A question of Bukh on sums of dilates Giorgis Petridis The - PowerPoint PPT Presentation

A question of Bukh on sums of dilates Giorgis Petridis The University of Georgia September 7th 2020

Introduction Joint work with Brandon Hanson. For subset A of a commutative group and a positive integer λ A + λ · A = { a + a ′ + · · · + a ′ : a , a ′ ∈ A } . � �� � λ times Bukh: In Z , | A + λ · A | ≥ ( λ + 1) | A | − o ( | A | ) (proved more). Conference participants with work on lower bounds: Nathanson, Plagne, Ru´ e, Serra, Shakan, and Tringali.

Upper bounds | A + λ · A | ≤ | A | 2 . Sharp when A is sparse in terms of λ . In theory and in applications interested in “additively structured” sets. A measure of additive structure is the doubling parameter K := | A + A | . | A | Have A + λ · A ⊆ the ( λ + 1)-fold sumset of A . An inequality of unnecke implies | A + λ · A | ≤ K λ +1 | A | . Pl¨ Bukh improved this to | A + λ · A | ≤ K C log( λ ) | A | (and proved much more). Bush and Zhao strengthened cases of Bukh’s general result. These methods do not work for small λ .

The question of Bukh Take λ = 2. We have seen | A + 2 · A | ≤ | A + A + A | ≤ K 3 | A | . The first inequality is sharp for arithmetic progressions and the second for some examples of Ruzsa. Ruzsa’s examples are the union of a 3-dimensional progression with 3 thin progressions. Question (Bukh, 2007) Does there exist a p < 3 such that for all K ∈ R and all finite A ⊂ Z such that | A + A | ≤ K | A | , we have | A + 2 · A | ≤ K p | A | ?

Affirmative answer to Bukh’s question Theorem (Hanson and P.) There exists p ≤ 2 . 95 such that for all K ∈ R and all finite subsets A of a commutative group that satisfy | A + A | ≤ K | A | , we have | A + 2 · A | ≤ K p | A | . Taking A = { 0 , 1 } n ⊂ Z n shows p ≥ log(2) / log(3 / 2) ≈ 1 . 7.

Motivation Key inequalities | A + 2 · A | ≤ | A + A + A | ≤ K 3 | A | . Aim: Understand structure of sets for which | A + A + A | ≤ K 3 | A | is nearly sharp. Differentiate between dilate sums and sumsets for sets not extremal to Pl¨ unnecke’s inequality. Proof suggests links between Pl¨ unnecke’s inequality and Freiman’s theorem.

Dilate sums vs sumsets k � Partition A = A i . i =1 k k � � | A + 2 · A | ≤ | A + 2 · A i | ≤ | A + A i + A i | . i =1 i =1 Contrast to k � | A + A + A | ≤ | A + A i + A j | . i , j =1

Key proposition Proposition There exists an absolute C > 0 such that for all finite subsets A of a commutative group with | A + A | ≤ K | A | , there exists a partition A = A 1 ∪ · · · ∪ A k with k � | A + A i + A i | ≤ C (log | A | ) 2 K 2 . 95 | A | . i =1 For | A + 2 · A | can ignore logarithms and constants because of the tensor power trick. The first attempt is to chose the A i greedily using averaging. When this does not work, use structural information.

A dichotomy Let M ≤ K . For every A ′ ⊆ A there exists a subset B ⊆ A ′ that satisfies one of the following two alternatives. 1 | B | ≥ | A ′ | / 3 and B = C 1 ∪ · · · ∪ C ℓ with ℓ | A + C i + C i | ≤ K 3 � M | A | , i =1 2 | B | ≥ | A ′ | / M 3 and | A + B + B | ≤ M 16 K 2 | A | .

Iteration Start with A ′ = A , find B ′ , set A ′ = A \ B ′ , find B ′′ , . . . The first alternative can occur at most log( | A | ) times. The contribution to � | A + A i + A i | is at most K 3 M | A | . The second alternative can occur at most log( | A | ) M 3 times. The contribution to � | A + A i + A i | is at most K 2 M 19 | A | . Choose M = K 1 / 20 .

Elements in the proof of the dichotomy The dichotomy is proved using: • tools of additive combinatorics popularised by Ruzsa, • considerations of Katz and Koester, • the Balog–Szemer´ edi–Gowers theorem, • considerations of Schoen and Shkredov. Delicate argument, does not work of A − 2 · A .

Rest of the talk Flesh out some details for A ′ = A . First explain how the first alternative might arise. Then what we learn when it doesn’t. Finally how to use this information to establish the second alternative.

Two basic tools Lemma (Pl¨ unnecke’s inequality) Let A be a finite subset of a commutative group such that | A + A | ≤ K | A | . There exists X ⊆ A such that for all sets C | X + C + A | ≤ K | X + C | . Lemma (Greedy covering lemma) A , X finite subsets of a commutative group and | X + A | ≤ K | X | . There exists a subset S such that A ⊆ S + X and | S | ≤ log( | A | ) K. We write � A = for A s = A ∩ ( s + X ) . A s s ∈ S

A hopeful calculation � Partition A into translates of X : A = A s , for A s = A ∩ ( s + X ). � � | A + A s + A s | ≤ | A + A s + X + s | s ∈ S s ∈ S � = | A + A s + X | s ∈ S � ≤ K | A s + X | s ∈ S ≤ K | S || A + X | ≤ K 3 | A | .

First alternative Obtain improvement to � | A + A s + A s | ≤ K 3 | A | when • either | S | smaller than K , • or most | X + A s | much smaller than K | A | . From now on focus on the first part A s 1 and assume • | A s 1 | = | A | / K and • all of | X + A s 1 | , | X + X | , | X + A | = K | A | .

Second alternative I A s 1 is selected greedily: using additive energy and by averaging can take s 1 to be any popular difference. D pop := { s ∈ A − X : | A s | ≥ | A | / K } � = ∅ . Since | A s 1 | = | A | / K also get | D pop | = K | A | . So have three sets of almost equal size K | A | : X + X , X + A and D pop .

Second alternative II By Katz-Koester X + A s ⊆ ( X + A ) ∩ ( s + ( X + X )) . Since | X + A s | = K | A | = | X + X | = | X + A | , every s ∈ D pop can be expressed as a difference of elements in X + A and X + X in an almost maximum number of ways. So have a nearly maximum number of solutions to the equation u − v = d with u ∈ X + A , v = X + X , d ∈ D pop .

Second alternative III Following Schoen-Shkredov, by Balog-Szemer´ edi-Gowers and unnecke there are subsets D ′ ⊆ D pop and V ⊆ X + X such that Pl¨ | V + D ′ + D ′ | comparable to | V | . By the greedy covering lemma, D ′ + D ′ is covered by few translates of X + X . Since D ′ ⊆ D pop , by averaging can find a translate of a large subset B ⊆ A inside D ′ . So B + B is covered by few translates of X + X . A + B + B is covered by few translates of A + X + X , which has size K 2 | A | , and | A + B + B | is much smaller than K 3 | A | .

Thank you.