A proof of Freiman s Theorem, continued. An analogue of Freiman s Theorem in a bounded torsion group

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1 A proof of Freiman s Theorem, continued Brad Hannigan-Daley University of Waterloo Freiman s Theorem Recall that a d-dimensional generalized arithmetic progression (GAP) in an abelian group G is a subset of the form Q = {a 0 + x i a i : 0 x i < n i } for some elements a 0,..., a k of G. Such a GAP is proper if each of those sums is distinct; equivalently, the size of the GAP is d n i. Given a finite set of integers A, recall that A + A 2 A 1 and that A is an arithmetic progression (that is, a 1-dimensional GAP) if and only if this bound is sharp. Freiman s Theorem similarly asserts that if A has a small sumset A + A, then A is contained in a GAP that isn t too big in the following sense: Freiman s Theorem. Let A be a finite subset of Z with A + A C A. Then there exist constants d and S depending only on C such that A is contained in a GAP of dimension at most d and size at most S A. Using some graph theory, particularly Menger s Theorem on counting vertex-disjoint paths, we have previously proved the following result which will be used several times in the following exposition: Plünnecke s Theorem A. If A and B are finite sets of integers for which A + B C B, then ka la C k+l B for all integers k and l. The particular case of this theorem that we find most useful is that if A + A C A, we have ka la C k+l A for all k and l. An analogue of Freiman s Theorem in a bounded torsion group Given a finite subset A of Z with small sumset A + A, Freiman s Theorem is about finding upper bounds for the smallest size and dimension of a GAP containing A. A naturally analogous problem to consider is, given a finite subset A of a bounded torsion group with small A + A, to find an upper bound for the size of the subgroup A that it generates (i.e. the smallest subgroup containing A). Suppose that G is an abelian group, written additively, such that each element of G has order at most r, and let A be a finite subset of G. Where A denotes the subgroup generated by A, we have A = { a A n aa : 0 n a < r} and hence A r A. This bound is not especially clever, though it is tight for general A. 1 However, if A + A is small, we can use one of Plünnecke s results to give what may be a much nicer bound: Ruzsa s version of Freiman s Theorem in a bounded torsion group. Let G be an abelian group in which every element has order at most r. Let A be a finite subset of G such that A + A C A for some C > 0. Then A C 2 r C4 A. 1 For example, suppose G = (Z/2Z) and A = {e 1,..., e n} for some n, where {e 1,..., e } is the usual basis for G. In this situation, we can take r = 2 and we have A = 2 n. 1

2 Proof. Suppose we had some X G such that A A A + X. We could then conclude that A A A X C 2 A r X where this final equality follows from Plünnecke s Theorem A (with k = l = 1). We proceed by looking for a small X which satisfies this condition. Since G is a torsion group, we have A = ja and similarly for each X G. We therefore want a subset X such that ja A A + X for every j 1. Take X to be a maximal subset of 2A A such that the translates {A + x : x X} are disjoint. (There exists such a set since any singleton {x} trivially satisfies this condition.) As each translate x + A is contained in 3A A, we then have X = 1 A (x + A) 1 A x X 3A A 1 A C4 A = C 4 where we have again used Plünnecke s Theorem A, with k = 3, l = 1. Let t 2A A. By maximality of X, there is some x X such that (A + t) (A + x), so t A A + X and thus 2A A A A + X. It follows that 3A A 2A A + X A A + 2X, and similarly we obtain (j + 1)A A A A + jx A A + X for each j 1. But observe that ja (j + 1)A A, since each element of ja is of the form a a j = (a a j + a j ) a j (j + 1)A A. We have therefore shown ja A A + X for each j 1, and thus A A A + X. Then from previous remarks, A C 2 A r X C 2 r C4 A. Freiman homomorphisms and Ruzsa s embedding lemma As we are interested in finding GAPs in abelian groups (Z, in particular), it would be useful to have an understanding of functions which preserve this structure. It is clear that any group homomorphism from G to another abelian group H preserves GAPs, but this is a severer restriction than we require. As a simple case, suppose φ : G H is a function which preserves 1-dimensional GAPs, i.e. arithmetic progressions. We must then require that if a, b, c G with b a = c b, we have φ(b) φ(a) = φ(c) φ(b) or, equivalently, a + c = b + b implies φ(a) + φ(c) = φ(b) + φ(b). We can consider this as motivating the following definition: Definition. Let A and B be subsets of abelian groups G and H respectively, written additively. A (Freiman) k-homomorphism from A to B is a map φ : A B such that whenever x x k = y y k holds for elements x i, y i A, we have φ(x 1 ) φ(x k ) = φ(y 1 ) φ(y k ). If such a φ is a bijection whose inverse is also a k-homomorphism, then φ is called a k-isomorphism and we shall say that A and B are k-isomorphic. A few simple remarks about Freiman homomorphisms: The restriction of a homomorphism of abelian groups to a subset of its domain is clearly a k-homomorphism for each k. If φ is a k-homomorphism then it is also a j-homomorphism for all 1 j k: choose any c A, and suppose x x j = y y j. Adding (k j)c to each side of this equation, we then have φ(x 1 ) φ(x j ) + (k j)φ(c) = φ(y 1 ) φ(y j ) + (k j)φ(c) and the result follows. The composition of two k-homomorphisms is again a k-homomorphism, and if they are both k-isomorphisms then so is their composition. If φ : A B is a k-homomorphism, then it induces a map from ka to kb by taking a a k to φ(a 1 ) φ(a k ). Hence if A and B are k-isomorphic then ka = kb. 2

3 Given coprime integers q and, the map from Z/ to itself which takes a to qa is a k-isomorphism for all k since it is a group automorphism. From previous discussion, we see that 2-homomorphisms preserve arithmetic progressions. Even more strongly, they preserve GAPs of all dimensions: Proposition. Let φ : A B be a 2-homomorphism, and suppose Q is a d-dimensional GAP in A. Then φ(q) is a d-dimensional GAP in B. Proof. Write Q = {a 0 + x i a i : 0 x i < n i }. We show that φ ( a 0 + ) x i a i = φ(a 0 ) + x i (φ(a 0 + a i ) φ(a 0 )) for all elements a 0 + d x ia i of Q, from which the result follows immediately. We proceed by induction on m = d x i. If m = 0 then each side of the equation is φ(a 0 ). If m = 1, then there is some j with x j = 1 and x i = 0 for all other i, in which case both sides of the equation are equal to φ(a 0 + a j ). ow, take any r = a 0 + d x ia i with d x i > 1. Choose an index j with x j 1. By induction, we then have φ (r a j ) = φ(a 0 ) + x i (φ(a 0 + a i ) φ(a 0 )) + (x j 1) (φ(a 0 + a j ) φ(a 0 )). i j Since φ is a 2-homomorphism and r + a 0 = (r a j ) + (a 0 + a j ) with each of these four terms lying in Q, we have φ(r) + φ(a 0 ) = φ(r a j ) + φ(a 0 + a j ) from which we recover as desired. φ(r) = φ(a 0 ) + x i (φ(a 0 + a i ) φ(a 0 )) Suppose we have A Z and a natural number. The map π : A Z/ given by reduction (mod ) is the restriction of a group homomorphism, and therefore it is a k-homomorphism for all k. It is injective if A lies in an interval of length less than, but in general its inverse may not be a k-homomorphism: as a simple example, suppose A = {1, 2} and we take π : A Z/2. This map is an injective k-homomorphism for all k, but its inverse is not a k-homomorphism for any k 2 since = but On the other hand, suppose we have defined ψ : Z/ Z by mapping each residue class to its unique representative in [1, ]. In( general this] is not a k-homomorphism, but suppose we restrict ψ to a subset B of Z/ such that ψ(b) j k, (j+1) k for some j. Using bars to denote residue classes, let a 1,..., a k, b 1,..., b k B with k a i = k b i. Then k ψ(a i) k ψ(b i) is a multiple of, but from the way we have restricted ψ we must have k ψ(a i) k ψ(b i) < and so we really have k ψ(a i) = k ψ(b i), thus ψ is a k-homomorphism. 3

4 Given a finite subset A of Z, we outline our strategy for finding a GAP containing A as follows: (Ruzsa s Embedding Lemma) Find sufficiently large such that A has a large subset A which is k-isomorphic to a subset A of Z/. (Bogolyubov s Lemma) Using a Fourier argument, find a highly structured Bohr set in 2A 2A. Using Minkowski s theorems from the geometry of numbers, show that such a Bohr set contains a large GAP Q. From Q, construct a GAP containing A as a large subset. Ruzsa s Embedding Lemma. Let A be a set of integers with ka ka C A. Then for any prime > 2C A we may find a subset A of A with A A /k such that A is k-isomorphic to a subset of Z/. Proof. Let p be a very large prime, and let 1 q p 1. Define the map φ q : A Z which takes each a to the reduction of qa (mod p) in [1, p] Z. We clearly have A = k 1 j=0 (( ]) jp φ 1 (j + 1)p q,. k k Then, (( by the pigeonhole ]) principle, we can choose some j such that the size of the preimage φ 1 jp q k, (j+1)p k is at least A /k. Denote this preimage by A q. Restrict φ q to A q ; from previous remarks, we see that φ q is then a k-isomorphism from A q to its image. Let > 0 be a parameter, and define φ q : A q Z/ by taking each a to the residue class of φ q (a) (mod ). As a composition of two k-homomorphisms, φ q is a k-homomorphism. Our goal is to show that for large enough, there is some q for which φ q is, in fact, a k-isomorphism. Suppose that φ q is not a k-isomorphism. Then there are a i, b i in A q with k a i k b i but k φ q(a i ) = k φ q(b i ). We will count the number of ( bad ) values of q for which this is possible, given fixed. Since φ q is a k-isomorphism, k φ q(a i ) k φ q(b i ), but these two quantities are congruent (mod ) from the previous inequation. That is, there is some nonzero integer l such that l = φ q (a i ) φ q (b i ). ote that since the image of φ q is contained in an interval of size p/k, we have l k(p/k) = p and hence 0 < l p/. From the definition of φ q, it follows that ( l q a i ) b i (mod p). Suppose we fix such nonzero k a i k b i and l. Assuming without loss of generality that p > max x,y ka x y k a i k b i, it follows that k a i k b i is not congruent to 0 (mod p) and hence there is exactly one q which satisfies the above congruence. As there are at most ka ka C A choices of k a i k b i and at most 2p/ 1 choices of l, there are at most C A (2p/ 1) bad values of q. To guarantee the existence of a good q, it therefore suffices to have ( ) 2p C A 1 < p 1 4

5 or, rearranged, from which the result follows. > 2C A 1 + C A 1 p Bohr sets and the geometry of numbers Definition. Let be a large prime, let K = {r 1,..., r k } be a set of distinct residue classes (mod ), and let δ [0, 1). The Bohr set B(K; δ) is the set of all residue classes s such that srj δ for each 1 j k. We refer to k as the dimension of this Bohr set. Bogolyubov s Lemma. Let X be a subset of Z/ with X δ. Then 2X 2X contains a Bohr set B(K; 1/4) of dimension k 1/δ 2. Proof. Assume without loss of generality that X = δ. Recall that, by Parseval s formula, we have r(mod ) ˆ1 X (r) 2 = X = δ 2. It follows that there are at most 1/δ 2 elements r of Z/ with ˆ1 X (r) δ X, since otherwise we would have ˆ1 X (r) 2 > 1 δ 2 ( δ X ) 2 = δ 2, r(mod ) a contradiction. One of these elements is 0, since ˆ1 X (0) = X > δ X. Let K = {r 1,..., r k } be the set of all nonzero r with ˆ1 X (r) δ X. We show that B(K; 1/4) 2X 2X. Let b be an element of that Bohr set. Consider the exponential sum S := = r(mod ) ˆ1 X (r) ( ) 4 br e ( (x3 + x 4 x 1 x 2 + b)r e r(mod ) x 1,x 2,x 3,x 4 X = # { (x 1, x 2, x 3, x 4 ) X 4 : x 1 + x 2 x 3 x 4 b(mod ) }. To show that b 2X 2X, then, it suffices to show that S is positive. Since S is real, we may rewrite it as r(mod ) ˆ1 X (r) 4 cos ( ) 2πbr. Clearly, the contribution from the term r = 0 is X 4. Suppose r K; then since b B(K; 1/4) we have br 1/4 and so ˆ1 X (r) 4 cos ( ) 2πbr ˆ1 X (r) 4 cos(π/2) = 0. It therefore suffices to show that the contribution to S from the nonzero r K that is, those r for which ˆ1 X (r) < δ X is less than X 4 in absolute value. Let R be the set of such r, so and the result follows. ˆ1 X (r) ( ) 4 2πbr cos r R ( δ X ) 2 r R < δ X 2 r(mod ) ) ˆ1 X (r) 2 ˆ1 X (r) 2 = δ X 2 ( X ) = X 4, 5

6 We shall need some results from the geometry of numbers. In particular, recall Minkowski s Second Theorem: Minkowski s Second Theorem. Let K be a symmetric convex body and Λ a lattice in R k. Let λ 1 λ k be the successive minima of C with respect to Λ. Then λ 1 λ k 2 k det(λ)/vol(k). Recall also the Volume Packing Lemma: Lemma (Volume Packing Lemma). Let Λ, Λ be lattices with Λ Λ R n. Then det(λ) = det(λ )[Λ : Λ]. We are now ready to prove that Bohr sets contain large GAPs. Proposition. Let be a large prime, let r 1,..., r k be distinct residue classes (mod ) with k 2, and let δ (0, 1). Then the Bohr set B(r 1,..., r k ; δ) in Z/ contains a proper GAP of dimension k and size at least (δ/k) k. Proof. Let r = (r 1,..., r k ), identifying each r j with its representative in [1, ], and define Λ to be the lattice generated by Z k and r, i.e. 1 Λ = (lr + (Z) k ). l=0 As is prime and the r j are distinct, at least one of the r j is coprime to. Assume without loss of generality that r 1 is coprime to. It follows that the cosets lr + (Z) k are disjoint for 0 l 1, since the values lr 1 are distinct (mod ), and hence [Λ : (Z) k ] =. We therefore have det(λ) = det((z) k )/ = k 1. ow, let C = {x R k : x i < 1, 1 i k}. Then clearly C is a convex symmetric body with vol(c) = 2 k. Taking λ 1... λ k to be the successive minima of C with respect to Λ, choose linearly independent lattice points b 1,..., b k such that b j λ j C. For 1 j k, write b j = (b j r 1 + q 1,j,..., b j r k + q k,j ) for b j {0,..., 1} and q i,j Z. Consider linear combinations of the form n j b j = n j (b j r 1 + q 1,j ),..., n j (b j r k + q k,j ) where the n j are integers with n j δ/(kλ j ). Since each b j λ j C, each coordinate k n j(b j r i + q i,j ) of this linear combination is bounded in absolute value by k n j λ j ( ) k δ kλ j λ j = δ. Then for each 1 i k, k r i n jb j = k n j(b j r i + q i,j ) k n j(b j r i + q i,j ) δ = δ and so k n jb j B(r 1,..., r k ; δ). This is the k-dimensional GAP that we seek. To see that it is a proper GAP, consider that if k n jb j = k n j b j, we then have k n jb j k n j b j (mod ) 6

7 coordinatewise, but we have seen that the coordinates of these vectors are bounded in absolute value by δ < and so the two vectors must actually be equal. Then n j = n j for all j since the vectors b j are linearly independent, so the GAP is proper. Since our n j run over n j δ kλ j, the size of the GAP is k ( ) δ kλ j k δ kλ j ( ) k δ k (λ 1 λ k ) 1. k From Minkowski s Second Theorem, we have λ 1 λ k 2 k det(λ) k 1 vol(c) = 2k = k 1, and thus the size of 2 k the GAP is at least ( ) δ k k as desired. ow we put everything together to prove Freiman s Theorem. The proof of Freiman s Theorem Freiman s Theorem. Let A be a finite subset of Z with A + A C A. Then there exist constants d and S depending only on C such that A is contained in a GAP of dimension at most d and size at most S A. Proof. By Plünnecke s inequality, 8A 8A C 16 A. Let be any prime with 2C 16 A < < 4C 16 A ; such exists by Bertrand s postulate. Then by Ruzsa s Embedding Lemma, there exists some A 1 A which is 8-isomorphic to a subset A 2 of Z/, with A 1 A /8. Thus A 2 A /8 /(32C 16 ). Taking δ = 1/(32C 16 ), Bogolyubov s Lemma implies that 2A 2 2A 2 contains a Bohr set B(K; 1/4) of dimension k 1/δ C 32. From the geometry of numbers, we conclude that 2A 2 2A 2 contains a proper GAP Q of dimension d about 1024C 32, and size about e C33 A. Let X be a maximal subset of A such that the translates {Q + x : x X} are all disjoint. Then Q + X 3A 2A, so from Plünnecke s inequality we have Q + X C 5 A. Clearly X = Q+X Q C5 A = e C33 A C5 e C33. ow, by maximality of X, for any a A we must have (Q + a) (Q + X), and hence a X + Q Q. Then A X + Q Q. If we write Q = {a 0 + d y ja j : 1 x j n j }, we see that Q Q is the GAP Q Q = { d y ja j : y j n j 1} of dimension d about 1024C 32 and size at most d d (2n j 1) 2 d = 2 d Q n j C32 e C33 A Clearly X + Q Q is contained in the GAP Z = { x X δ xx + d y ja j : δ x {0, 1}, y j n j 1} of dimension d + X 2 10 C 32 + C 5 e C33 A. and size Z 2 X Q Q 2 X A 2 C5 e C33 A. 7

8 This Z is the GAP we seek, and both its dimension and the density of A in Z are bounded by functions of C as desired. We note finally that [2] by using some results of Chang [3], the bounds on d and S may be improved to d C 2 (log C) 2 and S e 220 C 2 (log C) 2. References [1] M. B. athanson. Additive umber Theory: Inverse Problems and the Geometry of Sumsets. Springer-Verlag, GTM 165, [2] K. Soundararajan, Additive Combinatorics: Winter Stanford University, [3] M.-C. Chang, A polynomial bound in Freiman s theorem. Duke Mathematical Journal 113 (2002), no. 3,