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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
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
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
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
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.
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 2 / 20
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.
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 2 / 20
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.
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 2 / 20
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.
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 2 / 20
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.
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 2 / 20
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).
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 3 / 20
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).
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 3 / 20
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).
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 3 / 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
“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
3.
(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.
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 5 / 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
3.
(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.
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 5 / 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
3.
(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.
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 5 / 20
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?
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 6 / 20
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?
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 6 / 20
Large Sum-Free Sets in Fn3 (Continued)
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.
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 7 / 20
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 ).
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 8 / 20
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 ).
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 8 / 20
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 ).
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 8 / 20
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 ).
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 8 / 20
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 ).
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 8 / 20
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 ).
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 8 / 20
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.
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 9 / 20
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.
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 9 / 20
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?
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 10 / 20
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?
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 10 / 20
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?
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 10 / 20
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.)
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 11 / 20
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.)
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 11 / 20
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.)
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 11 / 20
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.)
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 11 / 20
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?
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 12 / 20
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?
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 12 / 20
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?
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 12 / 20
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?
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 12 / 20
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".
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 13 / 20
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".
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 13 / 20
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".
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 13 / 20
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".
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 13 / 20
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).)
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 14 / 20
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).)
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 14 / 20
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).)
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 14 / 20
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?
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 15 / 20
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?
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 15 / 20
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?
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 15 / 20
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
4
seems particularly unsatisfactory.)
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 16 / 20
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
4
seems particularly unsatisfactory.)
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 16 / 20
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
4
seems particularly unsatisfactory.)
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 16 / 20
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?
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 17 / 20
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?
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 17 / 20
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?
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 17 / 20
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?
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 17 / 20
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.
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 18 / 20
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.
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 18 / 20
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|.
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 19 / 20
An “Elementary” Kneser’s Theorem
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|.
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|.
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 19 / 20
An “Elementary” Kneser’s Theorem
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|.
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|.
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 19 / 20
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).
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 20 / 20
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).
Vsevolod F. Lev (The University of Haifa) Problems in Additive Combinatorics Graz, January 2015 20 / 20
Thank you!