Notes on Freiman s Theorem

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1 Notes on Freiman s Theorem Jacques Verstraëte Department of Mathematics University of California, San Diego La Jolla, CA, jacques@ucsd.edu. 1 Introduction Freiman s Theorem describes the structure of a set A under the condition that A + A has size close to that of A. If P is a generalized arithmetic progression, then P + P is close to P. Freiman s Theorem states the partial converse: if P + P is close to P then P must be contained in a small generalized arithmetic progression. The theorem may be stated as follows, and we will give the remarkable proof of this theorem due to Ruzsa. Theorem 1 Let A Z N be a set such that A + A C A. Then A is contained in a d-dimensional arithmetic progression P of cardinality at most k A where d and k depend on C only. The constants from this theorem can be chosen to be d = C 40C and k = C C40C. 2 Plünnecke s Theorem A layered graph is a directed graph G with vertex set comprising a union of directed paths (v 1, v 2,..., v n ) where v i V i for i [n] and the sets V i, called layers, are pairwise disjoint. A layered graph G is a Plünnecke Graph if for each arc u v and Γ + (v) = {w 1, w 2,..., w k } and Γ (u) = {t 1, t 2,..., t l } there exist vertices u 1, u 2,..., u l and v 1, v 2,..., v k and arcs v i w i for i [k] and arcs t j u j for i [l]. Given two layered graphs G, H with layers V i, W i for i [n], the product graph G H has layers V i W i for i [n] and the arc (v, w) (v, w ) is present if and only if v v E(G) or w w E(G). The reader is invited to check that G H is a Plünnecke Graph if G and H are. For i [n] and Z V 0 let Γ i (Z) be the ith neighborhood of Z, namely the set of vertices in V i joined 1

2 from Z by a directed path. The ith magnification ratio D i (G) of a layered graph G with layers V i : i [n] is The first lemma is { Γi (Z) min Z } : Z V 1, Z Lemma 1 Let G, H be layered graphs. Then D i (G H) = D i (G)D i (H). Proof. It is straightforward to check D i (G H) D i (G)D i (H). For the reverse inequality, let P = V 1 W 1, Q = V 1 W i and R = V i W i. Join (v, w) P to (v, w ) Q in F if v = v and w Γ i (w) and (v, w ) Q to (v, w ) R if w = w and v Γ i (v ). Then there exists a path from (v, w) P to (v, w ) Q if and only if there exists a path from (v, w) to (v, w ) in G H. Hence in the layered graph F with layers P, Q and R we have that the magnification D P R from P to R is exactly D i (G H). We next show that the magnification D P Q is D i (H) and D QR is D i (G). If C W 1 with Γ i (C) / C = D i (H), let v V 1 and C = {v} C. This shows D P Q D i (H). Conversely, let C P = V 1 W 1. For each x V 1, let C x = {y : (x, y) C}. Then Γ(C x ) Q / C x D i (H) hence D P Q D i (H). Therefore D P Q = D i (H) and similarly D QR = D i (G) from Q to R. Since D P Q D QR D P R, we have D i (G)D i (H) D i (G H). Lemma 2 Let G be a Plünnecke Graph with layers V i : i [n] such that D n (G) 1. Then there are V 1 disjoint paths from V 1 to V n and D i (G) 1 for all i [n]. Proof. Add a vertex a joined to all of V 1 and a vertex b joined to all of V n. Let m be the maximum number of disjoint a-b paths. There exists a set S = {s 1, s 2,..., s m } of size m separating a from b, by Menger s Theorem. Set S i = S V i. Choose S such that M = s i S i is a minimum. We claim that S V 1 V n. Suppose this is false; there exists i : 1 i < n such that S V i = {s 1, s 2,..., s q } =. Let P 1, P 2,..., P m be disjoint paths from V 1 to V n. Each P i contains exactly one s i, by the minimality of m. Let s i and s + i denote the predecessor and successor of s i on the path containing s i, oriented from V 1 to V n, 1 i q. By the minimality of M, we cannot replace any elements of S with predecessors on the paths. So we find a path P from V 1 to V n that misses {s 1, s 2,..., s q, s q+1,..., s m }. 2

3 This path must intersect S, as S is a separating set. Let {r} = P V i 1. Then the next vertex of P must be s i for some i : 1 i q. We claim that every path from {s 1, s 2,..., s q, r} to s + 1, s + 2,..., s + q passes through the vertices s 1, s 2,..., s q. Suppose that this claim is false. If there exists a path Q from s i to s + j missing s 1, s 2,..., s q, then the path comprises the segment of P i to s i, the segment of Q to s + j and the segment of P j onwards, misses S. This contradicts the fact that S is a separating set. Therefore the graph induced by {s 1,..., s q, r}, {s 1,..., s q } and {s + 1,..., s + q } is a Plünnecke Graph. In this subgraph, let d + (x) and d (X) be the in- and out-degrees of x. Since s i is joined to s i, d + (s i ) d+ (s i ). Similarly, d (s i ) d (s + i ). Also q i=1 d+ (s i ) = q i=1 d (s + i ) by counting edges, and d+ (r) + d + (s i ) = d (s i ). Since d + (r) > 0, we have a contradiction. Therefore S V 1 V n and, by minimality, S = (V 1 S) (Γ n (V 1 \S) and S = V 1 S + Γ n (V 1 \S) V 1 S + V 1 \S = V 1. Theorem 2 (Plünnecke s Theorem). Let G be a Plünnecke Graph with layers V 0, V i : i [n]. Then D 1 (G) D 2 (G) 1/2 D n (G) 1/n. Proof. The theorem follows from the last lemma when D n := D n (G) = 1. Consider first the case D n < 1. It is enough to show D 1/i i Dn 1/n for all i [n]. Let m, r N be chosen with m i and ( ) m + n 1 Dn r 1. n We suppose m is a minimum subject to this constraint. Let A = {a 1, a 2,..., a m } Z all of whose increasing sums a j1 + a j2 + + a jm j 1 j 2 j m are distinct. Let H be the layered graph whose layers are V 0 = {0}, V k = ka for k [n] and such that x V k is joined to y V k+1 if x + a = y for some a A. Then by the choice of m and a preceding lemma, ( ) m + n 1 D n (G r H) = DnD r n (H) = Dn r 1. n By the last lemma, it follows that D i (G r H) 1. Using the bound ( ) m n m + n 1 m n n! n we have the inequality D i ( m+i 1 i 1 ) 1/r 1 ( n! m i/r 3 D r n ) i/rn.

4 Letting r, this converges to D i Dn i/n, as required. If D n > 1, then we reverse H to obtain a layered graph F whose layers in order are V 0 = na, V 1 = (n 1)A,..., V n = {0}. Applying similar arguments to those above on the layered graph G r F gives the bound for D n > 1. 3 Ruzsa s covering lemma In this section we see that if A + A is small, then so is ka la, where ka := {a 1 + a a k : a 1, a 2,..., a k A}. Note that repetitions are allowed in this definition. Lemma 3 Let k N and let A and B be non-empty subsets of Z such that A+kB C A. For h k, there is a non-empty set X A such that X + hb C h/k X. Proof. Let G be the natural Plünnecke Graph. If the result were false, then D h (G) > C h/k so D k (G) > C which implies that A + kb > C A, a contradiction. Lemma 4 If A is a non-empty subset of Z and A + A C A, then ka C k A for each k 3. Proof. Take k = 1 and B = A in the preceding lemma. This implies that there exists a non-empty X A such that X + ka C k A C k A, but X + ka ka, so the result is proved. Lemma 5 Let U, V, W Z. Then U V W U + V U + W. Proof. Define, for x V W, ϕ(u, x) = (u + v(x), u + w(x)) where v(x) V, w(x) W satisfy v(x) w(x) = x. Then ϕ is an injection U (V W ) (U + V ) (U + W ). Theorem 3 (Covering lemma) Let A, B Z such that A + B C A and let k and l be natural numbers with l k. Then kb lb C k+l A. Proof. Suppose l k 1. By the first lemma, there exists A A with A + kb C k A. Again there exists A A with A + lb C l A. Using the last lemma, A kb lb A + kb A + lb C k+l A A and the result follows on dividing by A. 4

5 4 Freiman homomorphisms Let A Z s or A Z and B Z t. Then ϕ : A B is called a (Freiman) k-homomorphism if whenever x 1 + x x k = y 1 + y y k, with x i, y i A, ϕ(x i ) = ϕ(y i ). In addition, ϕ is called a k-isomorphism if ϕ is invertible and ϕ and ϕ 1 are k-homomorphisms. Note that ϕ is a k-homomorphism if the map ψ : (x 1,..., x k ) ϕ(x i ) induced by ϕ is a well defined map ka kb, and a k-isomorphism if ψ is a bijection. Our interest will be in 2-isomorphisms, as these preserve arithmetic progressions a set 2-isomorphic to an arithmetic progression is clearly an arithmetic progression. We use the following notation: if ϕ : A B and X A, then ϕ X denotes the restriction of ϕ to X. Lemma 6 Let A Z and suppose ka ka < N where N is prime. Then there exists A A with X A /k that is k-isomorphic to a subset of Z N. Proof. We may suppose A N and select a prime p > k max A. Then the quotient map ϕ 1 : Z Z p is a homomorphism of all orders, and ϕ 1 A is a k-isomorphism. Now let q be a random element of [p 1] and define ϕ 2 : Z p Z p by ϕ 2 (x) = qx. Then ϕ 2 is an isomorphism of all orders, and hence a k-isomorphism. Let ϕ 3 (x) = x where ϕ 3 : Z p Z. Then for any j, ϕ 3 Ij is a k-isomorphism where I j = {x Z p : j 1 k p x < j k p 1}. For, if k i=1 x i = k i=1 y i (mod p) with x i, y i I j, then k i=1 x i = k i=1 y i in Z. By the pigeonhole principle, there exist A A with A A /k (depending on q) and ϕ 2 ϕ 1 [A ] I j for some j. Restricted to A, ϕ 3 ϕ 2 ϕ 1 is a k-homomorphism. Finally, let ϕ 4 be the quotient map (a k-homomorphism) Z Z N. Then with ϕ = ϕ 4 ϕ 3 ϕ 2 ϕ 1, ϕ(x) = qx (mod p) (mod N) and ϕ A is a k-homomorphism, as it is the composition of k-homomorphisms. We prove that for some q, ϕ A is a a k-isomorphism. Now i a i i a i with a i, a i A implies i a i i a i (mod p) so we have q( i a i i a i) (mod p) is a multiple of N for some a i, a i A if ϕ A fails to be a k-isomorphism. The probability of this event is at most ka ka /N < 1. So for some q, ϕ A is a k-isomorphism. 5 Proof of Freiman s Theorem We now present Ruzsa s proof of Freiman s Theorem. We start with a set A Z such that A + A C A. The first step is to show 2A 2A contains an arithmetic progression of 5

6 small dimension and size relative to A. By the covering lemma, 8A 8A C 16 A. By the last lemma, A contains a subset A of cardinality at least A /8 which is 8-isomorphic to a a set B Z N with C 16 A < N 2C 16 A, where N is prime and C A < N 2C A, using Bertrand s Postulate. So B = αn with α (16C 16 ) 1. By Bogolyubov s Theorem, 2B 2B contains an arithmetic progression of dimension at most α 2 at least (α 2 /4) α 2 N = β A where β is a constant depending only on C. 8-isomorphic to A, 2B 2B is 2-isomorphic to 2A 2A. and cardinality Since B is Any set 2-isomorphic to a d-dimensional arithmetic progression is a d-dimensional arithmetic progression. Therefore 2A 2A, and hence 2A 2A, contains an arithmetic progression Q of dimension at most α 2 and cardinality β A. Now let X = {x 1, x 2,..., x k } A be maximal such that x, y X, x y imply x y Q Q. Equivalently, all the sets x + Q are disjoint, so X + Q = X Q. Since X is maximal, A X +(Q Q) and X is contained in the k-dimensional arithmetic progression { k } R = a i x i : 0 a i 1. i=1 Therefore A is contained in the arithmetic progression R + (Q Q), of dimension at most α 2 + X. We know that X + Q A + 2A 2A C 5 A, by the covering lemma and R + (Q Q) kx + Q Q ka + 2A 2A C k+4 A. Since X + Q = X Q, this shows X C 5 A / Q. Since Q β A, we have k = X C 5 /β and therefore R + (Q Q) has dimension at most α 2 + β 1 C 5 and cardinality at most C C5 /β+4 A. This completes the proof of Freiman s Theorem. 6