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=