Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Sum Product Theorem
Additive Combinatorics Seminr, TAU, 2016
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
Noam Parzenchevski
December 13, 2016
Arithmetic and Geometric Progressions
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
Recall that if A is an arithmetic progression then
|A + A| ≈ 2|A|
Likewise, for a geometric progression |A · A| ≈ 2|A|
Can we find a set A over some field F such that A behaves
like an arithmetic and geometric progression?
Sum-Product Theorems
Sum Product
Theorem
Noam
Parzenchevski
We will see two main results in the next two meetings:
Introduction
Theorem (Erdős–Szemerédi)
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
For A ⊂ R, max{|A + A|, |A · A|} ≥ c · |A|1+
(for absolute c, )
Theorem (Katz-Bourgain-Tao)
∀α > 0 ∃ > 0 s.t. ∀p ∈ P ∀A ⊂ Fp with |A| ≤ p 1−α :
max{|A + A|, |A · A|} ≥ |A|1+
Polynomial Doubling over Bounded Sets
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
In order to establish the thm. we will first acquire a similar
lemma for growth of bounded sets.
Lemma (4.3)
Let A ⊂ R be a set bounded between [m, 2m], then:
max{|A + A|, |A · A|} ≥ c · |A|1+
for absolute c, Growth of Bounded Sets - Proof
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
Order the set a1 < ... < aN and partition A into N 7/8
sets of size s = N 1/8
Denote Ai = {ais , ais+1 , ... , ais+s−1 }
Define the diameter of a set diam(A) = max(A) − min(A)
Let B = argminAi (diam(Ai ))
Growth of Bounded Sets - Proof
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
We claim that if i − j ≥ 10 then:
Ai + B ∩ Aj + B = ∅ and Ai · B ∩ Aj · B = ∅
To see this, let a ∈ Ai , a0 ∈ Aj , b, b 0 ∈ B and let
a = a0 + x, b = b 0 + y
Note that
x = a − a0 ≥ diam(Aj+1 ) + ... + diam(Ai−1 ) ≥ 9 · diam(B)
Additionally, |y | = b − b 0 ≤ diam(B)
Together, we get that x ≥ 9 · diam(B) ≥ 9 · |y |
Growth of Bounded Sets - Proof
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
Assume
a+b = a0 +b 0 → a0 +b 0 +x +y = a+b = a0 +b 0 → x = |y |
which is a contradiction
Assume ab = a0 b 0 → (a0 + x)b = ab = a0 b 0 =
0
a0 (b − y ) → bx = −ya0 → yx = − ab
However, we know that x/y ≥ 9 and 1/2 ≤ a/b ≤ 2
which asserts our claim.
Growth of Bounded Sets - Proof
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
By the claim we can clearly see that
|A + A| + |A · A| ≥ Σi |A10i + B| + |A10i · B|
Let I = {i | |Ai + B|, |Ai · B| ≤ s 1+α }
Hopefully |I | is small and we can “ignore” elements in
these sets
We will choose an α that ensures us that there are
2 )b4
b1 , b2 , b3 , b4 ∈ B such that (bb1 −b
∈ Ai
3 −b4
To get this, note that there are s 2 nondistinct sums a + b
for a ∈ Ai and b ∈ B
By assumption there are ≤ s 1+α distinct sums so by
pigeonholing:
∃σ s.t. aj + bj = σ for
s2
s 1+α
= s 1−α pairs
Growth of Bounded Sets - Proof
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
1−α Consider the s 2 ≈ s 2−2α products aj bk for
aj + bj = ak + bk = σ
These products also take at most s 1+α distinct values by
assumption
By letting α < 1/3 we get again that there is a pair with
the same product
This means that there are a1 , a2 ∈ Ai and
b1 , b2 , b3 , b4 ∈ B s.t. a1 + b1 = a2 + b2 and a1 b3 = a2 b4
We now conclude that a1 =
(b1 −b2 )b4
b3 −b4
as needed
Growth of Bounded Sets - Proof
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
Now, since |B|4 = s 4 the expression
take at most s 4 distinct values
(b1 −b2 )b4
b3 −b4
∈ Ai can
Since Ai ∩ Aj = ∅ we can see that |I | ≤ s 4 = N 1/2
We now want to get a buond on the sum and product of
sets not in I
Growth of Bounded Sets - Proof (conclusion)
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
We note that there are
1 7/8
10 N
sets of type A10i
Additionally,
1
∀j ∈
/ I : |Aj + B| + |Aj · B| ≥ s 1+α = N 8 (1+α)
We can now see that:
1
1 7/8
|A + A| + |A · A| ≥ ( 10
N
− N 1/2 ) · N 8 (1+α) =
1 1+ α
8
10 N
(5+α)
α
1 1+ 8
− N 8 ≥ 20
N
Holds for example for N ≥ 3000 and α < 1/3
The lemma follows by an averaging argument Erdős–Szemerédi Proof
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
We now prove the theorem using the lemma
First, we can assume wlog that the set is positive
If not, we split the set into a A+ and A−
We now take the one with higher cardinality and change
sign if necessary
This reduces the set by at most a factor of 2
Erdős–Szemerédi Proof
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
Let Ai = {a ∈ A | 2i < a ≤ 2i+1 }
Note: Ai is bounded as needed by the lemma
Obviously ∀i 6= j (Ai + Ai ) ∩ (Aj + Aj ) = ∅
The same holds for the product set
We now set I = {i | 0 < |Ai | < N 1/4 } and distinguish two
cases
Erdős–Szemerédi Proof
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
We first examine the case where Σi∈I |Ai | < N/2:
We “empty” all small subsets {Ai | i ∈ I }
If Ai is not empty, it follows that |Ai | ≥ N 1/4
Since the lemma holds for Ai we get that:
|Ai + Ai | + |Ai · Ai | ≥ c · |Ai |1+ = c · |Ai | · |Ai | ≥
c · |Ai | · (N 1/4 ) = c · |Ai | · N /4
Erdős–Szemerédi Proof
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
We now get that
/4 ·|A |
|A+A|+|A·A| ≥ Σi ∈I
i
/ |Ai +Ai |+|Ai ·Ai | ≥ Σi ∈I
/ c ·N
Since Σi∈I |Ai | < N/2 we can conclude:
/4 · 1 N = c N 1+/4
c · N /4 · Σi ∈I
/ |Ai | ≥ c · N
2
2
Erdős–Szemerédi Proof
Sum Product
Theorem
Noam
Parzenchevski
Introduction
We now examine the case where Σi∈I |Ai | ≥ N/2:
Bounded Sets
Lemma
We first note that |I | ≥ 12 N 3/4
Erdős
Szemerédi
Theorem
Pick I 0 ⊂ I such that ∀i 6= j |i − j| ≥ 2 and
|I 0 | = 12 |I | ≥ 14 N 3/4
Sum-Product
in Finite
Fields
We now pick a representative ai ∈ Ai for each i ∈ I 0 and
claim that their pairwise sums are distinct.
Growth of
Sets Under
Polynomial
Expressions
This claim will conclude the proof
Erdős–Szemerédi Proof
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
To see the claim we assume the opposite
That is, ai + aj = ak + al for distinct i, j, k, l
Say, wlog, i > max(j, k, l) → ai > 2i
However, since k ≤ i − 2 and l ≤ i − 4 we get
k + l ≤ 2i−1 + 2i−3 < 2i Erdős–Szemerédi Proof (conclusion)
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
By the above claim, clearly
1 3/4 0 ≈ c · N 3/2
|A + A| ≥ |I2 | ≥ 4 N2
By proving a polynomial growth for both cases the
theorem follows Erdős–Szemerédi Proof (final remarks)
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Corollary
Erdős
Szemerédi
Theorem
Én elmegyek te itt maradsz isten veled te kismalac
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
I’m leaving, you’re staying. May god be with you, little
piglet
Sum-Product in Finite Fields: The Final Goal
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
We now want to prove the KBT theorem, and in order to
do that we will require two lemmas
The rest of this talk will prove the first lemma
Next week we will prove the second lemma and the main
theorem will follow
Polynomial Sets
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
The lemma we will prove deals with “polynomial
expressions” on sets
That is R(A) which employ a fixed number of additions,
subtractions and multiplications
Specifically, we will look at the following elegant
expression:
R(A) = (A−A)·(A−A)+(A−A)·A+(A−A+A·A−A·A)·A
Our lemma will deal with the growth of A under this
operation
Growth of Sets Under R(·)
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
We now state our lemma:
Erdős
Szemerédi
Theorem
Lemma (4.5)
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
Let A ⊂ Fp be a set such that |A| ≤ p 0.9 then |R(A)| ≥ |A|1.01
Growth of Sets Under R(·) - proof
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
Consider the set of λ ∈ Fp for which |A + λA| < |A|2
That is, there exist a1 6= a3 and a2 6= a4 such that
a1 + λa2 = a3 + λa4
Equivalently, we see that λ =
a3 −a1
a2 −a4
Let A0 = (A − A)\{0} and
B = A0 /A0 = {a10 /a20 | a10 , a20 ∈ A0 } be the set of those λ0 s
We will consider two cases: B = F∗p and B 6= F∗p
Interlude - Ruzsa Triangle Inequality for
Multiplicative Groups
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
Recall that the Ruzsa triangle inequality guarantees that
for any subset of a group A, B, C ⊂ G:
|A| · |B − C | ≤ |A − B| · |A − C |
In this sense the substraction operation can be seen as
addition of the inverse:
B − C = {b + c 0 | b ∈ B and ∃c ∈ C s.t. c + c 0 = 0}
We can now employ this inequality for the multiplicative
group F∗p
Interlude - conclusion
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
Recall, F∗p = ({1, ..., p − 1}, ·, 1)
In this group the multiplication by the inverse gives us
division, that is:
B/C = {b · c −1 | b ∈ B and c ∈ C }
And so we get the analogous of the inequality over F∗p :
|A| · |B/C | ≤ |A/B| · |A/C |
Growth of Sets Under R(·) - B = F∗p
Sum Product
Theorem
Noam
Parzenchevski
We now look at the multiplicative group B = F∗p and
R0 (A) = (A − A) · (A − A)
Bounded Sets
Lemma
We set Atriangle := A0 , Btriangle , Ctriangle := (A0 )−1 =
{a−1 | a ∈ A0 }
Erdős
Szemerédi
Theorem
By Ruzsa:
|A0 | · |(A0 )−1 /(A0 )−1 | ≤ |A0 /(A0 )−1 | · |A0 /(A0 )−1 |
Introduction
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
We invert the operations to get |A0 | · |A0 /A0 | ≤ |A0 · A0 |2
For B = A0 /A0 this yields |A0 · A0 | ≥ (|B| · |A0 |)1/2
We now conclude:
|R0 (A)| = |(A − A) · (A − A)| ≥ |A0 · A0 | ≥ (|B| · |A0 |)1/2
Since |B| = p and |A| ≤ p 0.9 we get that the above
= p 1/2 · |A0 |1/2 ≥ |A|19/18
Growth of Sets Under R(·) - B 6= F∗p
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
Fix a nonzero a ∈ A, and we claim that there exists λ ∈ B
s.t. λ + a ∈
/B
To prove this assume the opposite, that is
∀λ ∈ B : λ + a ∈ B
This implies λ + a + a ∈ B which implies λ + 3a ∈ B and
so on...
Since a group of prime order is generated by any non zero
element, we have that a generates Fp in B
Therefore, such a λ exists
We now set x = λ + a
Growth of Sets Under R(·) - B 6= F∗p
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
To finish this case we need one more claim:
|A + xA| = |A|2 iff x ∈
/B
Proof: |A + xA| = |A|2
iff the map (a, b) 7→ a + xb is injective on A × A
iff a + xb 6= c + xd for all distinct (a, b), (c, d) ∈ A × A
iff x 6=
a−c
d−b
iff x ∈
/B where a 6= c and b 6= d
Growth of Sets Under R(·) - B 6= F∗p
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
To recap, we now have |A + (λ + a) · A| = |A|2
We plug λ =
a3 −a1
a2 −a4
−a1
and get |A + ( aa23 −a
+ a) · A| = |A|2
4
We can now multiply by a2 − a4 and get:
|(a2 − a4 ) · A + ((a3 − a1 ) + aa2 − aa4 ) · A| ≥ |A|2 which
implies:
|(A − A) · A + (A − A + A · A − A · A) · A| ≥ |A|2
We now set
R1 (A) = (A − A) · A + (A − A + A · A − A · A) · A
Growth of Sets Under R(·) - conclusion
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions
For each case (B = F∗p and B 6= F∗p ) we found a poly.
expression with poly. growth
Thus, we
conclude:|R(A)| = |R0 (A)| + |R1 (A)| ≥ |A|19/18 > A1.01 Thank You
Sum Product
Theorem
Noam
Parzenchevski
Introduction
Bounded Sets
Lemma
Erdős
Szemerédi
Theorem
Sum-Product
in Finite
Fields
Growth of
Sets Under
Polynomial
Expressions