A SUM-PRODUCT ESTIMATE IN ALGEBRAIC DIVISION ALGEBRAS OVER R. Department of Mathematics University of California Riverside, CA

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1 A SUM-PRODUCT ESTIMATE IN ALGEBRAIC DIVISION ALGEBRAS OVER R 1 Mei-Chu Chang Department of Mathematics University of California Riverside, CA 951 mcc@math.ucr.edu Let A be a finite subset of an integral domain, and let A denote the cardinality of the set A. The sum set of A is and the product set of A is A A + A {a 1 + a a i A}, (0.1) A A.A {a 1 a a i A}. (0.) In [7], Erdös and Szemerédi conjectured that A and A cannot both be small. More precisely, they made the following Conjecture. A + A > A ɛ. What they proved is the following Theorem (Erdős-Szemerédi). If A R is a finite set of real numbers, then where δ is a constant. A + A A 1+δ, (0.3) Notation 1. X Y means X > cy, for some nonzero constant c. Nathanson [9] showed that δ = partially supported by NSA grant No. MDA Mathematics Subject Classification. 05A99. 1 Typeset by AMS-TEX

2 MEI-CHU CHANG Ford [1] obtained that δ = 1 15, and Elekes [5] showed that δ = 1 4 by using Szemerédi- Trotter Theorem [11]. So far, the best bound is obtained by Solymosi [13] who showed that δ = 3 11+ɛ also by using Szemerédi-Trotter Theorem cleverly. In a different direction, Bourgain, Katz, and Tao [1] proved the theorem for the case that A is a subset of a finite field. For other related results, see [], [3], [4], [6], [8], [10]. In this paper, we generalize the Erdős-Szemerédi argument to sets of elements contained in C, the field of complex numbers, or K, the quaternions, or certain normed subspaces V of an R algebra (See Theorem 3). Theorem 1. Let A be a finite subset of C, or K. Then A + A A (0.4) The proof of Erdős-Szemerédi Theorem uses the order of real numbers and therefore it does not generalize trivially for the complex case. The problem for the complex case was brought up by I. Ruzsa (private communication). The theorem follows from the following special case. Theorem. Let R be the annulus, R {z r z r}, (0.5) contained in C, or K and let B R be a finite subset of R. Then B + B B 1+δ, (0.6) where δ = ( 1 3 ɛ)δ 1, δ 1 = 1 9, ɛ = (0.7) Our method to prove Theorem is to cover R with disjoint boxes Q of equal size such that the maximal number of elements of B contained in each of the boxes is B δ 1. We fix one such a maximal box called Q 0, assuming B B 1+δ and B B 1+δ, and show that most of the boxes have nearly as many elements as B Q 0. For each of these boxes, we find a large subset C Q such that we can define a map from (B Q 0 ) 8 to C Q C Q. The cardinalities of the domain and the range give us a bound on δ 1. The following theorem can be used to prove a sum-product theorem for symmetric matrices.

3 ERDÖS-SZEMERÉDI THEOREM 3 Theorem 3. Let {R m, +, } be an R-algebra with + the componentwise addition. For a = (a 1,, a m ), let a = ( a i ) be the Hilbert-Schmidt norm, and let V Rm such that 1. c = c(m), a, b V, a b = c a b. for any a V \ {0}, a 1 exists (in a possibly larger field). Then for any A V, A + A + AA > A 1+δ. The paper is organized as follows: In Section 1, we prove Theorem. In Section, we reduce Theorem 1 to Theorem. For simplicity, we prove the theorems for complex numbers, and describe the differences for the cases of the quaternions and the normed subspace in Remark (Section 1) and Remark 5 (Section ) for those who care about the constant c as in Notation 1. (The exponents are the same for both C and K.) Section 1. The Theorem for an annulus. Then Let r > 0 be as in (0.5). We define A = { 1 r a : a A}. A = A, A + A = A + A, A.A = A.A, (1.1) and we may rescale the annulus R and assume R {z C : 1 z }. (1.) For any ρ > 0, the complex plane is covered by disjoint squares O j,k = {pρ + 1 qρ : (j 1) p < j, (k 1) q < k}, (1.3) where j, k Z. (For convenience, sometimes we denote Q j,k by Q.) We choose ρ such that max Q B Q B δ 1. (1.4) (See (0.7) for δ 1.) Let and Q 0 be an element of P with P {Q : Q B }, B Q 0 B δ 1. (1.5) First, we will show that the number of the sets Q + Q 0 and QQ 0 containing a fixed point is bounded by an absolute constant.

4 4 MEI-CHU CHANG Claim 1. Fix Q P. 1. For any x Q + Q 0, {Q : x Q + Q 0 } 4.. For any x QQ 0, {Q : x QQ 0 } 5. Proof. (1) Let Q j,k be as in (1.3), and Q 0 be as in (1.5), Then Q j,k + Q 0 is the set Q j,k = {pρ + 1 qρ : (j 1) p < j, (k 1) q < k}, Q 0 = {p 0 ρ + 1 q 0 ρ : (j 0 1) p 0 < j 0, (k 0 1) q 0 < k 0 }. {(p+p 0 )ρ+ 1 (q +q 0 )ρ : (j +j 0 ) p+p 0 < (j +j 0 ), (k +k 0 ) q +q 0 < (k +k 0 )}, The only sums of squares intersecting Q j,k +Q 0 are Q j+1,k +Q 0,Q j,k+1 +Q 0, Q j+1,k+1 +Q 0. () To count the number of products of squares containing x, we enclose each square by a disc with the same center and count the number of discs containing the product of the discs. We will use the multiplicative property of the Hilbert-Schmidt norm. Let z Q be the center of Q. Then any z Q is contained in a disc centered at z Q with radius ρ. Therefore, we have z = z Q + d, where d < ρ. (1.6) Similarly, let z Q0 be the center of Q 0. Then for any z 0 Q 0, we have z 0 = z Q0 + d 0, where d 0 < ρ. (1.7) We see easily that zz 0 is contained in a disc centered at z Q z Q0 zz 0 z Q z Q0 dz Q0 + d 0 z Q + dd 0 ρ + ρ < 7 ρ. with radius 7 ρ. In fact, For the second inequality, we use the fact that z Q, z Q0. The point zz 0 is in a disc centered at z Q z Qo if and only if z Q z Q0 is in a disc centered at zz 0 with the same radius. Therefore, counting the number of discs containing zz 0 is the same as counting the number of z Q z Q0 in the disc centered at zz 0. Since z Q z Q0 z Q z Q0 ρ, (1.8) a disc of radius 7 ρ with center zz 0 contains at most 49 of the points z Q z Q0. Therefore, every point zz 0 in QQ 0 is contained in at most 49 discs centered at z Q z Q0, hence zz 0 is contained in at most 49 of the sets QQ 0 centered at z Q z Q0.

5 ERDÖS-SZEMERÉDI THEOREM 5 Remark. If R K, then we cover R with boxes Q of volume ρ 4. Therefore, every point is contained in at most 16 of the sets Q + Q 0, and at most 9 4 of the sets QQ 0. More generally, if V is as in Theorem 3, then a point in V is contained in at most m of the sets Q + Q 0 and at most [1 + ( m + m 4 )] of the sets QQ 0. Next, we show that most squares have almost the maximal number of elements of B, if B < B 1+δ. Claim 3. We assume and let Then B < B 1+δ, (1.9) P 0 {Q P : B δ 1 (1+ɛ)δ B Q B δ 1 }. (1.10) Proof. First, it follows from (1.9), Claim 1 and (1.5) that Therefore, P 0 B 1 δ 1. (1.11) B 1+δ > B > (B Q 0 ) + (B Q) > 1 4 Q P (B Q 0 ) + (B Q) Q P Next, we see that (1.10) and (1.13) imply P B Q 0 P B δ 1 (1.1) P B 1+δ δ 1. (1.13) Q P\P 0 B Q < P \ P 0 B δ1 (1+ɛ)δ P B δ1 (1+ɛ)δ B 1 ɛδ, (1.14) which is < 1 B, when B is sufficiently large. Hence which, together with (1.4) gives (1.11). B Q > 1 B, (1.15) Q P 0 The next claim shows that for almost all Q P 0, the sum sets of B Q and B Q 0 are not too big, if B < B 1+δ.

6 6 MEI-CHU CHANG Claim 4. We assume B < B 1+δ, (1.16) and let Then P 1 {Q P 0 : (B Q) + (B Q 0 ) < B δ 1+(1+ɛ)δ }. (1.17) P 1 B 1 δ 1. (1.18) Proof. By (1.16), Claim 1, and (1.17), we have B 1+δ > B > (B Q 0 ) + (B Q) Q P 0 \P 1 > 1 (B Q) + (B Q 0 ) 4 Q P 0 \P 1 P 0 \ P 1 B δ1+(1+ɛ)δ. (1.19) Hence, P 0 \ P 1 B 1 δ1 ɛδ. (1.0) Now, (1.18) follows from (1.11) and (1.0). Standard Construction Let Q P 1 be fixed, and let Y be a maximal subset of B Q, such that the sets y + (B Q 0 ) for y Y are all disjoint. The following gives an estimate of Y. Claim 5. Y < B (1+ɛ)δ. Proof. The claim follows from (1.5) and the following inequality Y B Q 0 = y Y y + (B Q 0 ) (B Q) + (B Q 0 ) < B δ 1+(1+ɛ)δ. The last inequality is by (1.17).

7 ERDÖS-SZEMERÉDI THEOREM 7 Claim 6. For z B Q, we define B z = (B Q) ( z + (B Q 0 ) (B Q 0 ) ). Then there is a point z Q B Q, such that B zq > B δ 1 (1+ɛ)δ. (1.1) Proof. The maximality of Y implies that B Q ( y + (B Q0 ) (B Q 0 ) ). (1.) Therefore, B Q = y Y and there is z Q Y, such that Hence, y Y ( (B Q) ( y + (B Q0 ) (B Q 0 ) )), (1.3) B zq = (B Q) ( z Q + (B Q 0 ) (B Q 0 ) ) because of (1.10) and Claim 5. Remark 6.1. Denoting B zq Hence B zq B δ 1 (1+ɛ)δ (1+ɛ)δ, by B Q, we have B Q. (1.4) Y B Q = (B Q) ( z + (B Q 0 ) (B Q 0 ) ). (1.5) B Q B Q (B Q 0 ) (B Q 0 ). (1.6) Now, we will do the same construction with the product sets. Claim 7. We assume and let Then B < B 1+δ, (1.7) P {Q P 1 : (B Q)(B Q 0 ) < B δ 1+(1+ɛ)δ }. (1.8) In particular, if B + B < B 1+δ, then Proof. The same as Claim 4. P > P 1 B 1 δ 1 ɛδ. P 0 P 1 P B 1 δ 1. (1.9) Fix Q P, and let Y be a maximal subset of B Q, such that the sets y (B Q 0 ) for y Y are all disjoint.

8 8 MEI-CHU CHANG Claim 8. Y < B (1+ɛ)δ. Proof. The same as Claim 5. Remark 8.1. Here we use the fact that B Q 0 = y (B Q 0 ). This property holds for V in Theorem 3, because (y ) 1 exists. Since our results hold for the case when multiplication is not commutative, we will use the notation of inverse set. Notation. A 1 = {a 1 : a A} Claim 9. For z B Q, we define C z = B Q z (B Q 0 )(B Q 0 ) 1. Then there is a point z Q B Q, such that Proof. The same as Claim 6. Remark 9.1. Denoting C z Q C z Q > B δ 1 3(1+ɛ)δ. (1.30) by C Q, we have C Q = B Q z (B Q 0 )(B Q 0 ) 1, (1.31) and C 1 Q C Q (B Q 0 )(B Q 0 ) 1 (B Q 0 )(B Q 0 ) 1. (1.3) Moreover, Remark 6.1 and (1.31) give C Q C Q (B Q 0 ) (B Q 0 ). (1.33) Proof of Theorem. For any Q P and for any z z C Q, (1.33) implies that there are z 1,..., z 4 B Q 0, such that and (1.3) implies that there are z 5,..., z 8, such that z z = z 1 + z z 3 z 4, (1.34) z 1 z = z 5 z 1 6 z 7 z 1 8. (1.35) Therefore, given (z 1,..., z 8 ) (B Q 0 ) (B Q 0 ), (1.34) and (1.35) determine z, z uniquely. z = (z 1 + z z 3 z 4 )(z 5 z 1 6 z 7 z 1 8 1) 1 z = (z 1 + z z 3 z 4 ) ( (z 5 z 1 6 z 7 z 1 8 1) ). (1.36) Namely, (1.34) and (1.36) define a map on (B Q 0 ) (B Q 0 ) whose image contains C Q C Q. Thus, we have B Q 0 8 Q P C Q B 1+δ1 6(1+ɛ)δ. (1.37) The last inequality follows from (1.9) and (1.30). Now, the theorem follows from (1.5) and (1.37).

9 Remark. To prove Theorem 3, we observe that ERDÖS-SZEMERÉDI THEOREM 9 z(z 5 z 1 6 z 7z 1 8 1) = z 1 + z z 3 z 4 V \ {0}. Hence (z 5 z 1 6 z 7z 1 8 1) 1 exists for z 5,, z 8 B Q 0. Section. The General Case. In this section we will use Theorem to prove Theorem 1. First, according to (1.1), we may assume that We define, for any k N, A { z : 1 < z }. A k = {z A : k 1 < z k }. (.1) Claim 1. There is a set K N such that for any k K, we have where log A log A. A k > Proof. For l = 1,, log A, we define A K log A, (.) K l = {k l 1 A k < l }. (.3) Therefore, and there is l such that Let K = K l. Then for any k K, l K l > A, l l K l > A log A. A k l 1 > A K log A.

10 10 MEI-CHU CHANG Claim. To estimate A k and A k, we may assume that A k is contained in one of the quadrants, and A A k > 3 K log A. (.4.) Furthermore, for any z 1, z A k, z 1 + z max{ z 1, z }. (.5) Proof. One of the four quadrants contains at least 1 4 A k elements of A k. To see (.5), we note that (Rz 1 + Rz ) + (Iz 1 + Iz ) (Rzi + Iz i ), because z 1, z are in the same quadrant implies that Rz 1 + Rz max{ Rz 1, Rz } and Iz 1 + Iz max{ Iz 1, Iz }. Claim 3. Every element of A lies in at most two of the sets A k, and every element of A lies in exactly one of the sets A k. Proof. Let z 1, z A k. Then and k 1 < z 1 + z k+1, (.6) 4 k 1 < z 1 z 4 k. (.7) The first inequality in (.6) follows from (.5). Let z be an element with norm between k 1 and k for some k. Then z can only be in A k 1 and A k. Claim 4. Let δ be the absolute constant given in Theorem. Then A + A > Proof. Claim 3 gives the following inequalities. A > k K A > A 1+δ 4+3δ K δ. (.8) (log A ) 1+δ A k 1 A k, (.9) k K k K A k k K A k. (.10) Adding (.9) and (.10) together, and applying Theorem to A k, we have A + A > 1 A k 1+δ > k K A 1+δ 4+3δ K δ. (.11) (log A ) 1+δ

11 The last inequality follows from (.4). We divide the proof into two cases. Case 1. K < A 1 +ɛ. Then from (.11) Case. K > A 1 +ɛ. ERDÖS-SZEMERÉDI THEOREM 11 A + A A 1+δ ( 1 +ɛ)δ (log A ) 1+δ A 1+( 1 ɛ)δ. (.1) For each k K, we pick z k A k. We may assume that there is K 1 K with K 1 > 1 K such that 1 k 1 < Rz k < k, for all k K 1. (.13) Hence, A { z k + z k : k, k K 1 } { Rz k + Rz k : k, k K 1 } > 1 3 K 1 A 1+ɛ > A 1+( 1 ɛ)δ. (.14) The third inequality is by the same reasoning as that in Claim 1 in Section 1. The last inequality follows from our assumption (0.7). Finally, (.1), (.14), and (0.7) conclude the proof of the theorem. Remark 5. If A K, then in (.4), (.8), and (.11), 3 and 4+3δ are replaced by 5 and 6+5δ 1. Also in (.13), k 1 and K 1 > 1 1 K are replaced by 4 k 1 and K 1 > 1 4 K. For the normed space V, we have m+1, (m+1)(δ+1)+1 1, m k 1, 1 m K instead. Acknowledgement. The author would like to thank J. Stafney for helpful discussions. References [1]. J. Bourgain, N. Katz, and T. Tao, A sum-product estimate in finite fields, and applications, preprint. []. M. Chang, Erdös-Szemeredi problem on sum set and product set, Annals of Math. 157 (003),

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