Exponential Bounds for the Erd\H{o}s-Ginzburg-Ziv Constant

0.23 MB·

by  Eric Naslund

EXPONENTIAL BOUNDS FOR THE ERDŐS-GINZBURG-ZIV
CONSTANT
ERIC NASLUND
Abstract. The Erdős-Ginzburg-Ziv constant of an abelian group G, denoted s(G),
is the smallest k N such that any sequence of length s(G) contains a zero-sum
7
∈
1 subsequenceoflength exp(G). Inthispaper,weusethe multi-slicerankmethodfrom
0 [12], which is a variant of Tao’s slice rank method [15], to prove that
2
s Fn 3(p 1)p!(J(p)p)n
n p ≤ −
a where 0.8414 < J(p) < .9183(cid:0)7 is(cid:1)a constant depending on p, equal to the constant
J appearing in Ellenberg and Gijswijt’s bounds for arithmetic progression-free subsets
8 inFn [4]. We alsoimproveupona recentresultofHegedűs,andshowthatif (Z/kZ)n
p
1 satisfies property D as defined in [7], then

O] s((Z/kZ)n)≤(n−1)γkn,q +1
where
C k 1 xq −q−1
h. γk,q = q 0<inxf<1 1− x x k ,
t andqisthelargestprimepowerdividingk. In−particular,if(Z/qZ)n satisfiesproperty
a
m D where q is a prime power, then
[ s((Z/qZ)n) (n 1)4n+1.
≤ −

1
v
2
4
9 1. Introduction
4
0 For an abelian group G, let exp(G) denote the exponent of G, which is the maximal
1. order of any element in G. The Erdős-Ginzburg-Ziv constant of G, denoted s(G), is
0 defined to be the smallest k N such that any sequence of elements of G of length
7 s(G) contains a zero-sum subs∈equence of length exp(G). In their original paper, Erdős,
1
: Ginzburg and Ziv [5] proved that
v
i
X s(Z/kZ) = 2k 1.
−
r
a That is, among any sequence of 2k 1 integers there is a subsequence of length k whose
−
sum is divisible by k, and furthermore this is not true if 2k 1 is replaced by 2k 2.
− −
In 2007 Reiher [14] proved that
s (Z/kZ)2 = 4k 3,
−
which resolved a longstanding co(cid:0)njecture o(cid:1)f Kemnitz [10], and more generally when
1 k k ,
1 2
≤ |
s((Z/k Z) (Z/k Z)) = 2k +2k 3,
1 2 1 2
⊕ −
see [11, Theorem 5.8.3] for a proof. In the case where G = (Z/kZ)n with n large, very
little is known. Heiko [8] gave the elementary bounds
(k 1)2n +1 s((Z/kZ)n) (k 1)kn +1,
− ≤ ≤ −
Date: January 19, 2017.
1
EXPONENTIAL BOUNDS FOR THE ERDŐS-GINZBURG-ZIV CONSTANT 2
and Alon and Dubiner [1] proved that
s((Z/kZ)n) (cnlogn)nk
≤
for some c > 0. Ellenberg and Gijswijt’s recent bounds for capsets in Fn [4] imply that
3
s((Z/3Z)n) 2(J(3)3)n
≤
where
1
J(3) = 3 207+33√33 = 0.9183....
8
q
In this paper, for p 5 we give an unconditional exponential improvement to the
upper bounds for s F≥n as n , where F = (Z/pZ) denotes the finite field with p
p → ∞ p
elements. Our main result is:
(cid:0) (cid:1)
Theorem 1. Suppose that A Fn has size
⊂ p
A > 3(p 1)p!(J(p)p)n
| | −
where
1 1 xp
J(p) = inf − .
p 0<x<1 (1 x)x2/3
−
Then A contains p distinct elements x ,...,x such that
1 p
x + +x = 0.
1 p
···
We immediately have the following corollary:
Corollary 1. The Erdős-Ginzburg-Ziv constant of Fn satisfies
p
s(Fn) 3(p 1)p!(J(p)p)n.
p ≤ −
For p 5, this implies the bound
≥
(1.1) s(Fn) (J(p)p)n+p,
p ≤
since 3(p 1)p! (J(p)p)p for p 5. J(p) is the same constant appearing in Ellenberg
− ≤ ≥
and Gijswijt’s bound for arithmetic progression-free sets, and in [2, prop. 4.12] it was
proven that J(p) is a decreasing function of p, and that
z z−2
lim J(p) = inf − = 0.8414....
p→∞ z>1 3logz
Since J(3) = 0.9183..., it follows that for p 3, J(p) lies in the range
≥
0.8414 J(p) 0.9184.
≤ ≤
The proof of theorem 1 uses the multi-slice rank method, given in [12], which is a
variant of slice rank method [15]. The multi-slice rank allows us to handle the indicator
tensor that results from forcing the variables to be distinct, which has large slice rank
for p 5. The slice rank method was introduced by Tao [15] following the work of
≥
Ellenberg and Gijswijt [4], and the breakthrough result of Croot, Lev and Pach [3]. The
slice rank method has seen numerous applications, such as Tao’s proof of Ellenberg and
Gijswijt’s upper bound for progression-free sets in Fn [15, 4], new bounds for the Erdős-
p
Szemerédi sunflower problem [13, 6], and disproving certain conjectures concerning fast
matrix multiplication [2]. We refer the reader to [2] for a more detailed discussion of
the properties of the slice rank.
EXPONENTIAL BOUNDS FOR THE ERDŐS-GINZBURG-ZIV CONSTANT 3
We say that a group G satisfies property D (see [7, Sec. 7]) if every sequence S of
length s(G) 1 that does not contain exp(G) elements summing to 0, when viewed as
−
a multi-set, takes the form
exp(G)−1
S = T
∪i=1
for some set T. That is, if S is a maximal sequence that does not contain exp(G)
elements summing to zero, then every element in S appears exactly exp(G) 1 times.
Gao and Geroldinger conjecture [7, Conj. 7.2] that (Z/kZ)n satisfies prope−rty D for
everyk andn, andunder thisconjecturewecanimprove upontheboundsfortheErdős-
Ginzburg-Ziv constant using the slice rank method. Recently, Hegedűs [9] showed that
if Fn satisfies property D, then
p
1− (p−2)2 n+1
s Fn (p 1)p(cid:18) 2p2logp(cid:19) +1,
p ≤ −
and we give the following im(cid:0) pr(cid:1)ovement:
Theorem 2. Assume that (Z/kZ)n satisfies property D. Then
s((Z/kZ)n) (n 1)γn +1
≤ − k,q
where
k 1 xq
γk,q = inf − x−q−k1.
q 0<x<1 1 x
−
In particular, if q is a prime power and (Z/qZ)n satisfies property D, then
s((Z/qZ)n) (n 1)4n +1.
≤ −
Remark 1. When k = q is a prime power, γ is at most 4 since
q,q
γq,q = q inf 1−xq x−q−q1 < inf 1 = 4.
q 0<x<1 (1 x) 0<x<1 (1 x)x
− −
For general k,q,
q−1
γk,q < k inf 1 x−q−k1 = k 1+ q −1 1+ k k ,
q 0<x<1 1 x q k q 1
− (cid:18) (cid:19)(cid:18) − (cid:19)
1
and as (1+x)x is decreasing on (1, ), for k q we have the bound
∞ ≥
k 1
γ < 2 1+ − .
k,q
q
(cid:18) (cid:19)
To prove theorem 2, we first prove the following result for subsets of (Z/kZ)n:
Theorem 3. Suppose that A (Z/kZ)n satisfies
⊂
A > γn
| | k,q
where
k 1 xq
γk,q = inf − x−q−k1,
q 0<x<1 1 x
−
and q is the largest prime power dividing k. Then A contains k not necessarily distinct,
but not all equal, elements x ,...,x which sum to 0.
1 k
Using this result, theorem 2 follows as a corollary.
EXPONENTIAL BOUNDS FOR THE ERDŐS-GINZBURG-ZIV CONSTANT 4
Proof. (Theorem 3 implies theorem 2) Let S be a sequence in (Z/kZ)n of length
s((Z/kZ)n) 1 for which no k elements that sum to 0. By property D, it follows
−
that as a multi-set
S = k−1T
∪i=1
for some T (Z/kZ)n. This implies that the only solution to
⊂
x + +x = 0
1 k
···
for x T is when x = x = = x , and so by theorem 3 T γn, and hence
s((Z/ik∈Z)n) (n 1)1γn +12. ··· k | | ≤ k (cid:3)
≤ − k
Theproofoftheorem3uses theslice rankmethodanddoesnotrequirethemulti-slice
rank. The key ideas originate in [2], where they extend Ellenberg and Gijswijt’s bounds
for progression-free sets in Fn to progression-free sets in (Z/kZ)n using a binomial
p
coefficient indicator as well as a lemma concerning the slice-rank of a tensor-product.
In particular, in [2] they proved the following theorem:
Theorem 4. ([2, Theorem 4.14]) Suppose that A (Z/kZ)n contains no non-trivial 3-
⊂
term arithmetic progression, that is it contains no non-trivial solutions to the equations
x+y = 2z. Then
A (k J(q))n
| | ≤ ·
where q is the largest prime power dividing k.
Theorem 3 extends the above result to a k-variable linear equation.
2. The Multi-Slice Rank
To motivate the definition of the slice rank and the multi-slice rank, we begin by
recalling the definition of the tensor rank. For finite sets X ,...,X , a function
1 n
h : X X F,
1 n
×···× →
h = 0, of the form
6
h(x ,...,x ) = f (x )f (x ) f (x )
1 n 1 1 2 2 n n
···
for some f ,...,f , is called a rank 1 function, and the tensor rank of
1 n
F : X X F
1 n
×···× →
is defined to be the minimal k such that
k
F = g
i
i=1
X
where the g are rank 1 functions. To define the multi-slice rank, we first need some
i
notation. Given variables x ,...,x and a set S 1,...,n , S = s ,...,s , let x
1 n 1 k −→S
⊂ { } { }
denote the k-tuple
x ,...,x ,
s1 sk
and so for a function g of k variables, we have that
g(x ) = g(x ,...,x ).
−→S s1 sk
EXPONENTIAL BOUNDS FOR THE ERDŐS-GINZBURG-ZIV CONSTANT 5
Definition 1. Let X ,...,X be finite sets. We say that the function
1 n
h : X X F,
1 n
×···× →
h = 0 is a b-slice for 1 b n if
6 ≤ ≤ 2
h(x ,...,x ) = f(x )g x
1 n −→S −→{1,...,n}\S
for some f,g and S 1,2,...,n of size S =(cid:0)b. More gen(cid:1)erally, we say that F is a
⊂ { } | |
slice if it is a b-slice for some 1 b n.
≤ ≤ 2
For example, the function
F : X X X X F
× × × →
defined by
1 if x = y and z = w
F (x,y,z,w) =
0 otherwise
(
takes the form F(x,y,z,w) = δ(x,y)δ(z,w) where δ(x,y) is 1 if x = y, and 0 otherwise,
and so F is a 2-slice.
Definition 2. Let X ,...,X be finite sets. The slice rank of F is the minimal k such
1 n
that
k
F = c h
i i
i=1
X
where each h is a 1-slice. The multi-slice rank of a function
i
F : X X F
1 n
×···× →
is the minimal k such that
k
F = c g
i i
i=1
X
where the g are slices, that is each g is a b-slice for some b n.
i i ≤ 2
We note that for two variables, the slice-rank, multi-slice rank and tensor rank are
equivalent. For three variables, the multi-slice rank and the slice rank are equivalent,
and for 4 or more variables, all three ranks are different. A key property of the multi-
slice rank is the following lemma, given in [12, Lemma 8], which generalizes [15, Lemma
1].
Lemma 1. Let X be a finite set, and let Xk denote the k-fold Cartesian product of X
with itself. Suppose that
F : Xk F
→
is a diagonal tensor, that is
F(x ,...,x ) = c δ (x ) δ (x )
1 n a a 1 a n
···
a∈A
X
for some A X where c = 0, and
a
⊂ 6
1 x = a
δ (x) = .
a
0 otherwise
(
Then
multi-slice-rank(F) = A .
| |
EXPONENTIAL BOUNDS FOR THE ERDŐS-GINZBURG-ZIV CONSTANT 6
Proof. See [12, Lemma 8]. (cid:3)
Thefollowinglemmaisproposition4.2in[2],anditplaysanimportantroleinproving
theorem 2.
Lemma 2. Let G,H be finite abelian groups, and suppose that u : Gk F and v :
Hk F where Gk denotes the k-fold product G G. Define the t→ensor on the
→ × ···×
product of the groups
u v : (G H)k F
⊗ × →
by writting each element t G H as a pair t = (g,h) where g G and h H, and
∈ × ∈ ∈
setting
u v(t ,...,t ) = u(g ,...,g )v(h ,...,h )
1 k 1 k 1 k
⊗
where t is given as the pair (g ,h ) for each i. Then
i i i
slice-rank(u v) slice-rank(u) H .
⊗ ≤ ·| |
Proof. Suppose that slice-rank(u) = k. Then we may write
k
u(g ,...,g ) = f (g )v g
1 k i ji i −→{1,...,k}\ji
i=1
X (cid:0) (cid:1)
for some choice of functions f ,v and indices j . For r ,...,r H we can split v based
i i i 1 k
∈
on its value at each point in Hk and write
v(r ,...,r ) = c δ(h ,r ) δ(h ,r )
1 k h1,...,hk 1 1 ··· k k
(h1,..X.,hk)∈Hk
where
1 if h = r
δ(h,r) = ,
0 otherwise
(
and c F are constants. Then
h1,...,hk ∈
k
u(g ,...,g )v(r ,...,r ) = f (g )δ(h ,r )v g R −→h
1 k 1 k i ji ji ji i −→{1,...,k}\ji {1,...,k}\ji
Xi=1 hXji∈H (cid:0) (cid:1) (cid:16) (cid:17)
where
 k 
R −→h = c δ(h ,r ) ,
{1,...,k}\ji h1,...,hk l l
 
−→
(cid:16) (cid:17) h{1,...,kX}\ji∈Hk−1 lY= 1 
 l = j 
i
 6 
 
and so
slice-rank(u v) k H .
⊗ ≤ | |
(cid:3)
EXPONENTIAL BOUNDS FOR THE ERDŐS-GINZBURG-ZIV CONSTANT 7
3. Unconditional Bounds for s Fn
p
Let p > 2 be a prime number. For x ,...,x Fn, define(cid:0) (cid:1)
1 p ∈ p
F : Fn p F
p p → p
by (cid:0) (cid:1)
n
(3.1) F (x ,...,x ) = 1 (x +x + +x )p−1 ,
p 1 p 1i 2i pi
− ···
i=1
Y(cid:0) (cid:1)
where x is understood to be the ith coordinate of the vector x Fn, and the notation
ji j ∈ p
Fn p is used to denote the p-fold Cartesian product of Fn with itself. Then
p p
(cid:0) (cid:1)
1 if x +x + +x = 0
1 2 p
F (x ,...,x ) = ··· .
p 1 p
0 otherwise
(
This tensor does not take into account whether or not the variables are distinct, and
so in particular for any x Fn
∈ p
F (x,...,x) = 1.
p
In order to modify this tensor so that it picks up only distinct k-tuples of elements
summing to zero, we use the technique in [12], and introduce an indicator which is a
sum over the permutations in the symmetric group S . For every σ S , define
p p
∈
(3.2) f : Xp F
σ
→
to be the function that is 1 if (x ,...,x ) is a fixed point of σ, and 0 otherwise. The
1 p
following is [12, Lemma 4]:
Lemma 3. We have the identity
1 if x ,...,x are distinct
1 k
sgn(σ)f (x ,...,x ) = ,
σ 1 k
0 otherwise
(
σX∈Sk
where sgn(σ) is the sign of the permutation.
Proof. By definition,
sgn(σ)f (x ,...,x ) = sgn(σ)
σ 1 k
σX∈Sk σ∈SXtab(−→x)
whereStab(x) S isthestabilizerof x. Sincethestabilizerisaproductofsymmetric
−→ k −→
⊂
groups, this will be non-zero precisely when Stab(x) is trivial, and hence x ,...,x
−→ 1 k
must be distinct. (cid:3)
Using this lemma, we give the following modification of [12, Lemma 5]:
Lemma 4. Let S denote the set all elements in S given as a product of i disjoint
i k k
C ⊂
cycles. Define
R (x ,...,x ) = sgn(σ)f (x ,...,x ).
k 1 k σ 1 k
σXSk
∈
σ / C ,C
1 2
∈
EXPONENTIAL BOUNDS FOR THE ERDŐS-GINZBURG-ZIV CONSTANT 8
Then for k 3,
≥
1 if x ,...,x are distinct
1 k
( 1)k−1(k 1)! k−1 1 if x = = x
R (x ,...,x ) =  − − j=2 j 1 ··· k ,
k 1 k
α(x1,...,xk) P if x1,...,xk take on 2 distinct values

0 otherwise


where 
α(x ,...,x ) = ( 1)k−1# σ that fix (x ,...,x ) .
1 k 2 1 k
− { ∈ C }
Proof. If there are 3 or more distinct elements among x ,...,x then f (x ,...,x ) = 0
1 k σ 1 k
for any σ , , and so the identity holds by lemma 3. When there are exactly two
2 1
∈ C C
distinct elements among x ,...,x , f (x ,...,x ) = 0 for any σ , and so
1 k σ 1 k 1
∈ C
R (x ,...,x ) = ( 1)k−1# σ that fix (x ,...,x )
k 1 k 2 1 k
− { ∈ C }
since sgn(σ) = ( 1)k for any σ . Lastly, when x = = x , we have that
2 1 k
− ∈ C ···
R (x ,...,x )+( 1)k +( 1)k−1 = 0
k 1 k 2 1
− |C | − |C |
by lemma 3 since sgnσ = ( 1)k−1 for σ . Since
1
− ∈ C
k−1
1
= (k 1)! and = (k 1)! ,
1 2
|C | − |C | − j
j=1
X
it follows that
k−1
1
R (x ,...,x ) = ( 1)k−1
l 1 k
− j
j=2
X
when x = = x . (cid:3)
1 k
···
Remark 2. Let
1 x = = x
1 k
δ(x ,...,x ) = ··· .
1 k
0 otherwise
(
We can calculate R (x ,...,x ) exactly as a linear combination of k! (k 1)! (k
k 1 k
− − − −
1)! k−1 1 delta functions, including the constant function. When k = 3
j=1 j
P R (x ,x ,x ) = 1,
3 1 2 3
when k = 4
R (x ,x ,x ,x ) =1 δ(x ,x ) δ (x ,x ) δ(x ,x )
4 1 2 3 4 1 2 2 2 3 3 4
− − −
δ(x ,x ) δ(x ,x ) δ(x ,x ),
4 1 1 3 2 4
− − −
and when k = 5
R (x ,x ,x ,x ,x ) =1 δ(x ,x )+2 δ(x ,x ,x )
5 1 2 3 4 5 i j i j l
−
i<j≤5 i<j<l≤5
X X
δ(x ,x ) δ(x ,x ).
i j l m
i<Xj≤5 l <Xm 5
≤
l,m = i,j
6
EXPONENTIAL BOUNDS FOR THE ERDŐS-GINZBURG-ZIV CONSTANT 9
Starting with k = 5, there will be terms that are the product of multiple delta functions
in multiple variables, and to handle these we need to use the multi-slice rank. Note,
for example, that as a function on X4,
δ(x ,x )δ(x ,x )
1 2 3 4
has slice rank equal to X (see [16]) but multi-slice rank equal to 1.
| |
We now use the function R (x ,...,x ) to modify F , and arrive at an appropriate
p 1 p p
tensor.
Lemma 5. For x ,...,x Fn define
1 p ∈ p
I (x ,...,x ) = R (x ,...,x )F (x ,...,x )
p 1 p p 1 p p 1 p
where F is defined in (3.1) and R is defined in 4. Let B Fn be any subset which
p p ⊂ p
has no two elements x,y B satisfying x = cy for some c F×. Then when restricted
∈ ∈ p
to Ap = A A, I is given by
p
×···×
1 if x ,...,x are distinct and sum to zero
1 p
I (x ,...,x ) = ( 1)p−1 if x = = x .
p 1 p  1 p
− ···
0 otherwise
Proof. By pairing every element in F× with it’s inverse we have that
 p
p−1 p−1
1 1
= 1+ = 1,
j − j −
j=2 j=1
X X
and so by Wilson’s theorem
p−1
1
( 1)p−1(p 1)! = ( 1)p−1,
− − j −
j=2
X
and hence if x = = x , I (x ,...,x ) = ( 1)p−1. If there are exactly two distinct
1 p p 1 p
elements among x·,·.·..,x , say x,y Fn, then−
1 p ∈ p
x + +x = cx+(p c)y
1 p
··· −
where 1 c p 1, and this can never equal 0 by the assumption on B, and so
≤ ≤ −
F (x ,...,x ) = 0 in this case. When x ,...,x are distinct, F (x ,...,x ) = 1, and
p 1 p 1 p p 1 p
so the result follows. (cid:3)
We are now ready to prove theorem 1.
Proof. Suppose that A Fn, and that there does not exist distinct x ,...,x A
⊂ p 1 p ∈
satifying
x + +x = 0.
1 p
···
Let v Fn be any vector. Then A cannot contain the entire line 0,v,2v,...,(p 1)v,
∈ p −
since this sums to 0, and so A contains at most p 1 elements along any line. Thus
−
there exists B A of size
⊂ 1
B A
| | ≥ p 1| |
−
such that no two elements in B lie on the same line. Since B is a subset of A, there
does not exist distinct x ,...,x B satisfying x + +x = 0. Letting I be defined
1 p 1 p p
∈ ···
EXPONENTIAL BOUNDS FOR THE ERDŐS-GINZBURG-ZIV CONSTANT 10
as in 5, when we restrict I to Bp it will be a diagonal tensor with ( 1)p−1 along the
p
−
diagonal, and so by 1
B multi-slice-rank(I ).
p
| | ≤
To bound the multi-slice rank of I (x ,...,x ), consider
p 1 p
(3.3) f (x ,...,x )F (x ,...,x )
σ 1 p p 1 p
where f was defined in (3.2) and F was defined in (3.1). We can decompose σ as a
σ p
product of disjoint cycles π π . Let r be the number of elements in the cycle π ,
1 t i i
so that t r = p. Write ea··ch· cycle explicitely as π = (j j j ) where j are
i=1 i i i1 i2 ··· iri il
indices, and i ranges from 1 to t, and l ranges from 1 to r . For f (x ,...,x ) to be
i σ 1 p
P
non-zero, we must have
x = x = = x .
ji1 ji2 ··· iri
Define the function
n t p−1
M (z ,...,z ) = 1 r z .
r1,...,rt 1 t  − i ij 
!
j=1 i=1
Y X
 
When f = 1, then we have the equality of functions
σ
f (x ,...,x )F (x ,...,x ) = f (x ,...,x )M x ,...,x
σ 1 p p 1 p σ 1 p r1,...,rt j1,1 jt,1
and so the mult-slice rank f f (x ,...,x )F (x ,...,x ) will b(cid:0)e at most the(cid:1)slice rank
σ 1 p p 1 p
of M (z ,...,z ). This is a polynomial of degree (p 1)n in t variables, and we
r1,...,rt 1 t −
may expand the product as a linear combination of monomials of the form
(ze11 ze1n)(ze21 ze2n) (zet1 zetn),
11 ··· 1n 21 ··· 2n ··· t1 ··· tn
where e p 1 and t n e (p 1)n. For each term, there will be some
ij ≤ − i=1 j=1 ij ≤ −
coordinate i such that the monomial
P P
zei1 zein
i1 ··· in
hasdegree at most (p−1)n. By always slicing offthe lowest degree piece of our monomial,
t
it follows that the slice rank of M will be at most t multiplied by the number of
r1,...,rt
n-variable monomials over F of degree at most n(p−1). In other words,
p t
n
n(p 1)
(3.4) slice-rank(M ) t # v 0,1,...,p 1 n : v − .
r1,...,rt ≤ · ∈ { − } i ≤ t
( )
i=1
X
Let n denote the unsigned Stirling numbers of the first kind, which count the number
k
of elements in S which are the product of exactly k disjoint cycles. Thus we have
n
(cid:2) (cid:3)
proven that
(3.5)
p n
n n(p 1)
multi-slice-rank(I ) t # v 0,1,...,p 1 n : v − .
p i
≤ t · ∈ { − } ≤ t
( )
Xt=3 h i Xi=1
We will use a slightly cruder bound, using the fact that the right hand side of (3.4) as
a function of t is largest when t = 3. Since there are p! terms, we have the bound
n
n(p 1)
multi-slice-rank(I ) 3p!# v 0,1,...,p 1 n : v − .
p i
≤ ∈ { − } ≤ 3
( )
i=1
X