₂|+|A₂+q·(A\affA2)|. (38) On the one hand,

|A₁+q·(A\affA1)|=|A₁+q·A₂|

by (35)

≥ |π1(q·A2)|(|A1| −1) +|q·A2|

by Cor. 1

=|π1(A2)|(|A1| −1) +|A2|

= 2(d−d₁)|A₁|+|A₂| −2(d−d₁).

by (36)

(39) On the other hand,

|A2+q·(A\affA2)| ≥ |π2(q·(A\affA2))||A2|

by Cor. 1

(Cont. next page)

4. Proof of Theorem 2

=|π₂(A\affA2)||A₂|

≥(|π2(A)| −1)|A2|

≥2(d−d₂)|A₂|.

by (26)

(40)

>From (37)-(40), note that

|A+q·A| ≥ |A₁+q·A|+|A₂+q·A₂|+ 2(d−d₂)|A₂|

|A+q·A| ≥ |A₂+q·A|+|A₁+q·A₁|+ 2(d−d₁)|A₁|+|A₂| −2(d−d₁), and

|A+q·A| ≥|A₁+q·A₁|+|A₂+q·A₂|+|A₂|

+ 2(d−d1)|A1|+ 2(d−d2)|A2| −2(d−d1),

and this concludes this case.

4 Proof of Theorem 2

In this section we complete the proof of Theorem 2.

Proof. (Theorem 2)We start the proof with three reductions. First, translating if necessary, we will assume that0∈A. Second, note that if dimA < d, then there is a basisB:={f₁,f₂, . . . ,f_d}ofZ^dsuch thatB⁰:={f₁,f₂, . . . ,f_dimA}is a basis of affA∩Z^d. Thusπ_B,B⁰ satisfies that dimπ_B,B⁰(A) = dimA= dimR^dimA,π_B,B⁰(A)⊆Z^dimA,

|π_B,B⁰(A)|=|A|and|π_B,B⁰(A+q·A)|=|A+q·A|; hence we will assume from now on that dimA=d. Third, since dimA=d, Remark 2 allows us to assume thatAis reduced.

Recall that for eachi, j∈Z, we have thatci,j= 4^j4i. The proof of Theorem 2 is done first by induction ondand then onm. First note that ifd= 0,AandA+q·A are singletons and hence

|A+q·A|= 1≥ m

|q| −1|A| −c0,m.

Thus, from now on, we assume that the statement is true for all 0≤d⁰ < d. Now assume thatm=|q|+ 2d−1. Then Theorem 3 leads to

|A+q·A| ≥2|A| −1≥|q|+ 2d−1

|q|+ 2d−1|A| −c_d,|q|+2d−1.

>From now on, we will assume that the statement holds for all|q|+ 2d−1≤m⁰< m.

To complete the induction, we will divide the proof into two cases.

• Assume that there is a low-dimensionalq-weak partition ofA. This case will also be divided into two subcases.

? Assume that there is a flatq-weak partition ofA. From Lemma 7, there is aq-partitionA=A1]A2such that, ifd1:= dimA1,d2:= dimA2and c1:= 2(1 +d−d1)(1 +d−d2), then

|A+q·A| ≥|A₁+q·A₁|+|A₂+q·A₂|+ 2δ(A₁, A₂)|A|

+ 2(d−d1)|A1|+ 2(d−d2)|A2| −c1. (41) Ifd1< dandd2< d, then the induction hypothesis yields that

|A₁+q·A₁| ≥(|q|+ 2d₁−1)|A₁| −c_d1,(|q|+2d₁−1)²

|A₂+q·A₂| ≥(|q|+ 2d₂−1)|A₂| −c_d2,(|q|+2d₂−1)² (42) and thereby

|A+q·A| ≥|A₁+q·A₁|+|A₂+q·A₂|

+ 2(d−d₁)|A₁|+ 2(d−d₂)|A₂| −c₁

by (41)

≥(|q|+ 2d−1)|A| −c_d1,(|q|+2d₁−1)²

−c_d2,(|q|+2d₂−1)²−c1

by (42)

≥ m

|q|+ 2d−1|A| −c_d,m.

Ifd₁=dord₂=d, we have thatδ(d₁, d₂) = 1. For anyi∈ {1,2}, we have by the hypothesis of induction,

|A_i+q·A_i| ≥











(|q|+ 2d_i−1)|A_i| −c_di,(|q|+2d−1)² ifd_i < d;

m−1

|q|+2d−1|A_i| −c_d,m−1, ifd_i =d.

and in any case

|A_i+q·A_i|+ 2(d−d_i)|A_i| ≥ m−1

|q|+ 2d−1|A_i| −c_d,m−1. (43) Thus

|A+q·A| ≥|A₁+q·A₁|+|A₂+q·A₂|+ 2|A| + 2(d−d₁)|A₁|+ 2(d−d₂)|A₂| −c₁

by (41)

≥ m−1

|q|+ 2d−1|A|+ 2|A| −2c_d,m−1−c₁

by (43)

≥ m

|q|+ 2d−1|A| −c_d,m.