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Sum-product theorems over finite fields
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Goal Last week we saw Theorem[Erdos-Szemeredi]: βπ π .π‘ πππ πππ¦ π΄βπ
, max π΄+π΄ , π΄β
π΄ β₯| π΄| 1+π We saw a proof last week of π= and itβs conjectured to be close to 1. Today we wish to extend this theorem to finite Fields Problem 1: If πΉ has a subfield πΉβ², then clearly for π΄= πΉ β² , |π΄+π΄|=|π΄β
π΄|=|π΄| . So this limits us to πΉ π where p is a prime. Problem 2: It always holds that π΄+π΄ , π΄β
π΄ β€π then clearly if π΄ is too close to π we canβt expect too much growth
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Goal General statement:
β πΌ>0 β π>0 π .π‘ πΉππ πππ¦ πππππ π πππ π ππ‘ π΄β πΉ π ππ π ππ§π π΄ β€ π 1βπΌ ππ‘ βππππ π‘βππ‘ max π΄+π΄ , π΄β
π΄ max π΄+π΄ , π΄β
π΄ β₯ π΄ 1+π Today we show: πΉππ πππ¦ πππππ π πππ π ππ‘ π΄β πΉ π ππ π ππ§π π΄ β€ π 0.9 ππ‘ βππππ π‘βππ‘ max π΄+π΄ , π΄β
π΄ β₯ π΄ π πππ π πππ ππππ π‘πππ‘ π
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Plan Lemma 1: πΉππ πππ¦ π΄β πΉ π , π΄ β€ π 0.9 , ππ‘ βππππ π
β² π΄ β₯ π΄ 1.01
πΉππ πππ¦ π΄β πΉ π , π΄ β€ π 0.9 , ππ‘ βππππ π
β² π΄ β₯ π΄ 1.01 Where Rβ² = π΄βπ΄ β
π΄βπ΄ + π΄βπ΄ β
π΄+ π΄βπ΄+π΄β
π΄βπ΄β
π΄ β
π΄ (We saw the proof of this lemma last week) Lemma 2: πΏππ‘ π΄β πΉ π ππ π .π‘ π΄+π΄ , π΄β
π΄ β€ π΄ 1+π . πΉππ πππ¦ ππππ¦ππππππ ππ₯ππππ π πππ π
(β
) π‘βπππ ππ₯ππ π‘ π π π’ππ ππ‘ π΅ π π’πβ π‘βππ‘ π
π΅ β€ π΅ 1+ππ π€βπππ π=π(π
) is π ππππ π‘πππ‘ π‘βππ‘ πππππππ ππππ¦ ππ π‘βπ ππππ¦ππππππ
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Proof of Theorem (Using lemmas)
Let π΄ β€ π 0.9 and let π be such that π΄+π΄ , π΄β
π΄ β€ |π΄| 1+π . Using lemma 2 with Rβ gives us that exists π΅βπ΄ such that π
β² π΅ β€|π΅| 1+π( π
β² )π But by lemma 1 we know that π
β² π΅ β₯|π΅| 1.01 Meaning πβ₯ 1 100π( π
β² ) as requested.
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Lemma 2 plan From now until the end of the talk, we aim to prove lemma 2: πΏππ‘ π΄β πΉ π ππ π .π‘ π΄+π΄ , π΄β
π΄ β€ π΄ 1+π . πΉππ πππ¦ ππππ¦ππππππ ππ₯ππππ π πππ π
(β
) π‘βπππ ππ₯ππ π‘ π π π’ππ ππ‘ π΅ π π’πβ π‘βππ‘ π
π΅ β€ π΅ 1+ππ π€βπππ π=π(π
) is π ππππ π‘πππ‘ π‘βππ‘ πππππππ ππππ¦ ππ π‘βπ ππππ¦ππππππ
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Proving Lemma 2 Lemma 3: Let A be a subset of an Abelian group s.t π΄+π΄ β€ πΎ|π΄|. Then exists π΅β π΄ of size π΅ β₯ πΎ βπ 1 |π΄| s.t any π 1 β π 2 βπ΅βπ΅ can be written as: π 1 β π 2 = π=1 12 π π , π π βAβͺβπ΄ In at least πΎ βπ 1 π΄ 11 distinct ways and any π 1 + π 2 βπ΅+π΅ can be written as π 1 + π 2 = π=1 12 π π , π π βAβͺβπ΄ Write statement on board
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Proving Lemma 3 (1/4) We will use the BSG theorem (distinct paths of length 3 in dense bipartite graph) Let π = |π΄| ; Let K s.t π΄+π΄ β€ πΎ|π΄| Let πΆβ π΄+π΄ be the set of elements that can be written as π = π 1 + π 2 in π 2π distinct ways. We note that πΆ β₯ π 2π (Shown on board) Denote H as the following bipartite graph: each element in aβ π΄ has a vertex in both the left and the right side. πΈ(π») = {( π 1 , π 2 )| π 1 + π 2 β πΆ}
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Proving Lemma 3 (2/4) Recall Sudakovβs Lemma - Let H = (A;B;E) be a bipartite graph with vertex sets A,B and edge set E. Assume that |π΄|=|π΄β²| =π and |πΈ| =π π 2 . Then there exist π΅βπ΄,π΅ββπ΄β² of size |π΅|,|π΅β|β₯ ( π )π such that for any πβπ΅, πβ²β π΅β there are at least Ξ©( π 5 π 2 ) paths of length 3 between b and bβ. Here E = πΆ β
π 2π β₯ π 2 4 π 2 , meaning we have |π΅|,|π΅β|β₯ πΎ βπ 1 π, where each b,bβ has at least πΎ βπ 1 π 2 paths of length 3.
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Proving Lemma 3 (3/4) |π΅|,|π΅β|β₯ πΎ βπ 1 π ; Each π,πβ has at least πΎ βπ 1 π 2 paths of length 3. A path like that looks like (b,a,aβ,bβ). Meaning we can write πβ+π= π+π β π+ π β² + π β² + π β² = π 1 β π 2 + π 3 In at least πΎ βπ 1 π 2 ways. Any element in C can be written as π= π 1 + π 2 at least π 2π distinct ways means that π+πβ= π=1 6 π π ; π π βπ΄βͺβπ΄ Can be written in πΎ βπ 1 π 5 distinct ways
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Proving Lemma 3 (4/4) So we can express any π 1 β π 2 βπ΅βπ΅ as ( π 1 +πβ) β ( π 2 +πβ) for any πββ π΅β² so overall we can express π 1 β π 2 = π=1 12 π π ; π π βπ΄βͺβπ΄ in πΎ βπ 1 π 5 β
πΎ βπ 1 π 5 β
πΎ βπ 1 π= πΎ βπ 1 π 11 ways. π΅+π΅ proven in similar way
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Lemma 4 We proceed to prove by induction that any polynomial expression does not grow (which is the original goal of lemma 2) Lemma 4: For any integers t,s there exists a constant c = c(t,s) > 0 such that the following holds. If π΄+π΄ , π΄β
π΄ β€ π΄ 1+π then there exists π΅β π΄ of size π΅ β₯ π΄ 1βππ such that | π΅ π‘ β π΅ π‘ π΅ π π΅ βπ | β€ π΅ 1+ππ
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Lemma 2 Proof From Lemma 4 Reminder: PlΓΌnnekeβRuzsa theorem
πΌπ π΄βπ΄ ππ π΄+π΄ β€ πΎ π΄ π‘βππ ππ΄βππ΄ β€ πΎ π+π π΄ for any π,πβπ Note that this is a stronger than lemma 2 for the case of linear functions
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Lemma 2 Proof From Lemma 4 If π΄+π΄ , π΄β
π΄ β€ π΄ 1+π then there exists π΅β π΄ of size π΅ β₯ π΄ 1βππ such that | π΅ π‘ β π΅ π‘ π΅ π π΅ βπ | β€ π΅ 1+ππ Plug π =0. we get that | π΅ π‘ β π΅ π‘ | β€ π΅ 1+ππ . By PlΓΌnnekeβRuzsa theorem, there exists cβ such that | π‘π΅ π‘ β π‘π΅ π‘ | β€ π΅ 1+πβ²π . So we get that any monomial any sum of monomials over degree t is not large. Take t to be the degree of R(B) and we get that R(B) isnβt too large
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Proving Lemma 4 (1/8) Proof by induction on t. Base case π‘=1
Apply lemma 3 to get a B of size π΅ β₯ π΄ 1βπ π such that π 1 β π 2 βπ΅βπ΅ can be written as: π 1 β π 2 = π=1 12 π π , π π βAβͺβπ΄ in at least π΄ 11βπ π distinct ways. So any π₯β π΅βπ΅ π΅ π π΅ βπ can be written as π₯= π=1 12 π₯ π , π₯ π β Β±A π +1 π΄ βπ in at least π΄ 11βπ π distinct ways.
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Proving Lemma 4 (2/8) Applying PlΓΌnnekeβRuzsa on the multiplicative group of πΉ π we get that π΄ π +1 π΄ βπ β€ π΄ 1+ 2π +1 π = π΄ 1+π(ππ ) (here m=s+1,l=s), therefore the number of choices for the π₯ π βs is a most ( π΄ 1+π ππ ) 12 = π΄ 12+π(ππ ) So π΅βπ΅ π΅ π π΅ βπ β€ π΄ 12+π ππ π΄ 11βπ π = π΄ 1+π ππ β€ π΅ 1+π ππ +π(π) Which concludes the case of t=1 (for any s).
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Proving Lemma 4 (3/8) Assume correctness for t, and weβll show that it holds for t+1. Let l(t,s)=t+s+12. By induction we assume that there exists π΅β π΄ of size π΅ β₯ π΄ 1βππ s.t | π΅ π‘ β π΅ π‘ π΅ π π΅ βπ | β€ π΅ 1+ππ . In particular π΅β
π΅ β€ π΅ 1+ππ , so applying lemma 3 to the multiplicative group of πΉ π , we get that there exists πΆβ π΅ of size πΆ β₯ π΅ 1βπ ππ s.t any π₯β πΆβ
C can be written as π₯= Ξ π=1 12 π π π π βπ΅βͺ π΅ β1 in at least π΅ 11βπ ππ distinct ways.
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Proving Lemma 4 (4/8) We can assume (and lose a constant factor) that for some 1β€πβ€ 12 , π 1 ,β¦, π π β π΅ and π π+1 ,β¦, π 12 β π΅ β1 (i.e elements in B are consecutive, and elements in π΅ β1 are consecutive) Multiplying by an element in πΆ π‘β1 we get that we can write any element π₯β πΆ π‘+1 as π₯= Ξ π=1 12 π π ; π 1 β π΅ π‘ , π 2 ,β¦, π π βπ΅, π π+1 ,β¦ π 12 β π΅ β1 in at least π΅ 11βπ ππ distinct ways. Now we look at π₯,π¦β πΆ π‘+1 , and look at π₯= π₯ 1 β
π₯ 2 β
β¦β
π₯ 12 , π¦= π¦ 1 β
π¦ 2 β
β¦β
π¦ 12 π₯ 1 , π¦ 1 β π΅ π‘ , π₯ π , π¦ π βπ΅ πππ 2β€πβ€π, π₯ π , π¦ π β π΅ β1 πππ π+1β€πβ€12
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Proving Lemma 4 (5/8) π₯βπ¦= π₯ 1 β
π₯ 2 β
β¦β
π₯ 12 β π¦ 1 β
π¦ 2 β
β¦β
π¦ 12 =
π₯ 1 β π¦ 1 π₯ 2 π₯ 3 β¦ π₯ 12 + π¦ 1 π₯ 2 β π¦ 2 π₯ 3 β¦ π₯ 12 + π¦ 1 π¦ 2 π₯ 3 β π¦ 3 π₯ 4 β¦ π₯ 12 + β¦ + π¦ 1 π¦ 2 β¦ π¦ 11 ( π₯ 12 β π¦ 12 ) (This is a telescopic sum)
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Proving Lemma 4 (6/8) π₯ 1 β π¦ 1 π₯ 2 π₯ 3 β¦ π₯ 12 +
π₯ 1 β π¦ 1 π₯ 2 π₯ 3 β¦ π₯ 12 + π¦ 1 π₯ 2 β π¦ 2 π₯ 3 β¦ π₯ 12 + π¦ 1 π¦ 2 π₯ 3 β π¦ 3 π₯ 4 β¦ π₯ 12 + β¦ + π¦ 1 π¦ 2 β¦ π¦ 11 ( π₯ 12 β π¦ 12 ) The first sum element is contained in π΅ π‘ β π΅ π‘ π΅ πβ1 π΅ 12βπ The next m-1 elements are contained in B π‘ π΅βπ΅ π΅ πβ2 π΅ β(12βπ) The next 12-m elements are contained in B π‘ π΅ β1 β π΅ β1 π΅ πβ1 π΅ β(11βπ)
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Proving Lemma 4 (7/8) Itβs easy to see that π΅ β1 β π΅ β1 β π΅βπ΅ π΅ β2 .
So π΅ π‘ β π΅ π‘ π΅ πβ1 π΅ 12βπ , B π‘ π΅βπ΅ π΅ πβ2 π΅ β(12βπ) , B π‘ π΅ β1 β π΅ β1 π΅ πβ1 π΅ β(11βπ) are all contained in π΅ π‘ β π΅ π‘ π΅ π‘ π΅ 12 π΅ β12 If we multiply by an element of π΅ π π΅ βπ we get that any element of πΆ π‘+1 β πΆ π‘+1 πΆ π πΆ βπ can be written as 12 terms of π΅ π‘ β π΅ π‘ π΅ π‘+π +12 π΅ β(π‘+π +12) = π΅ π‘ β π΅ π‘ π΅ π π΅ βπ in π΅ 11βππ ways
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Proving Lemma 4 (8/8) Any element of πΆ π‘+1 β πΆ π‘+1 πΆ π πΆ βπ can be written as 12 terms of π΅ π‘ β π΅ π‘ π΅ π‘+π +12 π΅ β(π‘+π +12) = π΅ π‘ β π΅ π‘ π΅ π π΅ βπ in π΅ 11βππ ways By induction on t we have that the number of xβs in π΅ π‘ β π΅ π‘ π΅ π π΅ βπ is at most π΅ 12βπ ππ | πΆ π‘+1 β πΆ π‘+1 πΆ π πΆ βπ |β€ π΅ 12βπ ππ π΅ 11βπ ππ = π΅ 1+π ππ Which concludes the reduction and the theorem.
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Considering the constants
We can get from the base case that π(1,π ) = π(π ) And the induction step gives us π(π‘+1,π ) = π(π(π‘,π‘+π +12)) [Here the O(.) is very important)] What weβve shown grows exponentially in t (and linearly in s) For our chosen Rβ from lemma 1 Rβ² = π΄βπ΄ β
π΄βπ΄ + π΄βπ΄ β
π΄+ π΄βπ΄+π΄β
π΄βπ΄β
π΄ β
π΄ we get a very large constant, meaning we shown over all that πβ₯ 1 100π
π β 1 100β
ππππ π‘ 12
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