The Cardinality of Sumsets: Different Summands

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1 Te Cardinality of Sumsets: Different Summands Brendan Murpy, Eyvindur Ari Palsson and Giorgis Petridis arxiv: v3 [mat.co] 17 Feb 2017 Abstract We offer a compete answer to te following question on te growt of sumsets in commutative groups. Let be a positive integer and A,B 1,B 2,...,B be finite sets in a commutative group. We bound A + B B from above in terms of A, A+B 1,..., A+B and. Extremal examples, wic demonstrate tat te bound is asymptotically sarp in all parameters, are furtermore provided. 1 Introduction Given non-empty) finite sets A,B 1,B 2,...,B in a commutative group, teir sumset also referred to as teir Minkowski sum) is A+B 1 + +B = {a+b 1 + +b : a A,b i B i for 1 i }. We obtain an upper bound on te cardinality of A + B B in terms of and te cardinalities of A and A+B 1,...,A+B. Note tat te question becomes trivial unless some constraints are put on te sets as A + B B A B 1... B ; and te bound is attained wen A,B 1,...,B are sets of distinct generators of a free commutative group. Te best known upper bound is as follows. Teorem 1.1. Let and m be positive integers and α 1,...,α be positive real numbers. Suppose tat A,B 1,...,B are finite sets in a commutative group tat satisfy A = m and A+B i α i m for all 1 i. Ten A+B 1 + +B α 1...α m 2 1/. Teorem 1.1 can be proved by different metods. It can be deduced from te work of Ruzsa in [10, 11]. It also follows by combining an inequality of Balister and Bollobás in [1] wit 1

2 Te Cardinality of Sumsets: Different Summands an inequality of Ruzsa [9]. Madiman, Marcus and Tetali ave given a different proof of te inequality of Balister and Bollobás in [5]. We discuss te various proofs in more detail in Section 2. It is wort pointing out ere tat te metods used by te tree groups of autors are different: Ruzsa relied on grap teory; Bollobás and Balister on projections; and Madiman, Marcus and Tetali on entropy. Te upper bound in Teorem 1.1 as te correct dependence on α and m. Te following example a modification of similar examples given by Ruzsa in [10, 11]) demonstrates tis. Example 1.2. Let be a positive integer. Tere exist infinitely many α 1,...,α ) Q + ) wit te following property. For eac suc -tuple α 1,...,α ) tere exist infinitely many m and sets A,B 1,...,B in a commutative group wit A = m, A+B i 1+o1))α i m and A+B 1 + +B 1+o1)) Te o1) term is o m 1). ) α 1...α m 2 1/. Wesow tattesets inexample 1.2areextremal to tisproblemby proving amatcing upper bound and so settle te question of bounding from above te cardinality of iger sumsets in commutative groups. Teorem 1.3. Let be a positive integer, α 1,...,α be positive real numbers and m an arbitrarily large integer. Suppose tat A,B 1,...,B are finite sets in a commutative group tat satisfy A = m and A+B i α i m for all 1 i. Ten ) A+B 1 + +B α 1...α m 2 1/ +O m 2 2/)) ) = 1+o1)) α 1...α m 2 1/. Te o1) term is o m 1). Note. For large te main term is rougly e 1 α 1...α m 2 1/. Te proof is a refinement of Ruzsa s grap teoretic approac. Te upper bound in Teorem 1.3 is submultiplicative wit respect to direct products. In oter words if one replaces A by, say, its Cartesian product A A = {a,a ) : a,a A} and te B i by teir Cartesian products B i B i, ten after some standard calculations of te form A A)+B B) = A+B) A+B) = A+B 2 ) one obtains α 1...α A 2 1/, 2

3 B. Murpy, E.A. Palsson and G. Petridis wic is weaker tan wat te teorem gives. Tis particular feature of te upper bound makes using one of te key ingredients in Ruzsa s metod, te product trick, more delicate. From a tecnical point of view tis is te greatest difficulty tat must be overcome. Te special case wen B 1 = B 2 = = B and α 1 = = α = α was considered in [7]. Te sumset A+B 1 + +B in tis case is abbreviated to A+B. Te upper bound obtained tere is sligtly stronger: A+B 1+o1)) C 2α m 2 1/ 1.1) for an absolute constant C > 0. Te extra factor of in te denominator can be accounted for by te fact tat wile S 1 + +S S 1... S olds for general sets S i, wen te same set S is added to itself one as te stronger inequality S + + S ) S + 1. Inequality 1.1) probably does not ave te correct dependence in as te largest value of A+B in examples is of te order 1 α m 2 1/. It would be of interest to bridge tat gap. Te proof of Teorem 1.3 is similar to tat of inequality 1.1). Tere are noneteless tecnical differences. Rougly speaking we combine ideas from te proof of inequality 1.1) wit a strategy used repeatedly in te literature for example in [3, 12]) to prove a generalisation of te afore mentioned result of Ruzsa. We could not find a result general enoug for our purposes in te literature and so give a detailed proof in Section 5. Te paper is organised as follows. In Section 2 we discuss te different proofs of Teorem 1.1. Te proof of Teorem 1.3 is done in Section 3. Example 1.2 is described in Section 4. In Section 5 te grap teoretic framework of te proof is developed. 2 Proof of Teorem 1.1 Teorem 1.1 follows by combining an inequality of Balister and Bollobás wit an inequality of Ruzsa. Teorem 2.1 Balister-Bollobás, [1]). Let and m be positive integers and α 1,...,α be positive real numbers. Suppose tat A,B 1,...,B are finite sets in a commutative group tat satisfy A = m and A+B i α i m for all 1 i. Ten for any subset C B 1 + +B A+C α 1...α ) 1/ m C 1 1/. Te proof given by Balister and Bollobás is sort and elegant. It combines an idea of Gyarmati, Matolcsi and Ruzsa in [4] wit te Box Teorem in [2]. Madiman, Marcus and Tetali gave a 3

4 Te Cardinality of Sumsets: Different Summands somewat different proof based on entropy [5]. Te teorem can also be proved by metods developed by Ruzsa for example in [10, 11]). To deduce Teorem 1.1 one naturally sets C = B 1 + +B. Tis gives A+B 1 + +B α 1...α ) 1/ m B 1 + +B 1 1/. 2.1) We are left wit bounding B B in terms of m and te α i. Ruzsa acieved tis by modifying a grap teoretic metod of Plünnecke in [8], a variant of wic we describe in Section 5. Teorem 2.2 Ruzsa, [9]). Let and m be positive integers and α 1,...,α be positive real numbers. Suppose tat A,B 1,...,B are finite sets in a commutative group tat satisfy A = m and A+B i α i m for all 1 i. Ten tere exists a non-empty subset X A suc tat X +B 1 + +B α 1...α X. In particular B 1 + +B X +B 1 + +B α 1...α X α 1...α m. Substituting te last inequality in inequality 2.1) gives te bound in Teorem 1.1. Teorems 1.1 and 2.2 differ crucially in te exponent of m. Ruzsa as sown in [12] tat if one is interested in bounding X +B 1 + +B for a suitably cosen large subset of A, ten te correct exponent of m is 1. Specifically e proved tat for any ε > 0 tere exists a non-empty subset X A suc tat X > 1 ε) A and ) ε 1 1 X +B 1 + +B α 1...α X 2ε 1 α 1...α X. 1 Te exponent of X in te upper bound remains 1 even wen X is required to ave very large density in A. Te nature of te upper bound canges wen te cardinality of te wole of A+B 1 + +B is bounded. 4

5 B. Murpy, E.A. Palsson and G. Petridis 3 Proof of Teorem 1.3 Te upper bound in Teorem 1.3 is an increasing function of te α i and te ratios A+B i / A are rational numbers so we may assume tat α i Q +. Te next step is to reduce to te special case were all te α i are equal. We prove te following. Proposition 3.1. Let be a positive integer, α be a positive rational number and m an arbitrarily large integer. Suppose tat A,B 1,...,B are finite sets in a commutative group tat satisfy A = m and A+B i αm for all 1 i. Ten A+B 1 + +B α m+ 1 1 = ) ) 1 α m 2 1/ +O m 2 2/)) α m 2 1/ +O m 2 2/)). Teorem 1.3 is deduced from te above proposition in a standard way by working in direct products of groups see for example [11, 12]). Deduction of Teorem 1.3 from Proposition 3.1. Let α i = p i /q i and set n = q 1...q. Furtermore, let T 1,...,T be pairwise disjoint sets of generators of a free abelian group F wit cardinality n i := T i = n j i α j; and let 0 denote te identity of F. Eac n i is cosen so tat ) α i n i is equal to n j α j. We apply Proposition 3.1 to te sets A = A {0}, B 1 = B 1 T 1,...,B = B T. As A +B i = T i A+B i mn α j = n for all i = 1,..., te proposition yields ) α j A ) ) m A +B 1 + +B n α 2 1/ j +Om 2 2/ ) ) ) ) m = α i n 2 1/ i +Om 2 2/ ) ). Teorem 1.3 follows by observing tat A +B 1 + +B = A+B 1 + +B {0}+T 1 + +T = A+B 1 + +B n i, 5

6 Te Cardinality of Sumsets: Different Summands and dividing by n 1 n. We next prove Proposition 3.1. Te roug strategy is as follows. We initially apply Teorem 2.2 to find a non-empty subset X 1 A wose growt under addition wit te B i can be bounded. We are left wit bounding A+B 1 + +B )\X 1 +B 1 + +B ). We would like to iterate tis process, wic requires a stronger statement tan Teorem 2.2. From a tecnical point of view, tis is te eart of te argument. It requires a detour in grapteoretic tecniques developed by Plünnecke and Ruzsa and so is left for Section 5. Te key result we employ in te proof of Proposition 3.1 is as follows. It will be proved in a sligtly stronger form as Corollary 5.18 in p.23. Lemma 3.2 Bound for sumsets wit a component removed). Let be a positive integer. Suppose tat A,B 1,...,B are finite sets in a commutative group and E A a subset of A. If X A\E is a subset of A\E tat minimises te quantity µz) := 1 Z +B i )\E +B i ) Z over all non-empty subsets Z A\E, ten X +B 1 + +B )\E +B 1 + +B ) µ X, were µ = µx) = 1 X +B i )\E +B i ). X Note tat setting E = gives Teorem 2.2 for te special case wen α 1 = = α = α, as 1 µ = min Z A Z +B i Z 1 A+B i A α. Te ultimate task for tis section is to deduce Proposition 3.1 from te above estimate. Proof of Proposition 3.1. Applying te bound stated above successively we partition A into X 1 X k for some finite k A is finite), wose exact value is irrelevant to te argument. 6

7 B. Murpy, E.A. Palsson and G. Petridis More precisely in te jt step set we set E = j 1 l=1 X l E = for j = 1) and cose X j to be te minimal non-empty subset of A\E tat minimises te quantity µ j := 1 Z +B i )\E +B i ), Z wic we set to be µ j. Te inequality we get is X j +B 1 + +B )\ j 1 l=1 X l +B 1 + +B ) µ j X j. 3.1) It is crucial to observe tat te defining properties and especially te minimality) of te X j imply tat te µ j form an increasing sequence. Indeed µ j µ j+1 as µ j X j +µ j X j+1 = µ j X j X j+1 1 j 1 ) X j X j+1 )+B i )\ X l +B i l=1 = 1 j 1 ) X j +B i )\ X l +B i l=1 + 1 j ) X j+1 +B i )\ X l +B i = µ j X j +µ j+1 X j+1. l=1 Wen te µ j are large it turns out tat replacing te estimate in 3.1) by a more elementary one is more economical. We ave j 1 ) X j +B 1 + +B )\ X l +B 1 + +B X j +B 1 + +B l=1 X j B 1 + +B. To bound B 1 + +B we adapt accordingly te argument in Teorem 2.2. B 1 + +B X 1 +B 1 + +B µ 1 X µ 1m. 7

8 Te Cardinality of Sumsets: Different Summands Combining 3.1) wit te last two inequalities gives X j +B 1 + +B )\ Summing over j = 1,...,k gives A+B 1 + +B = j 1 l=1 X l +B 1 + +B ) min { µ j,µ 1 m} X j. k X j +B 1 + +B )\ k min { µ j,µ 1m } X j. j 1 l=1 X l +B 1 + +B ) We are left wit bounding te sum k min{ µ j,µ 1 m} X j subject to two constraints: k X j = m and k µ j X j = = 1 = 1 k 1 k A+B i αm. X j +B i )\ X j +B i )\ j 1 l=1 j 1 l=1 X l +B i ) ) X l +B i ) ) Te two quantities inside te min are equal if µ j = µ := µ 1 m 1/. As µ j µ 1 for all 1 j, we can replace te min by te straigt line µ 1 +µ j µ 1 ) µ µ 1 µ µ 1, wic, tougt of as function of µ j, intersects te curve µ j at µ j = µ 1 and µ j = µ. Te slope 8

9 B. Murpy, E.A. Palsson and G. Petridis is bounded by µ 1 1 m 1 m 1/ 1 µ 1 1 m 1 1/ +2m 1 2/). Terefore A+B 1 + +B k = µ 1 µ 1 +µ 1 1 µ j µ 1 ) m 1 1/ +2m 1 2/)) X j k X j +µ 1 1 m 1 1/ +2m 1 2/) k µ j X j µ 1 m 1 1/ +2m 1 2/) k X j µ 1m+α µ 1 )µ 1 1 α m+α µ 1 )µ 1 1 m 2 1/ +2m 2 2/) m 2 1/ +2m 2 2/). Tefinaltaskistoselecttevalueof1 µ 1 αtatmaximises tisexpression. Differentiating te expression α µ 1 )µ 1 1 wit respect to µ 1 gives tat it is maximised wen 1)α µ 1 ) = µ 1 = µ 1 = 1 1 ) α or α µ 1 = α. Substituting above gives A+B 1 + +B α m+ = α m+ ) α m 2 1/ +2m 2 2/) ) α m 2 1/ +O m 2 2/)). Tis completes te proof of Proposition 3.1 modulo te proof of Lemma 3.2 on p. 6, wic as we ave seen implies Teorem 1.3. Te proof of te estimate is given in Section 5. We next provide examples wic sow tat te upper bound given by Teorem 1.3 is asymptotically sarp. 9

10 Te Cardinality of Sumsets: Different Summands 4 Examples We construct te sets in Example 1.2. To keep te notation simple we assume tat te α i are all equal: α 1 = = α = α. Once tese examples ave been constructed, it is straigtforward to construct examples for different α 1,...,α )byconsidering direct products. Verymuclikeintefirststepofteproof ofteorem1.3insection3onetenconsiderssetsa = A {0},B 1 = B 1 T 1,...,B = B T to get a different -tuple α 1,...,α ), were α i = α T i. Te T i are sets of distinct generators of a free commutative group. Te details are as follows, A +B i A = A+B i) T i A = A+B i T i 1+o1))α T i A and A +B 1 + +B = A+B 1 + +B ) T 1 + +T ) = A+B 1 + +B T 1 + +T 1+o1))α A 2 1/ T 1 T = 1+o1))α 1 α A 2 1/. To construct te sets for te special case wen α i = α for all i, we fix and let a and l be integers, wic we consider as variables wit a assumed to be large and divisible by 1. We set b = la and work in Z k b, were k = +a 1 / 1). We write x i for te it coordinate of te vector x. We consider A = A 1 A 2 were A 1 = {x : x i {0,l,2l,...,a 1)l} for 1 i and x i = 0 oterwise} and A 2 is a collection of a 1 / 1) independent points k A 2 = {x : x j = 1, x j = 0 oterwise}. j=+1 B i is taken to be a copy of Z b B i = {x : x i {0,...,b 1}, x j = 0 for all j i}. 10

11 B. Murpy, E.A. Palsson and G. Petridis We now estimate te cardinality of te sets tat interest us. A = A 1 + A 2 = a + a 1 1 = 1+o1))a. As is fixed different values of a result to different values of m. To bound A+B i we note tat A 1 +B i equals {x : x i {0,...,b 1},x j {0,l,2l,...,a 1)l},j i} = ba 1 and tat A 2 +B i = A 2 B i = ba 1 1. Tus A+B i A 1 +B i + A 2 +B i Terefore α = ba 1 + ba 1 1 = 1+ 1 ) la 1 = 1+o1)) 1+ 1 ) l ) lm. is fixed and so different values of l result to different values of α. To bound A+B 1 + +B from below observe tat B 1 + +B = Z b = b and tat for a,a A 2 te intersection a+b 1 + +B ) a +B 1 + +B ) is emty. Tus A+B 1 + +B A 2 +B 1 + +B = A 2 B 1 + +B = a 1 1 b = l 1 a2 1 = 1+o1)) = 1+o1)) ) α m 2 1/ ) α m 2 1/. Wearedone. Asisexpectedtestructureoftesetspresented ereissuctateveryinequality in Section 3 is more or less attained. 11

12 Te Cardinality of Sumsets: Different Summands 5 Grap Teory In tis section we develop te grap teoretic framework necessary for our proof of te estimate on p.6; te last step of te proof of Teorem 1.3. Te results and metods of tis section are influenced by te work of Ruzsa, c.f. [11, 12]. We define a type of layered commutative grap, called a commutative ypercube grap, tat generalises te addition grap associated to sumsets of te form A + B B, defined in te first example below. Te class of commutative ypercube graps includes graps tat result from removing a component from an addition grap. Te main result of tis section is an analog of Teorem 2.2 for commutative ypercube graps. Trougout tis section stands for disjoint union. 5.1 Hypercube graps and teir products Let Q denote te set of all subsets of {1,...,} and for I in Q, let I denote te cardinality of I. Given I and I in Q, we will write I I if I = I {i} for some i I. Definition 5.1 Hypercube Grap). Let G be a directed grap wit vertex set V and edge set E. We say tat G is a ypercube grap indexed by Q if it satisfies two conditions: i) For eac I in Q tere exists a set U I V suc tat V is te disjoint union of te U I s: V = I Q U I. ii) Tere is an edge u v in E only if u U I and v U I were I I. For sort, we may say G is a Q -ypercube grap. Note tat a Q -ypercube grap is a layered grap wit +1 layers: V = V 0 V, were V i = I =i U I. We give some examples of ypercube graps. Te most important example of a ypercube grap is an addition grap wit different summands, wic are featured in [12]. Example 5.2 Addition graps). Let A,B 1,...,B be finite subsets of a commutative group G. Teir addition grap G + A,B 1,...,B ) is defined as follows: for eac I in Q, let U I = A+ i I B i. We consider eac U I to be contained in a separate copy of G, and we let V = I Q U I. For eac vertex x in U I tere is an edge to y in U I if I = I {i} and y = x+b for some b in B i. Tus G + A,B 1,...,B ) is a ypercube grap indexed by Q. Note tat any subgrap of a Q -ypercube grap is automatically a Q -ypercube grap. For certain induced subgraps of a ypercube grap, we can say someting more. We recall from [11] a definition. 12

13 B. Murpy, E.A. Palsson and G. Petridis Definition 5.3 Cannels of directed graps). Given a directed grap G = GV,E) and two sets of vertices X,Y V, te cannel GX,Y) between X and Y is te subgrap of G induced by te set of vertices tat lie on a pat from X to Y including endpoints). Example 5.4 Cannels are ypercube graps). Let G be a ypercube grap indexed by Q and let I and I be elements of Q suc tat I I. Given subsets X U I and Y U I, te cannel GX,Y) is a ypercube grap indexed by Q j, were j = I \I. Proof. Since te edges of G are edges of G, condition 5.1 of Definition 5.1 is automatically satisfied. Tus it remains to be sown tat condition 5.1 is satisfied. Note tat te set of J in Q suc tat I J I is in one-to-one correspondence wit Q j. Fixing one suc correspondence, let J denote te element in Q j corresponding to J and set U JG) = VG) U J. Since any vertex in VG) must be an element of some U J wit I J I, we ave VG) = J Qj U JG), as desired. To prove te analog of Teorem 2.2, we must define a type of grapproduct between ypercube graps tat is motivated by addition graps of product sets. Definition 5.5 Hypercube Product). Let G and G be ypercube graps indexed by Q. We define a ypercube grap G = G G also indexed by Q as follows: for eac I Q, we define U I G) = U I G ) U I G ), and for u,v) U I G),u,v ) U I G), we ave u,v) u,v ) if and only if u u and v v. We call G te ypercube product of G and G. ItiseasytoseetatG + A A,B 1 B 1,...,B B )isteypercubeproductofg +A,B 1,...,B ) and G + A,B 1,...,B ). In tis sense, direct products in te group setting correspond to ypercube products in te grap setting and so ypercube products are natural objects. 5.2 Square commutativity Te key feature of addition graps tat makes tem useful in additive number teory is tat tey capture in a grap teoretic way te commutativity of addition. Tis particular feature was first exploited by Plünnecke in [8], wo worked wit a class of directed layered graps e called commutative. Te importance of commutative graps to additive number teory is detailed in [6, 14, 11]. We will only mention tem briefly as we need a stronger form of commutativity in order to prove Teorem 2.2, one tat works better for ypercube graps. Firstwemakeanauxiliarydefinition: givenindexsetsi,i,andi inq suctati I I, tereisauniqueindexseti c inq suctati c I andi I c I ; explicitlyi c = I I \I ). We will call I c te associate of I. 13

14 Te Cardinality of Sumsets: Different Summands Definition 5.6 Square Commutativity). Let G be a ypercube grap indexed by Q. We say tat G is square commutative if it satisfies two conditions: 1. Upward square commutativity: Given indices I,I and I in Q suc tat I I I, and vertices v U I, v U I, and v 1,...,v n U I suc tat v v v i for i = 1,...,n, tere exist distinct vertices v 1,...,v n U I c suc tat v v i v i for i = 1,...,n. 2. Downward square commutativity: Given indices I,I and I in Q suc tat I I I, and vertices v 1,...,v n U I, v U I, and v U I suc tat v i v v for i = 1,...,n, tere exist distinct vertices v 1,...,v n U I c suc tat v i v i v for i = 1,...,n. Square commutative graps are commutative in te sense defined by Plünnecke; square commutativity strengtens commutativity by requiring tat te alternate pats from v to v i or from v i to v ) go troug te associate vertex set. Tis is an important observation as later on we will need to apply Plünnecke s result. In our language, Ruzsa as already sown in pp of [12] tat addition graps are square commutative: Proposition 5.7 Ruzsa, [12]). Let A,B 1,...,B be subsets of a commutative group. Ten teir addition grap G + A,B 1,...,B ) is square commutative. Cannels of square commutative ypercube graps are also square commutative. Lemma 5.8. Let G be a square commutative ypercube grap indexed by Q, and let G = GX,Y) be a cannel of G. Ten G is square commutative. Additionally, if X U I and Y U I were I I Q, ten G is a Q j square commutative ypercube grap, were j = I \I. Proof. We ave already sown on p.13 tat G is a ypercube grap indexed by Q I \I. Tat it is square commutative follows from te fact tat G is square commutative combined wit te fact tat if x,z VG) and x y z ten y VG). Now tat we ave sown tat te main examples of ypercube graps are square commutative, we will prove tat square commutativity is inerited by products. Lemma 5.9. Let G 1 and G 2 be square commutative ypercube graps indexed by Q, and let G = G 1 G 2 be teir ypercube product. Ten G is square commutative. 14

15 B. Murpy, E.A. Palsson and G. Petridis Proof. Te proof is a straigtforward verification of square commutativity. We will prove only te upward condition, since te proof of te downward condition is similar. Let I,I, and I be indices in Q suc tat I I I, and suppose we ave vertices u,v) U I G), u,v ) U I G), andu 1,v 1 ),...,u n,v n ) U I G) suc tat u,v) u,v ) u i,v i ) for i = 1,...,n. We must find u 1,v 1 ),...,u n,v n ) U I c G) suc tat u,v) u i,v i ) u i,v i) for i = 1,...,n. Consider te sequences of vertices u u u i in G 1. Since G 1 is square commutative, tere exist distinct vertices u i U I c G 1 ) suc tat u u i u i for i = 1,...,n. Similarly tere exist distinct vertices v i U I c G 2 ) suc tat v v i v i for i = 1,...,n. Tus we ave distinct vertices u i,v i ) U I c G 1) U I c G 2 ) = U I c G) suc tat u,v) u i,v i ) u i,v i ) for i = 1,...,n, as desired. 5.3 A Plünnecke-type inequality for square commutative graps Te main goal in tis section is to extend Ruzsa s Teorem 2.2. Our result can furtermore be tougt of as an extension to square commutative graps of Plünnecke s inequality Teorem 5.10 below). Before we do tis we need to establis some notation and lemmas regarding magnification ratios. Given a directed grap G and subsets X,Y VG), we will use Im G X,Y) to denote te set of elements in Y tat can be reaced from X by pats in G. If G as layers V 0,...,V, we will use µ i G) to denote te it magnification ratio of G, wic is defined as Im G Z,V i ) µ i G) := min. Z V 0 Z We will say tat X V 0 acieves µ i G) wen µ i G) = Im GX,V i ). X Plünnecke bounded te growt of magnification ratios of commutative graps. We state a special case of is result tat will be applied later. Teorem 5.10 Plünnecke, [8]). Let 1 be a positive integer and G be a commutative grap. Ten µ G) µ 1 G). Square commutative graps are commutative, so Teorem 5.10 applies; owever, te bound on 15

16 Te Cardinality of Sumsets: Different Summands µ G) is not adequate for our purpose. Te goal of tis subsection is to improve it for square commutative graps. If G is a ypercube grap indexed by Q, ten te magnification of a subset X V 0 in U I, were I Q, is defined as β I X) := Im GX,U I ). X If I = {i}, ten we will use β i X) to denote β {i} X). Te following lemma relates te β I to te usual magnification ratio µ i. Lemma Let G be a ypercube grap indexed by Q. For any X V 0 G) we ave µ i G) I =iβ I X), wit equality if and only if X acieves µ i G) i.e., µ i G) = Im GX,V i ). X Proof. By te definition of µ i G) we ave µ i G) Im GX,V i ) X wit equality if and only if X acieves µ i G). Since V i is a disjoint union of te U I suc tat I = i, we ave Im G X,V i ) X = I =i Im GX,U I ) X = I =i Im G X,U I ) X = I =iβ I X). Combining tese two equations yields te desired inequality. Later we will also need te following elementary identity, wic asserts tat te β i are multiplicative. Lemma Let G,G be ypercube graps indexed by Q and G = G G. Ten for all i = 1,..., and Z V 0 G ), Z V 0 G ) we ave: β i Z Z ) = β i Z )β i Z ). Proof. We ave V 1 G ) = U {i} and V 1G ) = U {i}. Te way G is constructed gives V 1 G) = U {i} U {i}). Note tat Im G Z Z,U {i} U {i}) = Im G Z,U {i}) Im G Z,U {i}). 16

17 B. Murpy, E.A. Palsson and G. Petridis Te claim follows by taking cardinalities: β i Z Z ) = Im GZ Z,U{i} U {i} ) Z Z = Im G Z,U {i} ) Z = β i Z )β i Z ). Im G Z,U {i} ) Z Magnification ratio is multiplicative wit respect to tensor product of layered graps. However, for ypercube graps tis is only true for te top level magnification ratio, wic is multiplicative for square commutative ypercube graps. Square commutativity is not necessary, but it is sufficient logically and for our purposes). Lemma Let G 1 and G 2 be square commutative ypercube graps indexed by Q, and let G 3 be te ypercube product G 1 G 2. Ten µ G 3 ) = µ G 1 )µ G 2 ). Proof. For i = 1,2,3, we will define an auxiliary layered grap Ĝi as follows: V 0 Ĝi) := V 0 G i ), V 1 Ĝi) := V G i ), and v,v ) EĜi) if and only if tere is a pat from v to v in G i. Te proof rests on te following fact: Claim Ĝ3 = Ĝ1 Ĝ2. In words, Ĝ 3 is te directed layered tensor product of Ĝ1 and Ĝ2. It sould be noted ere tat tis would not be te case if we were working wit it magnification ratios for 1 i <, and tat square commutativity is essential for our proof. Proof of claim. It suffices to sow tat for any pair of vertices u 0,v 0 ) in V 0 G 1 ) V 0 G 2 ) and any pair of vertices u,v ) in V G 1 ) V G 2 ), we can find a sequence of index sets I 1 I = {1,...,}, and pats u 0 u 1 u in G 1 and v 0 v 1 v in G 2 suc tat u j U Ij G 1 ) and v j U Ij G 2 ). Tis guarantees tat te product pat u 0,v 0 ) u 1,v 1 ) u,v ) is contained in te ypercube product G 1 G 2, ence te edge u 0,v 0 ) u,v ) is contained in Ĝ3. Let u 0 u 1 u be any pat in G 1 from u 0 to u. We will use square commutativity to sow tat tere is a pat u 0 u 1 u 1 u suc tat u j U {1,...,j}. Applying te same argument for a pat v 0 v 1 v will prove te claim. For eac u j, let I j be te index set in Q suc tat u j U Ij G 1 ). By definition, u j u j+1 only if tere exists i j+1 suc tat I j+1 = I j {i j+1 }. Tus we may represent te sequence of index 17

18 Te Cardinality of Sumsets: Different Summands sets by a permutation: ) 1 2. i 1 i 2 i Applying upward square commutativity to te sequence I j 1 I j I j+1 is equivalent to switcing te pair i j and i j+1. An example tat illustrates tis, is tat by applying upward square commutativity to te layers V 0, V 1 and V 2 we transform te sequence V 0 = U U {i1 } U {i1,i 2 } to V 0 = U U {i2 } U {i1,i 2 } and so, in te permutation notation, we get ) ) i 1 i 2 i 3 i i 2 i 1 i 3 i Tus by repeated application of upward square commutativity, we can find a pat u 0 u 1 u 2 u suctatu 1 U {1} G 1 ). Againbyrepeatedapplicationofsquarecommutativity, we can find a pat u 0 u 1 u 2 u 3 u suc tat u 2 U {1,2}G 1 ), and so on. Now we continue wit te proof of te lemma. By definition of Ĝi, we ave µ 1 Ĝi) = µ G i ) for i = 1,2,3. Since Ĝ3 is te layered product of Ĝ1 and Ĝ2, by te multiplicativity of magnification ratios of directed layered graps e.g. Teorem 7.1 in [6]) we ave µ 1 Ĝ3) = µ 1 Ĝ1)µ 1 Ĝ2). Tus µ G 3 ) = µ G 1 )µ G 2 ), as desired. We are now ready to state and prove te teorem. Teorem 5.15 A Plünnecke-type inequality for square commutative graps). Let G be a square commutative grap indexed by Q. Ten for every Z V 0 we ave µ G) β 1 Z) β Z). Moreover, ) µ1 G) µ G). Proof. As usual i=0 V i is te vertex set of G and V 1 = U {i}. G is a square commutative grap and so in particular is commutative. Applying Teorem 5.10 and Lemma 5.11 successively gives: µ G) µ 1 G) β i Z)), 18

19 B. Murpy, E.A. Palsson and G. Petridis for all Z V 0. A first improvement is as follows. Claim For all Z V 0, we ave µ G) ) max β iz). 1 i 5.1) To prove Claim 5.16 we use te tensor product trick [12], see also [13]). Let n be any positive integer. We let G n = G G denote te n-fold ypercube product of G wit itself and S n V i G n ) te subset of V i G n ) tat is precisely te n-fold product of S wit itself. By Lemma 5.13 and Teorem 5.10 & Lemma 5.9 we get tat for all positive integers n µ G) n = µ G n ) µ 1 G n ). By Lemma 5.11 and Lemma 5.12 we ave µ 1 G n ) β iz n ) = β iz) n and so ) /n µ G) β i Z) n. Letting n go to infinity proves Claim To deduce te first inequality in te statement of te teorem we use a trick of Ruzsa e.g. [11]), wic appears in is proof of Teorem 2.2 and is similar to tat used in te deduction of Teorem 1.3 from Proposition 3.1. We begin by recalling tat Z is fixed. Let T 1,...,T be pairwise disjoint sets of generators of a free abelian group wit identity 0. For now we leave n i = T i undetermined, but note tat tey will depend on Z. Let T denote te addition grap G + {0},T 1,...,T ) and let G = G T. Te subsets of V 0 G ) are of te form S {0} for S V 0. Combining Claim 5.16 wit Lemma 5.12 gives µ G ) ) ) ) max β iz {0}) = max β iz)β i {0}) = max β iz)n i. 1 i 1 i 1 i We now cose te value of te n i. Te β i Z) are rational numbers so we set β i Z) = p i /q i and n = q 1 q. By coosing n i = n j i β jz) we ave β i Z)n i = n l=1 n l = β j Z)n j for all 19

20 Te Cardinality of Sumsets: Different Summands i,j = 1,...,n. Tus ) max β iz)n i = 1 i k β i Z)n i. On te oter and Lemma 5.13 gives µ G ) = µ G)µ T ) = µ G) T 1 + +T = µ G)n 1...n. Combining te above proves te first inequality in te statement of te teorem: µ G) = µ G ) n 1...n max 1 i β i Z)n i ) n 1...n = β iz)n i n 1...n = β i Z). To get te second inequality in te statement of te teorem we first apply te aritmetic mean - geometric mean inequality and get µ G) β 1 Z) β Z) 1 β i Z)). Te last step is to let X V 0 be te subset tat acieves te first magnification ratio µ 1 G) i.e., µ 1 G) = Im GX,V 1 ). Lemma 5.11 gives X β ix) = µ 1 G) and we are done. A couple remarks of some interest. Considering G = G + {0},T 1,...,T ) as constructed above wit T 1 = = T sows tat te upper bound cannot be trivially improved. Teorem 5.10 follows from Teorem Let G be a commutative grap wit vertex set V 0,V 1,...,V. We construct a ypercube grap H as follows: U I = V I and for every I I, u U I and v I, uv EH) if and only if uv EG). One may tink of of te it layer of H as consisting of i) copies of Vi and te set of edges between U I and U I is a copy of te set of edges between V I and V I, wenever I I. A routine calculation confirms tat H is square commutative, tat µ H) = µ G) and tat µ 1 H) = µ 1 G). Terefore, ) µ1 H) µ G) = µ H) = µ 1 G). 20

21 B. Murpy, E.A. Palsson and G. Petridis 5.4 A stronger Plünnecke-type inequality for square commutative graps Teorem 5.15 as one unsatisfactory aspect from a tecnical point of view: it does not provide any information on te subset Z V 0 tat acieves µ G) i.e., te subset tat satisfies µ G) = Im GZ,V ). We strengten Teorem 5.15 by proving tat te subset X V 0 Z tat acieves µ 1 G) as restricted growt and in fact satisfies te bound given in Teorem A similar result was proved for commutative graps in [7]. Teorem Let G be a square commutative grap wit vertex set V 0 V. Suppose tat X V 0 acieves µ 1 G) i.e., µ 1 G) = Im GX,V 1 ). Ten X ) µ1 G) Im G X,V ) X. Proof. We work in te cannel G = GX,V ) rater tan te original square commutative grap. IntiscontextwewillprovetatifG isacommutativegrapwitvertexsetv 0 V, wic satisfies µ 1 G ) = V 1 / V 0, ten V µ1 G ) ) V 0. Suppose not. Let G be a counterexample were V 0 is minimal. Teorem 5.15 implies tat te collection { ) Z V 0 : Im G Z,V ) µ1 G ) Z } is nonempty. Let S V 0 be a set of maximal cardinality in te collection S cannot equal V 0 \ Im G S,V assumed tat G is a counter example) and H = G V 0 \ S,V as we ave )). In words H is te cannel consisting of pats in G tat do not start in S and do not end in its image in V. Suppose tat W 0 W 1 W are te layers of H. Observe also tat for all Z W 0 and all i = 1,..., we ave Im H Z,W i ) = Im G Z,W i ). W 1 does not intersect Im G S,V 1 ) as tere would ten exist a pat in H leading to Im G S,V ). We terefore ave W 1 V 1 Im G S,V 1) 21

22 Te Cardinality of Sumsets: Different Summands V 1 µ 1 G ) S = V 1 V 1 V 0 S = V 1 V 0 S V 0 = W 0 V 1 V 0, as W 0 = V 0 S. Consequently µ 1 H) W 1 W 0 V 1 V 0 = µ 1G ). 5.2) Let T W 0 be any subset tat satisfies Im H T,W 1 ) = µ 1 H) T. Let us get a lower bound on Im H T,W ). We know from te maximality of S tat µ1 G ) Tis implies ) S T < Im G S T,V ) = Im G S,V ) + Im G T,V )\Im G S,V ) = Im G S,V ) + Im HT,W ) ) µ1 G ) S + Im H T,W ). ) µ1 G ) Im H T,W ) > T. 5.3) Finally we consider H = HT,W ), te cannel consisting of all pats in H starting at T. H is a square commutative grap wit layers T 0 T and magnification ratio µ 1 H ) = µ 1 H). By inequalities 5.3) and 5.2) we get: T = Im H T,W ) = Im H T,W ) ) µ1 G ) ) µ1 H) > T T 0 ) µ1 H ) = T 0. Tus H is anoter counterexample. However, T 0 = T W 0 = V 0 contradicts te minimality of V 0. \ S < V 0, wic 22

23 B. Murpy, E.A. Palsson and G. Petridis 5.5 Application to sumsets wit a component removed Our final task is to deduce from Teorem 5.17 te upper bound on sumsets wit a component removed, wic was used in Section 3. Corollary Let be a positive integer. Suppose tat A,B 1,...,B are finite sets in a commutative group and E A a subset of A. If X A\E is a subset of A\E tat minimises te quantity Z +B i )\E +B i ) Z over all non-empty subsets Z A\E, ten X +B 1 + +B )\E +B 1 + +B ) µ X, were µ := µx) = 1 X +B i )\E +B i ). X Proof. We work in to te ypercube grap G indexed by Q wit vertex set given by U I = A + i I B i) \ E + i I B i); and edge set determined as follows: an edge exists between u V I and v V I {j} if v u B j. G is square commutative by Lemma 5.8, because it is precisely te cannel ) G A\E,A+ B i )\E + B i ) in te square commutative addition grap G + A,B 1,...,B ). Identifying Z A\E wit te corresponding subset of V 0 G) gives Z +B i )\E +B i ) Z = Im GZ,V 1 ). Z In particular te defining property of X implies tat X acieves µ 1 G) and so µ 1 G) = 1 Im G X,V 1 ) X = 1 Im G X,U {i} ) X 23

24 Te Cardinality of Sumsets: Different Summands = 1 = µ. X +B i )\E +B i ) X Te condition in Teorem 5.17 is satisfied and so ) µ1 G) X +B 1 + +B )\E +B 1 + +B ) = Im G X,V ) = µ X, as claimed. Acknowledgements Te autors would like to tank te referee for a careful reading of te manuscript and elpful suggestions; and Dimitris Koukoulopoulos for spotting an omission in te statement of Lemma Te second autor would like to tank te University of Rocester for teir funding during is postdoc tere were te bulk of tis project was completed. Te tird autor would like to tank Imre Ruzsa for is elp in simplifying some of te proofs in te related paper [7], wic made te proof of te main result of te present paper simpler tan it would ave oterwise been. References [1] P. Balister and B. Bollobás, Projections, entropy and sumsets, Combinatorica ), [2] B. Bollobás and A. Tomason, Projections of bodies and ereditary properties of ypergraps, Bull. London Mat. Soc ), [3] K. Gyarmati, M. Matolcsi and I. Z. Ruzsa, Plünnecke s inequality for different summands, in: Building Bridges: between matematics and computer science, M. Grötscel and G.O.H. Katona. eds.), Bolyai Society Matematical Studies Vol. 19, New York: Springer, [4] K. Gyarmati, M. Matolcsi and I. Z. Ruzsa, A supperadditivity and submultiplicativity property for cardinalities of sumsets, Combinatorica ), [5] M. Madiman, and A. W. Marcus, and P. Tetali, Entropy and set cardinality inequalities for partition-determined functions, Random Structures Algoritms ),

25 B. Murpy, E.A. Palsson and G. Petridis [6] M. B. Natanson, Additive Number Teory: Inverse Problems and te Geometry of Sumsets. New York: Springer, [7] G. Petridis, Upper bounds on te cardinality of iger sumsets, Acta Arit, ), [8] H. Plünnecke, Eine zalenteoretisce Anwendung der Grapteorie, J. Reine Angew. Mat ), [9] I. Z. Ruzsa, An application of grap teory to additive number teory, Scientia, Ser. A ), [10] I. Z. Ruzsa, Cardinality questions about sumsets, in: Additive Combinatorics, A. Granville and M.B. Natanson and J. Solymosi eds.), CRM Proceedings & Lecture Notes, New York: American Matematical Society, [11] I. Z. Ruzsa, Sumsets and Structure, in: Combinatorial Number teory and Additive Group Teory. New York: Springer, [12] I. Z. Ruzsa, Towards a noncommutative Plünnecke-type inequality, in: An Irregular Mind: Szemerédi is 70 Bolyai Society Matematical Studies). New York: Springer, [13] T. Tao, Tricks Wiki article: Te tensor power trick, terrytao.wordpress.com/2008/08/25/tricks-wiki-article-te-tensor-product-trick/. [14] T. Tao and V. H. Vu, Additive Combinatorics. Cambridge: Cambridge University Press, Department Matematics, University of Rocester, Rocester, NY, USA addresses: murpy@mat.rocester.edu, palsson@mat.rocester.edu, giorgis@cantab.net 25

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