PROGRESSION-FREE SETS IN Z
n
4
ARE EXPONENTIALLY SMALL
ERNIE CROOT, VSEVOLOD F. LEV, AND P
´
ETER P
´
AL PACH
†
Abstract. We show that for integer n ≥ 1, any subset A ⊆ Z
n
4
free of three-term
arithmetic progressions has size |A| ≤ 4
γn
, with an absolute constant γ ≈ 0.926.
1. Background and Motivation
In his influential papers [R52, R53], Roth has shown that if a set A ⊆ {1, 2, . . . , N}
does not contain three elements in an arithmetic progression, then |A| = o(N) and in-
deed, |A| = O(N/ log log N) as N grows. Since then, estimating the largest possible
size of such a set has become one of the central problems in additive combinatorics.
Roth’s original results were improved by Heath-Brown [H87], Szemer´edi [S90], Bour-
gain [B99], Sanders [S12, S11], and Bloom [B], the current record due to Bloom being
|A| = O(N(log log N)
4
/ log N).
It is easily seen that Roth’s problem is essentially equivalent to estimating the largest
possible size of a subset of the cyclic group Z
N
, free of three-term arithmetic progressions.
This makes it natural to investigate other finite abelian groups.
We say that a subset A of an (additively written) abelian group G is progression-free
if there do not exist pairwise distinct a, b, c ∈ A with a + b = 2c, and we denote by r
3
(G)
the largest size of a progression-free subset A ⊆ G. For abelian groups G of odd order,
Brown and Buhler [BB82] and independently Frankl, Graham, and R¨odl [FGR87] proved
that r
3
(G) = o(|G|) as |G| grows. Meshulam [M95], following the general lines of Roth’s
argument, has shown that if G is an abelian group of odd order, then r
3
(G) ≤ 2|G|/ rk(G)
(where we use the standard notation rk(G) for the rank of G); in particular, r
3
(Z
n
m
) ≤
2m
n
/n. Despite many efforts, no further progress was made for over 15 years, till Bateman
and Katz in their ground-breaking paper [BK12] proved that r
3
(Z
n
3
) = O(3
n
/n
1+ε
) with
an absolute constant ε > 0.
Abelian groups of even order were first considered in [L04] where, as a further elabo-
ration on the Roth-Meshulam proof, it is shown that r
3
(G) < 2|G|/ rk(2G) for any finite
abelian group G; here 2G = {2g : g ∈ G}. For the homocyclic groups of exponent 4 this
†
Supported by the Hungarian Scientific Research Funds (Grant Nr. OTKA PD115978 and
OTKA K108947) and the J´anos Bolyai Research Scholarship of the Hungarian Academy of Sciences.
1
2 ERNIE CROOT, VSEVOLOD F. LEV, AND P
´
ETER P
´
AL PACH
result was improved by Sanders [S11] who proved that r
3
(Z
n
4
) = O(4
n
/n(log n)
ε
) with an
absolute constant ε > 0. The goal of this paper is to further improve Sanders’s result, as
follows.
Let H denote the binary entropy function; that is,
H(x) = −x log
2
x − (1 − x) log
2
(1 − x), x ∈ (0, 1),
where log
2
x is the base-2 logarithm of x. For the rest of the paper, we set
γ := max
n
1
2
H(0.5 − ε) + H(2ε)
: 0 < ε < 0.25
o
≈ 0.926.
Theorem 1. If n ≥ 1 and A ⊆ Z
n
4
is progression-free, then |A| ≤ 4
γn
.
The proof of Theorem 1 is presented in the next section.
We note that the exponential reduction in Theorem 1 is the first of its kind for problems
of this sort.
Starting from Roth, the standard way to obtain quantitative estimates for r
3
(G) in-
volves a combination of the Fourier analysis and the density increment technique; the
only exception is [L12] where for the groups G
∼
=
Z
n
q
with a prime power q, the above-
mentioned Meshulam’s result is recovered using a completely elementary argument. In
contrast, in the present paper we use the polynomial method, without resorting to the
familiar Fourier analysis – density increment strategy.
For a finite abelian group G
∼
=
Z
m
1
⊕ · · · ⊕ Z
m
k
with positive integer m
1
| · · · | m
k
,
denote by rk
4
(G) the number of indices i ∈ [1, k] with 4 | m
i
. Since, writing n := rk
4
(G),
the group G is a union of 4
−n
|G| cosets of a subgroup isomorphic to Z
n
4
, as a direct
consequence of Theorem 1 we get the following corollary.
Corollary 1. If A is a progression-free subset of a finite abelian group G then, writing
n := rk
4
(G), we have |A| ≤ 4
−(1−γ)n
|G|.
2. Proof of Theorem 1
We recall that the degree of a multivariate polynomial is the largest sum of the ex-
ponents of all of its monomials. The polynomial is multilinear if it is linear in every
individual variable.
The proof of Theorem 1 is based on the following lemma.
Lemma 1. Suppose that n ≥ 1 and d ≥ 0 are integers, P is a multilinear polynomial
in n variables of total degree at most d over a field F, and A ⊆ F
n
is a set with |A| >
2
P
0≤i≤d/2
n
i
. If P (a − b) = 0 for all a, b ∈ A with a 6= b, then also P (0) = 0.
PROGRESSION-FREE SETS IN Z
n
4
3
Proof. Let m :=
P
0≤i≤d/2
n
i
, and let K = {K
1
, . . . , K
m
} be the collection of all sets
K ⊆ [n] with |K| ≤ d/2. Writing for brevity
x
I
:=
Y
i∈I
x
i
, x = (x
1
, . . . , x
n
) ∈ F
n
, I ⊆ [n],
there exist coefficients C
I,J
∈ F (I, J ⊆ [n]) depending only on the polynomial P , such
that for all x, y ∈ F
n
we have
P (x − y) =
X
I,J⊆[n]
I∩J=∅
|I|+|J|≤d
C
I,J
x
I
y
J
=
X
I∈K
x
I
X
J⊆[n]\I
|J|≤d−|I|
C
I,J
y
J
+
X
J∈K
X
I⊆[n]\J
d/2<|I|≤d−|J|
C
I,J
x
I
!
y
J
.
The right-hand side can be interpreted as the scalar product of the vectors u(x), v(y) ∈
F
2m
defined by
u
i
(x) = x
K
i
, u
m+i
(x) =
X
I⊆[n]\K
i
d/2<|I|≤d−|K
i
|
C
I,K
i
x
I
and
v
i
(y) =
X
J⊆[n]\K
i
|J|≤d−|K
i
|
C
K
i
,J
y
J
, v
m+i
(y) = y
K
i
for all 1 ≤ i ≤ m. Consequently, if we had P (a − b) = 0 for all a, b ∈ A with a 6= b, while
P (0) 6= 0, this would imply that the vectors u(a) and v(b) are orthogonal if and only if
a 6= b. As a result, the vectors u(a) would be linearly independent (an equality of the
sort
P
a∈A
λ
a
u(a) = 0 with the coefficients λ
a
∈ F after a scalar multiplication by v(b)
yields λ
b
= 0, for any b ∈ A). Finally, the linear independence of {u(a): a ∈ A} ⊆ F
2m
implies |A| ≤ 2m, contrary to the assumptions of the lemma.
Remark. It is easy to extend the lemma relaxing the multilinearity assumption to the
assumption that P has bounded degree in each individual variable. Specifically, denoting
by f
δ
(n, d) the number of monomials x
i
1
1
. . . x
i
n
n
with 0 ≤ i
1
, . . . , i
n
≤ δ and i
1
+· · ·+i
n
≤ d,
if P has all individual degrees not exceeding δ, and the total degree not exceeding d, then
|A| > 2f
δ
(n, bd/2c) along with P (a − b) = 0 (a, b ∈ A, a 6= b) imply P (0) = 0. Moreover,
taking δ = d, or δ = |F| − 1 for F finite, one can drop the individual degree assumption
altogether.
4 ERNIE CROOT, VSEVOLOD F. LEV, AND P
´
ETER P
´
AL PACH
We will use the estimate
X
0≤i≤z
n
i
< 2
nH(z/n)
(1)
valid for all integer n ≥ 1 and real 0 < z ≤ n/2; see, for instance, [McWS77, Ch. 10, §11,
Lemma 8].
Recall, that for integer n ≥ d ≥ 0, the sum
P
d
i=0
n
i
is the dimension of the vector
space of all multilinear polynomials in n variables of total degree at most d over the
two-element field F
2
. In particular, the dimension of the vector space of all multilinear
polynomials in n variables over F
2
is equal to the dimension of the vector space of all F
2
-
valued functions on F
n
2
, and it follows that any non-zero multilinear polynomial represents
a non-zero function. These basic facts are used in the proof of Proposition 1 below.
For integer n ≥ 1, denote by F
n
the subgroup of the group Z
n
4
generated by its
involutions; thus, F
n
is both the image and the kernel of the doubling endomorphism of
Z
n
4
defined by g 7→ 2g (g ∈ Z
n
4
), and we have F
n
∼
=
Z
n
2
.
Proposition 1. Suppose that n ≥ 1 and A ⊆ Z
n
4
is progression-free. Then for every
0 < ε < 0.25, the number of F
n
-cosets containing at least 2
nH(0.5−ε)+1
elements of A is
less than 2
nH(2ε)
.
Proof. Let R be the set of all those F
n
-cosets containing at least 2
nH(0.5−ε)+1
elements of
A, and for each coset R ∈ R let A
R
:= A ∩ R; thus, ∪
R∈R
A
R
⊆ A (where the union is
disjoint), and
|A
R
| ≥ 2
nH(0.5−ε)+1
, R ∈ R. (2)
For a subset S ⊆ Z
n
4
, write
2 · S := {s
0
+ s
00
: (s
0
, s
00
) ∈ S × S, s
0
6= s
00
} and 2 ∗ S := {2s: s ∈ S}.
The assumption that A is progression-free implies that the sets
B := ∪
R∈R
(2 · A
R
) ⊆ F
n
and C := ∪
R∈R
(2 ∗ R) ⊆ F
n
are disjoint: this follows by observing that if 2r ∈ 2 · A with some r ∈ R, then for each
a ∈ r+F
n
we have 2a = 2r ∈ 2·A. Furthermore, the sets 2∗R are in fact pairwise distinct
singletons (for 2r
1
= 2r
2
is equivalent to r
1
− r
2
∈ F
n
and thus to r
1
+ F
n
= r
2
+ F
n
),
whence |C| = |R|.
Let d = n − d2εne so that, in view of (2) and (1),
2
X
0≤i≤d/2
n
i
< 2
nH(0.5−ε)+1
≤ |A
R
|, R ∈ R. (3)
PROGRESSION-FREE SETS IN Z
n
4
5
Denoting by C the complement of C in F
n
, and assuming, contrary to what we want to
prove, that |R| ≥ 2
nH(2ε)
, from (1) we get
d
X
i=0
n
i
= 2
n
−
d2εne−1
X
i=0
n
i
> 2
n
− 2
nH(2ε)
≥ 2
n
− |R| = 2
n
− |C| = |C|.
(This is the computation where the assumption ε < 0.25 is used.) Consequently, iden-
tifying F
n
with the additive group of the vector space F
n
2
, and accordingly considering
B and C as subsets of F
n
2
, we conclude that the dimension of the vector space of all
multilinear n-variate polynomials over the field F
2
exceeds the dimension of the vector
space of all F
2
-valued functions on C. Thus, the evaluation map, associating with every
polynomial the corresponding function, is degenerate. As a result, there exists a non-zero
multilinear polynomial P ∈ F
2
[x
1
, . . . , x
n
] of total degree deg P ≤ d such that P vanishes
on C. In particular, P vanishes on B ⊆ C, and therefore on each set 2·A
R
, for all R ∈ R.
Fixing arbitrarily an element r ∈ R, the polynomial P (2r + x) thus vanishes whenever
x ∈ 2 · (A
R
− r). Hence, also P (2r) = 0 by Lemma 1 (which is applicable in view of (3));
that is, P also vanishes on each singleton set 2 ∗ A
R
, for all R ∈ R. It follows that P
vanishes on C. However, P was chosen to vanish on C. Therefore, P vanishes on all of
F
n
2
, and it follows that P is the zero polynomial. This is a contradiction showing that
|R| < 2
nH(2ε)
, and thus completing the proof.
Proof of Theorem 1. For x ≥ 0, let N(x) denote the number of F
n
-cosets containing at
least x elements of A; thus N(x) = 0 for x > 2
n
, and we can write
|A| =
Z
2
n+1
0
N(x) dx. (4)
Trivially, we have N(x) ≤ 2
n
for all x ≥ 0, so that
Z
2
nH(1/4)+1
0
N(x) dx ≤ 2
(H(1/4)+1)n+1
< 2 · 4
γn
. (5)
On the other hand, the substitution x = 2
nH(0.5−ε)+1
gives
Z
2
n+1
2
nH(1/4)+1
N(x) dx = n
Z
1/4
0
2
nH(0.5−ε)+1
N(2
nH(0.5−ε)+1
) log
0.5 + ε
0.5 − ε
dε, (6)
and applying Proposition 1, the integral in the right-hand side can be estimated as
2n
Z
1/4
0
2
n(H(0.5−ε)+H(2ε))
log
0.5 + ε
0.5 − ε
dε < 3n
Z
1/4
0
2
n(H(0.5−ε)+H(2ε))
dε < n · 4
γn
. (7)
From (4)–(7) we get |A| < (n + 2) · 4
γn
, and to conclude the proof we use the tensor
power trick: for integer k ≥ 1, the set A × · · · × A ⊆ Z
kn
4
is progression-free and therefore
|A|
k
< (kn + 2) · 4
γkn