SUPER AWESOME STUFF · 3-Term Arithmetic Progressions in Groups A More General Question What is a 3-Term Arithmetic Progression? 3 numbers de ned by x 1 = a, x 2 = a + 2d, and x 3

IntroductionLemmas

Proof of TheoremThe End

SUPER AWESOME STUFFLinear Forms Over Finite Abelian Groups

Ran JiMentor: Craig V. Spencer

Kansas State UniversityHome Institution: Wellesley College

July 27, 2011

Ran Ji Mentor: Craig V. Spencer Super Awesome Stuff

IntroductionLemmas


3-Term Arithmetic Progressions in GroupsA More General Question

What is a 3-Term Arithmetic Progression?

3 numbers defined by x1 = a, x2 = a + 2d , and x3 = a + dwhere a, d ∈ Z

Formalized as a solution to x1 − x2 − 2x3 = ~r · ~x = 0 where~r = (1, 1,−2)

Ex. 1: 3, 7, 11 ∈ ZEx. 2: 5, 0, 2 ∈ Z7


IntroductionLemmas




3 numbers defined by x1 = a, x2 = a + 2d , and x3 = a + dwhere a, d ∈ ZFormalized as a solution to x1 − x2 − 2x3 = ~r · ~x = 0 where~r = (1, 1,−2)

Ex. 1: 3, 7, 11 ∈ ZEx. 2: 5, 0, 2 ∈ Z7


IntroductionLemmas





Ex. 1: 3, 7, 11 ∈ Z

Ex. 2: 5, 0, 2 ∈ Z7


IntroductionLemmas





Ex. 1: 3, 7, 11 ∈ ZEx. 2: 5, 0, 2 ∈ Z7


IntroductionLemmas



The Problem

Define ~x as a trivial solution to ~r · ~x = 0 if d = 0, i.e.x1 = x2 = x3

For a finite abelian group G , want to find the maximum sizeof A ⊆ G such that A contains no non-trivial solutions to~r · ~x = 0

Theorem (Meshulam, 1993)

Let G be a finite abelian group with 2 - G and let D(G ) denote themaximal cardinality of a set A ⊆ G such that A contains nonon-trivial 3-term arithmetic progressions. Then for someM, β > 0,

D(G ) ≤ M|G |

(ln |G |)β.


IntroductionLemmas



The Problem





D(G ) ≤ M|G |

(ln |G |)β.


IntroductionLemmas



The Problem





D(G ) ≤ M|G |

(ln |G |)β.


IntroductionLemmas



Now we want to generalize the question to finding solutions to~r · ~x = 0 for the general linear form ~r = (r1, · · · , rs).

Theorem (Liu and Spencer, 2009)

Let r1, . . . , rs be non-zero integers satisfying r1 + . . .+ rs = 0. Let

G ' Zk1 ⊕ · · · ⊕ Zkn , n is called the rank of G

be a finite abelian group with ki |ki−1 (2 ≤ i ≤ n), and supposethat (ri , k1) = 1 (1 ≤ i ≤ s). Let D(G ) denote the maximalcardinality of a set A ⊂ G containing no non-trivial solution ofr1x1 + · · ·+ rsxs = 0 with xi ∈ A (1 ≤ i ≤ s). Then for somecomputable constant M,

D(G ) ≤ M|G |

ns−2 .


IntroductionLemmas



Now we want to generalize the question to finding solutions to~r · ~x = 0 for the general linear form ~r = (r1, · · · , rs).

Theorem (Liu and Spencer, 2009)

Let r1, . . . , rs be non-zero integers satisfying r1 + . . .+ rs = 0. Let

G ' Zk1 ⊕ · · · ⊕ Zkn , n is called the rank of G

be a finite abelian group with ki |ki−1 (2 ≤ i ≤ n), and supposethat (ri , k1) = 1 (1 ≤ i ≤ s). Let D(G ) denote the maximalcardinality of a set A ⊂ G containing no non-trivial solution ofr1x1 + · · ·+ rsxs = 0 with xi ∈ A (1 ≤ i ≤ s). Then for somecomputable constant M,

D(G ) ≤ M|G |

ns−2 .


IntroductionLemmas



For this summer project, we used the less general linear form~r = (1, . . . , 1,−s) ∈ Zs+1. Note: ~r = (1, . . . , 1,−s) mimics theconstruction of (1, 1,−2).


IntroductionLemmas



Main Theorem

Theorem

Let ~r = (1, . . . , 1,−s), G be a finite abelian group, and D(G ) bethe maximal cardinality of a set A ⊆ G containing no non-trivialsolution to x1 + · · ·+ xs − sxs+1 = 0 with xi ∈ A (1 ≤ i ≤ s + 1).Define

d(m) = supc(H)≥m

D(H)

|H|

where c(H) is the rank of H. For any n ∈ N, d(n) ≤ Cns−2 , where

C = max

{1,√

s2 + s

√√√√(2s−4log 2

)2s−42

(2s−4log 2

) ,2s−1

2(s−2)2(2s−2 − 1)s−2

}.


IntroductionLemmas


First LemmaSecond Lemma

Lemma (Lemma 1)

For 1 ≤ i ≤ s + 1, let fi (χ) =∑x∈A

χ(rix) =∑x∈A

χri (x).

For ~r = (11, . . . , 1s ,−s),

(F) =∑χ∈G

f1(χ)f2(χ) · · · fs+1(χ) ≤ |G | |A|s−1(

s + 1

2

).

Let G be a finite abelian group with rank ≥ n. Suppose thatA ⊆ G contains no non-trivial solution to x1 + · · ·+ xs − sxs+1 = 0with xi ∈ A (1 ≤ i ≤ s + 1). G is the set of all homomorphismsfrom G to C×. Let χ ∈ G .


IntroductionLemmas



Lemma (Lemma 1)

For 1 ≤ i ≤ s + 1, let fi (χ) =∑x∈A

χ(rix) =∑x∈A

χri (x).

For ~r = (11, . . . , 1s ,−s),

(F) =∑χ∈G

f1(χ)f2(χ) · · · fs+1(χ) ≤ |G | |A|s−1(

s + 1

2

).

Let G be a finite abelian group with rank ≥ n. Suppose thatA ⊆ G contains no non-trivial solution to x1 + · · ·+ xs − sxs+1 = 0with xi ∈ A (1 ≤ i ≤ s + 1).

G is the set of all homomorphismsfrom G to C×. Let χ ∈ G .


IntroductionLemmas



Lemma (Lemma 1)

For 1 ≤ i ≤ s + 1, let fi (χ) =∑x∈A

χ(rix) =∑x∈A

χri (x).

For ~r = (11, . . . , 1s ,−s),

(F) =∑χ∈G

f1(χ)f2(χ) · · · fs+1(χ) ≤ |G | |A|s−1(

s + 1

2

).

Let G be a finite abelian group with rank ≥ n. Suppose thatA ⊆ G contains no non-trivial solution to x1 + · · ·+ xs − sxs+1 = 0with xi ∈ A (1 ≤ i ≤ s + 1). G is the set of all homomorphismsfrom G to C×. Let χ ∈ G .


IntroductionLemmas



Then

(F) =∑χ∈G

f1(χ)f2(χ) · · · fs+1(χ)

=∑χ∈G

(∑x1∈A

χ(x1)

)· · ·

(∑xs∈A

χ(xl)

)( ∑xs+1∈A

χ(−sxs+1)

)

=∑x1∈A· · ·

∑xs+1∈A

∑χ∈G

χ(x1 + x2 + · · ·+ xs − sxs+1)

Fact 1:∑χ∈G

χ(x) =

{|G | if x = 0

0 if x 6= 0


IntroductionLemmas



Then

(F) =∑χ∈G

f1(χ)f2(χ) · · · fs+1(χ)

=∑χ∈G

(∑x1∈A

χ(x1)

)· · ·

(∑xs∈A

χ(xl)

)( ∑xs+1∈A

χ(−sxs+1)

)

=∑x1∈A· · ·

∑xs+1∈A

∑χ∈G

χ(x1 + x2 + · · ·+ xs − sxs+1)

Fact 1:∑χ∈G

χ(x) =

{|G | if x = 0

0 if x 6= 0


IntroductionLemmas



According to Fact 1,∑χ∈G

χ(x1 + · · ·+ xs − sxs+1) = 1 only when

x1 + · · ·+ xs − sxs+1 = 0. So the above equation becomes

|G | · {~x ∈ As+1 : ~r · ~x = 0}.

Since A contains no non-trivial solutions, we are left with

|G | ·∑

1≤i<j≤s+1

# {~x ∈ As+1 : xi = xj ,~r · ~x = 0}.


IntroductionLemmas



According to Fact 1,∑χ∈G

χ(x1 + · · ·+ xs − sxs+1) = 1 only when

x1 + · · ·+ xs − sxs+1 = 0. So the above equation becomes

|G | · {~x ∈ As+1 : ~r · ~x = 0}.

Since A contains no non-trivial solutions, we are left with

|G | ·∑

1≤i<j≤s+1

# {~x ∈ As+1 : xi = xj ,~r · ~x = 0}.


IntroductionLemmas



We want to find upper bound on∑1≤i<j≤s+1

# {~x ∈ As+1 : xi = xj ,~r · ~x = 0}.

For some fixed i , 1 ≤ i ≤ s + 1, we want to look at all solutions ~xto ~r · ~x = 0 such that xi = xj for some i 6= j . l+1∑

w=1w 6=i , j

rwxw

+ (ri + rj)xi = 0

Rewrite as q1x1 + · · ·+ qsxs = 0.


IntroductionLemmas




# {~x ∈ As+1 : xi = xj ,~r · ~x = 0}.


w=1w 6=i , j

rwxw

+ (ri + rj)xi = 0



IntroductionLemmas




# {~x ∈ As+1 : xi = xj ,~r · ~x = 0}.


w=1w 6=i , j

rwxw

+ (ri + rj)xi = 0



IntroductionLemmas



WLOG, let q1 = · · · = qs−2 = 1. Then,

x1 =−q2x2 − q3x3 − · · · − qsxs

q1

= −q2x2 − q3x3 − · · · − qsxs

So we have that

∑1≤i<j≤s+1

#{~x ∈ As+1 : xi = xj , ~r · ~x = 0} ≤∑

1≤i<j≤s+1

|A|s−1.


IntroductionLemmas



WLOG, let q1 = · · · = qs−2 = 1. Then,

x1 =−q2x2 − q3x3 − · · · − qsxs

q1

= −q2x2 − q3x3 − · · · − qsxs

So we have that

∑1≤i<j≤s+1

#{~x ∈ As+1 : xi = xj , ~r · ~x = 0} ≤∑

1≤i<j≤s+1

|A|s−1.


IntroductionLemmas



Plugging this upper bound back into our original equation gives us

(F) =∑χ∈G

f1(χ)f2(χ) · · · fs+1(χ)

= |G |∑

1≤i<j≤s+1

#{~x ∈ As+1 : xi = xj ,~r · ~x = 0}

≤ |G |∑

1≤i<j≤s+1

|A|s−1

= |G | |A|s−1(

s + 1

2

).


IntroductionLemmas



Let W be the kernel of χ1, where

∣∣∣∣∑x∈A

χ1(x)

∣∣∣∣ = maxχ 6=χ0

∣∣∣∣∑x∈A

χ(x)

∣∣∣∣.

Lemma (Lemma 2)

For 1 ≤ i ≤ s + 1, let fi (χ) =∑x∈A

χ(rix) =∑x∈A

χri (x).

For ~r = (11, . . . , 1s ,−s),

(F) =∑χ∈G

f1(χ)f2(χ) · · · fs+1(χ)

≥ |A|s+1 − |A|2 |G |

(D(W )

|W ||G | − |A|

)s−2

.


IntroductionLemmas



Let W be the kernel of χ1, where

∣∣∣∣∑x∈A

χ1(x)

∣∣∣∣ = maxχ 6=χ0

∣∣∣∣∑x∈A

χ(x)

∣∣∣∣.Lemma (Lemma 2)

For 1 ≤ i ≤ s + 1, let fi (χ) =∑x∈A

χ(rix) =∑x∈A

χri (x).

For ~r = (11, . . . , 1s ,−s),

(F) =∑χ∈G

f1(χ)f2(χ) · · · fs+1(χ)

≥ |A|s+1 − |A|2 |G |

(D(W )

|W ||G | − |A|

)s−2

.


IntroductionLemmas



(F) =∑χ∈G

f1(χ)f2(χ) · · · fs+1(χ)

=∑χ∈G

(∑x1∈A

χ(x1)

)· · ·

(∑xs∈A

χ(xs)

)( ∑xs+1∈A

χ(−sxs+1)

)

Let ♥ =

(∑x1∈A

χ(x1)

)· · ·

(∑xs∈A

χ(xs)

)( ∑xs+1∈A

χ(−sxs+1)

).

We can split the sum over χ ∈ G into two parts, χ = χ0 andχ 6= χ0.


IntroductionLemmas



(F) =∑χ∈G

f1(χ)f2(χ) · · · fs+1(χ)

=∑χ∈G

(∑x1∈A

χ(x1)

)· · ·

(∑xs∈A

χ(xs)

)( ∑xs+1∈A

χ(−sxs+1)

)

Let ♥ =

(∑x1∈A

χ(x1)

)· · ·

(∑xs∈A

χ(xs)

)( ∑xs+1∈A

χ(−sxs+1)

).



IntroductionLemmas



(F) =∑χ∈G

f1(χ)f2(χ) · · · fs+1(χ)

=∑χ∈G

(∑x1∈A

χ(x1)

)· · ·

(∑xs∈A

χ(xs)

)( ∑xs+1∈A

χ(−sxs+1)

)

Let ♥ =

(∑x1∈A

χ(x1)

)· · ·

(∑xs∈A

χ(xs)

)( ∑xs+1∈A

χ(−sxs+1)

).



IntroductionLemmas



For all x ∈ G , χ0(x) = 1. Then∑x∈A

χ0(x) = |A|.

Using this fact,

we conclude that

(F) =∑χ∈G

♥

=∑χ=χ0

♥+∑χ6=χ0

♥

= |A|s+1 +∑χ 6=χ0

♥

We want to find an upper bound on

∣∣∣∣ ∑χ 6=χ0

♥∣∣∣∣ and subtract it from

|A|s+1 in order to find a lower bound on (F).


IntroductionLemmas




χ0(x) = |A|. Using this fact,

we conclude that

(F) =∑χ∈G

♥

=∑χ=χ0

♥+∑χ 6=χ0

♥

= |A|s+1 +∑χ 6=χ0

♥


∣∣∣∣ ∑χ 6=χ0




IntroductionLemmas




χ0(x) = |A|. Using this fact,

we conclude that

(F) =∑χ∈G

♥

=∑χ=χ0

♥+∑χ 6=χ0

♥

= |A|s+1 +∑χ 6=χ0

♥


∣∣∣∣ ∑χ 6=χ0




IntroductionLemmas



∣∣∣∣ ∑χ 6=χ0

♥∣∣∣∣

≤∑χ 6=χ0

∣∣∣∣ ∑x1∈A

χ(x1)

∣∣∣∣ . . . ∣∣∣∣ ∑xs∈A

χ(xs)

∣∣∣∣∣∣∣∣ ∑xs+1∈A

χ(−sxs+1)

∣∣∣∣=∑χ 6=χ0

∣∣∣∣∑x∈A

χ(x)

∣∣∣∣s ∣∣∣∣∑y∈A

χ(−sy)

∣∣∣∣≤[

maxχ 6=χ0

∣∣∑x∈A

χ(x)∣∣]s−2 ∑

χ 6=χ0

∣∣∣∣∑x∈A

χ(x)

∣∣∣∣2∣∣∣∣∑y∈A

χ(−sy)

∣∣∣∣≤[

maxχ 6=χ0

∣∣∑x∈A

χ(x)∣∣

︸︷︷︸A

]s−2 ∑χ 6=χ0

∣∣∣∣∑x∈A

χ(x)

∣∣∣∣2︸︷︷︸B

|A|


IntroductionLemmas



After some work, we obtain the results that

A ≤ D(W )

|W ||G | − |A|

B = |A| |G | .

This gives us

∣∣∣∣ ∑χ 6=χ0

♥∣∣∣∣

≤As−2B |A|

≤ |A|2 |G |

(D(W )

|W ||G | − |A|

)s−2

.


IntroductionLemmas



After some work, we obtain the results that

A ≤ D(W )

|W ||G | − |A|

B = |A| |G | .

This gives us

∣∣∣∣ ∑χ 6=χ0

♥∣∣∣∣

≤As−2B |A|

≤ |A|2 |G |

(D(W )

|W ||G | − |A|

)s−2

.


IntroductionLemmas



Then

(F)

≥ |A|s+1 −∣∣∣∣ ∑χ 6=χ0

♥∣∣∣∣

≥ |A|s+1 − |A|2 |G |

(D(W )

|W ||G | − |A|

)s−2

.


IntroductionLemmas


The Induction Proof

Proof of Theorem

If we put together Lemma 1 and Lemma 2, we obtain the followinginequality:

|A|s+1 − |A|2 |G |

(D(W )

|W ||G | − |A|

)s−2

≤(

s + 1

2

)|A|s−1 |G | .


IntroductionLemmas


The Induction Proof

The Induction

We will show that d(n) = maxc(H)≥n

D(H)

|H|≤ C

ns−2 , where

C = max

{1,√

s2 + s

√√√√(2s−4log 2

)2s−42

(2s−4log 2

) ,2s−1

2(s−2)2(2s−2 − 1)s−2

},

using a proof by induction on n.


IntroductionLemmas


The Induction Proof

For n = 1, the expression simplifies to d(n) ≤ Cns−2 = C. Since

D(G ) can be at most |G |, d(n) ≤ 1. Then as long as C ≥ 1, ourtheorem holds.

For n > 1, we will assume that d(n − 1) ≤ C(n−1)s−2 and prove that

d(n) ≤ Cns−2 .

In order to solve for C, we can split this into 2 cases:

Case 1: |A|s+1

2 ≤(

s + 1

2

)|G | |A|s−1

Case 2: |A|s+1

2 >

(s + 1

2

)|G | |A|s−1

For each case, we will show that |A||G | ≤C

ns−2 for all A, implying that

d(n) ≤ C1ns−2 .


IntroductionLemmas


The Induction Proof




d(n) ≤ Cns−2 .


Case 1: |A|s+1

2 ≤(

s + 1

2

)|G | |A|s−1

Case 2: |A|s+1

2 >

(s + 1

2

)|G | |A|s−1



d(n) ≤ C1ns−2 .


IntroductionLemmas


The Induction Proof




d(n) ≤ Cns−2 .


Case 1: |A|s+1

2 ≤(

s + 1

2

)|G | |A|s−1

Case 2: |A|s+1

2 >

(s + 1

2

)|G | |A|s−1



d(n) ≤ C1ns−2 .


IntroductionLemmas


The Induction Proof




d(n) ≤ Cns−2 .


Case 1: |A|s+1

2 ≤(

s + 1

2

)|G | |A|s−1

Case 2: |A|s+1

2 >

(s + 1

2

)|G | |A|s−1



d(n) ≤ C1ns−2 .


IntroductionLemmas


The Induction Proof

Case 1

Assume |A|s+1

2 ≤(

s + 1

2

)|G | |A|s−1. We must find

C ≥√

s2 + s√

n2s−4

2n .

√n2s−4

2n has a single maximum at n = 2s−4log 2 . Let

C ≥√

s2 + s

√√√√(2s−4log 2

)2s−42

(2s−4log 2

) .


IntroductionLemmas


The Induction Proof

Case 1

Assume |A|s+1

2 ≤(

s + 1

2


C ≥√

s2 + s√

n2s−4

2n .√n2s−4

2n has a single maximum at n = 2s−4log 2 . Let

C ≥√

s2 + s

√√√√(2s−4log 2

)2s−42

(2s−4log 2

) .


IntroductionLemmas


The Induction Proof

Case 2

Assume |A|s+1

2 >

(s + 1

2


C ≥ 1

2b2(nb+1)

(nb

(n−1)b − 1

)b

where b = s − 2.

1

2b2(nb+1)

(nb

(n−1)b − 1

)b

is a strictly decreasing function of n.

Take n = 2 and let

C ≥ 2b+1

2b2(2b − 1)b =

2s−1

2(s−2)2(2s−2 − 1)s−2.


IntroductionLemmas


The Induction Proof

Case 2

Assume |A|s+1

2 >

(s + 1

2


C ≥ 1

2b2(nb+1)

(nb

(n−1)b − 1

)b

where b = s − 2.

1

2b2(nb+1)

(nb

(n−1)b − 1

)b

is a strictly decreasing function of n.

Take n = 2 and let

C ≥ 2b+1

2b2(2b − 1)b =

2s−1

2(s−2)2(2s−2 − 1)s−2.


IntroductionLemmas


The Induction Proof

Let C = max

{1,√

s2 + s

√(2s−4log 2

)2s−4

2

(2s−4log 2

) , 2s−1

2(s−2)2(2s−2 − 1)s−2

}for

each s, and we have obtained the upper bound d(n) ≤ Cns−2 .


IntroductionLemmas


Thank you for your time and attention!


Documents

SUPER AWESOME STUFF · 3-Term Arithmetic Progressions in Groups A More General Question What is a 3-Term Arithmetic Progression? 3 numbers de ned by x 1 = a, x 2 = a + 2d, and x 3