ON FREIMAN S 2.4-THEOREM

ON FREIMAN S 2.4-THEOREM ØYSTEIN J. RØDSETH Abstract. Gregory Freiman s celebrated 2.4-Theorem says that if A is a set of residue classes modulo a rime satisfying 2A 2.4 A 3 and A < /35, then A is contained in an arithmetic rogression of length 2A A +1. Without much extra effort, the bound A < /35 can be relaxed to A /11.3. A result of Freiman on the distribution of oints on the circle lays an imortant rôle in the roof. This result was recently refined by Lev. By using Lev s result instead of Freiman s, we increase the bound on the size of A to A /10.7. 1. Introduction Let be a rime, and let A be a non-emty subset of one of the rings Z or Z/Z. We use A to denote the cardinality of A, and we write 2A for the set of elements a + a where a, a A. As usual, for ring elements a, d, the set {a + id i = 0, 1,..., 1} is an arithmetic rogression with difference d. The number of distinct elements is the length of the rogression. If A is contained in an arithmetic rogression, we say that A is covered by the rogression. If A can be covered by a rogression of length 2A A + 1, then A has the covering roerty. The following beautiful result dates bac to 1961. Freiman s 2.4-Theorem. Let A Z/Z. If 2A 2.4 A 3 and if A < /35, then A has the covering roerty. In this note we show that Freiman s Theorem still holds if we relace the condition A < /35 by the weaer A /10.7. We shall follow Freiman s original roof as resented in [4] and also in [9]). At some oints our estimates will be slightly more recise. In articular, we shall aly a recent result of Lev [8] on the distribution of oints on the circle, which is a refinement of the corresonding result of Freiman. We also aly the theorems of Cauchy-Davenort and of Voser. Theorem 1. If 2A 2.4 A 3 and if A /10.7, then A has the covering roerty. In addition, we rove the following result. Theorem 2. Let 2 < σ 2.392. Suose that 2A σ A 3 and A /4σ). Then A has the covering roerty. 2000 Mathematics Subject Classification. 11A07, 11P70. Key words and hrases. Sumsets mod, Freiman isomorhisms, rectification. 1

2 Ø. J. RØDSETH These two theorems are roved in Section 3. But first we consider briefly a secial tye of isomorhism. 2. Rectification The roofs are by rectification, which means that we transform roblems in Z/Z to questions in Z which are hoefully) easier to answer. Additive roblems in Z/Z are often more difficult than the corresonding roblems in Z. For examle, a wellnown conjecture states that if is sufficiently large and if A Z/Z satisfies 2A min{3 A 4, 1}, then A has the covering roerty without further restrictions on the size of A). The corresonding result for integers was roven true by Freiman [2] in 1959: Theorem 3. Let K be a finite set of integers satisfying 2K 3 K 4. Then K has the covering roerty. Suose that we have a subset K of Z such that the canonical homomorhism Z Z/Z induces a bijective) ma φ : K A. If, for any r 1, r 2, r 3, r 4 K, r 1 + r 2 = r 3 + r 4 if and only if φr 1 ) + φr 2 ) = φr 3 ) + φr 4 ), then the ma φ is called a Freiman isomorhism of order 2). Such a ma reserves additive roerties involving two summands, and various results for 2A can be roven by roving the corresonding results for 2K. More generally, if φ has the roerty that for any r 1,..., r 2s K, if and only if r 1 + + r s = r s+1 + + r 2s φr 1 ) + + φr s ) = φr s+1 ) + + φr 2s ), then φ is a Freiman isomorhism of order s, or F s -isomorhism for short. An F s - isomorhism is also an F s 1 -isomorhism. The basic result on rectification and F s -isomorhisms is given by Freiman [4, Section 3.12], with imortant contributions in [1] and [5]. Let A Z/Z with A =. For b, c Z/Z, b 0, let ψ : Z/Z Z/Z be given by ψx) = bx + c for x Z/Z. Then ψ is an affine transformation, and A = ψa) is an affine image of A. We also write ψ for the induced ma A A. For any ositive integer s, if x 1,..., x 2s Z/Z, then if and only if x 1 + + x s = x s+1 + + x 2s ψx 1 ) + + ψx s ) = ψx s+1 ) + + ψx 2s ). Thus, if for some set K of integers, φ : K A is an F s -isomorhism, then the comosite ma ψφ : K A is also an F s -isomorhism.

FREIMAN S 2.4-THEOREM 3 The quantities A and 2A are affine invariants, that is, A = A and 2A = 2A for any affine image A of A. Let K be a set of integers such that the canonical homomorhism Z Z/Z induces a bijection K A. Then K is a set of integer reresentatives for A. Suose that A has the covering roerty. Then any affine image of A also has the roerty. In articular, there is an affine image A with a set K of integer reresentatives, such that K [0, 2A A ]. Let s 2 be an integer. For 2A A < /s, there is an F s -isomorhism φ : K A. We also have an affine transformation ψ : A A. Then the comosite ma ψφ : K A is an F s -isomorhism. Proosition 1. Let s 2 be an integer, and suose that A has the covering roerty. If s < / 2A A ), there exists an F s -isomorhism K A. If we have 2A σ A 3 and A c 1, we can write the result as follows. Proosition 2. Let s 2 be an integer, and suose that A has the covering roerty. If 1 s, σ 1)c 1 then there exists an F s -isomorhism K A. It is well nown that imlicit in Freiman s roof of the 2.4-Theorem there is a roof of the following result: If 2 < σ 2.4 and if A < /35, then there exists an F 2 -isomorhism K A for some set K Z; cf. [1,. 346]. However, under these hyotheses, we have, by Proosition 2, an F 25 -isomorhism from some set K Z onto A. If σ 2.4 and A < /10.7, there is an F 7 -isomorhism K A. An F 2 -isomorhism is actually all we need. So this indicates that there is still much room for imrovement in the 2.4-Theorem, that is, the bound on the size of A is still too small. Perhas A < /4 suffices? 3. Proofs Let K be a set of integer reresentatives for some affine image A of A. Suose that K consists of the integers r i, i = 0, 1,..., 1, where 0 = r 0 < r 1 < < r 1 < r :=. Let h be defined by r h 1 < /2 < r h. Choose the affine image A of A such that h is maximal. Let l = 2A, and assume first that h > l + 1. 1) 3 Then h 2. We shall show that if the hyotheses of Theorem 1 or Theorem 2 are satisfied, then 1) imlies that h =. Let d = gcdr 1,..., r h 1 ). Then d 0 mod ), and the set {ad 1 a A }, which is an affine image of A, has a set of integer reresentatives containing 0 = r 0 /d < r 1 /d < < r h 1 /d. We can therefore assume that A is chosen such that d = 1.

4 Ø. J. RØDSETH Set H = {r 0, r 1,..., r h 1 }. The canonical homomorhism Z /Z induces an injection 2H 2A. Hence, using 1), we have 2H 2A = l < 3h 3. Thus, by Theorem 3, H is covered by an arithmetic rogression of length at most 2H H + 1. Since gcdr 1,..., r h 1 ) = 1, this rogression has difference 1. Hence, so that H [0, 2H H ] [0, l h], 2H [0, 2l 2h]. Suose that there is a t K in the interval 2l 2h + 1 t 1 l + h. 2) This interval is non-emty if the hyotheses of Theorem 1 or of Theorem 2 are satisfied.) Then H + {t} [2l 2h + 1, 1]. Now we have l = 2A 2H) H + {t}) = 2H + h 3h 1, which contradicts 1). Thus there is no t K in the interval 2). Since K [0, ), we have K [0, 2l 2h] [ l + h, 1]. As a set of integer reresentatives for A, we may relace K by a set K [ l + h, 1] [0, 2l 2h]. A translation of A gives us an A with a set of integer reresentatives K, where so, by 1), K [0, 3l 3h], K [0, 2l 4]. The hyotheses of Theorem 1, and also those of Theorem 2, imly l < /4 + 2. Therefore K [0, /2). Hence h = and H = K. Now we have a Freiman isomorhism K A of order 2. Thus, by 1), 2K = l < 3 K 3. By Theorem 3, K has the covering roerty, and hence, so has A. Since K = H, we actually have a stronger result.) Next, we assume that 1) is false, that is, we assume that h l + 1. 3) 3 We now rove Theorem 1 by showing that the hyotheses of the theorem are violated. We also rove Theorem 2 by deducing a contradiction.

FREIMAN S 2.4-THEOREM 5 We define the discrete Fourier transform of the indicator function of A at x Z by Âx) = a A ex2πiax/), and similarly for 2Ax). We ut θ = 1 max x 0 Âx), where the maximum is taen over a comlete set of non-zero residues mod ). Then θ is an affine invariant for A. We follow [4,. 48 49] and [9,. 69], using [9, Lemma 2.15], the triangle inequality, etc., and obtain 1 1 2 = Âx) 2 2A x) 2 l + θ Âx) 2Ax). x=0 x=1 The Cauchy-Schwarz inequality and Parseval s formula give 2 2 l + θ 1 1 Âx) 2 2Ax) 2 x=1 x=1 = 2 l + θ 2 l l 2. This yields l θ 1 1. 4) We shall use the following case of Corollary 2 of Lev [8]. Lemma 1. Let A Z/Z, A =. Then there exists an integer u such that the image of the interval [u, u + /2) under the canonical homomorhism Z Z/Z contains at least n elements of A, where ex2πia/) a A sin 2n )π/). sinπ/) There is an integer z 0 mod ) such that Âz) = θ. The affine image A = {az a A} of A satisfies θ = ex2πia/). a A Hence, by the definition of h and Lemma 1, θ sin 2h ) π/). 5) sinπ/)

6 Ø. J. RØDSETH Notice that the right hand side of 5) does not exceed 2h, so that θ 2h, which is a well-nown result of Freiman [3]; see [4, Lemma 2.2] or [9, Theorem 2.9]. By 4), 5), and 3), we have 1 l 1 sin π ) ) 2l π sin 3 + 2. 6) We now rove Theorem 1. Assume the hyotheses of the theorem.then 3. By the Cauchy-Davenort Theorem see [10, Theorem 1]), we have l 2 1 if l, and we obtain 5. By Voser s Theorem see [10, Theorem 4]), Theorem 1 holds for = 5, 6, and 7. We may therefore assume that 8. Then 89. If we also use Theorem 1 in [6], we obtain 107. If we, in addition, use Theorem 3 in [7], we get 139.) By the inequality l 2.4 3 and 4), we have 5 12 1 1 sin π < sin ) 3π. 7) 5 In articular, 5 1 12 1 < 3 5, which gives / < 11.3. Using the first few terms in the MacLaurin series for the sine function, we obtain by 7), 5 π 1 6 ) ) 3 π < 3π 5 1 6 1 12 1 We cancel π/ and use 89 to get f/) < g/) for fx) = gx) = 3 5 5 x 1 12 x 1 1 1 6 ) 3 3π + 1 5 120 1 1 ) π 2, 6 89) ) 2 3π + 1 5x 120 ) ) 4 3π. 5x ) 5 3π. 5 Both fx) and gx) are increasing for x > 2.4. We find that f10.8) < g11.3) is false; hence / < 10.8. One more ste gives / < 10.7. In fact, / < x 0 = 10.6837..., where fx 0 ) = gx 0 ). If we use the condition 139, we get / < 10.668.) Our result contradicts the hyotheses of Theorem 1. This comletes the roof of Theorem 1. Now to Theorem 2. Assume the hyotheses. Using the theorems of Cauchy-Davenort and of Voser we obtain 10, so that 83.

FREIMAN S 2.4-THEOREM 7 By 6) and the inequalities l σ 3, /4σ), we have 1 σ 1 sin π ) ) 2σ π < sin 3 1. 8) In articular, this imlies and for σ 2.392, we get We also have α := α σ 1 1 < 2σ 3 1, ) 2σ π 3 1 > 0.1862. ) 2σ π 3 1 4σ < 0.1953. Using the inequality / 4σ and the first few terms in the MacLaurin series for the sine function, we obtain by 8), ) 1/2 3 1 1 4σ 1 6 ) ) 2 ) π 2σ < 3 1 1 1 6 α2 + 1 ) 120 α4. Since σ 2.392, 83, 0.1862 < α < 0.1953, we arrive at the false result 0.5915 < 0.5913. This comletes the roof of Theorem 2. References [1] Y. F. Bilu, V. F. Lev, and I. Z. Ruzsa, Rectification rinciles in additive number theory, Discrete Comut. Geom., 19 1998), 343 353. [2] G. A. Freiman, The addition of finite sets. I. Russian), Izv. Vysš. Učebn. Zaved. Matematia, 13 6) 1959), 202 213. [3] G. A. Freiman, Inverse roblems of additive number theory, VII. On addition of finite sets, IV. Russian), Izv. Vysš. Učebn. Zaved. Matematia, 6 31) 1962), 131 144. [4] G. A. Freiman, Foundations of a Structural Theory of Set Addition, Translations of Mathematical Monograhs, Vol. 37, American Math. Soc., Providence, R. I. 1973). [5] B. J. Green and I. Z. Ruzsa, Sets with small sumset and rectification, Bull. London Math. Soc. 38 2006), 43 52. [6] Y. O. Hamidoune and Ø. J. Rødseth, An inverse theorem mod, Acta Arith., 92 3) 2000), 251 262. [7] Y. O. Hamidoune, O. Serra, and G. Zémor, On the critical air theory in Z/Z, Acta Arith. 121 2) 2006), 99 115. [8] V. F. Lev, Distribution of oints on arcs, Integers 5 2) 2005), #A11, 6. electronic). [9] M. B. Nathanson, Additive Number Theory: Inverse Problems and the Geometry of Sumsets, Graduate Texts in Mathematics, Vol. 165, Sringer, New Yor 1996). [10] Ø. J. Rødseth, Sumsets mod, Trans. R. Norw. Soc. Sci. Lett. to aear).

8 Ø. J. RØDSETH Deartment of Mathematics, University of Bergen, Johs. Brunsgt. 12, N-5008 Bergen, Norway E-mail address: Oystein.Rodseth@mi.uib.no URL: htt://www.uib.no/peole/nmaoy