Selected Problems in Additive Combinatorics...Selected Problems in Additive Combinatorics Vsevolod...

Selected Problemsin Additive Combinatorics

Vsevolod F. Lev

The University of Haifa

Graz, January 2015

Kneser’s Theorem for Restricted Addition

Theorem (Kneser)Suppose that A and B are finite, non-empty subsets of an abeliangroup. If |A + B| < |A|+ |B| − 1, then A + B is periodic.

What is the analogue of Kneser’s Theorem for the restrictedsumset

A+B := {a + b : a ∈ A, b ∈ B, a 6= b}?

We seek a result of the following sort:

If (A+B is small), then (something very special happens).

Suppose for simplicity that the underlying group has no involutions.

The (A+B is small) part: the natural bound is |A+B| < |A|+ |B| − 3.The (something special happens) part: not only is A+B periodic, butindeed we have A+B = A + B!

Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 1 / 20

A+B := {a + b : a ∈ A, b ∈ B, a 6= b}?

Kneser’s Theorem for Restricted Addition (Continued)Conjecture (J. de Théorie des Nombres de Bordeaux 2005)Suppose that A and B are finite, non-empty subsets of aninvolution-free abelian group. If |A+B| < |A|+ |B| − 3, thenA+B = A + B (whence A+B is periodic).

A+B = A + B implies periodicity of A+B in view of the assumption|A+B| < |A|+ |B| − 3, by Kneser’s theorem.Interestingly, periodicity suffices: if one could show that|A+B| < |A|+ |B| − 3 implies periodicity of A+B, this wouldgive the strong form of the conjecture (A+B = A + B).For the group Fp, the sumset A+B is periodic iff A+B = Fp;hence, for the prime-order groups, the conjecture is equivalent tothe Erdos-Heilbronn Conjecture.In the general case, the conjecture is both an analogue anda counterpart of Kneser’s Theorem: if A+B is small, thenit can be studied using KT.

Kneser’s Theorem for Restricted Addition (Continued)Kemperman and Scherk have shown that if there exists c ∈ A + Bwith a unique representation as c = a + b with a ∈ A and b ∈ B,then |A + B| ≥ |A|+ |B| − 1.

Conjecture (Restatement 1)Suppose that A and B are finite, non-empty subsets of an abeliangroup. If there exists c ∈ A + B with a unique representation asc = a + b with a ∈ A and b ∈ B, then |A+B| ≥ |A|+ |B| − 3.

Conjecture (Restatement 2)Suppose that A and B are finite, non-empty subsets of an abeliangroup. If |(A + B) \ (A+B)| ≥ 3, then |A+B| ≥ |A|+ |B| − 3.

The general case reduces to that where 0 ∈ B ⊆ A and 0 /∈ A+B.The conjecture is true in torsion-free groups, prime-order groups,groups of exponent 2, small-order cyclic groups (computationally).

“Small” Sets in the Freiman Isomorphism ClassesThe set {0,1,2,4, . . . ,2n−2} is linear (has Freiman’s rank 1); as such, itis not Freiman-isomorphic to any shorter set. Is this the extremal case?

Conjecture (Konyagin-Lev, Mathematika 2000)

Any n-element set of integers is isomorphic to a subset of [0,2n−2].

For sets of rank r , it is natural to seek “compact” isomorphic sets in Zr .

Conjecture (Konyagin-Lev, Mathematika 2000)For any n-element integer set A of rank r , there exist integerl1, . . . , lr ≥ 0 with l1 + · · ·+ lr ≤ n − r − 1 such that A is isomorphicto a subset of the parallelepiped [0, l1]× · · · × [0, lr ] ⊆ Zr .

Both conjectures are true in the extremal cases r = 1 (linear sets) andr = n − 1 (Sidon sets).

Grynkiewicz has some partial results on the second conjecture.Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 4 / 20

“Small” Sets in the Freiman Isomorphism ClassesThe set {0,1,2,4, . . . ,2n−2} is linear (has Freiman’s rank 1); as such, itis not Freiman-isomorphic to any shorter set. Is this the extremal case?

Conjecture (Konyagin-Lev, Mathematika 2000)

Any n-element set of integers is isomorphic to a subset of [0,2n−2].

For sets of rank r , it is natural to seek “compact” isomorphic sets in Zr .

Conjecture (Konyagin-Lev, Mathematika 2000)For any n-element integer set A of rank r , there exist integerl1, . . . , lr ≥ 0 with l1 + · · ·+ lr ≤ n − r − 1 such that A is isomorphicto a subset of the parallelepiped [0, l1]× · · · × [0, lr ] ⊆ Zr .

Both conjectures are true in the extremal cases r = 1 (linear sets) andr = n − 1 (Sidon sets).

Grynkiewicz has some partial results on the second conjecture.Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 4 / 20

Large Sum-Free Sets in Fn3

Any affine hyperplane in Fn3 which is not a linear hyperplane is a

sum-free subset of Fn3 of size 3n−1. Conversely, if A ⊆ Fn

3 is sum-free,then A, A− a, and A + a are pairwise disjoint for any fixed a ∈ A.Hence, 3n−1 is the largest possible size of a sum-free subset of Fn

(Notice that in a characteristic other than 3, one can have sum-freesets A with (A− a) ∩ (A + a) 6= ∅ for some a ∈ A.)

What do large sum-free subsets of Fn3 look like?

Theorem (J. Combinatorial Theory A, 2005)

If n ≥ 3 and A ⊆ Fn3 is sum-free of size |A| > 5

27 · 3n, then A is

contained in an affine hyperplane.

The coefficient 5/27 is best possible and cannot be replaced with asmaller number. Yet, there is a room for a significant improvement.

27 · 3n, then A is

Large Sum-Free Sets in Fn3 (Continued)

One can obtain sum-free sets in Fn3 by “lifting” sum-free sets

from lower dimensions: if m < n and ϕ : Fn3 → Fm

3 is a surjectivehomomorphism, then for any sum-free set A0 ⊆ Fm

3 , the fullinverse image A := ϕ−1(A0) ⊆ Fn

3 is sum-free and has the samedensity in Fn

3 as A0 has in Fm3 .

A sum-free set A ⊆ Fn3 can be obtained by lifting if and only if it is

periodic.One can also obtain sum-free sets by removing some elementsfrom larger sum-free sets.

Thus, of real interest are those sum-free sets which are aperiodic andmaximal (by inclusion). They are “building blocks” from which all othersum-free sets can be obtained by lifting / removing elements.

How large can a maximal, aperiodic sum-free set in Fn3 be?

One can obtain sum-free sets in Fn3 by “lifting” sum-free sets

from lower dimensions: if m < n and ϕ : Fn3 → Fm

3 is a surjectivehomomorphism, then for any sum-free set A0 ⊆ Fm

3 , the fullinverse image A := ϕ−1(A0) ⊆ Fn

3 is sum-free and has the samedensity in Fn

3 as A0 has in Fm3 .

A sum-free set A ⊆ Fn3 can be obtained by lifting if and only if it is

periodic.One can also obtain sum-free sets by removing some elementsfrom larger sum-free sets.

Thus, of real interest are those sum-free sets which are aperiodic andmaximal (by inclusion). They are “building blocks” from which all othersum-free sets can be obtained by lifting / removing elements.

How large can a maximal, aperiodic sum-free set in Fn3 be?

Conjecture (J. Combinatorial Theory A, 2005)The largest possible size of a maximal, aperiodic sum-free set in Fn

3is (3n−1 + 1)/2.

There is an explicit construction of a maximal, aperiodic sum-free setin Fn

3 of size (3n−1 + 1)/2; the conjecture thus says that one cannot gobeyond that.

Establishing this conjecture would allow one to classify all sum-freesubsets of Fn

3 of density larger than 16 + ε, for any fixed ε > 0.

When does P(a− b) = 0, a 6= b imply P(0) = 0?Suppose A ⊆ Fn

2. Given that P ∈ F2[x1, . . . , xn] vanishes onA+A = (A + A) \ {0}, can we conclude that also P(0) = 0?

Not necessarily: any function on Fn2 can be represented by a

polynomial. What if deg P is small, while A is large?

Given d ≥ 0, how large must A ⊆ Fn2 be to ensure that if deg P ≤ d

and P(a + b) = 0 for all a,b ∈ A, a 6= b, then also P(0) = 0?

P constant (d = 0): suffices to have |A| ≥ 2 (so that A+A 6= ∅).P linear (d = 1): |A| = 2 insufficient (take A = {0,e1},P = x1 + 1),while |A| ≥ 3 suffices (if {a,b, c} ⊆ A andP(a + b) = P(b + c) = P(c + a) = 0, then also P(0) = 0 as0 = (a + b) + (b + c) + (c + a) and P is linear).P quadratic (d = 2): |A| ≥ n + 3 suffices, |A| = n + 1 is not enough(consider A = {0,e1, . . . ,en}, P = 1 +

∑1≤i≤n xi +

∑1≤i<j≤n xixj ).

∑1≤i≤n xi +

When does P(a− b) = 0 imply P(0) = 0? (Continued)For cubic polynomials the problem is wide open: it is not even clearwhether |A| can be polynomial in n (or must be exponential).

ProblemHow large must A ⊆ Fn

2 be to ensure that if P is a cubic polynomialin n variables over F2 satisfying P(a + b) = 0 for all a,b ∈ A, a 6= b,then also P(0) = 0?

Generally, for d ≥ 3 given, how large must A ⊆ Fn2 be to ensure that if

P ∈ F2[x1, . . . , xn] is a polynomial of degree d satisfying P(a + b) = 0for all a,b ∈ A, a 6= b, then also P(0) = 0?

Motivation: an F2-analogue of the “Roth’ problem” by Ernie Croot.To make a progress in Ernie’s problem, one needs to show thatfor polynomials of degree about (0.5 + ε)n, it suffices to have,say, |A| > 2n/n2.

When does P(a− b) = 0 imply P(0) = 0? (Continued)For cubic polynomials the problem is wide open: it is not even clearwhether |A| can be polynomial in n (or must be exponential).

ProblemHow large must A ⊆ Fn

2 be to ensure that if P is a cubic polynomialin n variables over F2 satisfying P(a + b) = 0 for all a,b ∈ A, a 6= b,then also P(0) = 0?

Generally, for d ≥ 3 given, how large must A ⊆ Fn2 be to ensure that if

P ∈ F2[x1, . . . , xn] is a polynomial of degree d satisfying P(a + b) = 0for all a,b ∈ A, a 6= b, then also P(0) = 0?

Motivation: an F2-analogue of the “Roth’ problem” by Ernie Croot.To make a progress in Ernie’s problem, one needs to show thatfor polynomials of degree about (0.5 + ε)n, it suffices to have,say, |A| > 2n/n2.

Large Doubling-Critical Sets in Fn2

DefinitionWe say that a subset A of an abelian group is doubling-critical if,for any proper subset B ( A, we have 2B 6= 2A.

That is, A is doubling-critical if, for any given a ∈ A, there exists a′ ∈ Asuch that a + a′ has a unique representation in 2A (= A + A).

Every set A with |A| ≤ 2 is doubling-critical. (Three-elementsubgroups are not doubling-critical!)Every Sidon set is doubling-critical.

The ideology: small doubling-critical sets are common, largedoubling-critical sets are rare and must be structured.

What is the largest possible size of a doubling-critical subsetof a finite abelian group G? What is the structure

of large doubling-critical subsets of G?

Large Doubling-Critical Sets in Fn2 (Continued)

If A is a doubling-critical subset of a finite abelian group G, then|A| ≤ 1

2 |G|+ 1. (For if B ⊂ A, |B| > 12 |G|, then 2B = 2A = G.)

Equality is attained if |G| is even: if H < G with |H| = 12 |G| and

g ∈ G \ H, then A := (g + H) ∪ {0} is doubling-critical.

A generalization: if S ⊆ G is sum-free, then S ∪ {0} is doubling-critical— and so are its translates g + (S ∪ {0}).

Theorem (Grynkiewicz-Lev, SIDMA 2010)

Suppose that A ⊆ Fn2 is doubling critical. If |A| > 11

36 · 2n + 3, then

A = g + (S ∪ {0})with a sum-free set S ⊆ Fn

2 and an element g ∈ Fn2.

ProblemReplace the coefficient 11

36 with the best possible one.(At least 1

4 should be possible.)

36 · 2n + 3, then

Large Doubling-Critical Sets in Fp

For a prime p, let DC[Fp] denote the family of all doubling-criticalsubsets of Fp.Since Fp has sum-free subsets of size 1

3 p + O(1), we have13

p + O(1) ≤ max{|A| : A ∈ DC[Fp]} ≤ 12

p + O(1),

and I can improve this to13

p + O(1) ≤ max{|A| : A ∈ DC[Fp]} ≤ 25

p + O(1).

ProblemWhat is the largest possible size of a doubling-critical set in Fp?(It is tempting to conjecture 1

3 p + O(1).)

ProblemHow about difference-critical sets?

3 p + O(1), we have13

p + O(1) ≤ max{|A| : A ∈ DC[Fp]} ≤ 12

p + O(1),

p + O(1) ≤ max{|A| : A ∈ DC[Fp]} ≤ 25

p + O(1).

3 p + O(1).)

3 p + O(1), we have13

p + O(1) ≤ max{|A| : A ∈ DC[Fp]} ≤ 12

p + O(1),

p + O(1) ≤ max{|A| : A ∈ DC[Fp]} ≤ 25

p + O(1).

3 p + O(1).)

3 p + O(1), we have13

p + O(1) ≤ max{|A| : A ∈ DC[Fp]} ≤ 12

p + O(1),

p + O(1) ≤ max{|A| : A ∈ DC[Fp]} ≤ 25

p + O(1).

3 p + O(1).)

Quadratic Residues are not a Perfect Difference SetSárközy conjectured that the set Rp of all quadratic residues moduloa (sufficiently large) prime p is not a sumset:

Rp 6= A + B, min{|A|, |B|} > 1.

The case B = A was settled by Shkredov: Rp 6= 2A (p > 3, A ⊆ Fp).

For B = −A, Shkredov’s method does not work: it is believed thatRp ∪ {0} 6= A− A, but we cannot prove this. A more tractable version:

Conjecture (Lev-Sonn)For a prime p > 13, there does not exist a set A ⊆ Fp such that thedifferences a′ − a′′ (a′,a′′ ∈ A, a′ 6= a′′) list all elements of Rp, andevery element is listed exactly once.

With Jack Sonn, we have established a number of necessaryconditions, and used them to show that in the range 13 < p < 1020,there are no “exceptional primes".

Rp 6= A + B, min{|A|, |B|} > 1.

Dense Perfect Difference Sets in NA set A ⊆ N is a perfect difference set if every element of N has aunique representation as a difference of two elements of A.

A construction: start with A = ∅ and at each step find the smallestd /∈ A− A and add to A two elements u and u + d , where u is largeenough to avoid any element having two (or more) representations.

This yields a perfect difference set A with the counting functionA(x)� x1/3. On the other hand, for every perfect difference setA ⊆ N one has A(x)� x1/2.

Problem (E. Journal of Combinatorics, 2004)Do there exist perfect difference sets A ⊆ N with the counting functionsatisfying A(x)� x1/2−ε? If not, how large can lim infx→∞

ln A(x)ln x be?

(Nathanson and Cilleruelo (2008) constructed perfect difference setsA ⊆ N with the counting function A(x)� x

√2−1−o(1).)

ln A(x)ln x be?

√2−1−o(1).)

ln A(x)ln x be?

√2−1−o(1).)

Flat-full Sets in Fn2

For integer 1 ≤ d ≤ n, let γd (n) denote the smallest size of a subsetA ⊆ Fn

2 such that for every v ∈ Fn2, there is a d-dimensional subspace

L < Fn2 with v + L ⊆ A∪{v}. That is, through every point v ∈ Fn

2 passesa d-flat entirely contained in A, with the possible exception of v itself.

Equivalently, γd (n) is the smallest possible size of a union of the form⋃v∈Fn

2(v + Lv \ {0}),

for all families {Lv : v ∈ Fn2} of d-subspaces.

For instance, γ1(n) = 2, and it is easy to see that γ2(n) = Θ(2n/3).

Theorem (Blokhuis-Lev, MJCNT 2013)

We have 23n/8 � γ3(n)� 23n/7, with absolute implicit constants.

ProblemWhat is the true order of magnitude of γ3(n) as n grows?

2(v + Lv \ {0}),

Small Sums of Roots of UnityIn 1975, Gerry Myerson defined f (n,q) to be the smallest possible sumof n roots of unity of degree q (repeated summands allowed), andobtained some estimates for q even.

For q prime and repetitions forbidden, we are seeking lower bounds forSA :=

∑a∈A e2πia/p, A ⊆ Fp

in terms of p and n := |A|.

Theorem (Konyagin-Lev, INTEGERS 2000)

For any set A ⊆ Fp with n := |A| ∈ [3,p − 1], we have |SA| > n−p−1

4 .On the other hand, for any n = 2k < p/20, there exists A ⊆ Fp with|A| = n such that |SA| < n−c log p, with an absolute constant c > 0.

Problem

Narrow the gap between the estimates. (The lower bound n−p−1

seems particularly unsatisfactory.)

Problem

Popular Differences in FpFor a finite subset A and an element b of an abelian group G, let

∆A(b) := |(A + b) \ A|,

the Erdos – Heilbronn (1964) / Olson (1968) function. Basic properties:non-vanishing: ∆A(0) = 0, and ∆A(b) ≥ 1 unless A + b = A;symmetry: ∆A(−b) = ∆A(b);sub-additivity: ∆A(b1 + · · ·+ bk ) ≤ ∆A(b1) + · · ·+ ∆A(bk ).

In applications, one needs to know that ∆A does not assume too manysmall values; that is, any B ⊆ G contains some b ∈ B with ∆A(b) large.

Theorem (Konyagin-Lev, Israel J. Math. 2010)For all (finite) A ⊆ Z and B ⊆ N with |B| < c|A|/ log |A|, there existsb ∈ B with ∆A(b) ≥ |B|.

(The logarithmic factor cannot be dropped!)

What is the Fp-version of this result?

∆A(b) := |(A + b) \ A|,

Popular Differences in Fp (Continued)

Theorem (Lev, J. Number Theory 2011)For all A,B ⊆ Fp with B ∩ (−B) = ∅ and|B| < min{c|A|/ log |A|,

√p/8}, there exists b ∈ B with ∆A(b) ≥ |B|.

The logarithmic factor cannot be dropped, the term√

p/8 seems atechnical annoyance.

ProblemDetermine whether the

√p/8 –term can be dropped.

Popular Differences in Fp (Continued)

Theorem (Lev, J. Number Theory 2011)For all A,B ⊆ Fp with B ∩ (−B) = ∅ and|B| < min{c|A|/ log |A|,

√p/8}, there exists b ∈ B with ∆A(b) ≥ |B|.

The logarithmic factor cannot be dropped, the term√

p/8 seems atechnical annoyance.

ProblemDetermine whether the

√p/8 –term can be dropped.

TheoremSuppose that A,B, and C are subsets of the finite abelian group G.If A + B + C 6= G, then

|A|+ |B|+ |C|+ |A + B + C| ≤ 2|G|.

An “Elementary” Kneser’s Theorem

|A|+ |B|+ |C|+ |A + B + C| ≤ 2|G|.

This theorem is completely equivalent to Kneser’s Theorem!(Some adjustments are to be made if G is infinite.)

Besides being remarkably symmetric, this restatement avoids thenotion of a period, absolutely vital in the standard formulation of KT.

Deducing “Elementary KT” from the “Standard KT”Suppose that A + B + C 6= G. If H is the period of A + B + C, then

|G| − |H| ≥ |A + B + C| ≥ |A|+ |B|+ |C| − 2|H|whence|A|+ |B|+ |C|+ |A + B + C| ≤

(|G|+ |H|

)+(|G| − |H|

)= 2|G|.

An “Elementary” Kneser’s Theorem

|A|+ |B|+ |C|+ |A + B + C| ≤ 2|G|.

This theorem is completely equivalent to Kneser’s Theorem!(Some adjustments are to be made if G is infinite.)

Besides being remarkably symmetric, this restatement avoids thenotion of a period, absolutely vital in the standard formulation of KT.

Deducing “Elementary KT” from the “Standard KT”Suppose that A + B + C 6= G. If H is the period of A + B + C, then

|G| − |H| ≥ |A + B + C| ≥ |A|+ |B|+ |C| − 2|H|whence|A|+ |B|+ |C|+ |A + B + C| ≤

(|G|+ |H|

)+(|G| − |H|

)= 2|G|.

An “Elementary” Kneser’s Theorem (Continued)The “Elementary Kneser’s Theorem”Suppose that A,B, and C are subsets of the finite abelian group G.If A + B + C 6= G, then

|A|+ |B|+ |C|+ |A + B + C| ≤ 2|G|.

Deducing “Standard KT” from the “Elementary KT”

Given A,B ⊆ G, let C := −A + B. Then A + B + C = G \ H, whereH is the period of A + B (easy). Hence by the EKT,

2|G| ≥ |A|+ |B|+ |A + B|+ |G \ H|,implying

|A + B| ≥ |A|+ |B| − |H|.

ProblemGive the “Elementary Kneser’s Theorem” an independent, simple proof(preferably not appealing to the notion of a period).

An “Elementary” Kneser’s Theorem (Continued)The “Elementary Kneser’s Theorem”Suppose that A,B, and C are subsets of the finite abelian group G.If A + B + C 6= G, then

|A|+ |B|+ |C|+ |A + B + C| ≤ 2|G|.

Deducing “Standard KT” from the “Elementary KT”

Given A,B ⊆ G, let C := −A + B. Then A + B + C = G \ H, whereH is the period of A + B (easy). Hence by the EKT,

2|G| ≥ |A|+ |B|+ |A + B|+ |G \ H|,implying

|A + B| ≥ |A|+ |B| − |H|.

ProblemGive the “Elementary Kneser’s Theorem” an independent, simple proof(preferably not appealing to the notion of a period).

Thank you!

Selected Problems in Additive Combinatorics...Selected Problems in Additive Combinatorics Vsevolod...

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