Large Sieves and Exponential Sums Liangyi Zhao Thesis Director: Henryk Iwaniec
The large sieve was first intruded by Yuri Vladimirovich Linnik in 1941 and have been later refined by many, including Rényi, Barban, Davenport-Halberstam, Selberg, Bombieri, Gallagher, Hàlasz, Montgomery, Huxley, Elliott, Vaughan, etc.. The theorem, in the following form, was first introduced by Davenport and Halberstam. For x R, x = inf x z denotes the distance of a real number z Z to the integers. Given 1 > δ > 0, we say that, for a finite sequence of real numbers, {x k } is δ-spaced mod 1, if x k x l δ 2 for all x k x l. 1
Theorem 1 (Classical Large Sieve Inequality). Let {a n } be an arbitrary sequence of complex numbers and {x k } be a sequence of real numbers that are δ-spaced modulo 1. Also suppose N N and M Z. Then we have (1) k a n e(x k n) 2 (δ 1 + N) where the implied constant is absolute. a n 2, Save for the computation of the implied constant, the above inequality is the best possible. The best implied constant is 1, as Cohen and Selberg showed independently that k a n e(x k n) 2 (δ 1 1 + N) a n 2.
Theorem 1 admits the following corollaries. Corollary 1. Let {a n }, M, N be defined as in Theorem 1 and Q N. Then we have (2) Q q=1 a mod q gcd(a,q)=1 a n e ( a q n ) 2 (Q 2 + N) a n 2, where the implied constant is absolute. Corollary 2. Let {a n }, Q, M, N be defined as in Corollary 1. Then we have (3) Q q=1 q ϕ(q) χ mod q a n χ(n) 2 (Q 2 + N) a n 2, where the sum over primitive characters modulo q is denoted as. χ (mod q)
We are interested in having an estimate of the following kind: (4) Q q=1 a mod q 2 gcd(a,q)=1 a n e ( a q 2n ) 2 a n 2, where will depend on N and Q. Id est, we want to have a large sieve type inequality for square moduli. As one can tell from the statement of the classical large sieve inequality, the spacing properties of {x k } matter greatly. We need a result concerning the spacing properties of Farey fractions with square denominators.
It is a remark attributed to Borel that rational numbers in the real line are like stars in the heavens to illuminate the mystery of the continuum. Indeed, we must investigate the well-spacedness of some of these stars. Set S Q = a q 2 Q gcd(a, q) = 1, 1 a < q2, Q < q 2Q. If x x are two elements of S Q, then x x Q 4. By classical large sieve inequality, we may take, in (4), (5) = Q 4 + N.
a On the other hand, q 2 in S Q with a fixed denominator q 2 are clearly q 2 -spaced. Again by the virtue of the classical large sieve inequality, we may also take in (4) after summing over q, (6) = Q(Q 2 + N). When N Q 3, both (5) and (6) can be interpreted as Q 4. Neither (5) nor (6) exploits the fact that squares are so sparsely distributed among the integers. One deduces that there are about Q 3 rational numbers between 0 and 1 with square denominators and height at most Q 2. Hence, these rational numbers are on average Q 3 -spaced. So we should be able to do better.
We must investigate the spacing properties of the elements in S Q. Set (7) M(Q, N) = max x S Q { x S Q x x < 1 2N }, and we have Lemma 1. Given ǫ > 0 and N N, then we have the following lemma. (8) M(Q, N) Q3 N + ( Q + Q2 N ) N ǫ, where the implied constant in (8) depends on ǫ alone. Proof. Majorize M(Q, N) by an incomplete Gauss Sum, complete the sum via Fourier means, and then estimate using Weyl shift.
With Lemma 1 at our disposal, we can state the following. Theorem 2. With {a n }, Q, M, and N defined as before, we have (9) Q q=1 a mod q 2 gcd(a,q)=1 a n e log(2q) ( a q 2n ) 2 [ Q 3 + (N Q + ] NQ 2 )N ǫ a n 2, where the implied constant depends on ǫ alone. Proof. Apply duality principle and open the sum above and apply Lemma 1.
For N Q 3, our result gives the majorant of O(Q 7 2 +ǫ ) while both (5) and (6) give the majorant of O(Q 4 ). We also have corollary for multiplicative characters. Corollary 3. For any sequence of complex numbers {a n }, we have (10) Q q=1 q ϕ(q) χ mod q 2 log(2q) a n χ(n) 2 [ Q 3 + (N Q + ] NQ 2 )N ǫ a n 2, where the implied constant depends on ǫ alone. Proof. Use Gauss sums.
The ideas large sieve type inequalities with square moduli is generalizable to higher power moduli. The result of Weyl shift weakens as the polynomial in the amplitude is of high degree, we expect that our corresponding result also weakens. As before, let S Q,k = { a q k Q gcd(a, q) = 1, 1 a < qk Q k, Q < q 2Q }. We easily see that if x and x are two distinct elements of S Q, then x x 1 Q2k. Therefore, we may take (11) = Q 2k + N.
We may also take (12) = Q(Q k + N). As before, we set (13) M k (Q, N) = max x S Q,k { x S Q,k x x < 1 2N }, then we have Lemma 2. Given N N and set κ = 2 k 1, then we have (14) M k (Q, N) Qk+1 N + Q κ 1 κ + Qκ+k κ N 1 κ N ǫ, where the implied constant in (14) depends on ǫ and k. Proof. The result arrives similarly as before.
Given the result of Lemma 2, we have the following theorem. Theorem 3. With {a n }, Q, M, and N defined as before, we have (15) Q q=1 a mod q k gcd(a,q)=1 log(2q) a n e ( a q kn [ Q k+1 + (NQ κ 1 κ ) 2 + N 1 1 κq κ+k ] κ )N ǫ where the implied constant depends on ǫ and k. a n 2, Proof. The result arrives similarly as before.
For N Q k+1, our result gives the majorant of O(Q k+1+κ 1 κ +ǫ ) while (11) gives the extremely poor majorant of O(Q 2k ) and (12) give the majorant of O(Q k+2 ). Also, we have the following corollary. Corollary 4. For any sequence of complex numbers {a n }, we have (16) Q q=1 q ϕ(q) χ mod q k log(2q) a n χ(n) [ Q k+1 + (NQ κ 1 κ 2 + N 1 1 κq κ+k ] κ )N ǫ where the implied constant depends on ǫ and k. Proof. Use Gauss sums. a n 2,
Question: Are the result thus far the best possible? My Answer: Probably Not.
If we take N = Q 3 in Lemma 1, we have the following (17) M(Q) def x = max { x SQ x < (2Q) 3} 1 Q2 +ǫ, x S Q and the implied constant in (17) depends on ǫ alone. With this, we can arrive at something a slightly different. Proposition 1. With {a n }, Q, M, and N defined as before, we have (18) Q q=1 a mod q 2 gcd(a,q)=1 a n e ( a q 2n ) 2 Q 1 2 +ǫ (Q 3 + N) where the implied constant depends on ǫ alone. Proof. Similarly as before, but need to partition (0,1]. a n 2,
For higher power moduli, we have the following. Proposition 2. With {a n }, Q, M, and N defined as before, we have (19) Q q=1 a mod q k gcd(a,q)=1 a n e ( a q kn ) 2 Q κ 1 κ +ǫ (Q k+1 +N) where the implied constant depends on ǫ and k. a n 2, Proof. The proof of the present proposition goes analogously as that of the previous proposition. Moreover, both Propositions above admit obvious corollaries to multiplicative characters.
Q M(Q) Q M(Q) Q M(Q) Q M(Q) Q M(Q) 1 0 2 0 3 1 4 1 5 2 6 1 7 1 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 3 25 2 26 2 27 3 28 2 29 2 30 2 31 2 32 2 33 2 34 2 35 2 36 2 37 3 38 3 39 3 40 3 41 3 42 3 43 3 44 3 45 2 46 3 47 3 48 3 49 3 50 3 51 3 52 3 53 4 54 3 55 3 56 3 57 3 58 3 59 3 60 3 61 3 62 3 63 3 64 3 65 3 66 3 67 3 68 3 69 3 70 3 71 3 72 3 73 3 74 3 75 3 76 3 77 3 78 3 79 3 80 3 81 3 82 3 83 4 84 4 85 3 86 3 87 3 88 3 89 4 90 4 91 4 92 4 93 4 94 3 95 3
With the heuristics considered before and computations, we make the following conjectures. Conjecture 1. Set { a S Q = q 2 Q gcd(a, q) = 1, 1 a < q 2, Q < q 2Q Then we have (20) max { x x SQ x < Q 3 } Q ǫ, x S Q where the implied constant depends on ǫ alone. }.
In the same spirit, we also believe that analogous conjectures could be made for higher power moduli. Conjecture 2. Let S k,q denotes the rational numbers between 0 and 1 with k-th power denominator and height not exceeding Q k. In other words, { } a S k,q = q k Q gcd(a, q) = 1, 1 a < q k, Q < q 2Q. Then we have (21) max { x x Sk,Q x < Q k 1 } Q ǫ, x S k,q where the implied constant depends only on k and ǫ.
If we assume the truth of Conjecture 2, then we would have the following. Conjecture 3. Let {a n } be an arbitrary sequence of complex numbers, Q, N N and M Z. We have (22) Q q=1 a mod q k gcd(a,q)=1 a n e ( a q kn ) 2 Q ǫ ( Q k+1 + N ) M+N where the implied constant depends on ǫ and k. a n 2, Of course, it goes without saying that Conjecture 3 admits corollary for multiplicative characters, assuming its truth.
Moreover, we doubt that any improvement upon our present result is attainable via the means estimating of incomplete Gauss sums. Indeed, if one is a true believer of the so-called principle of square rooting, then we should believe that the saving attained by any estimation of an exponential sum cannot be good enough to yield the majorant in Conjecture 2. Consequently, we believe that any improvement upon our present result must come from a completely different avenue.
Next we consider a large sieve type inequality for special characters to prime-square moduli. We want to have a result of the following kind. (23) Q q=1 1 ϕ(q) χ mod q 2 a n χ(n) 2 a n 2, where the is run over some special Dirichlet characters to the specified modulus, and will be in terms of Q and N. The idea of studying characters sums to prime-power moduli was started in a paper by A. G. Postnikov. He gave formulas on the decomposition of groups of characters of powerful moduli. P. X. Gallagher also studied characters of this type. Iwaniec expanded the idea to composite moduli.
The group G = ( Z/q 2 Z ) contains the following subgroup H = {x G x 1 (mod q)}. If ξ (mod q 2 ) be a character on G. Then the induced character on H is an additive character, so for x H [ ] a(x 1) (24) ξ(x) = e, for some a (mod q). q 2 If ξ (mod q 2 ) is another character of the same kind. Then ξ ξ 1 is a character on G, trivial on H. Therefore, ξ = ξχ where χ is induced by a character on (Z/qZ). Let G a be the set of characters ξ (mod q 2 ) satisfying (24). G a = ϕ(q), and that every element of G a is obtained by multiplying a fixed character ξ G a by a character χ (mod q) on (Z/qZ).
Therefore, the context in which we shall consider the sum in the left-hand side of (23) is with a = 1 in (24). (25) Q q=1 1 ϕ(q) ξ G 1 a n ξ(n) The characters being summed are not necessarily primitive, a feature that the classical large sieve inequalities, Corollary 2, do not possess. 2. Estimating via Cauchy gives that (26) Q q=1 1 ϕ(q) ξ G 1 a n ξ(n) 2 = [Q + O(N)] n a n 2. Although (26) is obtained trivially, it is an asymptotic formula rather than simply an upper bound.
The following is a quote from M. N. Huxley: An analytic number theorist, it is said, is someone who is very good at using Cauchy s inequality. Indeed, we must demonstrate our worth in that regard. In this section, we shall restrict our attention to only moduli that are prime squares which has its advantages.
Theorem 4. Let ǫ > 0 be given. Suppose Q, N N, M Z and {a n } be a sequence of complex numbers. We have (27) Q q=1 q P 1 ϕ(q) ξ G 1 a n ξ(n) 2 Q ǫ (NQ 1 4 + N 1 4Q + NQ 1 4 + N 3 4Q 3 8 where the implied constant depends on ǫ alone. ) a n 2, Proof. Cauchy several times and then complete character sums and apply estimate for Kloosterman and Ramanujan sums. We need Weil s bound.
The asymptotic formula of (26) is useful when the error term does not exceed the main term or when N Q. Our Theorem 4 is better than (26) when Q 3 2 +ǫ N. We would certainly hope that our result is good whenever (26) is not, id est whenever Q N. Since we already have to resort to the strength of Weil bound for our present result, any desire for improvement upon it is perhaps too greedy, at least for the time being.
Next, we consider a large sieve type inequality with quadratic amplitude. We are interested in estimating the sum of the following kind. K k=1 a n e[x k f(n)] 2 a n 2, where {x k } are some well-spaced real numbers, and f(x) = αx 2 + βx + γ with α, β, γ R and α > 0. We employed the lemma, styled the double large sieve, of Bombieri and Iwaniec.
Lemma 3 (double large sieve). Suppose x (1),, x (M) and y (1),, y (N) are vectors in R k with X 1 2 x(m) 1 X i 2, Y i 2 y(m) i Y i 2, for i = 1,, k, m = 1,, M, n = 1,, N. Set ǫ i = X 1 δ i = X i, and Λ(x) = max(1 x,0). Then X i Y i + 1 M N m=1 n=1 M M m=1 r=1 a m b n e [ x (m) y (n)] a m a r K i=1 Λ 2 ( ) π 4k k 2 i=1 x(m) i x (r) N N i δ i n=1 r=1 (X i Y i + 1) b n b r k i=1 Λ i, 1 y (r) i y(n) ǫ i. Proof. Fourier and Cauchy.
By the virtue of the duality principle, it is equivalent to estimate for the following sum. (28) M+N K k=1 c k e(x k f(n)) where {c k } is an arbitrary sequence of complex numbers, as {a n }. Therefore, if f(n) = n 2, then the virtue of the regular large sieve inequality, in its dual form, would majorize (28) as follows. (29) K k=1 c k e ( x k n 2) 2 2 (δ 1 + N 2 ), K k=1 c k 2, where the N 2 in the (29) comes from changing the variables and then filling in all the non-squares n s.
If one consider the general case of f(n) = αn 2 + βn + γ, a quadratic polynomial with real coefficients, one could proceed á la Gallagher. More precisely, we apply the Sobolev-Gallagher s Lemma, to a n e[f(n)] 2 on the intervals [x k δ, x k + δ] and then sum over the x k s. Then apply Cauchy-Schwartz inequality and Parseval identity, we can take the majorant to be (30) [δ 1 + α (M + N) 2 ] a n 2. During this proof, there is hardly any waste at all in the estimates. The only inequalities that we applied are those of Hölder, and Sobolev-Gallagher, neither of which admits great waste. One must admit that it has certain virtues in (30).
( Set f(x) = α x + β ) 2 and F(Q) = 2α We need to consider the following. x F(Q) s t { a q c s c t e[αx(s t)(s + t + β/α)]. Q gcd(a, q) = 1, 1 a < q Lemma 4. Let S = [M + 1, M + N] Z, and ǫ > 0 be given, g(x, y) = (x y)(x + y + s/t), and m, n S with g(m, n) > 1. Also s, t Z with t > 0 and gcd(a, b) = 1. Let T denote the number of pairs of m, n S with g(m, n) g(m, n ) 1 2. Then we have (31) T t[t(m + N) + s ] ǫ, where the implied constant depends only on ǫ. }. Proof. Estimations and non-trivial counting.
From double large sieve, we have Theorem 5. Under the usual notations, and f(x) = αx 2 +βx+γ with β/α = u/v Q. We have (32) x F(Q) a n e[xf(n)] (α 1 + 1)(Q 2 + QN)v[v(M + N) + u ] ǫ where the implied constant depends on ǫ alone. 2 a n 2, Proof. Apply double large sieve and Lemma 4.
The above theorem may be extended to the case β/α R by approximating β/α with a rational number and the proof works almost exactly the same. More precisely, we have Proposition 3.The result of Theorem 5 still hold if the condition β α Q is replaced by β α R and there exists u Q with gcd(u, v) = v 1 and (33) β α u v < 1 4vN. Proof. The proof goes precisely the same as that of Theorem 5.
If we apply duality first and then double large sieve to the dual sum, we arrive at Theorem 6. Under the usual notations, we have (34) x F(Q) a n e[xf(n)] 2 N(1 + α 1 )(Q 2 + N) where the implied constant is absolute. a n 2. Proof. Apply duality and then double large sieve.
Last, we estimate an exponential sum over primes with square root amplitude twisted with Hecke eigenvalues. More precisely, we want to have an estimate for the following sum (35) S(N) = n N λ(n)λ(n)e(α n). Λ(n), as usual, denotes the von Mangoldt function and α is a non-zero positive real number. λ(n) are the normalized Fourier coefficients of Cusp forms for the full modular group SL 2 (Z) on the upper half plane H. They are also the Hecke eigenvalue, up to some normalization factor.
Regarding the sign changes of the Hecke eigenvalues, it was due to Hardy-Ramanujan, and Good, respectively that n N λ(n)e(αn) N 1 2 log(2n), and Moreover, M. R. Murty conjectured that p N n N λ(p) = Ω ± ( N logloglog N log N λ(n) N 1 3 +ǫ. and succeeded in proving it provided some L-function has no real zero between 1/2 and 1. Also, S. D. Adhikari proved essentially the same result for cusp forms for the group Γ 0 (N). ).
The method that we employ in estimating S(N) is that developed by I. M. Vinogradov. The best possible result one can hope for with Vinogradov s method is (36) S(N) N Θ+ǫ, with Θ = 3. However, if one is to believe the so-called principle 4 of square-rooting, then one may also be led to believe that S(N) N 1 2 +ǫ should hold. However, Iwaniec, Luo and Sarnak gave surprising heuristics that Θ = 3 is where truth actually lies. What they 4 were able to show, more precisely, (37) S(N) = ZN 3 4 + O(N 5 8 +ǫ ), where Z is some constant, non-zero of course, that depends on the cusp form.
We shall show the following. Theorem 7. With S(N) defined as in (35), we have (38) S(N) N 5 6[log(3N)] 36, where the implied constant depends on α in the definition of S(N) and the cusp form f(z). We shall need to treat two types of sums. One of them is a bilinear form, which we Cauchy and estimate. The second we estimate using the mean value theorem for automorphic L- functions and the integral form of the large sieve.
Lemma 5 (á la Vaughan). For N < n 2N, n y = N 1 3 and z = 2N, then Λ(n) = Λ 1 (n) + Λ 2 (n) + Λ 3 (n) + Λ 4 (n) + Λ 5 (n), with Λ 1 (n) = µ(b)log a, Λ 2 (n) = µ(b)λ(c), Λ 3 (n) = ab=n b y abc=n b,c y y<a<z µ(b)λ(c), Λ 4 (n) = Λ 5 (n) = abc=n b z y<c z abc=n b,c y a z abc=n c>y y<b<z µ(b)λ(c). Proof. This is done via combinatorial means. µ(b)λ(c), and
Thus the sum of our interest in (35) is decomposed and it suffices to estimate each component. We have (39) N<n 2N Λ(n)λ(n)e(α n) S i (N) = N<n 2N 5 i=1 S i (N), where Λ i (n)λ(n)e(α n) with the Λ i (n) s defined as in Lemma 5. We shall estimate the size of each of the S i (N) s individually.
For the bilinear form S 3, S 4 and S 5. We have the following. Lemma 6. With S i (N) defined as before, we have S 3 (N) + S 4 (N) + S 5 (N) N 5 6(log3N) 35, where the implied constant depends on α and the cusp form f(z). Proof. Cauchy and estimate the inner-most sums with the corresponding exponential integrals.
Lemma 7. With S 1 (N) defined as before and for any ǫ > 0 given, we have S 1 (N) N 3 4 +ǫ, where the implied constant depends on α, f(z) and ǫ. Proof. Stationary phase, Perron s formula, mean value theorem, large sieve. Lemma 8. With S 2 (N) defined as before and for ǫ > 0 given, we have S 2 (N) N 3 4 +ǫ, where the implied constant depends on α, f(z) and ǫ. Proof. As in Lemma 7.
Also, applying a result of M. Jutila s would give that S 1 (N) + S 2 (N) N 5 6 +ǫ. Indeed, this will not affect our final result of Theorem 7, but our results give better estimates. What stays us is the want of better means to estimate the bilinear forms. We could transform the bilinear forms into integrals then use the mean value theorems of L-function and Dirichlet series to estimate the resulting integrals. However, known results do not yield better estimates. If one assumes the truth of the Montgomery conjecture or the appropriate version of Lindelöf hypothesis, then we do obtain the expected majorant of Θ = 3 4, with Θ as in (36).
POST-DEFENSE BACCHANAL