Statistical Set Addition &
The Polynomial Freiman-Rusza Conjecture
Additive Combinatorics and its Applications in Theoretical CS
Sections 2.4, 2.6
Shachar Lovett
Presented by Guy Shalev
A Short Reminder

We discussed sets 𝐴 ⊆ 𝐺 with small doubling.

We bounded the size of subspaces containing such 𝐴
◦ Growth-factor was exponential in the doubling constant, 𝐾
◦ For example, we saw that ∃𝐻 < 𝐺, 𝐻 ≤

𝟒
2
𝑲
𝐾 𝒓
𝐴 s.t. 𝐴 ⊆ 𝐻
We bounded the size of sets of the form ℓ𝐴 − 𝑚𝐴
◦ ℓ𝐴 − 𝑚𝐴 ≤ 𝐾 𝑙+𝑚 𝐴
Today’s sections

(2.4) Statistical Set Addition
◦ The BSG theorem + lemma from graph theory.

(2.6) Polynomial Freiman-Rusza conjecture
◦ An important conjecture explaining the structure of a set with
small doubling.
◦ 5 equivalent conjectures over 𝔽𝑛2
◦ Versions for higher torsion groups
3
Statistical set addition – Motivation (1)


𝐴 + 𝐴 is much larger than 𝐴
But many pairs 𝑎1 + 𝑎2 lie in a small set (of size 𝐾|𝐴|)

For example: In 𝔽𝑛2 , take 𝐴 = 𝑉 ∪ 𝑢1 , 𝑢2 … , 𝑢2𝑑 , where 𝑉 is a
subspace of dimension 𝑑, and 𝑢𝑖 are random vectors.
𝐴 = 2𝑑+1
𝐴 + 𝐴 = 𝑉 ∪ 𝑉 + 𝑢1 , 𝑢2 … , 𝑢2𝑑 + 𝑢𝑖 + 𝑢𝑗




𝑑 2
𝐴+𝐴 ≈ 2
≈ 𝐴2
1
But for at least of the pairs 𝑎1 , 𝑎2 , we have 𝑎1 + 𝑎2 ∈ 𝑉
4
4
Statistical set addition – Motivation (2)



𝐴 = 𝑉 ∪ 𝑢1 , 𝑢2 … , 𝑢2𝑑
In this example, 𝐴 has large-doubling, but there is a large subset
(𝑉 ⊆ 𝐴) with small-doubling.
This is always the case. If a significant part of 𝐴 × 𝐴 “lands” in
a set of size 𝐾|𝐴|, then there is some large 𝐴′ ⊆ 𝐴 with smalldoubling.
◦ Another point of view – there is a “simple explanation” for such an 𝐴
◦ The BSG theorem is a generalization of this idea
5
BSG Theorem
 BSG Theorem: by Balog, Szemeredi & Gowers
1. Let 𝐴, 𝐵 ⊆ 𝐺, s.t. 𝐴 = 𝐵 = 𝑁
2. There exists 𝐶 ⊆ 𝐺 s.t. 𝐶 = 𝑐𝑁, and Pr [𝑎 + 𝑏 ∈ 𝐶] ≥ 𝜖
𝑎∈𝐴,𝑏∈𝐵
◦ Then there are subsets 𝐴 ⊆ 𝐴, 𝐵 ⊆ 𝐵 s.t. 𝐴 , 𝐵 ≥
′
◦ And

𝐴′
+ 𝐵′
≤
′
′
212 𝑐 3
𝑁
𝜖5
′
′
𝜖2
𝑁
16
A
′
′
a
′
By theorem 2.11, 𝐴 + 𝐴 , 𝐴 − 𝐴 ≤ 𝑝𝑜𝑙𝑦
1
𝑐, ∗
𝜖C
a+b
A+B
𝑁
b
B
6
The plan
 We
will state a lemma in graph theory
 Easily prove BSG using the lemma
 Work hard to prove the lemma
 Take a break before the second part
7
Lemma (by Sudakov)

The Lemma:
1. Let 𝐻 = (𝐴, 𝐵, 𝐸) be a bipartite graph
2. 𝐴 = 𝐵 = 𝑁 , 𝐸 = 𝜖𝑁 2
◦ Then there are subsets
𝐴′
⊆
𝐴, 𝐵′
“As many as we could expect”
⊆ 𝐵 s.t.
𝐴′
,
𝐵′
𝜖2
≥ 𝑁
16
−12 5 2
◦ Such that for any 𝑎 ∈ 𝐴′ , 𝑏 ∈ 𝐵′ there are at least 2
length 3 from 𝑎 to 𝑏.
a
𝜖 𝑁 paths of
𝑏
b
𝑎
8
Proof of BSG using the lemma (1/3)
We have sets 𝐴, 𝐵, 𝐶 ⊆ 𝐺 as given in the theorem.
 Build the “natural” bipartite graph 𝐻:

◦ The vertex sets are 𝐴, 𝐵
◦ 𝐸 = 𝑎, 𝑏 𝑎 + 𝑏 ∈ 𝐶 ⇒ 𝐸 = 𝜖𝑁 2

Obviously, the conditions of the lemma hold, so we get
vertex sets 𝐴′, 𝐵′ as stated, and we have “many paths” from
any 𝑎 ∈ 𝐴′ to 𝑏 ∈ 𝐵′
9
Proof of BSG using the lemma (2/3)
For any 𝑎 ∈ 𝐴′ , 𝑏 ∈ 𝐵′ , and a specific path (𝑎, 𝑏, 𝑎, 𝑏) we get
𝑦 =𝑎+𝑏 = 𝑎+𝑏 − 𝑎+𝑏 + 𝑎+𝑏
 Denote

◦ 𝑥1 = 𝑎 + 𝑏 , 𝑥2 = 𝑎 + 𝑏 , 𝑥3 = 𝑎 + 𝑏
From the definition of 𝐸, we know 𝑥1 , 𝑥2 , 𝑥3 ∈ 𝐶
 Any element 𝑦 ∈ 𝐴′ + 𝐵′ can be represented as
𝑦 = 𝑥1 − 𝑥2 + 𝑥3 in at least 2−12 𝜖 5 𝑁 2 ways.
 𝐶 = 𝑐𝑁 , so there are 𝑐𝑁 3 triples 𝑥1 , 𝑥2 , 𝑥3 ∈ 𝐶

10
Proof of BSG using the lemma (3/3)
Any element 𝑦 ∈ 𝐴′ + 𝐵′ can be represented as
𝑦 = 𝑥1 − 𝑥2 + 𝑥3 in at least 2−12 𝜖 5 𝑁 2 ways.
 There are 𝑐𝑁 3 triples 𝑥1 , 𝑥2 , 𝑥3 ∈ 𝐶

Splitting the triples between elements in 𝐴′ + 𝐵′ we get:
3𝑁3
12 3
𝑐
2
𝑐
′
′
𝐴 + 𝐵 ≤ −12 5 2 = 5 𝑁
2 𝜖 𝑁
𝜖
11
Half way there!
 We
will state a lemma in graph theory
 Easily prove BSG using the lemma
 Work hard to prove the lemma
 Take a break before the second part
12
High-level proof of lemma (1)
a1
a
a2
b1
a
b2
a1
a
a2
b1
a
b2
a1
a
a2
b1
a
b2
a3
a4
a’
a5
a6
b3
b4
a’
b5
b6
a3
a4
a’
a5
a6
b3
b4
a’
a3
a4
a’
a5
a6
b3
b4
a’
Define good pairs such as (𝑏1 , 𝑏2 ) and bad pairs such as (𝑏3 , 𝑏4 )
Define a good neighbor for vertex 𝑎 ∈ 𝐴:
𝑏 ∈ Γ 𝑎 forms few bad pairs inside Γ 𝑎
For a random 𝑎, we prove it has many good neighbors
Take specific 𝑎 with many good neighbors, and fix its good neighbors as 𝐵′
Define 𝐴′ as all 𝑎’s with many neighbors in 𝐵′
Show there are many paths of length 3 from 𝑎 ∈ 𝐴′to 𝑏 ∈ 𝐵′
13
High-level proof of lemma
a1
a
a2
b1
a
b2
a1
a
a2
b1
a
b2
a3
a4
a’
a5
a6
b3
b4
a’
a3
a4
a’
a5
a6
b3
b4
a’
Define good pairs such as (𝑏1 , 𝑏2 ) and bad pairs such as (𝑏3 , 𝑏4 )
Define a good neighbor for vertex 𝑎 ∈ 𝐴:
𝑏 ∈ Γ 𝑎 forms few bad pairs inside Γ 𝑎
For a random 𝑎, we prove it has many good neighbors
Take specific 𝑎 with many good neighbors, and fix its good neighbors as 𝐵′
Define 𝐴′ as all 𝑎’s with many neighbors in 𝐵′
Show there are many paths of length 3 from 𝑎 ∈ 𝐴′to 𝑏 ∈ 𝐵′
14
Proof of the lemma (1/6)
𝑏
a

The Lemma:
b
1. Let 𝐻 = (𝐴, 𝐵, 𝐸) be a bipartite graph
𝑎
2. 𝐴 = 𝐵 = 𝑁 , 𝐸 = 𝜖𝑁 2
◦ We want large subsets 𝐴′ , 𝐵′ s.t. for any 𝑎 ∈ 𝐴′ , 𝑏 ∈ 𝐵′ there are many
paths of length 3 from 𝑎 to 𝑏.


The average degree of a vertex is 𝜖𝑁. We start by removing
𝜖
vertices from 𝐵 with degree smaller than 2 𝑁
We deleted less than
𝜖 2
𝑁
2
edges, left with at least
𝜖 2
𝑁
2
edges
15
Proof of the lemma (2/6)

Define a bad pair 𝑏1 , 𝑏2 ∈ 𝐵 if they have less than
𝜖3
𝑁
128
common
neighbors in 𝐴
 For 𝑣 ∈ 𝐴, let 𝐵𝑃(𝑣) denote the number of bad pairs which belong
to Γ 𝑣
 For any bad pair 𝑏1 , 𝑏2 ∈ 𝐵, by definition:
𝜖3
𝑣 ∈ 𝐴 𝑏1 , 𝑏2 ∈ Γ 𝑣 ≤
𝑁
128
 So after “bucketing”, averaging over the vertices yields:
𝜖3
𝑁
3
⋅
𝑁
𝜖
128
𝐸𝑣 𝐵𝑃(𝑣) ≤ 2
≤
𝑁2
𝑁
256
16
Proof of the lemma (3/6)
For 𝑣 ∈ 𝐴, let 𝐵 ⊆ Γ2 𝑣 be the neighbors 𝑏 ∈ Γ 𝑣 which form a bad
𝜖
pair with at least 𝑁 other neighbors 𝑏′ ∈ Γ 𝑣
(𝐵 = “bad
32
neighbors”)
 Let 𝐵𝑁(𝑣) = |𝐵|. For any 𝑣 ∈ 𝐴 :

𝜖2
𝐵𝑁(𝑣) ⋅
𝑁 ≤ 2 ⋅ 𝐵𝑃(𝑣)
32
(Counting the bad pairs using the bad neighbors. Each pair is counted at most twice)
Taking expectation on both sides,
𝑬𝒗
𝑩𝑵(𝒗)
≤ 𝐸𝑣 𝐵𝑃(𝑣) ⋅
64
𝜖2𝑁
≤
𝜖
3
𝑁
256
2
64
⋅
𝜖2𝑁
𝝐
=
𝑵
𝟒
To conclude: we bounded the number of bad neighbors for an average 𝑣 ∈ 𝐴
17
Proof of the lemma (4/6)

So, for an “average” 𝑣 ∈ 𝐴,
◦ There are many neighbors (𝑬𝒗 𝒅𝒆𝒈(𝒗) ≥
𝝐
𝑵)
𝟐
◦ There are few bad neighbors (𝑬𝒗 𝑩𝑵(𝒗) ≤
𝝐
𝑵)
𝟒
◦ So there are many good neighbors! Denote 𝐵′ = Γ(𝑣)\𝐵
◦𝑬
𝑩′
= 𝑬 𝒅𝒆𝒈 − 𝑬 𝑩𝑵 ≥
◦ Take 𝑣 ∈ 𝐴 where
𝐵′
≥
𝝐
𝑵
𝟒
𝜖
𝑁. Fix
4
this to be 𝑩′ .
18
Proof of the lemma (5/6)

So we have our 𝐵′, and 𝐵 ≥
′
𝜖
𝑁
4
𝜖2
Define 𝐴′ = a ∈ 𝐴 | |Γ 𝑣 ∩ 𝐵′ | ≥ 𝑁
16
 Lets lower-bound |𝐴′ |:

◦ Intuition: If |𝐴′ | was small, 𝐸 𝐴, 𝐵′
𝐸 𝐴, 𝐵′
would have few edges (and it doesn’t)
2
𝟐
𝜖
𝝐
≤ 𝐴′ 𝐵′ + 𝐴 \ 𝐴′
𝑁 ≤ 𝑨′ 𝑵 +
𝑵𝟐
16
𝟏𝟔
𝟐
𝜖
𝝐
𝐸 𝐴, 𝐵′ ≥ 𝑁 ∗ 𝐵′ ≥ 𝑵𝟐
2
𝟖
Minimal degree in 𝐵′
𝜖2
Combining these we get 𝐴′ ≥ 16 𝑁
19
Proof of the lemma (6/6)
′
𝜖2
𝜖2
𝜖3
𝑁 −
𝑁 ∗
𝑁 = 2−12 𝜖 5 𝑁 2
16
32
128
𝜖2
𝑁
16

We have 𝐴 , 𝐵′, with 𝐴 , 𝐵 ≥

Last step: show that for any 𝑎 ∈ 𝐴′ , 𝑏 ∈ 𝐵′ there are many
paths of length 3 from 𝑎 to 𝑏.
′
′
Fix 𝑎, 𝑏. From 𝐴′ 𝑠 construction, 𝑎 has many neighbors 𝑏′ ∈ 𝐵′ .
2. Out of these neighbors, only few form a bad pair with 𝑏
3. For 𝑏′ ∈ 𝐵 ′ that don’t form a bad pair with 𝑏, there are many
common neighbors 𝑎′ ∈ 𝐴
4. So we have (many – few) * many paths 𝑎, 𝑏 ′ , 𝑎 ′ , 𝑏 of length 3
1.
20
To conclude the section
 We
will state a lemma in graph theory
 Easily prove BSG using the lemma
 Work hard to prove the lemma
 Take a break before the second part
Questions?
21