An Estimate For Heilbronn s Exponential Sum

Transcription

1 An Estimate For Heilbronn s Exonential Sum D.R. Heath-Brown Magdalen College, Oxford For Heini Halberstam, on his retirement Let be a rime, and set e(x) = ex(2πix). Heilbronn s exonential sum is defined by S(a) = e( an 2 ), n=1 for any integer a corime to. It is imortant to note here that if n n (mod ), then n n (mod 2 ). Thus the summand in S(a) has eriod with resect to n, so that S(a) is a comlete sum to modulus. It was a favourite roblem of Heilbronn, and later of Davenort, to show that S(a) = o() as. Odoni [3] examined sums related to S(a), and showed that they are O( 1/2 ) in a suitable average sense. However his argument fails to give a non-trivial uer bound for an individual value of S(a). Indeed he shows how Weil s aroach leads only to the estimate S(a) = O( 3/2 ), which is worse than the trivial bound! We can now answer Heilbronn s question with the following theorem. Theorem 1 If is a rime and / a then S(a) 11/12, uniformly in a. As a corollary one can handle sums over arbitrary intervals. Corollary If is a rime and / a then / n e( an 2 ) 11/12 log, uniformly in a, for all M and for all N. In articular one sees, by Weyl s criterion, that the numbers n are uniformly distributed modulo 2 for large. 1

2 Since the deduction of the corollary is very straightforward we shall resent it here. We have / n e( an 2 ) = 1 r=1 s=1 1 1 {N + r=1 e( as 2 ) cosec( πr ) } max r e( r(s n) ) s=1 e( as 2 )e(rs ), on using the estimate e( rn { ) N, r, cosec( πr ), / r. Since N and we deduce that / n 1 cosec( πr ) log, r=1 e( an ) (log ) max 2 r s=1 e( as 2 )e(rs ). However, since s s (mod ), we have whence s=1 e( as 2 )e(rs + r)s ) = e((a 2 ), e( as 2 )e(rs ) = S(a + r) 11/12, by Theorem 1. This bound holds uniformly in r, of course, so that the corollary follows. We take this oortunity to record an estimate for a second sum which involves the distribution modulo 2 of n 1 rather than n. For / n we shall set q(n) = n 1 1. This has sometimes been called the Fermat quotient. Our result shows that q(n) is uniformly distributed modulo for 1 n <. 2

3 Theorem 2 For any integer a corime to we have / n uniformly for M, N 1. In articular uniformly for / a. n=1 e( aq(n) ) N 1/2 3/8, 1 e( aq(n) ) 7/8, The roof of Theorem 2 is quite straightforward, and we resent it here. We observe that q(mn) = q(m)n 1 + q(n), whence q(mn) q(m) + q(n) (mod ), for / mn. Thus { 0, n, χ(n) = e( aq(n) ), / n, is a non-rincial character to modulus 2, of order. The sum in Theorem 2 can therefore be written as χ(n), and the required result follows by an estimate of Burgess [1]. (Actually the result stated by Burgess contains a factor k 3/16+ε for characters to modulus k, but it is easy to remove the exonent ε when k = 2.) We now return to Theorem 1, with which we shall be concerned for the remainder of the aer. Our treatment begins with some elementary maniulations of a tye which would have been quite familiar to Heilbronn, which lead to the following result. Lemma 1 Let and write f(x) = X + X2 2 + X X 1 1 ZZ [X], Then there is a value of r for which S r = {k ZZ {0, 1} : f(k) = r}, N r = #S r. S(a) 3/4 N 1/4 r. 3

4 The trivial bound N r leads to the estimate S(a), so that nothing has been lost u to this oint. On the other hand, it is not so clear how any non-trivial estimate for N r may be obtained. It turns out that ideas from the work of Steanov [4] rovide the necessary tool. Steanov was concerned with roving Weil s theorem on the number of oints on a curve over a finite field, rather than bounding the number of zeros of a olynomial in one variable. The method shows strong links with ideas from transcendence theory, where one constructs an auxilliary olynomial which vanishes to high order at the oints of interest. The resemblance between f(x) and the function log(1 X) = X + X2 2 + X Q[[X]] serves as a guide during the argument. To construct the auxilliary olynomial one uses the fact that f satisfies a simle differential equation. This is exressed by the following lemma. Lemma 2 For any ositive integer r there exist olynomials q r (X) and h r (X) in ZZ [X], of degrees at most r + 1 and r 1 resectively, such that {X(1 X)} r ( d dx )r f(x) = q r (X) + (X X)h r (X). In effect the lemma converts f(x), which has large degree, into q r (X), which has small degree. As is usual in transendence roofs, we have also to show that our auxilliary olynomial does not vanish identically. This is accomlished in our case via the observation that f(x) is almost equal to a transcendental function, log(1 X), so that it cannot satisfy an algebraic relation. The following lemma exresses this rincile. Lemma 3 Let F (X, Y ) ZZ [X, Y ] have degree less than A with resect to X, and degree less than B with resect to Y. Then if F does not vanish identically we will have X / F (X, f(x)), roviding only that AB. This result is remarkably shar. Indeed if AB >, the olynomial F will have enough coefficients to ensure that X F (X, f(x)) is ossible. Secifically, one may note that the coefficients of F (X, f(x)) are linear functions of the coefficients of F (X, Y ). To make X F (X, f(x)) we must arrange that such linear functions vanish, and we have AB variables at our disosal. Hence if AB > a suitable olynomial F (X, Y ) can be found. With the hel of Lemmas 2 and 3, Steanov s method enables us to give the following bound for N r. Lemma 4 We have N r = O( 2/3 ) uniformly in r. 4

5 Theorem 1 now follows, via Lemma 1. After the original version of this aer was submitted, the referee kindly ointed out that Lemma 4 aears already in a aer of Mit kin [2]. Indeed Mit kin shows an analogous result for the olynomial as well. f(x) = 1 + X + X2 2! + X3 3! X 1 ( 1)! ZZ [X], 1 Preliminaries In this section we shall rove Lemma 1. We begin by writing so that S(a) = 1 + S 0 (a). Then S 0 (a) 2 = 1 S 0 (a) = e( an 2 ) 1 m,n=1 n=1 e( a(m n ) 2 ). When m n we set m n b (mod ) and m kb, n (k 1)b (mod ), so that b runs over the residues 1,..., 1 modulo and k runs similarly over the residues 2,..., 1. This yields 1 1 S 0 (a) 2 = ( 1) + = ( 1) + k b=1 k=2 e( ab {k (k 1) } 2 ) S(a{k (k 1) }). We now observe that k (k 1) = whence ( 1) l 1 ( )k l 1 f(k) (mod 2 ), l l=1 S 0 (a) 2 = ( 1) + N r S(a(1 r)). r=1 Cauchy s inequality then leads to the estimate S 0 (a) { Nr 2 }{ S(a(1 r)) 2 }. r=1 r=1 5

6 Here r N r is just the number of available integers k, namely 2. Moreover S 0 (a(1 r)) 2 = r=1 1 m,n=1 e( a(m n ) 2 ) r=1 e( ar(n m ) ). The inner sum vanishes unless n m, but this imlies that n m (mod ). Thus S 0 (a(1 r)) 2 = ( 1), so that r=1 Since max N r 1 we deduce that S 0 (a) ( 2)( 1) max N r. r S(a) 1 + 1/2 + 3/4 (max r as required for Lemma 1. 2 Steanov s Method N r ) 1/4 3/4 (max N r ) 1/4, r We shall begin by taking a olynomial Φ(X, Y, Z) ZZ [X, Y, Z], for which deg X Φ < A, deg Y Φ < B, deg Z Φ < C. The underlying idea is to arrange that the olynomial Ψ(X) = Φ(X, f(x), X ) has a zero of order at least D, say, at each oint k S r. We will therefore be able to conclude that DN r deg Ψ(X), roviding that Ψ does not vanish identically. We note that deg Ψ (deg X Φ) + (deg f)(deg Y Φ) + (deg X )(deg Z Φ) < A + ( 1)B + C, whence DN r A + (B + C), (1) roviding that Ψ does not vanish. In order for Ψ to have a zero of order at least D at a oint k we need d n Ψ(X) dx n = 0 for n < D. X=k Since k 0, 1 in our alication, this will be equivalent to {X(1 X)} n dn Ψ(X) dx n = 0. (2) X=k 6

7 For any term X a f(x) b X c we have {X(1 X)} n ( d dx )n {X a f(x) b X c } = X c {X(1 X)} n ( d dx )n {X a f(x) b }. We may now aly Leibnitz formula, using Lemma 2 together with the fact that {X(1 X)} g ( d dx )g X a either vanishes (if g > a) or is a olynomial of degree a + g. It follows that {X(1 X)} n ( d dx )n {X a f(x) b } is a linear combination of terms modulo X X, where and ˆq(X)q g1 (X)... q gl (X)f(X) b l 0 l min(b, n), deg q gi (X) g i + 1, (1 i l), deg ˆq(X) = a + n g 1... g l. Since X c X c (mod X X) we can now write {X(1 X)} n ( d dx )n {X a f(x) b X c } 0 β<b with deg P β (X) < A + 2n + C. Hence if P (X; a, b, c, n, r) = P β (X; a, b, c, n)f(x) β (mod X X), 0 β<b P β (X; a, b, c, n)r β we deduce that {X(1 X)} n dn dx n {Xa f(x) b X c } = P (k; a, b, c, n, r) X=k for any k S r. Here we use the observation that k k = 0 for any such k. We now write Φ(X, Y, Z) = a,b,c λ a,b,c X a f(x) b X c 7

8 and P n (X) = a,b,c λ a,b,c P (X; a, b, c, n, r), so that deg P n (X) < A + 2n + C and {X(1 X)} n dn dx n Φ(X, f(x), X ) = P n (k) X=k for any k S r. We shall arrange, by aroriate choice of the coefficients λ a,b,c, that P n (X) vanishes identically for n < D. This will ensure that (2) holds for k S r. Each olynomial P n (X) has at most A + 2n + C A + 2D + C coefficients, which are linear forms in the original λ a,b,c. Thus if D(A + 2D + C) < ABC (3) there will be a set of coefficents λ a,b,c, not all zero, for which the olynomials P n (X) vanish for all n < D. We must now consider whether Φ(X, f(x), X ) can vanish if Φ(X, Y, Z) does not. We shall write Φ(X, Y, Z) = c F c (X, Y )Z c, and take c 0 to be the smallest value of c for which F c (X, Y ) is not identically zero. It follows that Φ(X, f(x), X ) = X c 0 F c (X, f(x))x (c c0), c 0 c<c so that if Φ(X, f(x), X ) is identically zero we must have F c0 (X, f(x)) 0 (mod X ). This will contradict Lemma 3, roviding that AB, (4) as we now assume. Finally we conclude that (1) holds, subject to the conditions (3) and (4). One readily sees that a suitable choice of arameters is A = [ 2/3 ], B = C = [ 1/3 ], D = [ 1 3 2/3 ]. These are satisfactory if is large enough, and then (1) yields N r 2/3 as required for Lemma 4. 8

9 3 Proof of Lemmas 2 and 3 For the roof of Lemma 2 a simle induction argument suffices. For r = 1 we have X(1 X) d dx f(x) = X X, so that we may take q 1 (X) = 0 and h 1 (X) = 1. For the general case we differentiate the formula {X(1 X)} r ( d dx )r f(x) = q r (X) + (X X)h r (X) and multily by X(1 X) to obtain {X(1 X)} r+1 ( d dx )r+1 f(x) + r(1 2X){X(1 X)} r ( d dx )r f(x) = X(1 X)q r(x) X(1 X)h r (X) + (X X)X(1 X)h r(x). This allows us to set and q r+1 (X) = X(1 X)q r(x) X(1 X)h r (X) r(1 2X)q r (X) h r+1 (X) = X(1 X)h r(x) r(1 2X)h r (X). The required bounds for deg q r (X) and deg h r (X) then follow by induction. The roof of Lemma 3 is more difficult. We shall use the following auxilliary result. Lemma 5 Let F (X, Y ) ZZ [X, Y ]. Let deg X F = m 0 and deg Y F = n 1 and suose that m, n <. Then (1 X) m+1 ( d dx )m+1 F (X, f(x)) G(X, f(x)) (mod X 1 m ), for some G(X, Y ) ZZ [X, Y ] with deg X G m and deg Y G = n 1. In articular G does not vanish identically. with To rove this result it will suffice to show that (1 X) m+1 ( d dx )m+1 X a f(x) b G(X, f(x); a, b) (mod X 1 m ), (a m, b n) (5) deg X G(X, Y ; a, b) a, deg Y G(X, Y ; a, b) b 1, and such that the coefficient of X a Y b 1 in G(X, Y ; a, b) is non-zero. 9

10 In order to establish (5) we aly Leibnitz formula, observing that (1 X) k ( d dx )k X a is either identically zero (if k > a) or is a olynomial of degree a, and that (1 X) l ( d dx )l f(x) = (1 X) l ( d dx )l 1 {1 + X X 2 } = (1 X) l ( d 1 dx )l 1 { 1 X + O(X 1 )} for l 1. One can therefore see that = (1 X) l (l 1)! { (1 X) l + O(X l )} = (l 1)! + O(X l ) (6) deg X G(X, Y ; a, b) a, deg Y G(X, Y ; a, b) b, in (5). To check that deg Y G(X, Y ; a, b) b we use Leibnitz formula to write ( d m+1 dx )m+1 X a f(x) b = ( m + 1 )( d k dx )k X a ( d dx )m+1 k f(x) b and ( d dx )m+1 f(x) b = ( m + 1 k k 1,..., k b k 1,...,k b where k k b = m + 1 k and ( m + 1 k k 1,..., k b ) = ){( d dx )k 1 f(x)}... {( d dx )k b f(x)}, (m + 1 k)! k 1!... k b! is a multinomial coefficient. We now see that terms with k > a vanish identically. Moreover, if k a then m + 1 k m + 1 a > 0, so that k i > 0 for some index i. It follows that there will be no term involving f(x) b in (5). Finally we have to examine the coefficient of X a f(x) b 1, which arises in the above formulae from those terms in which recisely one of the k i is non-zero. Since k can be at most a if we are to have a non-zero contribution, we see, using (6), that the required coefficient is b a ( m + 1 k )( 1) k a! (m k)! (a k)! = b(m a)!a! 10 a ( 1) k ( m + 1 k )( m k m a ). (7)

11 It remains to determine whether or not this exression vanishes in ZZ. To this end we consider the exansion 1 X m+1 1 X = (1 X) 1 1 ( 1 X 1)m+1 (1 X) m = m+1 (1 X) 1 + { ( 1) m k ( m )( k 1 X )k }(1 X) m = m ( 1) m k ( m + 1 k )(1 X) m k = m ( 1) m k ( m + 1 m k ) ( 1) j ( m k k j )X j. j=0 Now the sum over k on the right of (7) is just the coefficient of X m a in the final exression above, aart from a factor ( 1) a. Moreover 1 X m+1 1 X = 1 + X Xm so that the coefficient of X m a is just 1. We can therefore conclude that b(m a)!a! a ( 1) k ( m + 1 k )( m k m a ) = ( 1)a b(m a)!a!. Since this is non-zero in ZZ for b, m < the lemma now follows. It remains to derive Lemma 3 from Lemma 5. We shall take a non-zero olynomial F (X, Y ) ZZ [X, Y ] with deg X F = m and deg Y F = n. Using induction on n we shall show that X (m+1)(n+1) / F (X, f(x)) roviding that (m + 1)(n + 1). This clearly suffices for Lemma 3. When n = 0 the result is obvious. Suose now that the result has been roved when n is relaced by n 1. If F is as above but X (m+1)(n+1) F (X, f(x)) then X (m+1)n (1 X) m+1 ( d dx )m+1 F (X, f(x)). In the notation of Lemma 5 this yields X (m+1)n G(X, f(x)), in view of the fact that (m + 1)n (m + 1). Moreover we will have deg X G = m m, and deg Y G = n 1. However we now have X (m +1)n G(X, f(x)), contradicting the induction hyothesis. It therefore follows that X (m+1)(n+1) / F (X, f(x)) as required for the induction ste. This comletes the roof of our assertion, and with it the roof of Lemma 3. 11

12 4 Acknowledgement This aer was reared while the author was a guest of the Mathematics Deartment of Oklahoma State University. Their suort is gratefully acknowledged. References [1] D.A. Burgess, On character sums and L-functions. II., Proc. London Math. Soc. (3), 13 (1963), [2] D.A. Mit kin, An estimate for the number of roots of some comarisons by the Steanov method, Mat. Zametki, 51 (1992), 52-58, 157. (Translated as Math. Notes, 51 (1992), ) [3] R.W.K. Odoni, Trigonometric sums of Heilbronn s tye, Math. Proc. Camb. Phil. Soc., 98 (1985), [4] S.A. Steanov, The number of oints of a hyerellitic curve over a rime field, Izv. Akad. Nauk SSSR Ser. Mat., 33 (1969),