A structure theorem for product sets in extra special groups

Transcription

1 A structure theorem for roduct sets in extra secial grous Thang Pham Michael Tait Le Anh Vinh Robert Won arxiv: v1 [math.nt] 25 Ar 2017 Abstract HegyváriandHennecartshowedthatifB isasufficientlylargebrickofaheisenberg grou, then the roduct set B B contains many cosets of the center of the grou. We give a new, robust roof of this theorem that extends to all extra secial grous as well as to a large family of quasigrous. 1 Introduction Let be a rime. An extra secial grou G is a -grou whose center Z is cyclic of order such that G/Z is an elementary abelian -grou (nice treatments of extra secial grous can be found in [2, 6]). The extra secial grous have order 2n+1 for some n 1 and occur in two families. Denote by H n and M n the two non-isomorhic extra secial grous of order 2n+1. Presentations for these grous are given in [4]: H n = a 1,b 1,...,a n,b n,c [a i,a j ] = [b i,b j ] = 1,[a i,b j ] = 1 for i j, [a i,c] = [b i,c] = 1,[a i,b i ] = c,a i = b i = c i = 1 for 1 i n M n = a 1,b 1,...,a n,b n,c [a i,a j ] = [b i,b j ] = 1,[a i,b j ] = 1 for i j, [a i,c] = [b i,c] = 1,[a i,b i ] = c,a i = c i = 1,b i = c for 1 i n. Fromthese resentations, it isnot hardto see thatthecenter ofeach ofthese grous consists of the owers of c so are cyclic of order. It is also clear that the quotient of both grous by their centers yield elementary abelian -grous. In this aer we consider the structure of roducts of subsets of extra secial grous. The structure of sum or roduct sets of grous and other algebraic structures has a rich history in combinatorial number theory. One seminal result is Freiman s theorem [5], which asserts that if A is a subset of integers and A+A = O( A ), then A must be a subset of a small generalized arithmetic rogression. Green and Ruzsa [7] showed that the same result is true in any abelian grou. On the other hand, commutativity is imortant as the theorem is not true for general non-abelian grous [8]. With this in mind, Hegyvári and Hennecart were motivated to study what actually can be said about the structure of roduct sets in non-abelian grous. They roved the following theorem. Deartment of Mathematics, EPFL, Lausanne, Switzerland. thang.ham@efl.ch Deartment ofmathematicalsciences, CarnegieMellon University. Researchis suortedby NSF grant DMS mtait@cmu.edu. University of Education, Vietnam National University. vinhla@vnu.edu.vn Deartment of Mathematics, Wake Forest University. wonrj@wfu.edu 1

2 Theorem 1.1 (Hegyvári-Hennecart, [9]). For every ε > 0, there exists a ositive integer n 0 such that if n n 0, B H n is a brick and B > H n 3/4+ε then there exists a non trivial subgrou G of H n, namely its center [0,0,F ], such that B B contains at least B / cosets of G. The grou H 1 is the classical Heisenberg grou, so the grous H n form natural generalizations of the Heisenberg grou. Our main focus is on the second family of extra secial grous M n. The grou H n has a well-known reresentation as a subgrou of GL n+2 (F ) consisting of uer triangular matrices 1 x z [x,y,z] := 0 I n y where x,y F n, z F, and I n is the n n identity matrix. Let e i F n be the ith standard basis vector. In the resentation for H n, a i corresonds to [e i,0,0], b i corresonds to [0,e i,0] and c corresonds to [0,0,1]. By matrix multilication, we have [x,y,z] [x,y,z ] = [x+x,y +y,z +z + x,y ] where, denotes the usual dot roduct. A second focus of this aer is to consider generalizations of the higher dimensional Heisenberg grous where entries come from a quasifield Q rather than F. We recall the definition of a quasifield: A set L with a binary oeration is called a loo if 1. the equation a x = b has a unique solution in x for every a,b L, 2. the equation y a = b has a unique solution in y for every a,b L, and 3. there is an element e L such that e x = x e = x for all x L. A (left) quasifield Q is a set with two binary oerations + and such that (Q,+) is a grou with additive identity 0, (Q, ) is a loo where Q = Q\{0}, and the following three conditions hold: 1. a (b+c) = a b+a c for all a,b,c Q, 2. 0 x = 0 for all x Q, and 3. the equation a x = b x+c has exactly one solution for every a,b,c Q with a b. Given a quasifield Q, we define H n (Q) by the set of elements and a multilication oeration defined by {[x,y,z] : x Q n,y Q n,z Q} [x,y,z] [x,y,z ] = [x+x,y +y,z +z + x,y ]. Then H n (Q) is a quasigrou with an identity element (ie, a loo), and when Q = F we have that H n (Q) is the n-dimensional Heisenberg grou H n. 2

3 1.1 Statement of main results Let H n be a Heisenberg grou. A subset B of H n is said to be a brick if B = {[x,y,z] such that x X, y Y, z Z} where X = X 1 X n and Y = Y 1 Y n with non emty-subsets X i,y i,z F. Our theorems are analogs of Hegyvári and Hennecart s theorem for the grous M n and the quasigrous H n (Q). In articular, their structure result is true for all extra secial grous. We will define what it means for a subset B of M n to be a brick in Section 2.1. Theorem 1.2. For every ε > 0, there exists a ositive integer n 0 = n 0 (ε) such that if n n 0, B M n is a brick and B > M n 3/4+ε then there exists a non trivial subgrou G of M n, namely its center, such that B B contains at least B / cosets of G. Combining Theorem 1.1 and Theorem 1.2, we have Theorem 1.3. Let G be an extra secial grou. For every ε > 0, there exists a ositive integer n 0 = n 0 (ǫ) such that if n n 0, B G is a brick and B > G 3/4+ε then there exists a non trivial subgrou G of G, namely its center, such that B B contains at least B / cosets of G. For Q a finite quasifield, we similarly define a subset B H n (Q) to be a brick if B = {[x,y,z] such that x X, y Y, z Z} where X = X 1 X n and Y = Y 1 Y n with non emty-subsets X i,y i,z Q. Theorem 1.4. Let Q be a finite quasifield of order q. For every ε > 0, there exists an n 0 = n 0 (ε) such that if n n 0, B H n (Q) is a brick, and B > H n (Q) 3/4+ε, then there exists a non trivial subquasigrou G of H n (Q), namely its center [0,0,Q] such that B B contains at least B /q cosets of G. Taking Q = F gives Theorem 1.1 as a corollary. 2 Preliminaries 2.1 A descrition of M n We give a descrition of M n with which it is convenient to work. Define a grou G whose elements are triles [x,y,z] where x = (x 1,...,x n ), y = (y 1,...,y n ), with x i,y i,z F for 1 i n. The grou oeration in G is given by [x,y,z] [x,y,z ] = [x+x,y +y,z +z + x,y +f(y,y )] 3

4 where the function f : Z n Z n N is defined by n f ((y 1,...,y n ),(y 1,...,y n)) yi mod +y i = mod. Concretely, f counts the number of comonents where (after reducing mod ) y i +y i. This is slight abuse of notation, as y,y F n, but is well-defined if we regard them as elements of Z n. Lemma 2.1. With the oeration defined above, G is a grou isomorhic to M n. Proof. We first need to check associativity of the oeration. After cancellation, this reduces to checking the equality f(y +y,y )+f(y,y ) = f(y,y +y )+f(y,y ) which holds because (yi +y i ) mod +y i mod yi mod +y i + mod yi mod +y i mod +y i mod = (yi +y = i) mod +y i mod (yi +y + i) mod +y i mod, as all three of the exressions count the largest multile of dividing y i mod +y i mod +y i mod. Since G is generated {[e i,0,0],[0,e i,0],[0,0,1]}, we define a homomorhism ϕ : G M n by ϕ ( [e i,0,0] ) = a i, ϕ ( [0,e i,0] ) = b i, and ϕ([0,0,1]) = c. This ma is clearly surjective and it is easy to check that the generators of G satisfy the relations in M n. Since G = 2n+1, ϕ is an isomorhism and G = M n, as claimed. With this descrition, there is a natural way to define a brick in M n. A subset B of M n is said to be a brick if B = {[x,y,z] such that x X, y Y, z Z} where X = X 1 X n and Y = Y 1 Y n with non emty-subsets X i,y i,z F. 2.2 Tools from sectral grah theory For a grah G with vertex set {v 1,...,v n }, the adjacency matrix of G is the matrix with a 1 in row i and column j if v i v j and a 0 otherwise. Since this is a real, symmetric matrix, it has a full set of real eigenvalues. Let λ 1 λ 2 λ n be the eigenvalues of its adjacency matrix. If G is a d-regular grah, then its adjacency matrix has row sum d. In this case, λ 1 = d with the all-one eigenvector 1. Let v i denote the corresonding eigenvector for λ i. We will make use of the trick that for i 2, v i 1, so Jv i = 0 where J is the all-one matrix of size n n (see [3] for more background on sectral grah theory). 4

5 It is well-known (see [1, Chater 9] for more details) that if λ 2 is much smaller than the degree d, then G has certain random-like roerties. A grah is called biartite if its vertex set can be artitioned into two arts such that all edges have one endoint in each art. For G be a biartite grah with artite sets P 1 and P 2 and U P 1 and W P 2, let e(u,w) be the number of airs (u,w) such that u U, w W, and (u,w) is an edge of G. We recall the following well-known fact (see, for examle, [1]). Lemma 2.2 (Corollary9.2.5, [1]). Let G = (V,E) be d-regular biartite grah on 2n vertices with artite sets P 1 and P 2. For any two sets B P 1 and C P 2, we have d B C e(b,c) n λ 2 B C. 2.3 Sum-roduct grahs Let Q be a finite quasifield. The sum-roduct grah SP Q,n is defined as follows. SP Q,n is a biartite grah with its vertex set artitioned into artite sets X and Y, where X = Y = Q n Q. Two vertices U = (x,z) X and V = (y,z ) Y are connected by an edge, (U,V) E(SP Q,n ), if and only if x,y = z+z. We need information about the eigenvalues of SP Q,n. Lemma 2.3. If Q is a quasifield of order q, then the grah SP Q,n is q n regular and has λ 2 2 1/2 q n/2. We rovide a roofof Lemma 2.3 for comleteness inthe aendix, and we note that similar lemmas were roved in [11] and [10]. 3 Proof of Theorem 1.2 Lemma 3.1. Let B M n be a brick in M n with B = [X,Y,Z] where X = X 1 X n and Y = Y 1 Y n. For given a = (a 1,...,a n ),b = (b 1,...,b n ) F n, suose that Z 2 X i (a i X i ) Y i (b i Y i ) > 2 n+2, then we have B B [a,b,f ]. Proof. Let X i = X i (a i X i ), Y i = Y i (b i Y i ),X = (X 1,...,X n ), and Y = (Y 1,...,Y n ). We first have B B {[x,y,z] [a x,b y,z ] : x X,y Y,z,z Z}. On the other hand, it follows from the multilicative rule in M n that for [x,y,z] [a x,b y,z ] = [a,b,z +z + x,(b y) +f(y,b y)]. To conclude the roof of the lemma, it is enough to rove that { z +z + x,(b y) +f(y,b y): z,z Z,x X,y Y } = F 5

6 under the condition Z 2 X Y > 2 n+2. To rove this claim, let λ be an arbitrary element in F, we define two sets in the sumroduct grah SP F,n, E X and F Y as follows: E = X ( Z +λ), F = { (b y, z f(y,b y)): z Z,y Y }. It is clear that E = Z X and F = Z Y. It follows from Lemma 2.2 and Lemma 2.3 that if Z 2 X Y > 2 n+2, then e(e,f) > 0. It follows that there exist x X,y Y, and z,z Z such that z +z + x,(b y) +f(y,b y) = λ. Since λ is chosen arbitrarily, we have { z +z + x,(b y) +f(y,b y): z,z Z,x X,y Y } = F. Proof of Theorem 1.2. We follow the method of [9, Theorem 1.3]. First we note that if Z > /2, then we have Z +Z = F. This imlies that B B = [2X,2Y,F ]. Therefore, B B contains at least B B / B / cosets of the subgrou [0,0,F ]. Thus, in the rest of the roof, we may assume that Z /2. For 1 i n, we have a i F X i (a i X i ) = X i 2, b i F Y i (b i Y i ) = Y i 2, which imlies that X i (a i X i ) Y i (b i Y i ) = a i F b i F X i 2 Y i 2. Therefore we obtain X i (a i X i ) Y i (b i Y i ) = a,b F n X i 2 Y i 2. (1) Let N be the number of airs (a,b) F n Fn such that Z 2 X i (a i X i ) Y i (b i Y i ) > 2 n+2. It follows from Lemma 3.1 that [a,b,f ] B B for such airs (a,b). Then by equation (1) ( n ) ( n 2 X i Y i N +2 n+2 ( 2n N) > X i Y i ), and so N > n X i 2 Y i 2 2 3n+2 n X i Y i 2 n+2. 6

7 By the assumtion of Theorem 1.2, we have ( ) B = Z X i Y i > M n 3/4+ε = 3n/2+3/4+ε(2n+1). (2) Thus when n > 1/ǫ, we have since Z. In other words, X i Y i > 3n/2+7/4, N (1 2 3/2 ) X i Y i = (1 2 3/2 ) B Z B, since Z /2. 4 Proof of Theorem 1.4 Lemma 4.1. Let Q be a quasifield of order q and let [X,Y,Z] = B H n (Q) be a brick. For a given a = (a 1,...,a n ), b = (b 1,...,b n ) Q n, suose that Z 2 X i (a i X i ) Y i (b i Y i ) > 2q n+2, then we have B B [a,b,q]. Proof. The roof is similar to that of Lemma 3.1, so we leave some details to the reader. Let X = (X 1 (a 1 X 1 ),...,X n (a n X n )), Y = (Y 1 (b 1 Y 1 ),...,Y n (b n Y n )) and E X, F Y in SP Q,n where E = X ( Z +λ), F = { (b y, z): z Z,y Y }, and λ Q is arbitrary. Then e(e,f) > 0 which imlies that there exist x X, y Y, and z,z Z such that z +z + x,(b y) = λ. This imlies that [a,b,q] B B. The rest of the roof of Theorem 1.4 is identical to that of Theorem 1.2. We need only to show that if Z Q and Z > Q /2, then Z + Z = Q. However, this follows since the additive structure of Q is a grou. 7

8 References [1] N. Alon and J. H. Sencer, The Probabilistic Method, 2nd ed., Wiley-Interscience, [2] M. Aschbacher, Finite Grou Theory, Vol. 10. Cambridge University Press, [3] A. Brouwer and W. Haemers, Sectra of Grahs, Sringer, New York, etc., [4] P. Diaconis, Threads through grou theory, Character Theory of Finite grous, Contemorary Mathematics, 524 (2010): [5] G. A. Freiman, Addition of finite sets, Sov. Math. Dokl. 5 (1964) [6] D. Gorenstein, Finite Grous, Vol American Mathematical Soc., [7] B. Green and I. Ruzsa, Freiman s theorem in an arbitrary abelian grou, J. Lond. Math. Soc. 75 (2007), [8] N Hegyvári and F. Hennecart, A note on Freiman models in Heisenberg grous, Israel J. of Math. 189 (2012), [9] N. Hegyvári and F. Hennecart, A structure result for bricks in Heisenberg grous, Journal of Number Theory 133(9) (2013): [10] T. Pham, M. Tait, C. Timmons, and L. A. Vinh, A Szemerédi-Trotter tye theorem, sum-roduct estimates in finite quasifields, and related results, J. Combin. Theory Ser. A 147 (2017) [11] L. A. Vinh, The solvability of norm, bilinear and quadratic equations over finite fields via sectra of grahs, Forum Mathematicum, Vol. 26 (2014), No. 1, Aendix Proof of Lemma 2.3. Let Q be a finite quasifield of order q and let SP Q,n be the biartite grah with artite sets X = Y = Q n Q where (x 1,...,x n,z x ) (y 1,...,y n,z y ) if and only if z x +z y = x 1 y 1 + +x n y n. (3) First we show that SP Q,n is q n regular. Let (x 1,...,x n,z x ) be an arbitrary element of X. Choose y 1,...,y n Q arbitrarily. Then there is a unique choice for z y that makes (3) hold, and so the degree of (x 1,...,x n,z x ) is q n. A similar argument shows the degree of each vertex in Y is q n. Next we show that λ 2 is small. Let M be the adjacency matrix for SP Q,n where the first q n+1 rows and columns are indexed by X. We can write ( ) 0 N M = N T 0 where N is the q n+1 q n+1 matrix whose (x 1,...,x n,x z ) X (y 1,...,y n,y z ) Y entry is 1 if (3) holds and 0 otherwise. The matrix M 2 counts the number of walks of length 2 between vertices. Since SP Q,n is q n regular, the diagonal entries of M 2 are all q n. Since SP Q,n is biartite, there are no 8

9 walks of length 2 from a vertex in X to a vertex in Y. Now let x = (x 1,...,x n,x z ) and x = (x 1,...,x n,x z ) be two distinct vertices in X. To count the walks of length 2 between them is equivalent to counting their common neighbors in Y. That is, we must count solutions (y 1,...,y n,z y ) to the system of equations x z +y z = x 1 y 1 + +x n y n (4) and x z +y z = x 1 y 1 + +x n y n. (5) Case 1: For i 1 n we have x i = x i : In this case we must have x z x z. Subtracting (4) from (5) shows that the system has no solutions and so x and x have no common neighbors. Case 2: There is an i such that x i x i: Subtracting (5) from (4) gives x z x z = x 1 y 1 + +x n y n x 1 y 1 x n y n. (6) There are q n 1 choices for y 1,...,y i 1,y i+1,...y n. Since x i x i 0, these choices determine y i uniquely, which then determines y z uniquely. Therefore, in this case x and x have exactly q n 1 common neighbors. A similar argument shows that for y = (y 1,...,y n,y z ) and y = (y 1,...,y n,y z), then either y and y have either no common neighbors or exactly q n 1 common neighbors. Now let H be the grah whose vertex set is X Y and two vertices are adjacent if and only if they are either both in X or both in Y, and they have no common neighbors. For this to occur, we must be in Case 1, and therefore we must have either x z x z or y z y z and all of the other coordinates equal. Therefore, this grah is q 1 regular, as for each fixed vertex there are exactly q 1 vertices with a different last coordinate and the same entries on the first n coordinates. Let E be the adjacency matrix of H and note that since H is q 1 regular, all of the eigenvalues of E are at most q 1 in absolute value. Let J be the q n+1 by q n+1 all ones matrix. By the above case analysis, it follows that M 2 = q n 1 ( J 0 0 J ) +(q n q n 1 )I q n 1 E (7) Now let v 2 be an eigenvector of M for λ 2. For a set of vertices Z let χ Z denote the vector which is 1 if a vertex is in Z and 0 otherwise (ie it is the characteristic vector for Z). Note that since SP Q,n is a regular biartite grah, we have that λ 1 = q n with corresonding eigenvector χ X + χ Y and λ n = q n with corresonding eigenvector χ X χ Y. Also note that v 2 is erendicular to both of these eigenvectors and therefore is also erendicular to both χ X and χ Y. This imlies that ( ) J 0 v 0 J 2 = 0. Now by (7), we have λ 2 2 v 2 = (q n q n 1 )v 2 q n 1 Ev 2. Therefore q 1 λ2 2 q n 1 is an eigenvalue of E and is therefore at most q 1 in absolute value, imlying that λ 2 2 1/2 q n/2. 9