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Chapter 30. A cross intersection problem with measures

Chapter 30. A cross intersection problem with measures

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196



30. A cross intersection problem with measures



proving the following result, which is an extension of Theorem 12.7

to cross intersecting families.

Theorem 30.1 ([105]). For i = 1, 2, let μi be a probability measure

(1)

(n)

on Ωi = 2[n] with respect to a probability vector pi = (pi , . . . , pi ),

and let Ui ⊂ Ωi . Suppose that U1 and U2 are cross intersecting. Then

we have the following.

(1)



(l)



(i) If pi = max{pi : 1 ≤ l ≤ n} < 1/2 for i = 1, 2, then

(1) (1)

μ1 (U1 )μ2 (U2 ) ≤ p1 p2 .

(1) (1)



(l) (l)



(l)



(ii) If p1 p2 = max{p1 p2 : 1 ≤ l ≤ n} and pi ≤ 1/3 for

(1) (1)

all i = 1, 2 and 1 ≤ l ≤ n, then μ1 (U1 )μ2 (U2 ) ≤ p1 p2 .

Borg considered a problem concerning cross intersecting integer

sequences in [14] and obtained similar results to (i) of Theorem 30.1

using a shifting technique. On the other hand, we note that in (ii),

(1)

(l)

pi is not necessarily the maximum of pi . This is a less convenient

situation for shifting, but the SDP approach works well even in such

settings.



30.2. Proof of the theorem

The proof goes as follows. We translate our problem into the setting

of Theorem 29.6 and then we construct a feasible optimal solution to

the dual problem (D). For the construction we follow Friedgut [68]

and first consider the case n = 1 carefully, and then we extend it to

the general case using a tensor product construction.

(1)



30.2.1. Setup. To simplify notation we write pi instead of pi . Let

( )

( )

qi = 1 − pi for i = 1, 2 and ∈ [n]. We also write qi instead of

(1)

qi . By symmetry we may assume

p1 ≥ p2 .

Let G be the bipartite Kneser graph. That is, V (G) = Ω1 Ω2

with Ωi = 2[n] (i = 1, 2), and x ∈ Ω1 and y ∈ Ω2 are adjacent if and

only if x ∩ y = ∅. We apply Theorem 29.6 to this graph by finding a



feasible solution to (D) with α = β = p1 p2 /2, which would yield

(30.1)



μ1 (U1 )μ2 (U2 ) ≤ p1 p2 .



30.2. Proof of the theorem



197



30.2.2. Proof of (i).

The case n = 1. We prove (30.1) by constructing a feasible solution,

which looks more complicated than we actually need to deal with this

trivial case. The point is that the construction can be easily extended

to the general case n ≥ 2 later.

Let ci =



pi /qi , and define

p



pj

qi



1 − qji

1



Ai,j =



Di,j =



,



0



1 ci

,

1 − c1i



Vi =



Δi =



1

0

,

0 −ci cj

qi

0



0

,

pi



where the rows and columns are indexed in the order ∅, {1}. A direct

computation shows that

(30.2)



ViT Δi Vi = I,



Ai,j Vj = Vi Di,j ,



(Δi Ai,j )T = Δj Aj,i .



Using the first two equalities, we get

ViT (Δi Ai,j )Vj = Di,j .



(30.3)



To construct a feasible solution to (D) in (29.1), let us find a PSD

matrix S and a symmetric matrix Z with Z ≥ 0. To this end, let

θ=





p1 p2 /2



and let

(30.4)

γx,y Ex,y = ηΔ1 A1,2 ,



α = β = θ,

x∼y



Z=



1 Δ1 A1,1



0



0



2 Δ2 A2,2



where η, 1 , 2 ∈ R. By definition Z is symmetric. Also Z ≥ 0 is

equivalent to i (1 − 2pi ) ≥ 0 for i = 1, 2. Since pi < 1/2, we need

i ≥ 0 for i = 1, 2. By (29.1) we define a symmetric matrix

(30.5)



S=



θΔ1 − 1 Δ1 A1,1

ηΔ2 A2,1 − 12 Δ2 J2,1 Δ1



We need to choose



1



≥ 0,



2



ηΔ1 A1,2 − 12 Δ1 J1,2 Δ2

.

θΔ2 − 2 Δ2 A2,2



≥ 0, and η ∈ R so that S



0.



198



30. A cross intersection problem with measures



By (30.2) and (30.3), we have



θ− 1

T



0

0

V

V1

0

S 1

=⎢

⎣η − 1

0 V2T

0 V2

2

0

Thus, S



0

θ + 1 c21

0

−ηc1 c2



η−

0

θ−

0



1

2



2





0

−ηc1 c2 ⎥

⎥.

0 ⎦

θ + 2 c22



0 if and only if

θ−

η−



1

1

2



η−

θ−



1

2



θ + 1 c21

−ηc1 c2



0,



2



−ηc1 c2

θ + 2 c22



0,



if and only if

0≤



(30.6)



1, 2



(θ −

θ+



≤ θ,



1 ) (θ

2

1 c1







2)



θ+





2

2 c2



η−



1

2



2



,



≥ η 2 c21 c22 .



By the above three conditions we obtain a one-parameter family of

solutions

p2

(p1 − p2 )p2

p2

q2

(30.7) 1 =

, 0 ≤ 2 ≤ θ.

, η=√



2+

2+

p1

2 p1 p2

p1 p2

2



This gives feasible solutions to (D) with objective value p1 p2 and

proves the bound (30.1) for the case n = 1.

The case n ≥ 2. The construction in this case is really similar to

that in the previous case. We replace the 2 × 2 matrices used in the

previous case with new n × n matrices by taking tensor products as

explained below.

(l)



Let ci =



(l)



(l)



pi /qi < 1, and let



(l)

pj

1 − (l)

(l)



q

Ai,j =

i

1



(l)



pj



(l)



qi





⎦,



0



where the rows and columns are indexed in the order ∅, { }. Then

we redefine

[n]



(n)



(n−1)



Ai,j = Ai,j := Ai,j ⊗ Ai,j



(1)



⊗ · · · ⊗ Ai,j .



We identify 2{1} × · · · × 2{n} with 2[n] , and thus we view the rows

(resp. columns) of Ai,j as indexed by Ωi (resp. Ωj ). Similarly we



30.2. Proof of the theorem

[n]



199



[n]



[n]



redefine Di,j = Di,j , Vi = Vi , and Δi = Δi . Note that Δi is the

diagonal matrix whose (x, x)-entry is μi ({x}) for x ∈ Ωi . These new

matrices satisfy (30.2) and (30.3).

We choose the variables by (30.4), where 1 , 2 , and η are given

by (30.7). Note that for x ∈ Ωi and y ∈ Ωj , the (x, y)-entry of Ai,j is

0 if x ∩ y = ∅ and positive otherwise. Thus Z is a symmetric matrix

with Z ≥ 0. Define the symmetric matrix S by (30.5). Then we have

V1T

0



(30.8)



0

θI1 − 1 D1,1

=

ηD2,1 − 12 E∅,∅

V2



0

V

S 1

V2T

0



ηD1,2 − 12 E∅,∅

,

θI2 − 2 D2,2



where Ii ∈ RΩi ×Ωi denotes the identity matrix. The RHS of (30.8) is

isomorphic to the direct sum z∈2[n] S (z) , where

S (∅) =



θ−

η−



1

1

2



η−

θ−



(z)



(z)



1

2



,



(z)



θ − 1 c1,1

(z)

ηc2,1



S (z) =



2

( ) ( )



ηc1,2

(z)

θ − 2 c2,2



(z = ∅),



(z)



. Note that |ci,j | < 1. We need to choose

with ci,j =

∈z − ci cj



0,



0,

and

η



R

so

that S 0, that is, for all z ∈ 2[n] ,

1

2

S (z)



(30.9)



0.



We have already verified (30.9) for z = ∅ and {1} in the previous

(z)

case. So let z = ∅, {1}. By (30.6) with |ci,j | < 1 the diagonal entries

of S (z) are nonnegative. Thus, S (z) 0 if and only if

θ−



(30.10)



(z)

1 c1,1



(z)

2 c2,2



θ−



(z) 2



≥ ηc1,2 .



Let us verify that (30.10) holds if 2 = 0. In this case, by (30.7),

1 = (p1 − p2 )p2 /(4θ) and η = q2 /2, and then (30.10) is rewritten as

(30.11)



(z)



p1 p2 ≥ c1,1



q22 c2,2 + (−1)|z| (p1 − p2 )p2 .

(z)



We also note that

(l)



pi



(z)



|ci,i | =



(l)



l∈z



1 − pi







pi

1 − pi



|z|



because p/(1 − p) is increasing in p ∈ (0, 1/2). Thus, if |z| is odd,

(z)

then |ci,i | ≤ pi /qi and the RHS of (30.11) is

≤ (p1 /q1 ) q22 (p2 /q2 ) − (p1 − p2 )p2 = p1 p2 .



200



30. A cross intersection problem with measures

(z)



If |z| is even, then |ci,i | ≤ (pi /qi )2 and the RHS of (30.11) is

≤ (p1 /q1 )2 q22 (p2 /q2 )2 + (p1 − p2 )p2 = p1 p2 (p1 /q1 )2 ≤ p1 p2 .

Thus in both cases we have (30.11) and S

0. This means that



a feasible solution to (D) with objective value p1 p2 is obtained by

setting 2 = 0. This completes the proof of (i).

30.2.3. Proof of (ii). If n = 1 then the result follows from (i). Let

n ≥ 2. We proceed as in the proof of (i), but this time we choose

2 = θ. Then 1 = θ and η = 1/2 by (30.7). We need to show (30.10)

for z = ∅, {1}. In this case it reads

(z)



(z)



1 − c2,2



1 − c1,1



p1 p2



(30.12)



(z)



≥ 1.



(z)



|c1,1 |



|c2,2 |



We distinguish the cases according to the parity of |z|.

(z)



If |z| is odd, then ci,i < 0, and

(z)



1 − ci,i

(z)



|ci,i |



=1+



1

(z)



|ci,i |



(l)



qi



=1+



.



(l)



l∈z



pi



The RHS is minimized when |z| = 1, say z = {l}. Then the LHS of

(30.12) is

(l)



(l)



(l)



(l)



(l) (l)



≥ p1 p2 (1 + q1 /p1 )(1 + q2 /p2 ) = p1 p2 /(p1 p2 ) ≥ 1,

as required.

(z)



If |z| ≥ 2 is even, then ci,i > 0, and

(z)



1 − ci,i

(z)



|ci,i |



= −1 +



1

(z)



|ci,i |



(l)



qi



= −1 +



(l)



l∈z



.



pi



The RHS is minimized when |z| = 2, say z = {k, l}, and the RHS is

(k) (l)



≥ −1 +



q1 q1



(k) (l)

p1 p1

(k)



(k)



=



(l)



1 − p1 − p1

(k) (l)

p1 p1



(l)



(k)



(k)







p1



(k) (l)

p1 p1



=



1

(l)



,



p1



where we used 1 − p1 − p1 ≥ 1/3 ≥ p1 . Thus the LHS of (30.12)

is

(l)

(l)

(l) (l)

≥ p1 p2 (1/p1 )(1/p2 ) = p1 p2 /(p1 p2 ) ≥ 1.

This completes the proof of (ii).



Chapter 31



Capsets and sunflowers



In this chapter we present a recent development of a polynomial

method (the slice rank method). We give two applications. One

is that a set containing no 3-term arithmetic progressions in Fn3 has

size less than (2.8)n for n sufficiently large, and the other is that a

family of subsets of [n] containing no sunflowers has size less than

(1.9)n for n sufficiently large.



31.1. Things happened in May 2016

Finding large subsets of an abelian group G with no 3-term arithmetic

progressions is one of the central problems in additive number theory.

In May of 2016 Croot, Lev, and Pach [20] proved a remarkable result

stating that the size of such subsets for the case G = (N/4N)n is

at most cn for some c < 4. This was a starting point of successive

breakthrough results all obtained in the same month. Soon after

they put the preprint on arXiv, Ellenberg and Gijswijt independently

found how to use the idea of [20] for the case G = Fn3 and obtained

the following result.

Theorem 31.1 (Ellenberg–Gijswijt [27]). Let X ⊂ Fn3 . Suppose that

for x, y, z ∈ X,

(31.1)



x + y + z = 0 if and only if x = y = z.



Then |X| < cn for some c < 3.

201



202



31. Capsets and sunflowers



Exercise 31.2. We say that three elements x, x + a, x + 2a ∈ Fn3

form a 3-term arithmetic progression if a = 0. Show that X ⊂ Fn3

contains no 3-term arithmetic progressions if and only if (31.1) holds.

A subset X ⊂ Fn3 satisfying (31.1) is called a capset. Bounding

the size of capsets is important in itself, and it is also related to the following two problems. One is a problem of determining the exponent ω

of matrix multiplication; see [12]. The other is a conjecture concerning the size of families without sunflowers. Recall that three distinct

subsets A, B, C ⊂ [n] form a sunflower if A ∩ B = B ∩ C = C ∩ A.

The following result was conjectured by Erd˝os and Szemer´edi [37].

Theorem 31.3. Let F ⊂ 2[n] . If F contains no sunflower, then there

exist c < 2 and n0 such that |F| < cn for all n > n0 .

Alon, Shpilka, and Umans [2] had proved that if Theorem 31.1 were

true (and we now know that it is indeed true), then it would imply

Theorem 31.3.

Tao reformulated the proof of Theorem 31.1 in his blog [106] by

introducing a slice rank. Then using this slice rank method Naslund

and Sawin [94] gave a direct proof of Theorem 31.3 with a better

constant c than the constant obtained from [2] with [27]. In the

following sections we explain the slice rank and its basic property,

and then we apply the property to prove Theorems 31.1 and 31.3.

Our presentation is partially based on a set of lecture notes by Zeeuw

[17], which is also recommended for other topics including the density

Hales–Jewett Theorem.



31.2. Slice rank

A diagonal matrix without zero diagonal entries is of full rank. We

will extend this fact to a “hypermatrix” in some sense. For this

purpose let us introduce the slice rank of a function.

Let X be a finite set, and let F be a field. Let f : X k → F be a

function in k variables. We say that f is sliced if f can be written as

a product of two functions a and b, where a is in one variable and b

is in the other k − 1 variables, for example,

f (x1 , x2 , . . . , xk ) = a(x3 ) b(x1 , x2 , x4 , . . . , xk ).



31.2. Slice rank



203



We can always write f as a sum of at most |X| sliced functions;

indeed,

f (x1 , . . . , xk ) =



az (x1 )bz (x2 , . . . , xk ),

z∈X



where az (x) = δz,x and bz (x2 , . . . , xk ) = f (z, x2 , . . . , xk ). Now the

slice rank of f , denoted by sr(f ), is the least number of sliced functions

such that f is represented as the sum of them, that is,

r



gi , where g1 , . . . , gr are sliced functions .



sr(f ) = min r : f =

i=1



As we just noted, sr(f ) ≤ |X|.

Lemma 31.4 (Tao [106]). Let X be a finite set, F be a field, and

k ≥ 2 be an integer. Let f : X k → F be a function such that

f (x1 , . . . , xk ) = 0 if and only if x1 = · · · = xk . Then sr(f ) = |X|.

Proof. We prove the statement by induction on k. Let k = 2. It

suffices to show that sr(f ) ≥ |X|. Suppose, to the contrary, that

r := sr(f ) < |X|. Then we can write

r



ai (x)bi (y)



f (x, y) =

i=1



for some nonzero functions ai and bi . Let F (resp. Fi ) be an |X| × |X|

matrix whose (x, y)-entry is f (x, y) (resp. ai (x)bi (y)). Then

r



F =



Fi .

i=1



Since the usual matrix rank of Fi is one, it follows that

r



rank F ≤



rank(Fi ) = r.

i=1



On the other hand, the assumption that f (x, y) = 0 if and only if

x = y implies that F is a diagonal matrix without zero diagonal

entries. Thus rank F = |X| > r, a contradiction.

Next we move on to the induction step, but to make things notationally easier we consider the case k = 3. Let f : X 3 → F and



204



31. Capsets and sunflowers



suppose that r := sr(f ) < |X|. Then we can write

(31.2)

ai (x)bi (y, z) +

ai (y)bi (x, z) +

f (x, y, z) =

i∈I



i∈J



ai (z)bi (x, y)

i∈K



for some nonzero functions ai and bi , where I

loss of generality we may assume that I = ∅.



J



K = [r]. Without



Let V be a vector space consisting of all functions v : X → F

such that x∈X v(x)ai (x) = 0 for all i ∈ I. Thus V is defined by |I|

linear equations in |X| variables, and

(31.3)



dim V ≥ |X| − |I| > r − |I|.



Choose v ∈ V so that the support S = {x ∈ X : v(x) = 0} is maximal.

We claim that

|S| ≥ dim V.



(31.4)



In fact, if |S| < dim V then we can find a nonzero w ∈ V such that

w(x) = 0 for all x ∈ S. But then v + w has a strictly larger support

than S, and this contradicts the maximality of the choice of S.

Define a function g : S 2 → F by g(y, z) = x∈X v(x)f (x, y, z).

We estimate the slice rank of g in two ways, which will give rise to

a contradiction, showing that the earlier assumption sr(f ) < |X| is

false. Compute g from the RHS of (31.2). For the first term we have

v(x)ai (x) bi (y, z) = 0.

i∈I



x∈X



The second and third terms we can rewrite as

ai (y)ci (z) +

i∈J



ai (z)ci (y),

i∈K



where

v(x)bi (x, z) for i ∈ J,



ci (z) =

x∈X



v(x)bi (x, y) for i ∈ K.



ci (y) =

x∈X



Thus we get

ai (y)ci (z) +



g(y, z) =

i∈J



ai (z)ci (y),

i∈K



31.2. Slice rank



205



that is, g is represented by |J| + |K| sliced functions, and

(31.5)



sr(g) ≤ |J| + |K| = r − |I|.



Finally we claim that g(y, z) = 0 if and only if y = z. In fact,

if y = z then f (x, y, z) = 0 and g(y, z) = x v(x)f (x, y, z) = 0. If

y = z, then g(y, y) = x v(x)f (x, y, y) = v(y)f (y, y, y) = 0 for y ∈ S,

as needed. Thus by the induction hypothesis with (31.3) and (31.4)

it follows that

sr(g) = |S| ≥ dim V > r − |I|,

which contradicts (31.5).

The next lemma gives us an upper bound for the slice rank of a

function. We will use the lemma to solve the capset problem and the

sunflower problem.

Lemma 31.5. Let F be a field, let Y ⊂ F be a finite set, and let

X ⊂ Y n , where n ∈ Z>0 . Define a function f : X 3 → F by

n



(31.6)



(xi + yi + zi )s − t ,



f (x, y, z) =

i=1



where t ∈ F and s ∈ Z>0 . Let g(x) = x− 3 (1 + x + · · · + xs ) be a realvalued function defined on the interval (0, 1), and let α be a unique

root of g (x) = 0. Then

s



sr(f ) < 3(g(α))n .

In particular the following hold.

(i) If s = 2, t = 1, and F = F3 , then sr(f ) < 3(2.76)n .

(ii) If s = 1, t = 2, and F = R, then sr(f ) < 3(1.89)n .

Proof. By expanding the RHS of (31.6) we can write f as a sum of

monomials in the form

xi11 · · · xinn y1j1 · · · ynjn z1k1 · · · znkn ,

with I + J + K ≤ sn, where I = nl=1 il , J = nl=1 jl ,

and K = nl=1 kl . For simplicity we write xI to mean xi11 · · · xinn

and define y J and z K similarly. Since one of the I , J , and K



206



31. Capsets and sunflowers



is at most sn/3, we can divide f into three parts f = fx + fy + fz as

follows. First we collect all monomials in f with I ≤ sn/3 and let

xI



fx =

I ≤sn/3



cIJK y J z K .

J,K



Next we collect all monomials in f − fx with J ≤ sn/3 and let

yJ



fy =

J ≤sn/3



cIJK xI z K .

J,K



Finally let fz = f − fx − fy , which is represented as

zK



fz =

K ≤sn/3



cIJK xI y J .

I,J



We note that fx is a sum of #I sliced functions with I ≤

sn/3. Moreover, #I ≤ N , where N is the number of integer solutions

(i1 , . . . , in ) satisfying

sn

i1 + · · · + in ≤

3

with 0 ≤ il ≤ s for all 1 ≤ l ≤ n. Suppose that (i1 , . . . , in ) is one of

the solutions, and let au = |{l : il = u, 1 ≤ l ≤ n}| for 0 ≤ u ≤ s.

Then

(P) a0 + · · · + as = n,

sn

.

3

On the other hand, for a fixed feasible (a0 , . . . , as ) the number of the

corresponding solutions (i1 , . . . , in ) is

(Q) a1 + 2a2 + · · · + sas ≤



n

a0



n − a0

n − (a0 + · · · + as−1 )

···

as

a1



Thus we have



=



n!

.

a0 !a1 ! · · · as !



n!

,

a0 !a1 ! · · · as !

where the sum is taken under the conditions of (P) and (Q). Clearly

N is also an upper bound for the number of sliced functions in fy as

well as in fz . Consequently f is a sum of at most 3N sliced functions,

and

sr(f ) ≤ 3N.

N=



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