On Linnik and Selberg s Conjecture about Sums of Kloosterman Sums

Transcription

1 On Linnik and Selberg s Conjeture about Sums of Kloosterman Sums Peter Sarnak 1,2 and Jaob Tsimerman 1 1 Department of Mathematis, Prineton University, Prineton, NJ 2 Institute for Advaned Study, Prineton, NJ Dediated to Y. Manin on the Oasion of his 70 th Birthday 1 Statements In [Li] Linnik and later Selberg [Se] put forth far reahing onjetures onerning anellations due to signs of Kloosterman Sums. Both point to the onnetion between this and the theory of modular forms. For eample, the generalization of their onjetures to sums over arithmeti progressions imply the general Ramanujan Conjetures for GL 2 /Q. Selberg eploited this onnetion to give a nontrivial bound towards his well known λ onjeture. Later Kuznetzov [Ku] used his summation formula, whih gives an epliit relation between these sums and the spetrum of GL 2 /Q modular forms, to prove a partial result towards Linnik s Conjeture. Given that we now have rather strong bounds towards the Ramanaujan Conjetures for GL 2 /Q [L-R-S], [Ki-Sh], [Ki-Sa] it seems timely to revisit the Linnik and Selberg Conjetures. by The Kloosterman Sum S(m, n;, for 1, m 1 and n 0, is defined ( m + n S(m, n; = e mod 1( (1

2 2 Peter Sarnak and Jaob Tsimerman (here e(z = e 2πiz. It follows from Weil s bound [Wei] for S(m, n; p, p prime, and elementary onsiderations that S(m, n; τ( (m, n, 1/2 (2 where τ( is the number of divisors of. Linnik s Conjeture asserts that for ɛ > 0 and n S(1, n; ɛ. (3 ɛ Note that from (2 we have that S(m, n; ((m, n ɛ. (4 ɛ On the other hand for m, n fied, Mihel [Mi] shows that S(m, n; 1 2 ɛ, (5 ɛ and hene if (3 is true it represents full square-root anellation due to the signs of the Kloosterman Sums. Selberg puts forth the muh stronger onjeture, whih has been repliated in many plaes and whih reads; For (m, n 1/2, S(m, n; ɛ. (6 ɛ

3 On Linnik and Selberg s Conjeture about Sums of Kloosterman Sums 3 As stated, this is false (see Setion 2. It needs to be modified to inorporate an ɛ-safety valve in all parameters and not only in and (m, n. For eample, the following modifiation seems okay S(m, n; ( mn ɛ. (7 ɛ We all the range mn the Linnik range and mn the Selberg range. Obtaining nontrivial bounds (i.e., a power saving beyond (4 for the sums (3 and (7 in the Selberg range is quite a bit more diffiult. By a smooth dyadi sum of the type (7 we mean S(m, n; ( F (8 where F C0 (R + is of ompat support and where the estimates for (8 as, m, n vary, are allowed to depend on F. Summation by parts shows that an estimate for the left hand side of (7 will give a similar one for (8, but not onversely. In partiular, estimates for (7 imply bounds for non-dyadi S(m, n; sums with β h and β < 1. We note that for +h many appliations understanding smooth dyadi sums is good enough and in the Linnik range these smooth dyadi sums are diretly onneted to the Ramanujan Conjetures, see Deshouilliers and Iwanie [D-I] and Iwanie and Kowalski [I-K pp ]. Let π = v π v be an automorphi uspidal representation of GL 2 (Q\GL 2 (A with a unitary entral harater. Here v runs over all primes p and. Let µ 1 (π v, µ 2 (π v denote the Satake parameters of π v at an unramified plae v of π. We normalize as in [Sa] and use the results desribed there. For 0 θ < 1 2 we denote by H θ the hypothesis; If any π as above R (µ j (π v θ, j = 1, 2. (9 Thus H 0 is the Ramanujan-Selberg Conjeture and H θ is known for θ = 7 64 ([Ki-Sa].

4 4 Peter Sarnak and Jaob Tsimerman Conerning (7 the main result is still that of Kuznetzov [Ku] who using his formula together with the elementary fat that λ 1 (SL 2 (Z\H 1 4 showed (the barrier 1 6 is disussed at the end of this paper: Theorem 1 (Kuznetzov Fi m and n then S(m, n; m,n 1/6 (log 1/3. One an ask for similar bounds when varies over an arithmeti progression. For this one applies the Kuznetzov formula for Γ = Γ 0 (q in plae of SL 2 (Z (see [D-I] or one an use the softer method in [G-S]. What is needed is the v = version of H θ with θ 1 6. This was first established in [Ki-Sh] and leads to Theorem 1. Fi m, n, q, then 0(q S(m, n; m,n,q 1 6 (log 1/3. Consider now what happens if m and n are allowed to be large as asked by Linnik and Selberg. We eamine the ase q = 1, the general ase with q also varying deserves a similar investigation. We also assume that mn > 0, the mn < 0 ase is similar eept that some results suh as Theorem 4 are slightly weaker sine when mn > 0 we an eploit H 0 whih is known for the Fourier oeffiients of holomorphi-usp forms [De]. The analogue of Theorem 1 that we seek is S(m, n; ( (mn 6 (mn ɛ. (10 ɛ As with Theorem 1 we show below that the eponent 1 6 in the mn aspet onstitutes a natural barrier. We will establish (10 assuming the general Ramanujan Conjetures for GL 2 /Q (i.e assuming H 0 and unonditionally we ome lose to proving it. Theorem 2 Assuming H θ we have

5 On Linnik and Selberg s Conjeture about Sums of Kloosterman Sums 5 S(m, n; ɛ ( (mn (m + n 1/8 (mn θ/2 (mn ɛ. Corollary 3 Assuming H 1 12 (and afortiori H 0, (10 is true. Unonditionally Theorem 2 is true with θ = 7/64 and in partiular (10 is true if m and n are lose in the sense that n 5/43 m n or m 5/43 n m. The proof of Theorem 2 uses Kuznetzov s formula to study the dyadi sums 2 S(m, n; see Setion 2. The dyadi piees with = (mn 1/3 or smaller are estimated trivially using (2 and give the (mn 1/6 term in (10. For (mn 1/3 we don t have any nontrivial bound for the sum even in smooth dyadi form. Indeed applying the Kuznetzov formula to suh smooth sums in range (mn 1/2 δ for some δ > 0, one finds that the main terms on the spetral side loalize in the transitional range for the Bessel funtions of large order and argument. The analysis beomes a deliate one with the Airy funtion. In this mn > 0 ase the main terms only involve Fourier oeffiients of holomorphi modular forms. This indiates that one should be able to obtain the result bakwards using the Petersson formula [I-K] together with the smooth k-averaging tehnique [I-L-S pp. 102]. Indeed this is so and we give the details of both methods in Setion 2. Another bonus that omes from this holomorphi loalization is that we an appeal to Deligne s Theorem, that is H 0, for these forms., Theorem 4 Fi F C 0 (0, and δ > 0. For mn > 0 and (mn 1 2 δ S(m, n; ( F F,δ mn. The bound in Theorem 4 is nontrivial only in the range (mn 1/3. To etend this to smaller requires establishing anellations in short spetral sums for Fourier oeffiients of modular forms (see equation (34 whih is an interesting hallenge. In any ase this analysis of the smooth dyadi sums eplains the (mn 1/6 barrier in (10.

6 6 Peter Sarnak and Jaob Tsimerman 2 Proofs As we remarked at the beginning, the randomness in the signs of the Kloosterman sums in the form (7 implies the general Ramanujan Conjetures for GL 2 /Q. Sine there seems to be no omplete proof of this in the literature we give one here. For the arhimedian q-aspet, that is the λ 1 1/4 Conjeture, this follows from Selberg [Se] sine his zeta funtion Z(m, n, s has poles with nonzero residues in R(s > 1 2 if there are eeptional eigenvalues (i.e. λ < 1/4. For the ase of Fourier oeffiients of holomorphi modular forms the impliation is derived in [Mu pp ]. So we fous on the Fourier oeffiients of Maass forms and restrit to level q = 1 (the ase of general q is similar. We apply Kuznetzov s formula with m = n for Γ = SL 2 (Z, see [I-K pp. 409] for the notation; j = 1 ρ j (n 2 h(t j osh(πt j + 1 4π = g 0 + = 1 τ(n, t 2 S(m, n; h(t osh(πt dt ( 4πn g. (11 Here h is entire and deays faster than (1 + t 2 δ, with δ > 0 as t ±, and and g 0 = 1 π 2 g( = 2i π r h(r tan h(πr dr J 2ir ( r h(r osh(πr dr. We want to prove that for a fied j = j 0, we have ρ j0 (n = O ɛ ( n ɛ as n. (12 It is well known how to dedue the various forms of H 0 for φ j0 from (12. We hoose h with h(t 0 for t R and h(t j0 1 and so that g C (0, is rapidly dereasing at and g vanishes at = 0 to order 1. Now let n in (11. Sine τ(n, t = O ɛ ( n ɛ we have

7 On Linnik and Selberg s Conjeture about Sums of Kloosterman Sums 7 ρ j0 (n 2 n ɛ + = 1 S(n, n; ( 4πn g. If we assume (7 then we an estimate the sum on by summing by parts and we obtain the desired bound (12. We turn net to (6 and show that as stated it is false. Let be a large integer. For eah prime p in (, 2 there is an integer a p suh that S(1, a p ; p p 1 10 (13 This follows from the easily verified identities 1 p 1 p a(p a(p S(1, a; p p = 0, ( 2 S(1, a; p = 1 1 p p. (14 We note in passing that the asymptotis of any moment 1 p a ( S(1,a;p m, p in the limit as p, have been determined by Katz [Ka]. The m th moment being that of the Sato-Tate measure. Now let n be an integer satisfying n 0 (mod (! 2 and n a p (mod p for < p < 2. Suh an n an be hosen sine (! 2 and the p s are all relatively prime to eah other. For 0 <, n 0 (mod and hene S(1, n; = µ( the Mobius funtion. For < < 2 and not a prime, we have S(1, n; = S(1, n; 1 2 with 1 < 1, 2 <. Hene n 0(mod and S(1, n; = µ(. It follows that

8 8 Peter Sarnak and Jaob Tsimerman S(1, n; = µ( + < p < 2 < 2 not prime µ( 1 p + O(log S(1, a p ; + < p < 2 Hene (6 is false. 1/2 log + O(log. (15 Of ourse the n onstruted above is very large and the right hand side in (12 is ertainly O ɛ (n ɛ for any ɛ > 0. Thus with the n ɛ in (7 this is no longer a ounter eample. We turn net to Theorem 4, that is the analysis of the smooth dyadi sums in the Selberg range (mn 1/3. We need Kuznetzov s formula [I-K]. Let f C 0 (R + and M f (t = πi sinh(2πt 0 (J 2it ( J 2it ( f( d (16 and N f (k = 4(k 1! (4πi k 0 J k 1 ( f( d. (17 Let f j,k (z, j = 1,..., dim S k (Γ, be an orthonormal basis of Heke eigenforms for the spae of holomorphi usp forms of weight k for Γ = SL 2 (Z. Let ψ j,k (n denote the orresponding Fourier oeffiients normalized by f j,k (z = k=1 ψ j,k (m m k 1 2 e(mz. (18 Let τ(m, t be the n th Fourier oeffiient of the unitary Eisenstein series E(z, it. Then for mn > 0

9 On Linnik and Selberg s Conjeture about Sums of Kloosterman Sums 9 ( S(m, n; 4π mn f = M f (t j ρ j (m ρ j (n + 1 4π j + k 0(2 N f (k 1 j dim S k (Γ M f (t τ(m, t τ(n, t dt ψ j,k (m ψ j,k (n. (19 For (mn δ Y (mn 1/2 we apply (19 with f Y ( = f 0 ( Y (20 and f 0 C 0 (R > 0 fied. The left hand side of the formula is S(m, n; ( 4π mn f 0 = Y S(m, n; F 0 ( X where F 0 (ξ = f 0 (1/ξ and X = 4π mn Y, whih is what we are interested in when Y is large. One an analyze N fy (t and M fy (t as Y using the known asymptotis of the Bessel funtions J it ( and J k 1 ( for and t large. Using repeated integrations by parts one an show that the main term omes from the holomorphi forms only, with their ontribution oming from the transition range; J k (Y, with Y k k 1/3. (21 Using the approimations [Wa pp. 249] by the Airy funtion; J k ( 1 π ( 2(k 3 1/2 K 1 3 ( 2 3/2 (k 3/2 3 1/2 for k k 1/3 < k (22 and

10 10 Peter Sarnak and Jaob Tsimerman ( 1/2 2( k ( ( 2 3/2 ( k 3/2 J1/3 + J 1/3 J k ( /2 for k < k + k 1/3 (23 and keeping only the leading terms of all asymptotis one finds that 1 π S(m, n; Ai(ξ dξ k ( 4π mn f 0 Y ( 1 k 4(k 1! k f 0 Y (4πi k 1 j dim S Γ (Γ ψ j,k (m ψ j,k (n. (24 Here Ai(ξ = os (t 3 + ξt dt, is the Airy funtion. The above derivation is tedious and ompliated but one we see the form of the answer and espeially that it involves only holomorphi modular forms, another approah suggests itself and fortunately it is simpler. We start with the Peterson formula [I-K]. Γ (k 1 (4π k 1 j=1,...,dim S k (Γ ψ j (m ψ j (n = δ(m, n + 2π i k = 1 S(m, n; J k 1 ( 4π mn. Setting ρ f (n = ( 1/2 Γ (k 1 (4π k 1 ψ f (n (25 we have f H k (Γ ρ f (m ρ f (n = δ(m, n = 2πi k = 1 S(m, n; ( 4π mn J k 1 (26

11 On Linnik and Selberg s Conjeture about Sums of Kloosterman Sums 11 where H k (Γ denotes any orthonormal basis of S k (Γ and in partiular the Heke basis whih we hoose. In this ase ρ f (m = ρ f (1 λ f (m (27 where λ f (m τ(m, by Deligne s proof of the holomorphi Ramanujan Conjetures [De]. With this normalization (24 reads S(m, n; ( 4π mn f 0 (onst Y k ( k f 0 ρ f (m ρ f (n. Y f H k (Γ (28 From (26 with m = n = 1 we have that for k large f H k (Γ ρ f (1 2 = 1 + small. (29 We use the k-averaging formula in [I-L-S pp.102] whih gives for K 1, > 0 and h C 0 (R > 0 that h k ( := k 0(2 ( k 1 h J k 1 ( = ik ĥ(tk sin ( sin 2πt dt. K (30 From this it follows easily that if K 1+ɛ0 then for any fied N 1, k 0(2 ( k 1 h J k 1 ( K N. (31 K N While for 0 K 1+ɛ0 and for any fied N

12 12 Peter Sarnak and Jaob Tsimerman h k ( = h 0 ( K + 1 ( K 2 h K K 2N h N ( ( + O N K 2N 1 K (32 where h 0 (ξ = h(ξ and h 1,..., h N involve derivatives of h(ξ and are also in C 0 (R > 0. From (26 we have that k 0(2 = k 0(2 ( k 1 i k h K f H k (Γ ρ f (m ρ f (n ( k 1 (i k h δ(m, n 2π K =1 S(m, n; ( 4π mn h K. (33 We assume that (mn δ K mn with δ > 0 and fied. Then for N large enough (depending on δ we have from (27 (28 and (29 that N j = 0 K 2j = k 0(2 S(m, n; ( 4π mn h j K ( k 1 i k h K f H k (Γ ρ f (m ρ f (n + O(1. (34 Thus the lead term j = 0 in (34 reovers the asymptotis (24 and in a muh more preise form. We an estimate the right hand side of (34 as being at most RHS k ( k 1 h K f H k (Γ λ f (mλ f (n ρ f (1 2 K τ(m τ(n, (35 by (27 and (29. It follows that N j = 0 K 2j S(m, n; ( 4π mn h j K τ(m τ(n. (36 K

13 On Linnik and Selberg s Conjeture about Sums of Kloosterman Sums 13 From this it follows by estimating the j 1 sums trivially that S(m, n; ( 4π mn h K K τ(m τ(n + (mn1/4 K 5/2 (37 Using this bound whih is now valid for h 1,..., h N we an feed it bak into (36 to get S(m, n; ( 4π mn h K ( K τ(m τ(n + K 2 K τ(m τ(n + (mn1/4 K τ(m τ(n + (mn1/4. K 5/2 K 5/2 (38 Repliating this iteration a finite number of times yields that for (mn δ K mn, S(m, n; ( 4π mn h K K τ(m τ(n (39 or what is the same thing S(m, n; ( F ɛ (mn 1/2 τ(m τ(n, for (mn 1/2 δ (40 This ompletes the proof of Theorem 4. Finally we turn to the proof of Theorem 2. The dependene on mn in Kuznetzov s argument was eamined briefly in Huley [Hu]. In order for us to bring in H θ effetively we first eamine the dyadi sums for in various ranges. 2 S(m,n; Proposition 5 Assume H θ, then for ɛ > 0 S(m, n; mn (mn ( ɛ 1/6 + ɛ 2 + (m + n 1/8 (mn θ/2.

14 14 Peter Sarnak and Jaob Tsimerman Theorem 2 follows from this Proposition by breaking the sum 1 into at most log dyadi piees y 2y with (mn 1/3 y and estimating the initial segment 1 min((mn 1/3, using (4. We onlude by proving Proposition 5. In the formula (19 we hoose the test funtion f to be φ(t depending on mn, and a parameter 1/3 T 2/3 to be hosen later. φ is smooth on (0, taking values in [0, 1] and satisfying (i φ(t = 1 for a 2 (ii φ(t = 0 for t ( (iii φ (t a T t a where a = 4π mn. a a 2 + 2T and t a T. 1 (iv φ and φ are pieewise monotone on a fied number of intervals. For φ hosen this way we have that ( S(m, n; 4π mn φ = 1 ɛ (mn ɛ T T 2 T T S(m, n;, S(m, n; τ( (mnɛ T log. (41 where we have used (2 and the mean value bound for the divisor funtion. We estimate φ using (19 and to this end, aording to = 1 S(m,n; ( 4π mn (16 we first estimate ˆφ(r = osh(πr M φ (r. We follow Kuznetzov keeping trak of the dependene on mn. The following asymptoti epansion of the Bessel funtion is uniform see [Du]; J ir (y = πr + eirw(y/r 2 (1 + y2 + r 2 a 1 r2 + y... 2 where, a 1 are onstants and w(s = 1 + s 2 + log ( 1 1 s s (42

15 On Linnik and Selberg s Conjeture about Sums of Kloosterman Sums 15 We analyze the leading term, the lower order terms are treated similarly and their ontribution is smaller. For r 1 we have ˆφ(r r 2 (43 as is lear from the Taylor epansion for J ir when a 1 and from (42 otherwise. So assume that r 1 (or 1 and making the substitution y = rs we are redued to bounding = r 3/2 r 1/2 0 0 e irw(s ds φ(rs (s /4 s ( e irw(s rw (s (44 φ(rs. (45 w (s s(s /4 Now w (s is bounded away from zero uniformly on (0, and approahes 2 as s and behaves like 1 s as s 0. We may then apply the following easily proven mean value estimate: If F and G are defined on [A, B] with G monotoni and taking values in [0, 1] then B A F ( G( d 2 sup A C B C A F (d. (46 This yields that the quantity in (45 is at most O(r 3/2 and hene that ˆφ(r r 3/2 for r 1. (47 For r large we seek a better bound whih one gets by integration by parts in (45. This gives r 3/2 e irw(s 0 d ds ( φ(rs w ds (ss = O(r 5/2 + r 3/2 0 e irw(s θ (s (s /4 sw (s ds

16 16 Peter Sarnak and Jaob Tsimerman where the first term follows as in (47 and θ(s = φ(rs. Applying the mean value estimate to the last integral and using property (iii of φ yields ˆφ(r T r 5/2. (48 The bounds (47 and (48 are the same as those obtained by Kuznetzov only now they are uniform in and nm. For the term involving τ(m, t in (19 and our hoie of test funtion we have whih follows immediately from (43. ˆφ(r dr 1 (49 ρ(1 + 2ir 2 The term in (19 involving the k-sum over Fourier oeffiients of holomorphi forms is handled by using the Ramanujan Conjeture for these. We find that this sum is bounded by (mn ɛ 4(k 1 N φ (k. (50 k 0(2 k > 2 Now for mn it is immediate from the deay of the Bessel funtion at small argument that N φ (k 1/k!. Hene the sum in (50 is O ɛ ((mn ɛ whih is a lot smaller than the upper bounds that we derive for the right hand side of (19. For mn we need to investigate the transition ranges for the Bessel funtions J k (y. That is, the ranges y k k 1/3, k k 1/3 y k + k 1/3 and y k + k 1/3. We invoke the formula for the leading term asymptoti behavior these in eah region, see ([ Ol ] for uniform asymptotis whih allows one to onnet the ranges. For our φ we break the integral (16 defining N φ (k into the orresponding ranges. In the range (0, k k 1/3 the Bessel funtion is eponentially small and so the ontribution is negligible. On the transitional range we use (22 and bound the integrand in absolute value. The ( ontribution mn from this part to (k 1 N φ (k is O(1 and sine there are O values of k for whih the transitional range is present we onlude that: The ontribution to the sum (50 from the transitional range is ( mn O (mn ɛ. (51

17 On Linnik and Selberg s Conjeture about Sums of Kloosterman Sums 17 We are left with the ontribution to (50 from the range y k + k 1/3 in the integral defining N φ (k. In this range we have J k (ks ikw (s e (52 k(s2 1 1/4 where W ((s = s 2 1 artan s 2 1. (53 In partiular W (s = s2 1. (54 s Changing variables for the integral in the range in question leads one to onsidering 1+k 2/3 ikw (s φ(ks e ds. (55 k s(s2 1 1/4 We argue as with ˆφ(r and the elementary mean-value estimate. It is enough to bound 1+k 2/3 ikw (s e s ds (56 k(k 2 1 1/4 independent of. Multiply and divide by kw (s and integrate by parts. The boundary terms are e ikw (s/ ( k 3/2 (s 2 1 3/4 evaluated at 1 + k 2/3 and. Hene they are O(1/k. The resulting new integral is k 3/2 1+k 2/3 ikw (s 3s e ds (57 2(s 2 1 7/4 Now bound this trivially by estimating the integrand in absolute value. This also gives a ontribution of O(1/k. ( The number of k s for whih this range mn intersets the support of φ is again O. It follows that; the ontribution to the k sum from this range is

18 18 Peter Sarnak and Jaob Tsimerman ( mn O (mn ɛ. (58 That is we have shown that N φ (k k 0(2 1 j dim S k (Γ ψ j,k (m ψ j,k(n (mn ɛ (1 + mn. (59 (49 and (59 give us the desired bounds for the last two terms in (19. In order to omplete the analysis we must estimate the first term on the right hand side of (19. It is here that we invoke H θ. Consider the dyadi sums A t j 2A ρ j (n ρ j (m osh πt j ˆφ(tj (60 One an treat these in two ways. Firstly we an use H θ diretly from whih it follows (for a Heke basis of Maass forms Hene ρ j (n τ(n n θ ρ j (1. (61 A t j 2A ρ j (n ρ j (m osh πt j τ(n τ(m(nmθ A t j 2A ρ j (1 2 osh πt j (62 We reall Kuznetzov s meal value estimate: t j y ρ j (n 2 osh πt j ( = y2 π + O ɛ y log y + yn ɛ + n 1 +ɛ 2 (63 Applying (63 with n = 1 in (62 yields A t j 2A ρ j (n ρ j (m osh πt j ɛ (nmθ+ɛ A 2. (64 Alternatively we an estimate (60 diretly using (63 via Cauhy-Shwarz and obtain

19 On Linnik and Selberg s Conjeture about Sums of Kloosterman Sums 19 A t j 2A ρ j (n ρ j (m osh (πt j ( 2A A 1/2 ( 2A ρ j (n 2 osh πt j A 1/2 ρ j (m 2 osh πt j ɛ (A + m 1/4+ɛ (A + n 1/4+ɛ. (65 With these we have j ˆφ(t j ρ j(m ρ j (n osh πt j j ˆφ(t j ρ j (n ρ j (m osh πt j (66 and breaking this into dyadi piees applying (47, (48, (64 or (65 one obtains 2A ˆφ(t j ρ j(n ρ j (m osh πt j (mnɛ A ( ( min 1, min A(nm θ, A + (n m 1/4 A 1/2 + (mn 1 4 A 3/2 T A (67 Hene, ( (mn ɛ ( ( min 1, A + m 1/8 + n 1/8 (mn θ/2. (68 T A 2A A ˆφ(t j ρ j(nρ j (m osh πt j (mnɛ Combining the dyadi piees yields ( ( m 1/8 + n 1/8 ( A, (mn θ/2 + min T A (69 j ˆφ(t j ρ j(n ρ j (m osh πt j ( ( (mn ɛ m 1/8 + n 1/8 (mn θ/2 +. (70 T Putting this together with (41 and (51 in (19 yields

20 20 Peter Sarnak and Jaob Tsimerman ( S(m, n; T mn (mn ɛ + 2 ( + m 1/8 + n 1/8 (nm θ/2 +. T Finally hoosing T = 2/3 yields 2 S(m, n; (mn ɛ ( 1/6 + mn + (m 1/8 + n 1/8 (mn θ/2. (71 This ompletes the proof of Proposition 5. In Theorem 4 we eplained the (mn 1/6 in Theorem 2. The 1/6 barrier is similar, that is in the proof of Proposition 5 if we want to go beyond the eponent 1/6 (ignoring the mn dependene we would need to apture anellations in sums of the type ρ j (1 2 itj. (72 osh πt j t j 1/3 This appears to be quite diffiult. A similar feature appears with the eponent of 1/3 in the remainder term in the hyperboli irle problem whih has resisted improvements (see [L-P] and [Iw]. Aknowledgement: We thank D. Hejhal for his omments on an earlier draft of this paper. 3 Referenes [D-I] [De] [Du] J. Deshouillers and H. Iwanie, Kloosterman sums and Fourier oeffiients of usp forms, Invent. Math., 70, (1982/83, no. 2, P. Deligne, La onjeture de Weil. I. (Frenh, Inst. Hautes Études Si. Publ. Math, 43, (1974, T.M. Dunster, Bessel funtions of purely imaginary order with appliations to seond order linear differential equations, Siam Jnl. Math. Anal., 21, No. 4, (1990.

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