INTRODUCTION TO SUM-PRODUCT PHENOMENON AND ARITHMETICAL APPLICATIONS

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1 INTRODUCTION TO SUM-PRODUCT PHENOMENON AND ARITHMETICAL APPLICATIONS JIYOU LI Abstract. This exploratory note is an introduction to additive combinatorics, with an emphasis on sum-product phenomenon and some arithmetical applications. Most results and proofs can be found in [3], [5], [44], [62]. 1. Introduction: From Additive Number Theory to Additive Combinatorics Through out this paper, Z always denotes the set of integers, Z + denotes the set of positive integers and [n] denotes the subset of integers {1, 2,, n}. G denotes an abelian group, R denotes a ring and F p denotes the prime field of p elements. A sequence a, a + d, a + 2d,..., a + (k 1)d is called an arithmetic progression of length k, and is denoted by k-ap for abbreviation. An AP has the simplest additive structure. The concept of arithmetical progression plays an important role in understanding the additive structure of an additive set. We will briefly review some important theorems in additive number theory. Theorem 1.1 (van der Waerden, 1927). For any positive integers k and r, if we put all the integers into r boxes, then there is at least one box containing a k-ap. There was a stronger conjecture: Conjecture 1.2 (Erdős and Turán). For any positive integers k and positive constant δ, there is an integer N = N(k, δ) such that any subset in [N] of cardinality greater that δn must contain a k-ap. Roth settled this conjecture in 1952 for k = 3. Theorem 1.3 (Roth, 1952, [52]). Let δ be a positive constant. For N large enough, any subset of cardinality greater that δn in [N] must contain a 3-AP. In this case much progress has been made. Theorem 1.4 (Roth, 1953, [53]). For some constant C, one can take δ = C log log N. Theorem 1.5 (Heath-Brown, 1987, [40]). There are constants C, c such that δ = C (log N) c. Szemerédi showed in 1990 that one can take c = 1/4 [59]. Bourgain showed that one can take c = 1/2 ɛ in 1999 [13] and c = 2/3 ɛ in 2008 [14]. The current best bound is c = 1 ɛ by Sanders in

2 2 JIYOU LI Theorem 1.6 (Tom Sanders, 2012, [63]). There is a constant C such that δ = C log log N log N. What is the smallest δ one can take? Behrend gave a construction showing that Theorem 1.7 (Behrend, 1945, [2]). There is a subset of size e c log N N containing no 3-APs. Note that this is a big gap. Szemerédi solved the Erdős-Turán Conjecture by ingenious and elementary methods. Theorem 1.8 (Szemerédi, 1975, [58]). Let δ > 0. For N large enough, any subset of cardinality greater that δn in [N] must contain a k-ap. Furstenberg [33] gave an ergodic proof for the above theorem in Gowers gave another proof by Fourier analysis in C Theorem 1.9 (Gowers, 2001, [34, 35, 36]). Let δ = log log N, where C is a constant. For N large enough, any subset of cardinality greater that δn in [N] must contain a k-ap. Gowers proof combines new ideas. He defined Gowers norm for uniformity and developed higher order Fourier analysis. All of the above results and methods, together with some new ideas, bourn into a breakthrough. Theorem 1.10 (Green-Tao, 2004, [37]). For any k, there is a k-ap in the set of prime numbers. In the last two theorems, additive combinatorics plays an important role. The set {a + a 1 d 1 + a 2 d a k d k, 1 a i N i, 1 i k} is called a generalized arithmetic progression of dimension k and of order N 1 N 2... N k, and is denoted by GAP for short. For two sets A, B Z, define the sumset of A and B and the product set of A and B A + B = {a + b, a A, b B}, AB = {ab, a A, b B}. Example If A Z, A + A = 2 A 1, then A is an AP. Interestingly, if A + A 3 A 3, one can show that A must be contained in an AP(see [50]). How about the general case, see if A + A C A for some constant C? Theorem 1.12 (Freiman, [50])). If A + A C A, then A is contained in a generalized progression of dimension d = d(c) and size s = s(c). Roughly speaking, Freiman s theorem says that any set with a small sumset is close to an generalized arithmetic progression. Similarly in a commutative ring R, a set with a small product set is close to a geometric progression. Can a set in R be simultaneously closed to both an arithmetic and a geometric progression? It sounds not possible, at least for real numbers, as showed by Erdős and Szemeredi: either the sumset or the product set must be large.

3 LECTURE NOTES ON GAUSS SUMS AND ARITHMETICAL APPLICATIONS 3 Theorem 1.13 (Erdős and Szemerédi, 1983, [32]). A R, then there is an absolute constant ɛ > 0 such that max{ A + A, AA } A 1+ɛ. Currently the best bound is due to Solynomsi. Theorem 1.14 (Solymosi, 2008, [57]). A R, max{ A + A, AA } A 4/3. Question Can we replace R by a finite ring? Obviously, if A is a subring of R, then it follows that A + A = A and AA = A. Furthermore, if A is close to a sub ring of R, then A + A A and AA A. Since F p has no nontrivial subring, we may expect there should be a sum-product phenomenon in F p, which means no subset A in F p is roughly closed under both addition and multiplication. This leads to many striking results in number theory and applications in theoretical computer science. Theorem 1.16 (Bourgain-Katz-Tao, 2004, [18]). Let A F p with p δ < A < p 1 δ. Then there is an absolute constant ɛ = ɛ(δ) > 0 such that max{ A + A, AA } A 1+ɛ. Bourgain-Glibchuk-Konyagin removed the condition p δ < A and obtained Theorem 1.17 (Bourgain-Glibchuk-Konyagin, 2006, [17]). Let A F p with A < p 1 δ. Then there is an absolute constant ɛ = ɛ(δ) > 0 such that max{ A + A, AA } A 1+ɛ. Remarks This constant ɛ can be computed explicitly. Much progress were made towards a large ɛ. The bound ɛ = 1 11 was given by Rudnev [54] and in [48] the authors gave a slightly improvement and generalization. The current best record is ɛ = 1 5 by Rudnev [55]. Theorem 1.18 (M. Z. Garev; Katz-Shen, [44]). There is a constant C such that for any A F p, max{ A + A, AA } C min{ A 14 13, p 1 12 A }. 2. Tools from Additive Combinatorics 2.1. Basic Terminology. Let G be an abelian group and R be a commutative ring. For two sets A, B G, define the sumset of A and B A + B = {a + b, a A, b B}, and ka = A + A + + A. }{{} k We also denotes k A = {ka, a A}. Similarly, in a ring we can also define and AA = {ab, a A, b B}, A k = AA }{{... A }. k

4 4 JIYOU LI Define A A = {a + b, a A, b B, a b} and k A = A A A. }{{} k Example 2.1. Take A = {1, 3, 5, 7}, then A + A = {2, 4, 6, 8, 10, 12, 14}. Example 2.2. Take A = {1, 3, 9, 27}, then A+A = {2, 4, 6, 10, 12, 18, 28, 30, 36, 54}. Example 2.3 (Goldbach Conjecture). Let P be the set of all odd prime numbers, then P + P = 2 Z + {2, 4}. Example 2.4 (Lagrange Theorem). Let S be the set of all squares, then S + S + S + S = Z + {0}. Example 2.5 (Waring s Problem). Let S k be the set of all k-th powers, then what is the minimal integer m such that ms k = Z + {0}? These problems show that the questions such as understanding the structure of A + A and the converse problem, understanding the structure of A if we have some information on A + A, are fundamental problems in number theory Some Basic Results. In this note we will focus on the finite group case. From now we assume all sets are finite and let A denote the cardinality of A. A B means there is a constant α such that A α B. We also define the additive energy E(A, B) of two sets A and B by E(A, B) = #{(a, b, c, d) A B A B, a + b = c + d}. We list some basic properties about the sumsets without proofs. Details of the proof can be found in [62]. Proposition 2.6. We have E(A, B) I. max A (x B) A B x A±B A B A ± B ; A + B 3 II. A B ; A B A C C B III. A B. C The last one means that if we define d(a, B) by A B d(a, B) = log, A 1 2 B 1 2 then d(a, B) satisfies the triangle inequality and thus can be viewed as a distance. Example 2.7. If we choose B = A and C = A in III of the Proposition 2.6, then A A A + A 2. A

5 LECTURE NOTES ON GAUSS SUMS AND ARITHMETICAL APPLICATIONS 5 Example 2.8 (Small energy implies large sumset). From I of the Proposition 2.6, Example 2.9. In particular, E(A, B) A 2 B 2 A ± B. E(A, A) A 4 A ± A. It is obvious that A+A = A implies that A must be a subgroup. And similarly A + B = A implies that B is contained in a subgroup H of G and A is the union of cosets of H (Left as an exercise). Just like the Freiman theorem, can we say anything about the structure of A and B if A + B is small compare to A? If A + B α A, then what about A + B + B + + B? The following useful theorem is a partial result on this direction and plays an important role in additive combinatorics. Theorem 2.10 (Plünnecke-Ruzsa inequality). If A, B satisfy A + B α A, then there exists A 1 A such that A 1 + kb α k A 1. Outline of the proof. Step 1 : Define the magnification ratio G of a bipartite graph G(A, B) by N(X) G = min, X A,A X where N(X) is the neighborhood of X. Step 2: For graphs G : A B and H : B C, define the composition H G : A C of G and H naturally. One checks that H G H G. (2.1) Step 3: Define the commutativity of two graphs and the concept of Plünnecke graphs. G : A B and H : B C are commutative if for any edge a a + b, a + b a + b + c, there are two edges a a + c a + b + c. A sequence of graphs (G 0, G 1,, G k ) is called a Plunnecke graph of order k + 1 if G i and G i+1 are commutative for all 0 i k 1. Step 4: (Most interesting part) Using Menger Theorem to prove: If then for all 1 i k, G k G k 1 G 1 1, G i G i 1 G 1 1. We can easily verify that this is indeed a statement in graph theory. Step 5: Using the technique of tensor product of graphs to prove that the statement in Step 4 is equivalent to Plünnecke Theorem, which states that the reverse inequality of 2.1 still holds in some sense for commutative graphs: If (G 1, G 2,, G k ) is a Plünnecke graph of order k. Then the sequence of magnification ratios G i G i 1 G 1 1 i, 1 i k are decreasing. For instance, and G 2 G 1 G 1 2, G k G k 1 G 1 G 1 k.

6 6 JIYOU LI Step 6: Define the sumset graph (G 1, G 2,, G k ) of order k by G 1 : A A + B,..., G i : A + (i 1)B A + ib, 1 i k 1. One checks that it is a Plünnecke graph of order k. The magnification ratio for this sumset graph is exactly what we want since for sumset graph (G 1, G 2,, G k ) of order k, G 1 = N(X) A + B min α, X A,X X A and thus G k G k 1 G 1 α k. It then follows by the definition G k G k 1 G 1 = there is a subset X in A such that N(X) min = min X A,X X X A,X X + B + B + + B α k X. X + B + B + + B, X Theorem 2.11 (Plunnecke-Ruzsa inequality: Distinct Summands). Let X, A i G satisfy X + A i α i X, then there is a X 1 X with X 1 + A 1 + A A k α 1 α 2 α k X 1. Proof. The proof uses techniques of tensor product. Corollary A 1 + A A k X + A 1 X + A 2 X + A k X k 1. Setting all the sets A 1, A 2,, A k, X to be A we get a useful corollary: if A has small doubling constant, then ka will be small. Corollary If A + A = α A, then Similarly, ka α k A. Corollary 2.14 (Ruzsa). If A + A = α A, then ma na α m+n A. Corollary There exists X 2 X, X 2 > 1 2 X such that X 2 + A 1 + A A k X + A 1 X + A 2 X + A k X k 1. Remark. Here the constant 1 2 can be replaced by 1 ɛ Basic Sum-Product Theorem Over F p. We first state the sum product theorem for prime fields. Theorem 2.16 (BKT, [19] BGK, [17]). For every ɛ > 0, there exists a δ = δ(ɛ) > 0 and a constant c = c(ɛ) such that if A F p and A > p ɛ, then A + A + AA > c A 1+δ. We also have the following quantitative statement.

7 LECTURE NOTES ON GAUSS SUMS AND ARITHMETICAL APPLICATIONS 7 Theorem 2.17 (Ga, KaS). A + A + AA min { } A , p 11 A 12. Define the multiplicative energy E (A, B) of two sets A and B by E (A, B) = #{(a, b, c, d) A B A B, ab = cd}. Theorem 2.17 then follows directly by Lemma E (A, A) 4 A + A 9 A p A + A 8 A 5. Outline of the proof: energy estimate by Plünnecke inequality. Step 1: Since E (A, A) = aa ba, a A it follows by the principle of pigeonhole that there is a subset A 1 in A with almost the same size as A and a positive integer N such that b A A 1 N E (A, A). A Step 2: Assume A1 A1 A 1 A 1 = F p, by the principle of pigeonhole again there is x = a1 a2 a 3 a 4 A 1 such that p = A 1 A 1 A 1 4 A 1 A 1 Definition E(xA 1, A 1 ) = A 1 2 xa 1 2 E(xA 1, A 1 ) Thus Example 2.8 xa 1 + A 1 = (a 1 a 2 )A 1 + (a 3 a 4 )A 1 a 1 A 1 a 2 A 1 + a 3 A 1 a 4 A 1 4 P-R A 3 (a i A ± b 0 A) i=1 Triangle A 3 Step 3: For the case A1 A1 A 1 3 A 1+A 1 4 A + A 2. N i=1 E (A, A) 4 1 p A + A 8 A 5. A 1 A 1 F p, the proof is similar. What we need to do is to replace p by in the first inequality of Step 2. Corollary 2.15 guarantees that we can give a lower bound on a 1 A 1 a 2 A 1 + a 3 A 1 a 4 A 1 for some a i A 1. Thus we also get E (A, A) 4 A + A 9 A 2. Roughly speaking, Balog-Szemeredi-Gowers Theorem asserts if E(A, B) A 3 2 B 3 2, then there is A 1 A, B 1 B and A 1 A, B 1 B such that A 1 + B 1 A B1 1 2.

8 8 JIYOU LI Theorem 2.19 (Balog-Szemeredi-Gowers Theorem: Special case, Large energy implies small sumset). Let G be an additive group. There is an absolute constant c such that the following holds. For any A G satisfying there is a subset A 1 in A satisfying and E + (A, A) > c 1 A 3, A 1 > c k A A 1 + A 1 < c k A Generalizations: Polynomials and Expanders. Theorem 2.20 (Bourgain, [11]). For every ɛ > 0, there exists a δ > 0 such that if A, B, C F p and A > B, C > p ɛ, then A + B + AC > c A 1+δ. Based on this theorem and the Szemerédi-Trotter theorem in finite fields, Bourgain proved the function f(x, y) = x(x + y) from F p F p F p satisfies for some α if A B p α. polynomial is an expander. f(a, B) p α+ɛ It is then an challenging task to determine if a 2.5. Some Quantitative Results. In general, few quantitative results are known in additive combinatorics. We just list some basic ones. Theorem 2.21 (Cauchy and Davenport). For any A, B F p, we have A + B min{p, A + B 1}. Conjecture 2.22 (Erdős and Heilbronn). For any A F p, A A min{p, 2 A 3}. Theorem 2.23 (Dias and Hamidoune, 1994). For any A F p, A A A min{p, k A k }{{} 2 + 1}. k 3. Applications to Gauss sum 3.1. Historical results on Gauss sum. Let p be a prime number and F p be the prime field of order p. Denote e p (x) = e 2πix/p. Let f(x) be a polynomial with integer coefficients. We consider the exponential sum S f = e p (af(x)). x mod p We are particularly interested in the special case when f(x) = x m, called the Gauss period of order m, G(m) = e p (ax m ). x mod p If χ is a multiplicative character of order m, then we also define the Gauss sum G(χ) of order m by G(χ) = χ(x)e p (x). x mod p

9 LECTURE NOTES ON GAUSS SUMS AND ARITHMETICAL APPLICATIONS 9 The two kinds of Gauss sum are related by G(m) = x mod p (1 + χ(x) + + χ(x) m 1 )e p (x) = Since G(χ)G(χ) = χ( 1)p for nonprincipal character χ, and one then has G(χ) = p, G(m) (m 1) p. m 1 j=1 G(χ j ). For two multiplicative characters χ and φ, define the Jacobi sum by J(χ, φ) = x χ(x)φ(1 x) and define J(χ) = J(χ, χ). It is known that G(χ)m = ω(χ) where ω(χ) = χ( 1)p m 2 j=1 J(χ, χj ) Q(e 2πi/m ), and the central problem in evaluating G(χ) is to obtain a simple criterion for determining which m-th root of ω(χ) equals G(χ). Gauss settled the problem for quadratic sums (m = 2). This is the first remarkable square-root bound. Theorem 3.1 (Gauss). For quadratic character χ, { p, p 1(mod 4); G(χ) = G(2) = i p, p 3(mod 4). Remark 1 There are many proofs for this result: Gauss: Gauss coefficient; Dirichlet: Poisson summation formula; Cauchy: classical θ function, contour integration and Abel-Plana summation formula; Mertens: Trigonometric proofs; Algebraic proofs. Remark 2 This sum was used by Gauss to prove the quadratic reciprocity, one of the deepest result in number theory. The problem is much deeper for cubic and quartic Gauss sums and involves elliptic functions. For example, when p 1( mod 3), G(3) satisfies a cubic equation x 3 3px pr = 0 where integer r satisfies 4p = r b 2 and r 1(mod 3). For m 5 it is unsolved in general, although some progress has been made on sums of orders 5, 6, 8, 12, 16 and 24. For details about G(m) for small m, please refer to Berndt and Evans [3]. Remark: Generally we can define the Gauss sum by G(ψ, χ) = x χ(x)ψ(x), where ψ, χ are additive character and multiplicative character over F q respectively. Then Gauss sum is a bridge between them. χ(x) = 1 q 1 G(ψ, χ)ψ( x); ψ ψ(x) = 1 G(ψ, χ)χ 1 (x). q Theorem 3.2 (Davenport-Hasse). If both ψ and χ are not principal, then χ G(ψ, χ ) = ( 1) s 1 G(ψ, χ) [E:Fq], where E/F q is a finite field extension and ψ, χ are induced character respectively.

10 10 JIYOU LI Remark 3 The problem of evaluating the Gauss sum G(χ) is extremely deep only when q is a prime. Remark 4 Many generalizations: multi variables; incomplete sum; general polynomials and rational functions such as Kl. sums; special rings, etc. Remark 5 For a multiplicative character χ, the type of sum x χ(f(x)) is much harder. For instance, even for the case deg f(x) = 3, understanding the sum ( ) f(x) x F p as a function of p is intrinsically related with the problem of the modularity of elliptic curves, which is one of the deepest parts of Wiless proof of the Fermat Last Theorem. Motivations: Arithmetic applications; additive combinatorics; analytic number theory; coding theory; algebraic geometry; algebraic combinatoric; theoretical computer science. Theorem 3.3 (Hasse-Weil). Let f(x) be a nonconstant rational function. Suppose that all the pole orders d P of f(x) are not divisible by p and let m = (d P + 1) deg P 1. Then P :prime χ(f(x)) (m 1)q 1/2. x F q Theorem 3.4 (Hasse-Weil). Suppose f(x) F q [x] has degree m and is not of the form of g(x) p g(x). Then the exponential sum admits the estimate χ(f(x)) (m 1)q 1/2. Furthermore, we have x F q Theorem 3.5 (Weil). Let f(x) be a nonconstant rational function. Suppose that all the pole orders d P of f(x) are not divisible by p and let m = (d P + 1) deg P 1. P :prime Suppose g(x) is a polynomial of degree n and (n, q) = 1. Then χ(f(x))ψ(g(x)) (m + n 1)q 1/2. x F q Thus for any a 0(mod p), the bound x F e p (ax m ) (m 1) p can be p viewed as the simplest case of the Weil bound. For polynomial cases, all above bounds are nontrivial only when the degrees satisfying m < p 1/2 ɛ. Heath-Brown, Konyagin and Shparlinski [39, 44] improved this restriction to m < p 2 3 ɛ. Precisely, they obtain the monomial exponential sum bound e p (ax m ) mp 1/2, m p 1/3 ; m 5/8 p 5/8, p 1/3 m p 1/2 ; x F p m 3/8 p 3/4, p 1/2 m p 2/3 ; p

11 LECTURE NOTES ON GAUSS SUMS AND ARITHMETICAL APPLICATIONS 11 for any integer a with p a. Define E (k) (A) = #{x 1 + x x k = y 1 + y y k, x i, y i A}. Generally for any A F p we have Theorem 3.6. If then max e p (ax) p ɛ A, x A a F p E (k) (A) = ( 1 p + O(p 2kɛ )) A 2k. Proof. By the circle method. Converse when A is a subgroup in F p, we have Theorem 3.7. If Then E (k) (H) p 1 2 δ H 2k, max a F p x H e p (ax) p δ k 2 H. Proof. Denote ω(z) = #{x 1 + x x k y 1 y 2 y k = z, x i, y i A} and note that Thus e p (ax) 4k2 = x H z ω(z) 2 = E (2k) (H). z e p (az)ω(z) 2k = 1 H z,h H e p (ahz)ω(z) 2k H 2k( ω(z) e p (ahz) ) 2k z h H Holder Inequality H 2k ( z ω(z)) 2k 1 z ω(z) h H e p (ahz) 2k H 4k2 4k z,z ω(z)ω(z )e p (azz ) Hadamard Inequality < H 4k2 δ. H 4k2 4k E (2k) (H) p

12 12 JIYOU LI Similarly for any integer k, we also have have max e p (ax) (pe(k) (H)/ H ) 1/2k, x H a F p and max e p (ax) p 1/2k 2 E (k) (H) 1/k2 H 1 2/k, x H a F p Proposition 3.8 (Konyagin and Shparlinski). For H < p 2/3, Proof. Stepanov s method. E (2) (H) H 5 2. Cochrane and Pinner [26] made explicit this bound to that e p (ax m ) mp 1/2, m 3p 1/3 ; λm 5/8 p 5/8, 3p 1/3 m < p 1/2 ; x F p λm 3/8 p 3/4, p 1/2 m < 1 3 p2/3 ; where λ can be chosen to be 2/ Konyagin released the gap to H p 1/ New results from sum-product estimate. When m is large, Bourgain and Konyagin [7, 17, 19] obtained a celebrated nontrivial bound for a large kind of subgroups. Let H be a subgroup of F p. Suppose H > p δ, then there exits a constant δ > 0 such that for any integer a with p a, x H e p (ax) < H 1 δ. (3.1) For instance, Bourgain and Garaev proved in [15] that if δ > 1/4, then one can take δ = o(1). Ideals of the proof: Additive combinatorics on the spectrum of the set H. Suppose H is a subgroup of F p, where H > p δ. Since H H = H and thus by the sum-product theorem H + H H 1+δ. Thus for any ɛ > 0, there is k = k(ɛ, δ) such that kh = H + H + + H > p 1 ɛ Introduction to Fourier Analysis. For a complex function f(x) from F p to C, define its Fourier transform by Define the L r norm of f(x) by and the l r norm of f(ξ) by ˆf(ξ) = x e p (ξx)f(x). f Lr = (E x f(x) r ) 1/r f lr = ( ξ f(ξ) r ) 1/r. The convolution of f and g is defined by f g(x) = 1 f(y)g(x y). p y

13 LECTURE NOTES ON GAUSS SUMS AND ARITHMETICAL APPLICATIONS 13 The support of a function f(x) is defined as Supp(f) = {x, f(x) 0}. It is direct to check Supp(1 A 1 B ) Supp(1 A ) + Supp(1 B ). Example 3.9. Let 1 A, 1 B be the normalized indicator function of A and B respectively. Then A + B = Supp(1 A 1 B ). Proposition Proposition 3.11 (Parsavel Equality). f g = ˆfĝ. p f L2 = f l2 ; Example Let 1 A, 1 B be the normalized indicator functions (thus both with a probabilistic measure) of A and B respectively. Then E(A, B) = p A 2 B 2 χ Fp 1 A (χ) 2 1 B (χ) 2. Proof. E(A, B) = A (x B) 2 x A+B = p 2 A 2 B 2 1 A 1 B (x) 2 x A+B = p 2 A 2 B 2 x F p 1 A 1 B (x) 2 = p A 2 B 2 χ Fp 1 A 1 B (χ) 2 = p A 2 B 2 χ Fp 1 A (χ) 2 1 B (χ) 2. Example E(A, A) = p A 4 χ Fp 1 A (χ) 4. Example E (k) (A) = p 2k 3 A 2k χ Fp 1 A (χ) 2k. Proof. E (2k) (A) = p 2k 2 A 2k 1 A 1 A 1 A (x) 2 x A+A+ +A = p 2k 2 A 2k x F p 1 A 1 A 1 A (x) 2 = p 2k 3 A 2k χ Fp 1 A 1 A 1 A (χ) 2

14 14 JIYOU LI = p 2k 3 A 2k χ Fp 1 A (χ) 2k. Theorem Suppose H is a subgroup of F p wit H > p δ. Define µ(x) to be the normalized indicator function on H. Then for all ɛ > 0, there is a k = k(ɛ, δ) such that max x µ(k) (x) < p 1+ɛ, where µ(x) = 1 H x H δ x, and µ (k) (x) is the k-convolution of µ(x) First proof of the main result (3.1). Proof 1. By Theorem 3.15 and we have E (2k) (H) p 2+2kɛ H 2k. The main result (4.1) then follows by Theorem 3.7. We can also give a direct proof avoiding Theorem 3.7. Proof 1. pf (2k) (0) = ξ ˆF = ξ ˆF < p ɛ. Since H is a subgroup, by the invariance we have x F e p (xξ) = x F 2k ˆf(ξ) H 2k x F e p (xξ) 2k e p (xhξ), h H H x F e p (xξ) 2k < H 2k p ɛ. This is equivalent to x F e p (xξ) < p δ/4k H and this completes the proof Second proof of the main result (3.1), from Notes of Bourgain. Let µ : F p {0, 1} be a probability measure. For any δ > 0 denote the spectrum δ by δ = {ξ F p, ˆµ(ξ) > p δ }. Lemma Corollary (x, y) δ δ x y 2δ > p 2δ δ 2. E + ( δ, δ ) > p 4δ 4 δ 2δ.

15 LECTURE NOTES ON GAUSS SUMS AND ARITHMETICAL APPLICATIONS 15 Corollary Let k > δ > 0. Then either or there is a subset δ such that 2δ > p k δ, > p Ck δ ; + < p Ck. Proposition 3.19 (Main Proposition). Suppose µ is a H-invariant measure (In our case, we usually take 1 H, the normalized indicator function of H). Denote δ = δ (µ). For any ρ, δ > 0, there is always a δ such that if δ {0}, then δ > p ρ. Proof. The statement certainly holds for ρ = α since µ is a H-invariant measure. Suppose we have established the statement for some fixed ρ < 1. We want to prove the statement for ρ + c min{ρ, 1 ρ}. Given a small enough δ, since the statement is true for ρ, there exists δ < δ such that δ {0} δ > p ρ. Also there exists δ < δ 2 such that δ {0} δ > p ρ. 2 By Corollary 3.18, for δ < k < ρ, either or there is a subset δ 2 δ > p k δ, 2 such that > p Ck δ > p ρ ck ; + < p Ck. In the first case we have done since δ > δ > p ρ+k. Our remaining task is then to prove δ > p ρ+c for the second case. By the sum-product Proposition 2.18, E ( ) p 3Ck p Define a new probability measure η on F p by η(x) = 1 µ µ ( x y ). One computes that It follows y ˆη(ξ) = 1 ˆµ(ξy) 2. y ˆη(1) = 1 ˆµ(y) 2 > p δ y δ (η) {0} δ (η) > p ρ.

16 16 JIYOU LI Define ω( δ ) = #{(y, ξ), ξy δ }, where = δ (η). By a double counting argument, on one hand ω( δ ) = #{(y, ξ), ξy δ } δ 1 2 E (, ) 1 2 δ 1 2 E ( ) 1 4 E ( ) 1 4 δ 1 2 E ( ) On the other hand, ω( δ ) = z δ #{(y, ξ), ξy = z} = z δ ω(z) > z δ ω(z) ˆµ(z) 2 = z ω(z) ˆµ(z) 2 > p δ p 2δ > (1 p δ )p δ. z δ ω(z) ˆµ(z) 2 Finally we have 1 2 p δ < δ 1 2 E ( ) and it then follows that either 4, or This completes the proof. δ > p 9 8 ρ 8kC, δ > p ρ 8kC. Proof-2. Take ρ = 1 α 3, δ = α 4. Apply the Main Proposition, we have a δ and δ {0}, thus But we then have δ > p ρ. p ρ 2δ < ξ ˆµ(ξ) 2 = p x µ(x) 2 = p 1 α. This leads a contradiction.

17 LECTURE NOTES ON GAUSS SUMS AND ARITHMETICAL APPLICATIONS Applications to Waring s Problems over Finite Fields 4.1. History of Classical Waring s Problem. Question 4.1 (Edward Waring, 1770). Let S k be the set of all k-th powers. What is g(k), the smallest positive integer m such that ms k = Z + {0}? Equivalently, g(k) is the least positive integer m such that the equation x k 1 + x k m + + x k m = b always has a integral solution for each b Z +. Conjecture 4.2 (Waring, 1770). g(3) 9; g(4) 9; g(k). Theorem 4.3 (Lagrange, 1770). All positive integers are the sum of at most four integral squares. Equivalently, let S be the set of all squares, then 4S = Z + {0}. Example 4.4. Lagrange s Theorem implies g(2) 4. Since 7 can not be written to be a sum of 3 squares, we have g(2) = 4. Conjecture 4.5 (J. A. Euler, 1772). g(k) = 2 k + [1.5 k ] 2. Theorem 4.6 (Dickson, Pillai,..., Chen...). g(k) = 2 k + [1.5 k ] 2. if 2 k {1.5 k } + [1.5 k ] 2 k. Definition 4.7. Let G(k) denote the least number m such that Z + \ms k is finite. Equivalently, G(k) denotes the least number m such that every large natural number is the sum of at most m k-th powers of natural numbers. We know less about G(k) compare to g(k). Example 4.8 (All known bounds). G(2) = 4(Lagrange, 1770) G(3) 7(Linnik, 1943) G(4) = 16(Davenport, 1939) G(5) 17(Vaughan & Wooley, 1995) G(6) 24(Vaughan & Wooley, 1994) G(k) k(log k + log log k o(1))(wooley, 1992, 1995). G(k) O(k).(Maillet, Hurwitz, Hardy and Littlewood, 1930s).

18 18 JIYOU LI 4.2. Known Results of Waring s Problems over Prime Fields. Definition 4.9. Assume k p 1. Let γ(k, p) denote Waring s number (mod p), the smallest positive integer k such that every integer is a sum of m-th power of k integers (mod p). Definition 4.10 (Covering Number). For S F p, let γ(s, p) denote the least integer m such that ms = F p. Example Take H = {x k, x F p }. Then Theorem 4.12 (Cauchy, 1813). γ(k, p) = γ(h, p). γ(k, p) k. This bound is sharp for k = p 1 or k = p 1 2. Proof. By Theorem Generally, it follows from Theorem 2.21 that Theorem γ(a, p) p 1. A 1 Theorem 4.14 (Chowla, Mann and Strauss, [28], 1959). For k < p 1 2, γ(k, p) k Proof. Vosper s theorem, a generalization of Cauchy-Davenport when A has some very special structure. Example 4.15 (Some special values of γ(k, p)). γ(2, p) = 2; γ(3, p) = 2; γ(4, p) 3 for p > 7. Conjecture 4.16 (Heilbronn, [40]). For k < p 1 2, γ(k, p) k 1 2. For any positive ɛ and sufficiently small k(k p C ɛ ), γ(k, p) k ɛ. Theorem 4.17 (Heilbronn (1964), Bovey (1977))). Let t = p 1 k. Then Proof. Geometric lattice method. γ(k, p) C(t)k 1/φ(t). Remark: This bound is valid only for k p 1 log p. The exact size of C(t) is not known. Many bounds wre established. Theorem 4.18 (Some progress). I. Chowla [27] showed in 1943 that if k < p/3; γ(k, p) k 0.88

19 LECTURE NOTES ON GAUSS SUMS AND ARITHMETICAL APPLICATIONS 19 and II. Dodson [29, 30] sharpened this to γ(k, p) k 7/8 for sufficiently lare k; III. Dodson and Tietäväinen [31] obtained that γ(k, p) k 1/2+ɛ The second Heilbronn Conjecture was proven by Konyagin (1992) and the first was proved by Cipra, Pinner, Cochrane (2007). Theorem 4.19 (Konyagin, [43]). if p k log k(log log k + 1) 1+ɛ γ(k, p) (log k) 2+ɛ Proof. Let φ(h) = max a 0 x H e p(ax). By the circle method one has log p γ(k, p). log p log k log(φ(h)) Conjecture 4.20 (Montgomery, Vaughan, Wooley, 1995). φ(h) p/k log(kp). Conjecture 4.21 (Companion Conjecture). If p k log p, then log p γ(k, p). log p log k log log p He also conjectured that the exponent of the logatirhm can be reduced to 1 + ɛ. Theorem 4.22 (Ciper, Cocharane and Pinner, 2008 [22]). For any ɛ > 0 there is a constant c(ɛ) such that if φ(t) 1/ɛ then γ(k, p) c(ɛ)k ɛ, where t = (p 1)/k and φ is the Euler s totient function. By the bound (4.1) of Bourgain and Konyagin, and by a similar argument of Konyagin and Shparlinski [44], one can easily get Proposition There is an absolute constant C > 0 such that for m < p 1 δ, γ(m, p) C 1/δ Subset version of Waring s number. We now consider a stronger version of Waring s number, namely, the distinct or subset version of Waring s number. Definition Let γ (k, p) denote the distinct Waring s number (mod p), the smallest positive integer n such that every integer is a sum of k-th power of n distinct elements (mod p). We notice the difference between the two Waring s numbers γ(k, p) and γ (k, p). For example, γ (k, p) does not exist if (k, p 1) is large enough. Theorem 4.25 (L-). There is a constant ɛ(δ) > 0 such that for any prime p and any m < p 1 δ, if ɛ 1 < (e 1)p δ ɛ, then we have γ (m, p) < ɛ 1.

20 20 JIYOU LI Proof. Li-Wan s sieve ([46, 47])+(3.1)+some combinatorial arguments. Obviously γ(m, p) γ (m, p) and thus this bound implies Corollary 4.23, the known constant bound for ordinary Waring s number Odlyzko-Stanley Counting Problem. This topic is close related to the Waring s problem. Theorem 4.26 (A.M. Odlyzko and R.P. Stanley, 1978, [51]). Let Nm(b) be the number of subsets S Z p such that x m = b. Then x S N m (b) 2p 1 exp O(m p log p). p Theorem 4.27 (Konyagin and Shparlinski, [44]). Let N m(b) be the number of subsets S Z p such that x S xm = b. Then N m (b) 2p 1 p exp O(mp 1/2 log p), m p 1/3 ; exp O(m 5/8 p 5/8 log p), p 1/3 m p 1/2 ; exp O(m 3/8 p 3/4 log p), p 1/2 m p 2/3. Theorem 4.28 (Zhu-Wan, 2012, [65]). Let Nm(k, b) be the number of k-subsets S Z p such that x S xm = b. Then N m (k, b) (q 1) ( k m q + k + q ). q p k When q = p is a prime number, one gets One then has N m (k, b) (p 1) k (m p + k)k. p N m (b) 2q 1 4p exp (m q + q/p) log q. q 2πq Theorem 4.29 (Bourgain, [6]). Let N m(b) be the number of subsets S F p such that x S xm = b. If m < p 1 δ, then there is a constant 0 < ɛ = ɛ(δ) < δ such that N m(b) p 1 2 p 1 e O(p1 ɛ). (4.1) Theorem 4.30 (L-, [45]). Let Nm(k, b) be the number of k-subsets S F p such that x S xm = b. If m < p 1 δ, then there is a constant 0 < ɛ = ɛ(δ) < δ such that ( ) ( p 1 p N m(k, b) p 1 1 ɛ ) + mk m. (4.2) k k

21 LECTURE NOTES ON GAUSS SUMS AND ARITHMETICAL APPLICATIONS Further Applications of Sum-Product Phenomenon 5.1. Review for Gauss Sums. Theorem 5.1. e p (ax m ) (m 1) p. x F p Theorem 5.2 (Heath-Brown, Konyagin, Shparlinski). For m p 2 3, e p (ax m ) p 1 ɛ. x F p Theorem 5.3 (Bourgain, Konyagin, 2003). For m p 1 δ, e p (ax m ) p 1 ɛ. x F p Conjecture 5.4 (Montgomery, Vaugham and Wooley). max e p (ay) min{p , C(log p) 2 H 2 }. a 0 y H 5.2. Mordell type exponential sum. Let f(x) = m i=1 a ix i F p [x], such that Then (a i, p 1) < p 1 δ, 1 i d; (a i a j, p 1) < p 1 δ, 1 i j d. x e p (f(x)) p 1 ɛ Multilinear Exponential Sums. Let k = 2 t 4 and A i F p, 1 i k.suppose k A i > p ( k 2 )0.968, then there is δ = δ(k) such that x i A i,1 i k i=1 e p (x 1 x 2 x k ) p δ k A i Theoretical Computer Science and Cryptography. 1. Cayley Graphs: Let G = Sl 2 (F p ), S G be a subset. Define the Cay(G; S) to be the graph as follow: the set of vertices is G and there is an arrow from g 1 to g 2 if for some s S, g 2 = g 1 s. Let λ(g; S) be the second largest eigenvalue of Cay(G; S). We say Cay(G; S) is an ɛ-expander if λ(g; S) (1 ɛ) S. If Cay(G; S) is an ɛ-expander, we have d(g; S) log G. This follows by a well known result: Theorem 5.5 (Fan Chung). For a directed k-regular graph G of order n, log n 1 d(g) log k, λ where λ is the second largest eigenvalue of G. i=1

22 22 JIYOU LI Theorem 5.6 (Selberg, Lubotzky, Sarnak and Margulis,1988, [49]). For some constructions S G, Cay(G; S) is an expander. Theorem 5.7 (Helfgott, 2008, [41]). For all S G, S is a generator of G, d(g; S) poly(log G ). Theorem 5.8 (Bourgain and Gamburd, 2008 [15]). For random generating set S G, Cay(G; S) is an expander with high probability. 2. Randomness Extractors; Theorem 5.9 (Barak, Impagliazzo and Wigderson, 2006, [1]). For each δ > 0 there is an explicit construction for extracting randomness from a constant (depending polynomially on 1 δ ) number of distributions over {0, 1}n, each having min-entropy δn. These extractors output n bits that are 2 n close to the uniform distribution. Remark This is the first deterministic extractors from a constant number of weak sources whose entropy rate is less than 1/2. Proof. Additive number theory+sum product theorem. 3. Diffie-Hellman Key exchange protocal. Theorem 5.10 (Bourgain, [6, 9, 12]). Let θ F p of order t > p ɛ. Then for t 1 t, there is δ = δ(ɛ) t 1 max e p (aθ i ) t1 p δ. a 0 i=1 Corollary 5.11 (Bourgain, [6, 9, 12]). Let θ F p of order t > p ɛ. Then t t max e p (aθ i + bθ j + cθ ij ) t 2 p δ. (a,b,c,p)=1 i=1 j=1 It then follows that the Diffie-Hellman triples (θ a, θ b, θ ab ) has very small discrepancy. This means the binary strings of a random Diffie-Hellman triple has extremely small distance from the uniform distribution. Acknowledgements. The author would like to thank the kind hospitality of SMS, PKU and BICMR, and especially Professor Rongquan Feng. The author would also like to thank Professor Keqin Feng and Professor Daqing Wan for helpful discussions. References [1] B. Barak, R. Impagliazzo and A. Wigderson, Extracting randomness using few independent sources, SIAM J. Comput. 36 (2006), no. 4, [2] F. Behrend, On sets of integers which contain no three terms in arithmetical progression, Proc. Nat. Acad. Sci. U. S. A. 32, (1946) [3] B.C. Berndt and R.J. Evans, The determination of Gauss sums, Bull. Amer. Math. Soc. (N.S.) 5 (1981), [4] B.C. Berndt, R.J. Evans and K.S. Williams, Gauss and Jacobi sums, Canadian Mathematical Society Series of Monographs and Advanced Texts, John Wiley & Sons, Inc., New York, [5] J. Bourgain, Sum-Product Theorems and Applications, Lecture Notes, [6] J. Bourgain, Estimates on exponential sums related to the Diffie-Hellman distributions, Geom. Funct. Anal. 15 (2005) [7] J. Bourgain, Mordell type exponential sum estimates in fields of prime order, C. R. Math. Acad. Sci. Paris 339 (2004)

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