Additive Combinatorics and its Applications in Theoretical CS By Shachar Lovett Chapter 2: Set addition, Sections 2.1-2.3, pp. 3-8. Presented by Tomer Bincovich

Set addition 𝐺 an abelian group (think about 𝐺= 𝔽 𝑛 ). 𝐴⊆𝐺. The sumset of 𝐴 is 2𝐴=𝐴+𝐴= 𝑎+ 𝑎 ′ 𝑎, 𝑎 ′ ∈𝐴 Always 2𝐴 ≥ 𝐴 . When does equality hold?

When does 2𝐴 =|𝐴|? Equality holds if 𝐴 is empty, a subgroup of 𝐺 or a coset of a subgroup. For the other direction: If 𝐴≠∅, we can assume w.l.o.g. 0∈𝐴 by shifting. Then 𝐴⊆2𝐴, and since 2𝐴 = 𝐴 we have that 2𝐴=𝐴. 𝐴 is a nonempty finite subset that is closed under addition, hence is a subgroup.

Planned topics Ruzsa calculus The span of sets of small doubling The growth of sets of small doubling

Ruzsa calculus A set of basic inequalities between sizes of sets and their sumsets. For 𝐴,𝐵⊆𝐺, define: Sumset: 𝐴+𝐵= 𝑎+𝑏 𝑎∈𝐴,𝑏∈𝐵 Difference set: 𝐴−𝐵= 𝑎−𝑏 𝑎∈𝐴,𝑏∈𝐵 Claim 2.1 (Ruzsa triangle inequality): 𝐴 𝐵−𝐶 ≤ 𝐴−𝐵 𝐴−𝐶

Proof of Claim 2.1 Claim 2.1 (Ruzsa triangle inequality): 𝐴 𝐵−𝐶 ≤ 𝐴−𝐵 𝐴−𝐶 Define a map 𝑓:𝐴× 𝐵−𝐶 → 𝐴−𝐵 × 𝐴−𝐶 : for any 𝑥∈𝐵−𝐶 fix 𝑏∈𝐵,𝑐∈𝐶 s.t. 𝑥=𝑏−𝑐, and define 𝑓 𝑎,𝑥 = 𝑎−𝑏,𝑎−𝑐 . 𝑓 is injective, hence 𝐴 𝐵−𝐶 ≤ 𝐴−𝐵 𝐴−𝐶 .

The Ruzsa distance 𝑑 𝐴,𝐵 = log 𝐴−𝐵 𝐴 1 2 𝐵 1 2 Not formally a distance function. Why? Claim 2.3: The Ruzsa distance is symmetric and obeys the triangle inequality.

Proof of Claim 2.3 Symmetry: since 𝐵−𝐴 = 𝐴−𝐵 . 𝑑 𝐴,𝐵 = log 𝐴−𝐵 𝐴 1 2 𝐵 1 2 Symmetry: since 𝐵−𝐴 = 𝐴−𝐵 . Moreover, 𝑑 𝐴,𝐶 ≤𝑑 𝐴,𝐵 +𝑑 𝐵,𝐶 is equivalent to log 𝐴−𝐶 𝐴 1 2 𝐶 1 2 ≤ log 𝐴−𝐵 𝐴 1 2 𝐵 1 2 + log 𝐵−𝐶 𝐵 1 2 𝐶 1 2 𝐴−𝐶 𝐴 1 2 𝐶 1 2 ≤ 𝐴−𝐵 𝐴 1 2 𝐵 1 2 𝐵−𝐶 𝐵 1 2 𝐶 1 2 𝐵 𝐴−𝐶 ≤ 𝐵−𝐴 𝐵−𝐶 Which follows from Claim 2.1.

Corollary 2.4 If 𝐴−𝐵 ≤𝐾 𝐴 1 2 𝐵 1 2 then 𝐴−𝐴 ≤ 𝐾 2 𝐴 . For 𝐵=−𝐴 we get that if 𝐴+𝐴 ≤𝐾 𝐴 then 𝐴−𝐴 ≤ 𝐾 2 𝐴 .

Proof of Corollary 2.4 If 𝐴−𝐵 ≤𝐾 𝐴 1 2 𝐵 1 2 then 𝐴−𝐴 ≤ 𝐾 2 𝐴 . 𝐴−𝐴 𝐴 = 𝑒 𝑑 𝐴,𝐴 ≤ 𝑒 𝑑 𝐴,𝐵 +𝑑 𝐵,𝐴 = 𝐴−𝐵 𝐴 1 2 𝐵 1 2 2 ≤ 𝐾 2 𝑑 𝐴,𝐵 = log 𝐴−𝐵 𝐴 1 2 𝐵 1 2

The span of sets of small doubling

The doubling constant The doubling constant of 𝐴 is 𝐾= 𝐴+𝐴 𝐴 . We saw that 𝐾=1 corresponds to subgroups. Can we use 𝐾 to say something about the size of the smallest subgroup containing 𝐴? (or, in the case of 𝐺= 𝔽 𝑛 , the size of the span of 𝐴)

Lemma 2.5 (Laba) If 𝐴−𝐴 < 3 2 𝐴 then 𝐴−𝐴 is a subgroup. The constant 3/2 is tight: take 𝐴= 0,1 ⊆ℤ. 𝐴−𝐴= −1,0,1 is not a subgroup.

Proof of Lemma 2.5 We will show that for any 𝑥∈𝐴−𝐴, 𝐴∩ 𝐴+𝑥 > 𝐴 /2. This implies that for any 𝑥,𝑦∈𝐴−𝐴, 𝐴+𝑥 ∩ 𝐴+𝑦 ≠∅: Denoting 𝐴 𝑥 =𝐴∩ 𝐴+𝑥 , by the inclusion exclusion principle 𝐴+𝑥 ∩ 𝐴+𝑦 ≥ 𝐴 𝑥 ∩ 𝐴 𝑦 = 𝐴 𝑥 + 𝐴 𝑦 − 𝐴 𝑥 ∪ A 𝑦 > 𝐴 /2+ 𝐴 /2− 𝐴 =0 There are 𝑎 1 , 𝑎 2 ∈𝐴 s.t. 𝑎 1 +𝑥= 𝑎 2 +𝑦, thus 𝑥−𝑦= 𝑎 2 − 𝑎 1 ∈𝐴−𝐴. 𝐴−𝐴 is closed under taking difference, hence must be a subgroup.

Proof of Lemma 2.5 cont. We will show that for any 𝑥∈𝐴−𝐴, 𝐴∩ 𝐴+𝑥 > 𝐴 /2. Let 𝑥=𝑎− 𝑎 ′ ∈𝐴−𝐴. Then 𝐴∩ 𝐴+𝑥 = 𝐴−𝑎 ∩ 𝐴− 𝑎 ′ . Similarly to before: 𝐴−𝑎 ∩ 𝐴− 𝑎 ′ = 𝐴−𝑎 + 𝐴− 𝑎 ′ − 𝐴−𝑎 ∪ 𝐴− 𝑎 ′ ≥ 𝐴 + 𝐴 − 𝐴−𝐴 >|𝐴|/2 As needed.

Lemma 2.6 (Freiman) Let 𝐴⊆ ℝ 𝑛 with 𝐴+𝐴 ≤𝐾 𝐴 . Then 𝐴 lies in an affine subspace of dimension at most 2𝐾−1.

Proof of Lemma 2.6 We will prove that if 𝐴 has affine dimension 𝑑 then 𝐴+𝐴 ≥ 𝑑+1 𝐴 − 𝑑+1 2 Thus 𝑑+1 𝐴 − 𝑑+1 2 ≤ 𝐴+𝐴 ≤𝐾 𝐴 , and since 𝐴 ≥𝑑, 𝑑+1−𝐾 𝑑≤ 𝑑+1−𝐾 𝐴 ≤ 𝑑+1 2 = 𝑑+1 2 𝑑 Hence 𝐾≥ 𝑑+1 2 , or 𝑑≤2𝐾−1.

Proof of Lemma 2.6 cont. We prove by induction on 𝐴 that if 𝐴 has affine dimension 𝑑 then 𝐴+𝐴 ≥ 𝑑+1 𝐴 − 𝑑+1 2 If 𝐴 =1 then 𝐴+𝐴 =1, 𝑑=0 and the claim follows. Assume 𝐴 >1. Let 𝑀 𝐴 = 𝑎+ 𝑎 ′ 2 𝑎, 𝑎 ′ ∈𝐴 . Clearly 𝑀 𝐴 = 𝐴+𝐴 . Let 𝐶(𝐴) be the convex hull of 𝐴, which is a 𝑑-dimensional polytope by assumption. Its vertices are the points in 𝐴 that cannot be obtained as a convex combination of the other points.

Inductive proof, case (1) Fix a vertex 𝑎 ∗ ∈𝐴 of 𝐶 𝐴 , and let 𝐴 ′ =𝐴∖ 𝑎 ∗ . We have two cases: The affine dimension of 𝐴 ′ is 𝑑−1. Then the mid-points 𝑎+ 𝑎 ∗ 2 for all 𝑎∈𝐴 are outside 𝐶( 𝐴 ′ ). 𝑀 𝐴 ≥ 𝐴 + 𝑀 𝐴 ′ ≥ 𝐴 +𝑑 𝐴 ′ − 𝑑 2 = 𝑑+1 𝐴 −𝑑− 𝑑 2 = 𝑑+1 𝐴 − 𝑑+1 2 𝐴+𝐴 ≥ ? 𝑑+1 𝐴 − 𝑑+1 2

Inductive proof, case (2) Fix a vertex 𝑎 ∗ ∈𝐴 of 𝐶 𝐴 , and let 𝐴 ′ =𝐴∖ 𝑎 ∗ . We have two cases: The affine dimension of 𝐴 ′ is 𝑑. Let 𝑎 1 ,…, 𝑎 𝑡 be the neighboring vertices to 𝑎 ∗ in 𝐶 𝐴 (two vertices are neighbors if the segment connecting them is a one-dim. face of 𝐶 𝐴 ). Note that 𝑡≥𝑑 since 𝐶 𝐴 is 𝑑-dim. The points 𝑎 ∗ and 𝑎 ∗ + 𝑎 𝑖 2 for 1≤𝑖≤𝑡 are mid-points outside 𝐶 𝐴 ′ . 𝑀 𝐴 ≥ 𝑡+1 + 𝑀 𝐴 ′ ≥𝑑+1+ 𝑑+1 𝐴 ′ − 𝑑+1 2 = 𝑑+1 𝐴 − 𝑑+1 2 𝐴+𝐴 ≥ ? 𝑑+1 𝐴 − 𝑑+1 2

Lemma 2.6 – tightness Let 𝐴⊆ ℝ 𝑛 with 𝐴+𝐴 ≤𝐾 𝐴 . Then 𝐴 lies in an affine subspace of dimension at most 2𝐾−1. The dimension is tight up to lower order terms. Take 𝐴= 𝑣 1 ,…, 𝑣 2𝐾 a set of linearly independent vectors, then 𝐴+𝐴 = 2𝐾 2 +2𝐾 =𝐾(2𝐾+1) and 𝐴+𝐴 𝐴 =𝐾+ 1 2 .

Theorem 2.7 (Ruzsa) Let 𝐺 be an Abelian group of torsion 𝑟, 𝐴⊆𝐺 with 𝐴+𝐴 ≤𝐾 𝐴 . Then there exists a subgroup 𝐻<𝐺, 𝐻 ≤ 𝐾 2 𝑟 𝐾 4 𝐴 s.t. 𝐴⊆𝐻. A group has torsion 𝑟≥1 if 𝑟∙𝑔=0 for all 𝑔∈𝐺.

Example 𝐺= 𝔽 𝑝 𝑛 has torsion 𝑟=𝑝. Let 𝑈,𝑉⊆ 𝔽 𝑝 𝑛 two subspaces with 𝑈∩𝑉= 0 , and set 𝐴=𝑈+{ 𝑣 1 ,…, 𝑣 2𝐾 } where 𝑣 1 ,…, 𝑣 2𝐾 ∈𝑉 are linearly independent. 𝐴 = 𝑈 ∙2𝐾. Since 𝐴+𝐴=𝑈+ 𝑣 𝑖 + 𝑣 𝑗 1≤𝑖≤𝑗≤2𝐾 , 𝐴+𝐴 = 𝑈 ∙𝐾 2𝐾+1 and 𝐴+𝐴 𝐴 =𝐾+ 1 2 ≈𝐾. However, the size of the minimal subspace containing 𝐴 is 𝑝 2𝐾 𝑈 = 𝑝 2𝐾 2𝐾 𝐴 Thus an exponential dependency on 𝐾 is unavoidable.

Conjecture 2.8 (Ruzsa) There exists an absolute constant 𝐶≥2 s.t. for any Abelian group 𝐺 of torsion 𝑟, and any 𝐴⊆𝐺 with 𝐴+𝐴 ≤𝐾|𝐴|, the subgroup generated by 𝐴 has order ≤ 𝑟 𝐶𝐾 𝐴 . Was established with 𝐶=2 for 𝑟=2 and then extended for prime 𝑟: Theorem 2.9: Let 𝑝 be a prime, 𝐴⊆ 𝔽 𝑝 𝑛 with 𝐴+𝐴 ≤𝐾 𝐴 . Then there exists a subspace 𝐻⊆ 𝔽 𝑝 𝑛 s.t. 𝐴⊆𝐻 and 𝐻 ≤ 𝑝 2𝐾 2𝐾−1 𝐴 .

Theorem 2.10 If 2𝐴 ≤𝐾 𝐴 then 2𝐴−2𝐴 ≤ 𝐾 4 𝐴 . Used to prove Theorem 2.7. We will prove a more general result later today.

Proof of Theorem 2.7 - Sketch Let 𝐴⊆𝐺 with 2𝐴 ≤𝐾 𝐴 . We can assume 0∈𝐴 by possibly replacing 𝐴 with 𝐴−𝑎 for some 𝑎∈𝐴. Let 𝐵= 𝑏 1 ,…, 𝑏 𝑚 ⊆2𝐴−𝐴 be a maximal collection of elements s.t. 𝑏 𝑖 −𝐴 are all disjoint (“Ruzsa covering”). 𝐴 has small doubling, thus Theorem 2.10 implies that 𝐵 is bounded. We will show that ℓ𝐴−𝐴⊆ ℓ−1 𝐵+𝐴−𝐴 for all ℓ≥1. Together the theorem follows.

Proof of Theorem 2.7 cont. Since 𝑏 𝑖 −𝐴⊆2𝐴−2𝐴, 𝑚≤ 2𝐴−2𝐴 𝐴 ≤ 𝐾 4 𝑚≤ 2𝐴−2𝐴 𝐴 ≤ 𝐾 4 Therefore 𝐵 ≤ 𝐾 4 . 2𝐴 ≤𝐾 𝐴 ⇒ 2𝐴−2𝐴 ≤ 𝐾 4 𝐴

ℓ𝐴−𝐴⊆ ℓ−1 𝐵+𝐴−𝐴 ℓ=1 is trivial. For ℓ=2, we need to show 2𝐴−𝐴⊆𝐵+𝐴−𝐴. Take 𝑥∈2𝐴−𝐴, by construction there exists 𝑏∈𝐵 s.t. 𝑥−𝐴 ∩ 𝑏−𝐴 ≠∅, i.e. 𝑥−𝑎=𝑏− 𝑎 ′ for some 𝑎, 𝑎 ′ ∈𝐴. Hence 𝑥=𝑏+𝑎− 𝑎 ′ ∈𝐵+𝐴−𝐴. By induction on ℓ, ℓ𝐴−𝐴=𝐴+ ℓ−1 𝐴−𝐴 ⊆𝐴+ ℓ−2 𝐵+𝐴−𝐴 ⊆ ℓ−2 𝐵+𝐵+𝐴−𝐴= ℓ−1 𝐵+𝐴−𝐴

Concluding the proof ℓ𝐴−𝐴⊆ ℓ−1 𝐵+𝐴−𝐴 Let 𝐴 be the subgroup spanned by 𝐴, and similarly for 𝐵. 𝐴 = ℓ≥1 ℓ𝐴 ⊆ ℓ≥1 ℓ𝐴−𝐴 ⊆ ℓ≥1 ℓ−1 𝐵+𝐴−𝐴 = 𝐵 +𝐴−𝐴 Hence, using Corollary 2.4 (which implies 𝐴−𝐴 ≤ 𝐾 2 𝐴 ), 𝐴 ≤ 𝐵 ∙ 𝐴−𝐴 ≤ 𝐵 ∙ 𝐾 2 𝐴 We conclude by bounding 𝐵 . As 𝐺 has torsion 𝑟 and we saw that 𝐵 ≤ 𝐾 4 we have 𝐵 ≤ 𝑟 𝐾 4 . Finally, 𝐴 ≤ 𝐾 2 𝑟 𝐾 4 𝐴 .

The growth of sets of small doubling

Iterated sumsets ℓ𝐴= 𝑎 1 +…+ 𝑎 ℓ 𝑎 1 ,…, 𝑎 ℓ ∈𝐴 ℓ𝐴= 𝑎 1 +…+ 𝑎 ℓ 𝑎 1 ,…, 𝑎 ℓ ∈𝐴 ℓ𝐴−𝑚𝐴= 𝑎 1 +…+ 𝑎 ℓ − 𝑎 ℓ+1 −…− 𝑎 ℓ+𝑚 𝑎 1 ,…, 𝑎 ℓ+𝑚 ∈𝐴 Theorem 2.11 (Plünneke-Ruzsa): If 𝐴 = 𝐵 and 𝐴+𝐵 ≤𝐾 𝐴 then ℓ𝐴−𝑚𝐴 ≤ 𝐾 ℓ+𝑚 𝐴 . For 𝐵=𝐴 or 𝐵=−𝐴 we obtain the following: Corollary 2.12: If 𝐴+𝐴 ≤𝐾 𝐴 or 𝐴−𝐴 ≤𝐾 𝐴 then ℓ𝐴−𝑚𝐴 ≤ 𝐾 ℓ+𝑚 𝐴 .

Lemma 2.13 Let 𝐴,𝐵⊆𝐺 with 𝐴 = 𝐵 and 𝐴+𝐵 ≤𝐾 𝐴 . Let 𝐵 0 ⊆𝐵 a nonempty set minimizing the ratio 𝐾 0 = 𝐴+ 𝐵 0 𝐵 0 . Then for any 𝐶⊆𝐺, 𝐴+ 𝐵 0 +𝐶 ≤ 𝐾 0 𝐵 0 +𝐶 Notice that 𝐾 0 ≤𝐾.

Proof of Theorem 2.11 ∀𝐶 𝐴+ 𝐵 0 +𝐶 ≤ 𝐾 0 𝐵 0 +𝐶 ∀𝐶 𝐴+ 𝐵 0 +𝐶 ≤ 𝐾 0 𝐵 0 +𝐶 Let 𝐵 0 ⊆𝐵 as in Lemma 2.13. We will prove by induction on ℓ that 𝐵 0 +ℓ𝐴 ≤ 𝐾 0 ℓ 𝐵 0 ≤ 𝐾 ℓ 𝐵 0 . ℓ=0 is easy: 𝐵 0 +0𝐴 =1∙ 𝐵 0 . For ℓ>0: 𝐵 0 +ℓ𝐴 = 𝐴+ 𝐵 0 +(ℓ−1)𝐴 ≤ 𝐾 0 𝐵 0 + ℓ−1 𝐴 ≤ 𝐾 0 ∙ 𝐾 0 ℓ−1 𝐵 0 = 𝐾 0 ℓ 𝐵 0 By the Ruzsa triangle inequality (Claim 2.1), applied to − 𝐵 0 ,ℓ𝐴,𝑚𝐴: −𝐵 0 ℓ𝐴−𝑚𝐴 ≤ 𝐵 0 +ℓ𝐴 𝐵 0 +𝑚𝐴 ≤ 𝐾 ℓ+𝑚 𝐵 0 2 Hence ℓ𝐴−𝑚𝐴 ≤ 𝐾 ℓ+𝑚 𝐵 0 ≤ 𝐾 ℓ+𝑚 𝐴 𝐴 𝐵−𝐶 ≤ 𝐴−𝐵 𝐴−𝐶

Lemma 2.13 (Reminder) Let 𝐴,𝐵⊆𝐺 with 𝐴 = 𝐵 and 𝐴+𝐵 ≤𝐾 𝐴 . Let 𝐵 0 ⊆𝐵 a nonempty set minimizing the ratio 𝐾 0 = 𝐴+ 𝐵 0 𝐵 0 . Then for any 𝐶⊆𝐺, 𝐴+ 𝐵 0 +𝐶 ≤ 𝐾 0 𝐵 0 +𝐶

Proof of Lemma 2.13 Observe that by minimality of 𝐾 0 , for any 𝐵 ′ ⊆𝐵 we have that 𝐴+ 𝐵 ′ ≥ 𝐾 0 𝐵 ′ . The proof is by induction on 𝐶 . If 𝐶= 𝑥 , 𝐴+ 𝐵 0 +𝑥 = 𝐴+ 𝐵 0 = 𝐾 0 𝐵 0 = 𝐾 0 𝐵 0 +𝑥 If 𝐶 >1, write 𝐶= 𝐶 ′ ∪ 𝑥 . Define 𝐵 ′ ⊆ 𝐵 0 as 𝐵 ′ = 𝑏∈ 𝐵 0 𝐴+𝑏+𝑥⊆𝐴+ 𝐵 0 + 𝐶 ′ . We can write 𝐴+ 𝐵 0 +𝐶= 𝐴+ 𝐵 0 + 𝐶 ′ ∪ 𝐴+ 𝐵 0 +𝑥 ∖ 𝐴+ 𝐵 ′ +𝑥

Proof of Lemma 2.13 cont. 𝐴+ 𝐵 0 +𝐶= 𝐴+ 𝐵 0 + 𝐶 ′ ∪ 𝐴+ 𝐵 0 +𝑥 ∖ 𝐴+ 𝐵 ′ +𝑥 𝐴+ 𝐵 0 +𝐶 ≤ 𝐴+ 𝐵 0 + 𝐶 ′ + 𝐴+ 𝐵 0 +𝑥 ∖ 𝐴+ 𝐵 ′ +𝑥 = 𝐴+ 𝐵 0 + 𝐶 ′ + 𝐴+ 𝐵 0 +𝑥 − 𝐴+ 𝐵 ′ +𝑥 ≤ 𝐾 0 𝐵 0 + 𝐶 ′ + 𝐴+ 𝐵 0 − 𝐴+ 𝐵 ′ ≤ 𝐾 0 𝐵 0 + 𝐶 ′ + 𝐾 0 𝐵 0 − 𝐾 0 𝐵 ′ = 𝐾 0 𝐵 0 + 𝐶 ′ + 𝐵 0 − 𝐵 ′ It remains to show that: 𝐵 0 + 𝐶 ′ + 𝐵 0 − 𝐵 ′ ≤ 𝐵 0 +𝐶 ∀ 𝐵 ′ ⊆𝐵 𝐴+ 𝐵 ′ ≥ 𝐾 0 𝐵 ′

Concluding the proof It remains to show that: 𝐵 ′ = 𝑏∈ 𝐵 0 𝐴+𝑏+𝑥⊆𝐴+ 𝐵 0 + 𝐶 ′ It remains to show that: 𝐵 0 + 𝐶 ′ + 𝐵 0 − 𝐵 ′ ≤ 𝐵 0 +𝐶 Define 𝐵 ′′ = 𝑏∈ 𝐵 0 𝑏+𝑥∈ 𝐵 0 + 𝐶 ′ . Note that 𝐵 ′′ ⊆ 𝐵 ′ . We decompose 𝐵 0 +𝐶 as a disjoint union: 𝐵 0 +𝐶= 𝐵 0 + 𝐶 ′ ∪ 𝐵 0 +𝑥 ∖ 𝐵 ′′ +𝑥 Thus 𝐵 0 +𝐶 = 𝐵 0 + 𝐶 ′ + 𝐵 0 − 𝐵 ′′ ≥ 𝐵 0 + 𝐶 ′ + 𝐵 0 − 𝐵 ′ Which is what we needed to show.