On mean values for Dirichlet's polynomials and L-functions

ON MEAN VALUES FOR DIRICHLET’S POLYNOMIALS AND L-FUNCTIONS

arXiv:1609.03738v1 [math.NT] 13 Sep 2016

¨ PATRICK KUHN, NICOLAS ROBLES, AND DIRK ZEINDLER

Abstract. Using a new family of Dirichlet polynomials we extend the current results of mollified integral moments of modular forms. As an application, we improve the current lower bound on critical zeros of holomorphic primitive cusp forms.

1. Introduction 1.1. Review of L-functions. Let H = {x + iy, x ∈ R, y > 0}. A modular form of weight k for the congruence subgroup of a square-free integer N , a b Γ0 (N ) = ∈ SL(2, Z) c ≡ 0 mod N , c d is a complex valued function f : H → C such that: • f is holomorphic; • (f |k γ)(z) := (cz + d)−k f (γz) = f (z) for each γ ∈ Γ0 (N ); • f is holomorphic at all cusps of Γ0 (N ) (meaning that the Fourier series at those cusps is a Taylor series in q := e2πiz ). The cusps are given by γ(∞) = ac where a b γ= c d is an element of Γ0 (N ) \ SL(2, Z). Additionally, f is a cusp form if it is a modular form and if it vanishes at all cusps of Γ0 (N ). Let f denote a primitive cusp form of even weight k. The Fourier expansion of f at the cusp ∞ is given by X f (z) = (1.1) λf (n)n(k−1)/2 e2πinz , n≥1

for every complex number z in the upper half-plane H. The arithmetic normalization is λf (1) = 1. The Fourier coefficients λf (n) satisfy the multiplicative relations X X m n mn (1.2) λf (n)λf (m) = λf and λf (mn) = µ(d)λf λf , d2 d d d|(m,n) (d,N )=1

d|(m,n) (d,N )=1

for all positive integers m and n. Here µ(n) is the usual M¨obius function. For σ := Re(s) > 1, we consider (1.3) X λf (n) Y λf (p) αf (p) −1 βf (p) −1 1 −1 Y L(f, s) = = 1− + χ0 (p) 2s = 1− 1− , ns ps p ps ps p p n≥1

2010 Mathematics Subject Classification. Primary: 11M26; Secondary: 11M06, 11N64. Keywords and phrases. Dirichlet polynomial, mollifier, zeros on the critical line, ratios conjecture technique, autocorrelation, holomorphic cusp form, modular forms, generalized M¨ obius functions. 1


2

which is an absolutely convergent and non-vanishing Dirichlet series. In the Euler product χ0 denotes the trivial character modulo N . Here αf (p), βf (p) are the complex roots of the equation X 2 − λf (p)X + χ0 (p) = 0 and they are called Satake parameters. The function √ s N k−1 L(f, s) = L∞ (f, s)L(f, s) Λ(f, s) = Γ s+ 2π 2 is called the completed L-function of L(f, s). It can be extended to a holomorphic function on C and it satisfies the functional equation Λ(f, s) = ε(f )Λ(f, 1 − s), where ε(f ) = ±1 is the sign of the L-function. The sign is real because the L-function is self-dual. We also use in the following pages the following function χf (s) := ε(f )

L∞ (f, 1 − s) , L∞ (f, s)

so that χf (s)L(f, 1 − s) = L(f, s). The duplication formula for the gamma function allows us to write k 1/2 √ s 2 N s k−1 s k+1 L∞ (f, s) = Γ + Γ + . 8π 2π 2 4 2 4 It is well-known, following analogies of the Riemann zeta-function, that the non-trivial zeros ρf = βf + iγf of L(f, s) are located inside the critical strip 0 < βf < 1. The Rankin-Selberg convolution of two L-functions coming from a primitive cusp forms f and g is the L-function defined as ∞ 2 2 X λf (n)λg (n) Y Y Y αf,i (p)αg,j (p) −1 L(f ⊗ g, s) = L(χ, 2s) = . 1− ns ps p n=1

i=1 j=1

This is an L-function of degree 4. For each prime p, αf,1 (p), αf,2 (p) and αg,1 (p), αg,2 (p) are the roots of the quadratic equations X 2 − λf (p)X + χq (p) = 0,

and X 2 − λg (p)X + χN (p) = 0.

We may also write Y χ0 (p) −1 Y 1 −1 Y 1 L(χ0 , s) = 1− = 1− s = 1 − s ζ(s) =: ζ (N ) (s), s p p p p p6 |N

p|N

so that (1.4)

L(f ⊗ f, s) = ζ (N ) (2s)

∞ X λ2f (n) n=1

ns

for Re(s) > 1. For an L-function of degree 2 we have the unconditional bound (1.5)

|λf (n)| ≤ τ (n)nθ ,

where

θ = 7/64,

where τ (n) is the divisor function, which satisfies τ (n) nε for each ε > 0. This bound, which currently holds the world record, is due to Kim and Sarnak [22]. The Ramanujan hypothesis, on the contrary, states that (1.6)

|λf (n)| ≤ τ (n),

and it is a much stronger bound. It is still unproven for many L-functions of degree 2. However, for primitive cusp forms the Ramanujan hypothesis is proved and the result is due to Deligne [13].

ON MEAN VALUES OF DIRICHLET POLYNOMIALS AND L-FUNCTIONS

3

If Nf (T ) denotes the number of critical (or non-trivial) zeros of L(f, s) up to height 0 < γf < T , then one can show by the argument principle that [21, § 5] T NTd log + O(log q(f, iT )), π (2πe)d for T ≥ 1 and where d denotes the degree of L. Here q denotes the analytic conductor Nf (T ) =

q(f, s) = N q∞ (s) = N

d Y

(|s + κj | + 3),

j=1

where N ≥ 1 is the conductor or level of L(f, s), see [21, pp. 93-95]. Lastly, we will need to know a zero-free region [21, Theorem 5.10]. Specifically, we know that provided the Rankin-Selberg convolutions L(f ⊗ f, s) and L(f ⊗ f¯, s) exist with the latter having a simple pole at s = 1 and the former being entire if f 6= f¯, then there exists an absolute constant c > 0 such that L(f, s) has no zeros in the region c (1.7) , σ ≥1− 4 d log(N (|t| + 3)) except possibly for one simple real zero βf < 1, in which case f is self-dual. 1.2. Motivation. Let Q be a polynomial with complex coefficients satisfying Q(0) = 1 and Q(x)+ Q(1 − x) = constant. Set 1 d (1.8) V (s) = Q − L(f, s), 2 log T ds P where, for large T , we set L = log T . Moreover, let P (x) = j aj xj be a polynomial satisfying P (0) = 0 and P (1) = 1, and let M = T ν/2− where 1 − 2θ 0 1 and σ0 = 1/2 − R/L. Here R is a bounded positive number to be chosen later and M is the length of the mollifier. It is well-known that to replicate the behavior of 1/L(f, s) in the mean value integral Z T (1.11) I= |V ψ(σ0 + it)|2 dt, 1

one chooses (1.10). In [8], Bui, Conrey and Young attached a second piece to this mollifier, i.e. they worked with ψ(s) = ψ1 (s) + ψ2 (s),

4


where ψ1 (s) is now given by (1.12) below X µf (h)hσ0 −1/2 P1 [h], hs

ψ1 (s) =

(1.12)

h≤M1

with M1 = T ν1 , 0 < ν1 < ν/2 as well as P1 (0) = 0 and P1 (1) = 1. Here ψ2 (s) has the shape ψ2 (s) = χf (s + 1/2 − σ0 )

X µf,2 (h)hσ0 −1/2 k 1/2−σ0 P2 [hk], hs k 1−s

hk≤M2

with M2 = T ν2 where 0 < ν2 ≤ ν1 . In this case P2 is some other polynomial such that P2 (0) = P20 (0) = P200 (0) = 0. The terms are µf,2 (h) are given by the Dirichlet convolution (µf ∗ µf )(h). By convention µf,1 = µf . The reasoning behind this choice comes from the formal calculation ∞ X χf (s)L(f, 1 − s) µf,2 (h) 1 = = . χf (s) hs k 1−s L2 (f, s) L(f, s) h,k=1

This indicates that, up some extend, the second piece ψ2 (s) also replicates the behavior of 1/L(f, s). Set λ∗2 f (k) = (λf ∗ λf )(k) and µf,3 (h) = (µf ∗ µf ∗ µf )(h). With this in mind, we can also claim that σ0 −1/2 k 1/2−σ0 X µf,3 (h)λ∗2 f (k)h 2 ψ3 (s) = χf (s + 1/2 − σ0 ) P3 [hk], hs k 1−s hk≤M3

for an appropriate P3 , is a suitable mollifier since (formally) ∞ X µf,3 (h)λ∗2 f (k)

χ2f (s)

hs k 1−s

h,k=1

=

χ2f (s)L2 (f, 1 − s) L3 (f, s)

=

1 . L(f, s)

Naturally, this welcomes a higher order generalization. Suppose that ` ∈ N. This idea may be extended by taking (1.13)

ψ` (s) = χ`−1 f (s +

1 2

X µf,` (h)λ∗`−1 (k)hσ0 −1/2 k 1/2−σ0 f

− σ0 )

hs k 1−s

hk≤M`

P` [hk],

where µf,` (n) is given by ∞

X µf,` (n) 1 = ns L` (f, s)

(1.14)

for

Re(s) > 1,

n=1

∗k and λ∗k f stands for convolving λf with itself exactly k times; in other words λf (n) = (λf ∗· · ·∗λf )(n). The conditions on P` are

P` (0) = 0 (1.15)

and

P` (1) = 1,

when

P` (0) = P`0 (0) = P`00 (0) = · · · =

` = 1,

(`(`−1)) P` (0)

= 0,

when

` > 1,

where P (m) denotes the m-th derivative of P . Moreover M` = T ν` , where 0 < ν` ≤ ν1 . Another formal calculation shows that indeed one has χ`−1 f (s)

∞ X µf,` (h)λ∗`−1 (k) f h,k=1

hs k 1−s

=

`−1 (f, 1 − s) χ`−1 f (s)L

L` (f, s)

=

1 . L(f, s)

Clearly, when ` = 1 (by the use of (4.7) below) and ` = 2, the pieces of Conrey and Levinson and of Bui, Conrey and Young follow as special cases, respectively. Consequently, the mollifier we will


5

be working with is given by ψ(s) =

(1.16)

L X

ψ` (s),

`=1

where L is a positive integer of our choice. Although we will establish our results for general L, when it comes to numerical purposes it is quite sufficient to take L = 2. In this paper we revise the techniques of [8] for a mollifier consisting of several pieces. This approach is extremely general. In fact, previous results in the literature due to Bernard [4] and Bui, Conrey and Young [8] follow as special cases. As an application, we modestly increase the current proportion of critical zeros of L(f, s) and clarify the situation of simple critical zeros. Our technique is based on developments by Conrey and Snaith [12] on ratios conjectures, and by Conrey, Farmer and Zirnbauer [11] on autocorrelation of ratios of L-functions. This generalization shows a very rich structure in the twisted second moments of modular Lfunctions and it is plausible that these new mean value integrals may have potential applications in other ares of number theory. 2. Results The method sketched in [8, 27] to deal with multiple piece mollifiers in the mean value integral (1.11) carries through and our main results are as follows. Theorem 2.1. Suppose that ν` = ν/2 − ε where ν = 1−2θ 4+2θ for ε > 0 small, and that ν` + ν`+1 < 1. Then Z 1 T |V ψ(σ0 + it)|2 dt = c(PL , Q, R, νL ) + o(1), T 1 where Y X 2 (c`,`+k` )k` c(PL , Q, R, νL ) = k1 , k2 , · · · , kL 1≤`≤L

k1 +k2 +···+kL =2

and the different ci,j are given by (2.1) and (2.2). Remark 2.1. There is no need to explicitly compute c`,`+j where j = 2, 3, · · · since the contribution of the associated integral is O(T L−1+ε ), see Theorem 2.5 below. 2.1. The smoothing argument. The idea of smoothing the mean value integrals was worked out in [4, 8, 29] and it makes the following computations more convenient. Let w(t) be a smooth function satisfying the following properties: (a) 0 ≤ w(t) ≤ 1 for all t ∈ R, (b) w has compact support in [T /4, 2T ], (c) w(j) (t) j ∆−j , for each j = 0, 1, 2, · · · and where ∆ = T /L. Note that for the Fourier transform of w, we have w(0) b = T /2 + O(T /L). This allows us to re-write Theorem 2.1 as follows. 1−2θ Theorem 2.2. Suppose that ν` = ν/2 − ε where ν = 4+2θ for ε > 0 small, and that ν` + ν`+1 < 1. For any w satisfying conditions (a), (b) and (c) and σ0 = 1/2 − R/L, Z ∞ w(t)|V ψ(σ0 + it)|2 dt = c(PL , Q, R, νL )w(0) b + O(T /L), −∞

uniformly for R 1, where c(PL , Q, R, νL ) =

X k1 +k2 +···+kL =2

2 k1 , k2 , · · · , kL

Y 1≤`≤L

(c`,`+k` )k` ,


6

where the different ci,j given by (2.1) and (2.2). The technique of a multi-piece mollifier was developed in [8, 20]. In [27] a 4-piece mollifier was handled. The idea is to open the square in the integrand Z Z Z Z Z 2 2 2 2 |V ψ| = |V ψ1 | + |V | ψ1 ψ2 + |V | ψ1 ψ2 + · · · + |V ψL |2 X {I`,` + I`,`+1 + I`+1,` + I`,`+j + I`+j,` }, = 1≤`≤L

for j = 2, 3, 4, · · · . We will compute these integrals in the next sections. The integral I`,`+1 is asymptotically real, thus I`+1,` follows from I`,`+1 , i.e. I`,`+1 ∼ I`+1,` . 2.2. The main terms. The main terms coming from integrals I`,` , I`,`+1 and I`,`+j where j = 2, 3, 4, · · · are now stated as theorems. 1−2θ Theorem 2.3. Let ` ∈ N. Suppose ν` = ν/2 − ε, where ν = 4+2θ for ε > 0 small. Then Z ∞ w(t)|V ψ` (σ0 + it)|2 dt ∼ c`,` (P` , Q, R, ν` )w(0) b + O(T /L) −∞

uniformly for R 1, where c`,` =

d2` 22`(`−1) 1 2 2 2 ((` − 2)!) (` + (` − 1) − 1)! dx` dy ` Z 1Z 1Z 1Z 1 2 1 2 × + (x + y − v(y + r) − u(x + r)) (1 − r)` +(`−1) −1 ν ` 0 0 0 0 × e−ν` R[x+y−v(y+r)−u(x+r)] e2Rt[1+ν` (x+y−v(y+r)−u(x+r))] × Q(ν` (−x + v(y + r)) + t(1 + ν` (x + y − v(y + r) − u(x + r)))) × Q(ν` (−y + u(x + r)) + t(1 + ν` (x + y − v(y + r) − u(x + r)))) × (x + r)`−1 (y + r)`−1 u`−2 v `−2

(2.1)

×

(`(`−1)) P` ((1

− u)(x +

(`(`−1)) r))P` ((1

− v)(y + r))dtdrdudv

. x=y=0

Remark 2.2. The case ` = 1 has to be handled with a certain amount of care as it superficially seems divergent due to the presence of u`−2 and v `−2 in the integrands. This is taken care of by (1/(` − 1)!)2 in the denominator of the first line. By the use of (4.11) below, the term (1/(` − 1)!)2 cancels out the integrals with respect to u and with respect to v, leaving us with 2 Z Z 1 1 1 2Rv d Rνx c1,1 (P, Q, R, ν1 ) = 1 + e (e Q(v + νx)P (x + u))|x=0 dudv, ν1 0 0 dx which is precisely the term recovered by Conrey [10] and Bernard [4], see also (4.6) and [29, p. 544]. Moreover, this interesting phenomenon illustrates the idea that the mollifiers ψ1 and ψ2 are not of ”quite different shape” as argued in [8, p. 37]. Theorem 2.4. Let ` ∈ N. Suppose ν` + ν`+1 < 1. Then Z ∞ w(t)V ψ` ψ`+1 (σ0 + it)dt ∼ c`,`+1 (P` , P`+1 , Q, R, ν` , ν`+1 )w(0) b + O(T /L) −∞

uniformly for R 1, where ZZ Z 1 2 2 22` ν`+1 `(`+1) R d2` c`,`+1 = e u2` (1 − u)2` −1 eR[ν` (y−x)+uν`+1 (a−b)] 2 ` ` a+b≤1 (2` − 1)! ν` dx dy 0 a,b≥0


(2.2)

7

× Q(−xν` + auν`+1 )Q(1 + yν` − buν`+1 ) ν`+1 (`(`+1)) (`(`−1)) `−1 P`+1 ((1 − a − b)u)(ab) dudadb × P` x + y + 1 − (1 − u) . ν` x=y=0

Theorem 2.5. Let ` ∈ N. Suppose ν` + ν`+j < 2 for j = 2, 3, 4, · · · . Then Z ∞ w(t)V ψ` ψ`+j (σ0 + it)dt ` T L−1+ε −∞

uniformly for α, β

L−1 .

The smoothing argument is helpful because we can easily deduce Theorem 2.1 from Theorem 2.2 and so on. By having chosen w(t) to satisfy conditions (a), (b) and (c) and in addition to being an upper bound for the characteristic function of the interval [T /2, T ], and with support [T /2 − ∆, T + ∆], we get Z T |V ψ(σ0 + it)|2 dt ≤ c(P1 , P` , Q, R, ν1 , ν2 )w(0) b + O(T /L). T /2

Note that w(0) b = T /2 + O(T /L). We similarly get a lower bound. Summing over dyadic segments gives the full result. 2.3. The shift parameters α and β. Rather than working directly with V (s), we shall instead consider the following three general shifted integrals Z ∞ I`,` (α, β) = w(t)L(f, 12 + α + it)L(f, 12 + β − it)ψ` ψ` (σ0 + it)dt, −∞ Z ∞ I`,`+1 (α, β) = w(t)L(f, 12 + α + it)L(f, 12 + β − it)ψ` ψ`+1 (σ0 + it)dt, Z−∞ ∞ I`,`+j (α, β) = w(t)L(f, 12 + α + it)L(f, 12 + β − it)ψ` ψ`+j (σ0 + it)dt, −∞

for j = 2, 3, 4, · · · . The computation is now reduced to proving the following three lemmas. Lemma 2.1. We have uniformly for α, β c`,` (α, β) =

L−1 ,

I`,` = c`,` (α, β)w(0) b + O(T /L), where

d2` 1 22`(`−1) ((` − 2)!)2 (`2 + (` − 1)2 − 1)! dx` dy ` Z 1Z 1Z 1Z 1 2 2 × (1 − r)` +(`−1) −1 u`−2 v `−2 0 0 0 0 ν` (β(x−v(y+r))+α(y−u(x+r)))

×T (T 1+ν` (x+y−v(y+r)−u(x+r)) )−t(α+β) 1 × + (x + y − v(y + r) − u(x + r)) (x + r)`−1 (y + r)`−1 ν` (`(`−1)) (`(`−1)) × P` ((1 − u)(x + r))P` ((1 − v)(y + r))dtdrdudv x=y=0

Lemma 2.2. We have I`,`+1 = c`,`+1 (α, β)w(0) b + O(T /L), uniformly for α, β where 2 22` ν`+1 `(`+1) d2` c`,`+1 (α, β) = (2`2 − 1)! ν` dx` dy ` L−1 ,

.


8

ZZ ×

Z a+b≤1 a,b≥0

1

2 −1

u2` (1 − u)2`

0

au (M`−x M`+1 )

−α

−bu (M`y M`+1 T)

−β

ν`+1 (`(`−1)) (`(`+1)) `−1 × P` x + y + 1 − (1 − u) P`+1 ((1 − a − b)u)(ab) dudadb . ν` x=y=0 Lemma 2.3. For j = 2, 3, 4, · · · , we have Z ∞ w(t)V ψ` ψ`+1 (σ0 + it)dt T L−1+ε −∞

uniformly for α, β

L−1 .

To get Theorems 2.3 and 2.4 we use the following technique. Let I? denote either of the integrals in questions, and note that 1 d 1 d I? = Q − Q − I? (α, β) . log T dα log T dβ α=β=R/L Since I? (α, β) and c? (α, β) are holomorphic with respect to α, β small, the derivatives appearing in the equation above can be obtained as integrals of radii L−1 around the points −R/L, using Cauchy’s integral formula. Since the error terms hold uniformly on these contours, the same error terms that hold for I? (α, β) also hold for I? . That the above differential operator on c? (α, β) does indeed give c? follows from log X −1 d −α X =Q X −α . Q log T dα log T Note that from the above equation we get −1 d −1 d −bu au −α T )−β ) (M`y M`+1 Q Q (M`−x M`+1 log T dα log T dβ au log M y M −bu T log M`−x M`+1 ` `+1 −bu au −α =Q T )−β ) (M`y M`+1 Q (M`−x M`+1 log T log T −bu au −α = Q(−xν` + auν`+1 )Q(1 + ν` y − buν`+1 )(M`−x M`+1 ) (M`y M`+1 T )−β ,

as well as −1 d −1 d Q Q T ν` (β(x−v(y+r))+α(y−u(x+r))) (T 1+ν` (x+y−v(y+r)−u(x+r)) )−t(α+β) log T dα log T dβ = Q(ν` (−x + v(y + r)) + t(1 + ν` (x + y − v(y + r) − u(x + r)))) × Q(ν` (−y + u(x + r)) + t(1 + ν` (x + y − v(y + r) − u(x + r)))) × T ν` (β(x−v(y+r))+α(y−u(x+r))) (T 1+ν` (x+y−v(y+r)−u(x+r)) )−t(α+β) . Hence using the differential operators Q((−1/ log T )d/dα) and Q((−1/ log T )d/dβ) on the last line of c`,`+1 (α, β) we get in the integrand eR eR[ν` (y−x)+uν`+1 (a−b)] Q(−xν` + auν`+1 )Q(1 + yν` − buν`+1 ), by setting α = β = −R/L and using T z/L = T z/ log T = ez . Hence Theorem 2.4 follows. Similarly, when we use the differential operators Q((−1/ log T )d/dα) and Q((−1/ log T )d/dβ) on the integrand of c`,` (α, β) it becomes e−ν` R[x+y−v(y+r)−u(x+r)] e2Rt[1+ν` (x+y−v(y+r)−u(x+r))] × Q(ν` (−x + v(y + r)) + t(1 + ν` (x + y − v(y + r) − u(x + r)))) × Q(ν` (−y + u(x + r)) + t(1 + ν` (x + y − v(y + r) − u(x + r)))),


9

by the same substitutions. This proves Theorem 2.3. 3. Preliminary tools The following results are needed throughout the paper. The first lemma is used to compute the ”square” terms I`,` . We start by quoting a result from Bernard’s paper [4, Lemma 1], who in turn quoted it with small modifications from [21, Theorem 5.3]. Lemma 3.1. Let G be any entire function which decays exponentially fast in vertical strips, is even and normalised by G(0) = 1. Then for any α, β ∈ C such that 0 ≤ | Re(α)|, | Re(β)| ≤ 12 , one has X X λf (m)λf (n) m −it Vα,β (mn, t) L(f, 21 + α + it)L(f, 12 + β − it) = m1/2+α n1/2+β n m≥1 n≥1 X X λf (m)λf (n) m −it V−β,−α (mn, t), + Xα,β,t (t) m1/2−β n1/2−α n m≥1 n≥1

where gα,β (s, t) =

L∞ (f, 21 + α + s + it)L∞ (f, 12 + β + s − it) , L∞ (f, 12 + α + it)L∞ (f, 12 + β − it)

and Vα,β (x, t) = as well as Xα,β,t (t) =

1 2πi

Z (1)

G(s) gα,β (s, t)x−s ds, s

L∞ (f, 12 − α − it)L∞ (f, 12 − β + it) . L∞ (f, 12 + α + it)L∞ (f, 12 + β − it)

Lemma 3.2. Suppose that w satisfies the three conditions (a), (b), (c), and suppose that h, k are positive integers with hk ≤ T ν . Then one has −it Z ∞ X λf (m)λf (n) Z ∞ h 1 1 w(t) L(f, 2 + α + it)L(f, 2 + β − it)dt = w(t)Vα,β (mn, t)dt k m1/2+α n1/2+β −∞ −∞ hm=kn X λf (m)λf (n) Z ∞ + w(t)V−β,−α (mn, t)Xα,β,t dt 1/2−β n1/2−α m −∞ hm=kn + Oε ((hk)(1+θ)/2 T 1/2+θ+ε ), for α, β L−1 . Proof. This is proved by applying Lemma 3.1 to the right-hand side above −it Z ∞ h L(f, 12 + α + it)L(f, 12 + β − it)dt w(t) k −∞ X X λf (m)λf (n) Z ∞ hm −it = w(t)Vα,β (mn, t)dt m1/2+α n1/2+β −∞ kn m≥1 n≥1 X X λf (m)λf (n) Z ∞ hm −it + w(t)Xα,β,t (t)V−β,−α (mn, t)dt. m1/2−β n1/2−α −∞ kn m≥1 n≥1

Clearly the main terms appearing in the statement of the lemma are given by the diagonal case hm = kn. Let us now look at the off-diagonal terms. Following [4] we set X λf (m)λf (n) Z ∞ hm −it ND1 Ih,k (α, β) = w(t)Vα,β (mn, t)dt, m1/2+α n1/2+β −∞ kn hm6=kn


10

and ND2 (α, β) Ih,k

λf (m)λf (n) = m1/2−β n1/2−α hm6=kn X

Z

∞

−∞

hm kn

−it w(t)Xα,β,t (t)Vα,β (mn, t)dt.

By [4, Lemma 3], we have that for any ε > 0, 0 < γ < 1 and for any real A > 0 Z ∞ X λf (m)λf (n) hm −it ND1 Vα,β (mn, t)dt + O(T −A ), Ih,k (α, β) = w(t) 1/2+α n1/2+β kn m −∞ km6=hn mnT 2+ε hm | kn −1|T −γ ≤ T ν and α, β

provided that hk L−1 . Now fix an arbitrary smooth function ρ :]0, ∞[7→ R, compactly supported in [1, 2] and with ∞ X ρ(2−l/2 x) = 1. l=−∞

For each integer l, we define ρl (x) = ρ(x/Al ) By [4, Lemma 4] one has X ND1 Ih,k (α, β) =

X

Al1 Al2 hkT 2+ε 0 0 such that 1 A H , max{P1 , P2 } (l1 l2 M1 M2 P1 P2 )ε then one has H X X a(h) λf (m1 )λf (m2 )gh (m1 , m2 ) h=1

m1 ,m2 ≥1

p 1/2 θ 3/2 A h1 ||a||2 (P1 + P2 ) P1 + P2 + × (l1 l2 M1 M2 P1 P2 H)ε , for all ε > 0.

A max{P1 , P2 }

s θ (h1 , l1 l2 )H 1+ h1 l1 l2


11

Interestingly, as remarked in [4, Remark 8], Theorem 1.3 of [5] would not have been enough to circumvent the technicalities behind estimate (3.1). See [4, pp. 215-217] for further details. The required bounds for the test function were established in [4, Lemma 5]. Specifically, let α, β L−1 be complex numbers and let σ be any positive real number. For all non-negative integers i and j, we have i+j a 1/2+Re(α)+σ b 1/2+Re(β)+σ 1+2σ j i j ∂ xy Fq;l1 ,l2 (x, y) i,j T log T, ∂xi ∂y j Al1 Al2 p where the implicit constant does not depend on q. Using Lemma 3.3 with H = T −γ Al1 Al2 , h1 = 1 and ( a(h) =

1, 0,

if h ≤ H, otherwise,

as well as the previous results on Fq;l1 ,l2 yields ND1 (α, β) ε (hk)(1+θ)/2 T 1/2+θ+ε . Ih,k

(3.1) Similarly, one has

ND2 Ih,k (α, β) ε (hk)(1+θ)/2 T 1/2+θ+ε ,

and this can be shown by a similar argument.

Remark 3.1. One can replace in Lemma 3.1 the exponential decay of G(s) by a weaker decay 2 rate. However, we will use only G(s) = es p(s) with p a polynomial and we thus will not go into further detail. Specifically we shall use the pole annihilator 2

G(s) = es p(s),

and p(s) =

(α + β)2 − (2s)2 , (α + β)2

The following lemma, which is an adaption of the approximate functional equation, is needed for the computation of the term ”crossterms” I`,`+1 . P Lemma 3.4. Let σα,−β (f, l) = ab=l a−α bβ λf (a)λf (b). For L2 ≤ |t| ≤ 2T one has L(f, 21 + α + it)L(f, 12 − β + it) =

∞ X σα,−β (f, l) l=1

l1/2+it

3

e−l/T + O(T −1+ε ),

uniformly for α, β < L−1 . Proof. See Lemma 4.1 of [8].

Lemma 3.5. Let {f (m)}m∈N be a sequence of complex numbers and suppose that X (3.2) f (m) = cM + O(M 3/5 ) m≤M

as M → ∞ for some c ∈ C. We then have for k ∈ N (3.3)

X

(f ∗k )(m) = ck M

m≤M

logk−1 M + O(M logk−2 M ) k!

as M → ∞. Furthermore, we have (3.4)

X (f ∗k )(m) logk M = ck + O(logk−1 M ) m k!k

m≤M


12

as M → ∞. Let n ≥ 1 and M 0 ≥ 1 with log M 0 log M be given. We then have uniformly in γ for all |γ| ≤ 1/2 0 Z n X (log M M 0 −γ ck M −γ logk M 1 γr k−1 ∗k m ) = M r (log(M 0 /M r ))n dr (f )(m) m m k! 0 m≤M

+ O((log M )k+n−1 ),

(3.5)

as M → ∞, and the expression in (3.5) has order of magnitude (log M )k+n if c 6= 0. Proof. We first prove (3.3). We do this by induction over k. For k = 1, this is trivial. We thus assume (3.3) is true for k − 1. To simplify the notation, we write g(m) := (f ∗k−1 )(m) and get X X X X X g(d)f (m/d) = g(a)f (b) (f ∗k )(m) = (g ∗ f )(m) = m≤M

m≤M

m≤M d|n

ab≤M

3/5 X X X cM M f (b) = g(a) = g(a) . +O a a

(3.6)

a≤M

b≤M/a

a≤M

We first consider the main term. We use partial summation and get X X 1 X X X cM 1 1 g(a) = cM g(a) = cM 1/M g(a) − g(a) − a a m+1 m a≤M a≤M a≤M m≤M −1 a≤m k−2 X m 1 k−2 k−1 log k−3 = O(M log M ) + cM c m + O(m log m) (k − 1)! m(m + 1) m≤M −1 X logk−2 m X logk−3 m ck M k−2 + O M log M +M = (k − 1)! m+1 m m≤M −1

= ck M

log

k−1

m≤M −1

M

+ O(M logk−2 M ). k! It remains to show that the error term in (3.6) is of lower order. We have 3/5 X X 1 M 3/5 g(a)O =O M g(a) a a3/5 a≤M a≤M X X 1 X 1 1 3/5 =M O g(a) g(a) − − M 3/5 a≤M (m + 1)3/5 m3/5 m≤M −1 a≤m X 1 k−2 k−2 = O(M log M) + O m log m · 6/5 = O(M logk−2 M ). m m≤M −1 This completes the proof of (3.3). The proof of (3.4) is almost the same as (3.5) and we thus prove only (3.5). We first consider the case γ = 0 and get (3.7) n n X (log M 0 )n X X log (M 0 /(m + 1)) log (M 0 /m)) ∗k ∗k m (f )(m) = − (f )(a) − . m m+1 m m≤M

m≤M −1

a≤m

Note that logn (M 0 /(m + 1)) (log(M 0 ) − log(m + 1))n 1 = m+1 m 1 + 1/m 0 (log(M ) − log(m) − log(1 + 1/m))n 1 1 = 1− +O m m m2


(log(M 0 /m) − 1/m + O(1/m2 ))n = m

1 1− +O m

1 m2

13

(log(M 0 /m))n 1 n = − 2 (log(M 0 /m))n − 2 (log(M 0 /m))n−1 + O m m m

logn−1 M 0 m3

.

Inserting this and (3.3) in (3.7) gives k−1 X m k log k−2 c m + O(m log m) k! m≤M −1 n−1 0 n n−1 n log M 1 0 0 . log(M /m) + 2 log(M /m) +O × 2 3 m m m Applying Euler-Maclaurin summation to the leading term gives Z ck X ck M 1 k−1 0 n (log m) (log(M /m)) = (log y)k−1 (log(M 0 /y))n dy + O(logk+n−1 M ) k! k! 1 y m≤M −1 Z 1 ck k (log y)k−1 (log(M 0 /M r ))n dr + O(logk+n−1 M ). = (log M ) k! 0 Using the variable substitution y = M r gives the main term in (3.5). Applying Euler-Maclaurin summation to the remaining terms, one sees immediately that they are O(logk+n−1 M ). This completes the proof of (3.5). The argumentation for γ 6= 0 is almost identical and has to use that 1 1 1 1 1 2 = 1−γ = 1−γ 1 − (1 − γ) + O(1/m ) , (m + 1)1−γ m (1 + 1/m)1−γ m m with O(1/m2 ) uniform in γ for |γ| ≤ 1/2. This completes the proof of the lemma.

This lemma can be upgraded to read like Lemma 3.3 of [27] and Lemma 4.4 of [8] by incorporating smooth functions F and H. Our choice of c will be ress=1 L(f ⊗ f, s) (3.8) , c= ζ (N ) (2) as per the asymptotic behavior X ress=1 L(f ⊗ f, s) (3.9) + O(x3/5 ), λ2f (n) = x (N ) (2) ζ n≤x as x → ∞, found by Rankin [25]. 4. Evaluation of the shifted mean value integrals I? (α, β) 4.1. The mean value integral I`,` (α, β). The strategy is to insert the definition of ψ` into the mean value integral I`,` and then compute this integral by using the tools we have developed. One key aspect will be the p-adic evaluation of a certain arithmetical sum into a ratio of L-functions. This has the effect of transforming the problem from an arithmetical one to an analytic counterpart. Using (1.13) on I`,` , we get Z ∞ I`,` (α, β) = w(t)L(f, 21 + α + it)L(f, 12 + β − it)ψ` ψ` (σ0 + it)dt Z−∞ ∞ `−1 1 1 1 1 w(t)χ`−1 = f ( 2 − it)χf ( 2 + it)L(f, 2 + α + it)L(f, 2 + β − it) −∞

×

X h1 k1 ≤M`

µf,` (h1 )λ∗`−1 (k1 ) f 1/2−it 1/2+it k1

h1

P` [h1 k1 ]

X h2 k2 ≤M`

µf,` (h2 )λ∗`−1 (k2 ) f 1/2+it 1/2−it k2

h2

P` [h2 k2 ]dt


14

X

=

µf,` (h1 )µf,` (h2 )λ∗`−1 (k1 )λ∗`−1 (k2 ) f f

X

(h1 h2 k1 k2 )1/2

h1 k1 ≤M` h2 k2 ≤M`

where Z

∞

−it

h2 k1 = w(t) h 1 k2 −∞

J2,f

P` [h1 k1 ]P` [h2 k2 ]J2,f ,

L(f, 12 + α + it)L(f, 12 + β − it)dt,

since χf ( 12 + it)χf ( 12 − it) = 1 for all values of t. We now use Lemma 3.2 and rewrite J2,f as Z ∞ X λf (m)λf (n) J2,f = (4.1) w(t)Vα,β (mn, t)dt m1/2+α n1/2+β −∞ h2 k1 m=h1 k2 n Z ∞ X λf (m)λf (n) w(t)V−β,−α (mn, t)Xα,β,t dt + m1/2−β n1/2−α −∞ h k m=h k n 2 1

1 2

+ Oε ((h2 k1 h1 k2 )(1+θ)/2 T 1/2+θ+ε ), 0 (α, β)+ with Vα,β , V−β,−α and Xα,β,t as in Lemma 3.2. This means that we can write I`,` (α, β) = I`,` 00 (α, β) + E(h , h , k , k ), where E is the error term above. Note that I 00 (α, β) can be obtained I`,` 1 2 1 2 `,` 0 (α, β) by switching α by −β and multiplying by from I`,` √ −2(α+β) t N i(α2 − β 2 ) −2 Xα,β,t = (4.2) 1+ + O(t ) . 2π t √ See [4, Lemma 2 and p. 229] to verify that the effect of N is immaterial. Furthermore −α−β t = T −α−β + O(L−1 ), 2π 00 (α, β) = T −α−β I 0 (−β, −α) + O(T /L). To estimate the error for t T . More precisely, we have I`,` `,` term, we use the bounds ε and λ∗` f (k) ` k .

µf,` (h) ` 1,

(4.3)

Note that we are working with cusp forms and we thus can use the bound (1.6) for λf . We get X

X

µf,` (h1 )λ∗`−1 (k1 )µf,` (h2 )λ∗`−1 (k2 ) f f (h1 k1 h2 k2 )1/2

h1 k1 ≤M` h2 k2 ≤M`

X

=

X

(k1 )µf,` (h2 )λ∗`−1 (k2 ) µf,` (h1 )λ∗`−1 f f (h1 k1 h2 k2 )

h1 k1 ≤M` h2 k2 ≤M`

= T 1/2+θ+ε

X

1/2+θ+ε

X

E(h1 , h2 , k1 , k2 )

X

1/2

T 1/2+θ+ε (h1 k1 h2 k2 )(1+θ)/2

µf,` (h1 )λ∗`−1 (k1 )µf,` (h2 )λ∗`−1 (k2 )(h1 k1 h2 k2 )θ/2 f f

h1 k1 ≤M` h2 k2 ≤M`

=T

µf,` (h)λ∗`−1 (k)(hk)θ/2 f

2

hk≤M`

` T

1/2+θ+ε

X

θ/2 θ/2+ε

h

k

2 =T

θ/2+1+2ε 2

τ (n)n

) = T 1/2+θ+ε+ T 2ν` (θ/2+1+2ε) ,

where we have used M` = T ν` . Hence, if (4.4)

X n≤M`

hk≤M`

T 1/2+θ+ε (M`

1/2+θ+ε

ν
−1/2, Re(z) > −1/4 − Re(α)/2 − Re(β)/2, Re(s) + Re(z) > −1/2 − Re(α), Re(u) + Re(z) > −1/2 − Re(β). Then one has X

µf,` (h1 )µf,` (h2 )λ∗`−1 (k1 )λ∗`−1 (k2 )λf (m)λf (n) f f

h2 k1 n=h1 k2 m

1/2+s 1/2+u 1/2+s 1/2+u 1/2+α+z 1/2+β+z h2 k1 k2 m n

h1 = ×

L`

2 +(`−1)2

(f ⊗ f, 1 + s + u)L(f ⊗ f, 1 + α + β + 2z) ⊗ f, 1 + 2s)L`(`−1) (f ⊗ f, 1 + 2u)

L`(`−1) (f

L`−1 (f ⊗ f, 1 + α + s + z)L`−1 (f ⊗ f, 1 + β + u + z) Aα,β (s, u, z), L` (f ⊗ f, 1 + β + s + z)L` (f ⊗ f, 1 + α + u + z)

where Aα,β (s, u, z) is given by an absolutely convergent Euler product on Ωα,β . Proof. Let us set (k2 )λf (m)λf (n) (k1 )λ∗`−1 µf,` (h1 )µf,` (h2 )λ∗`−1 f f

X

S`,` =

h2 k1 n=h1 k2 m

1/2+s 1/2+u 1/2+s 1/2+u 1/2+α+z 1/2+β+z h2 k1 k2 m n

.

h1

We now write this as an Euler product over primes so that S`,` =

Y

µf,` (p`1 )µf,` (p`2 )λ∗`−1 (p`3 )λ∗`−1 (p`4 )λf (p`5 )λf (p`6 ) f f

X

p `2 +`3 +`6 =`1 +`4 +`5

(p`1 )

1/2+s

(p`2 )

1/2+u

(p`3 )

1/2+s

(p`4 )

1/2+u

(p`5 )

1/2+α+z

1/2+β+z

(p`6 )

,

where we have employed the substitutions h1 = p`1 , h2 = p`2 , k1 = p`3 , k2 = p`4 and m = p`5 , n = p`6 . Using the fact µ`+1 (p) = −(` + 1)λf (p) and λ∗`−1 (p) = (` − 1)λf (p) we have f Y λf (p)2 (`2 + (` − 1)2 ) λf (p)2 `(` − 1) λf (p)2 ` λf (p)2 (` − 1)` S`,` = 1+ − − − p1+s+u p1+2s p1+2u p1+β+s+z p λf (p)2 (` − 1) λf (p)2 ` λf (p)2 (` − 1) λf (p)2 −2+δ(s,u,z,α,β) + − 1+α+u+z + + 1+α+β+2z + O(p ) p p1+α+s+z p1+β+u+z p =

L`

2 +(`−1)2


L`(`−1) (f


16

×

L`−1 (f ⊗ f, 1 + α + s + z)L`−1 (f ⊗ f, 1 + β + u + z) Aα,β (s, u, z), L` (f ⊗ f, 1 + β + s + z)L` (f ⊗ f, 1 + α + u + z)

where δ(s, u, z, α, β) ∈ Ωα,β and Aα,β (s, u, z) =

Y

1+

X

p

r,l

. (s,u,z,α,β)

ap,l,` (p) pr+Xr,l,`

Here |ap,l,` | `2 and Xr,l,` (s, u, z, α, β) are linear forms in s, u, z, α, β and the sum over r, l is absolutely convergent in Ωα,β . Note that when ` = 1, the above reduces to X µf (h1 )µf (h2 )λf (m)λf (n) 1/2+s 1/2+u 1/2+α+z 1/2+β+z h2 m n h2 n=h1 m h1

(4.6)

=

L(f ⊗ f, 1 + s + u)L(f ⊗ f, 1 + α + β + 2s) Aα,β (s, u, z), L(f ⊗ f, 1 + β + s + z)L(f ⊗ f, 1 + α + u + z)

since lim λ∗`−1 (k) `→1 f `∈N

(4.7)

( 1, = 0,

if k = 1, otherwise.

Consequently, we have z Z ∞ X X a`,i i!a`,j j! 1 3 Z Z Z t G(z) s+u I 0`,` (α, β) = w(t) M ` i+j 2πi 2π z log M` −∞ (1) (1) (1) i j × ×

L`

2 +(`−1)2


L`(`−1) (f

L`−1 (f ⊗ f, 1 + α + s + z)L`−1 (f ⊗ f, 1 + β + u + z) ds du Aα,β (s, u, z)dz i+1 j+1 dt. ` ` s u L (f ⊗ f, 1 + β + s + z)L (f ⊗ f, 1 + α + u + z)

Now that we have transformed the arithmetic part of the problem into its analytic counterpart, we can proceed to compute these integrals. To do so, we move the s-, u- and z-contours of integration to δ > 0 small. This is then followed by deforming the z-contour to −δ + ε, thereby crossing the simple pole of 1/z at z = 0. Recall that G(z) vanishes at the pole of ζ(1 + α + β + 2z). The new contour of integration yields a contribution of size z Z Z Z ∞ X X a`,i i!a`,j j! 1 3 Z t G(z) s+u M w(t) ` i+j 2πi 2π z log M` Re(s)=δ Re(u)=δ Re(z)=−δ+ε −∞ i j 2 +(`−1)2

L`

×


L`(`−1) (f

× Z

L`−1 (f ⊗ f, 1 + α + s + z)L`−1 (f ⊗ f, 1 + β + u + z) ds du Aα,β (s, u, z)dz i+1 j+1 dt ` ` s u L (f ⊗ f, 1 + β + s + z)L (f ⊗ f, 1 + α + u + z) ∞

−∞

|w(t)|dtT 2(−δ+ε) M`2δ T 1−(2−2ν` )δ+ε T 1−ε

0 (α, β) as I 0 (α, β) = I 0 1−ε ), where for sufficiently small ε. Let us now write I`,` `,` `,`,0 (α, β) + O(T 0 I`,`,0 (α, β) corresponds to the residue at z = 0, i.e. z Z ∞ X X a`,i i!a`,j j! 1 2 Z Z t G(z) s+u 0 res M` I `,`,0 (α, β) = w(t) i+j z=0 2πi 2π z log M` −∞ (δ) (δ) i j


L`

×

2 +(`−1)2

17


L`(`−1) (f

ds du L`−1 (f ⊗ f, 1 + α + s + z)L`−1 (f ⊗ f, 1 + β + u + z) Aα,β (s, u, z) i+1 j+1 dt ` ` s u L (f ⊗ f, 1 + β + s + z)L (f ⊗ f, 1 + α + u + z) X X a`,i i!a`,j j! = w(0)L(f b ⊗ f, 1 + α + β) K`,` , i+j log M ` i j ×

where K`,` =

1 2πi

2 Z

2

Z

(δ)

(δ)

M`s+u

2

L` +(`−1) (f ⊗ f, 1 + s + u) L`(`−1) (f ⊗ f, 1 + 2s)L`(`−1) (f ⊗ f, 1 + 2u)

L`−1 (f ⊗ f, 1 + α + s)L`−1 (f ⊗ f, 1 + β + u) ds du Aα,β (s, u, 0) i+1 j+1 . ` ` s u L (f ⊗ f, 1 + β + s)L (f ⊗ f, 1 + α + u)

×

Before we compute K`,` , we need to sort out the situation with Aα,β . One can see that

=

=

µf,` (h1 )µf,` (h2 )λ∗`−1 (k1 )λ∗`−1 (k2 )λf (m)λf (n) f f

X

A0,0 (s, s, s) =

(h1 h2 k1 k2 mn)1/2+s X ∗`−1 µf,` (h2 )λf (k1 )λf (n)

h2 k1 n=h1 k2 m ∞ X −1−s

j

j=1 ∞ X

X

(k2 )λf (m) µf,` (h1 )λ∗`−1 f

h1 k2 m=j

h2 k1 n=j

2 ∗`−1 −1−s j (µf,` ∗ λf ∗ λf )(j) .

j=1

It now follows by the definition of λf and µf,` , see (1.3) and (1.14), that A0,0 (s, s, s) = 1, for all values of s. We now use the Rankin-Selberg convolution L-function given by (1.4) to reverse the order of summation and integration and then move the contour integral to the far right. In other words, ∗`2 +(`−1)2

K`,` =

X (λ2f (n))

m

n≤M`

×

M` m

s+u

1 2πi

2 Z (δ)

Z (δ)

`2 +(`−1)2

{ζ (N ) (2(1 + s + u))} L`(`−1) (f ⊗ f, 1 + 2s)L`(`−1) (f ⊗ f, 1 + 2u)

L`−1 (f ⊗ f, 1 + α + s)L`−1 (f ⊗ f, 1 + β + u) ds du Aα,β (s, u, 0) i+1 j+1 . ` ` s u L (f ⊗ f, 1 + β + s)L (f ⊗ f, 1 + α + u)

To simplify the calculations that will follow shortly, we will set the integrand to be `2 +(`−1)2

(M` /m)s+u {ζ (N ) (2(1 + s + u))} r`,` (α, β, i, j, s, u) = i+1 j+1 s u L`(`−1) (f ⊗ f, 1 + 2s)L`(`−1) (f ⊗ f, 1 + 2u) ×

L`−1 (f ⊗ f, 1 + α + s)L`−1 (f ⊗ f, 1 + β + u) Aα,β (s, u, 0). L` (f ⊗ f, 1 + β + s)L` (f ⊗ f, 1 + α + u)

We are going to follow a reasoning analogous to [8] and [23] by using the zero-free region of L(f ⊗ f, s), see [21, Theorem 5.10]. More precisely, by taking (1.7) into account, we consider the contour γ = γ1 ∪ γ2 ∪ γ3 given by γ1 = {iτ : |τ | ≥ Y }, γ2 = {σ ± iY : −c/ log Y ≤ σ ≤ 0}, γ3 = {−c/ log Y + iτ : |τ | ≤ Y },


18

with c > 0 and Y ≥ 1 large, where c is so chosen that there are no zeros between the curve γ and Re = δ. Since L(f ⊗ f, s) does not vanish, we replace the double integrals of Re(u) = Re(v) = δ by the contour of integration γ so that by the Cauchy residue theorem we have Z Z 1 2 r`,` (α, β, i, j, s, u)dsdu 2πi (δ) (δ) Z Z Z 1 1 2 = res r`,` (α, β, i, j, s, u)du + r`,` (α, β, i, j, s, u)dsdu s=0 2πi Re(u)=δ 2πi s∈γ Re(u)=δ Z 1 r`,` (α, β, i, j, s, u)du = res r`,` (α, β, i, j, s, u) + res s=u=0 s=0 2πi u∈γ Z Z Z 1 1 2 + res r`,` (α, β, i, j, s, u)ds + (4.8) r`,` (α, β, i, j, s, u)dsdu. u=0 2πi s∈γ 2πi s∈γ u∈γ R 1 The first estimation will be that of ress=0 2πi s∈γ r`,` (α, β, i, j, s, u)ds. To estimate this, we will first write the residue as a contour integral over a small circle of radius 1/L centered at 0, i.e. Z Z 1 1 2 (M` /m)u L`−1 (f ⊗ f, 1 + β + u) res r`,` (α, β, i, j, s, u)du = `(`−1) (f ⊗ f, 1 + 2u)L` (f ⊗ f, 1 + α + u) s=0 2πi u∈γ 2πi u∈γ L I M` s (N ) `2 +(`−1)2 × {ζ (2(1 + s + u))} Aα,β (s, u, 0) m D(0,L−1 ) ×

L`−1 (f ⊗ f, 1 + α + s) ds du . `(`−1) ` L (f ⊗ f, 1 + 2s)L (f ⊗ f, 1 + β + s) si+1 uj+1

We also have the bound [4, 21] 1 log |τ |. L(f ⊗ f, σ + iτ )

(4.9) Next we use the fact that

ζ (N ) (2(1 + s + u))Aα,β (s, u, 0) 1 in this contour of integration, as well as the bound ` L`−1 (f ⊗ f, 1 + α + s) `(`−1)−i−1 (β + s) (2s) Li−`(`−1) , si+1 L`(`−1) (f ⊗ f, 1 + 2s)L` (f ⊗ f, 1 + β + s) (α + s)`−1

1

since s 1/L. Using the fact that the arclength of the curve is 1/L, we obtain Z 1 res r`,` (α, β, i, j, s, u)du s=0 2πi u∈γ Z du (M` /m)Re(u) L`−1 (f ⊗ f, 1 + β + u) i−1−`(`−1) L `(`−1) ` (f ⊗ f, 1 + 2u)L (f ⊗ f, 1 + α + u) |u|j+1 u∈γ L Z Z 0 (log τ )`(`−1)+` dσ i−1−`(`−1) i−1−`(`−1) `(`−1)+` L dτ + L (log Y ) j+1 |τ | |σ + iY |j+1 |τ |≥Y −c/ log Y −c/ log Y Z M` dτ + Li−1−`(`−1) (log Y )`(`−1)+` j+1 m |τ |≤Y |τ − ic/ log Y | 1 M` −c/ log Y `(`−1)+` i−1−`(`−1) j L (log Y ) + (log Y ) . Yj m


19

Consequently, we get −c/ log Y Z 1 1 j M` i−1−`(`−1) `(`−1)+` res r`,` (α, β, i, j, s, u)du L (log Y ) + (log Y ) . s=0 2πi u∈γ Yj m For reasons of symmetry, i.e. r(α, β, i, j, s, u) = r(β, α, j, i, u, s), we also get −c/ log Y Z 1 1 j−1−`(`−1) `(`−1)+` i M` res r`,` (α, β, i, j, s, u)du L (log Y ) . + (log Y ) u=0 2πi s∈γ Yi m Keeping this bound in mind, we can bound the double integrals over γ as Z Z 1 2 r`,` (α, β, i, j, s, u)dsdu 2πi s∈γ u∈γ Z (M` /m)Re(s) L`−1 (f ⊗ f, 1 + α + s) ds `(`−1) ` (f ⊗ f, 1 + 2s)L (f ⊗ f, 1 + β + s) |s|i+1 s∈γ L Z (M` /m)u L`−1 (f ⊗ f, 1 + β + u) du × `(`−1) ` (f ⊗ f, 1 + 2u)L (f ⊗ f, 1 + α + u) |u|j+1 u∈γ L −c/ log Y −c/ log Y 1 1 i M` j M` (log Y )2(`(`−1)+`) + (log Y ) + (log Y ) Yi m Yj m −c/ log Y 1 M` (log Y )2(`(`−1)+`) + (log Y )i+j . i+j Y m Let us now set 2 2 −c/ log Y X (λ2f (n))∗` +(`−1) 1 q M` Ω(`, q) := + (log Y ) . n Yq m n≤M`

Using (3.9) and Lemma 3.5, we can bound Ω(`, q) by 1 2 2 2 2 −c/ log Y c/ log Y Ω(`, q) q (log M` )` +(`−1) + M` (log Y )q M` (log M` )` +(`−1) Y 2 2 (log T )` +(`−1) 2 2 (4.10) + (log Y )q (log T )` +(`−1) , q Y 2 2 since log T log M` . Choosing Y = log T , we obtain Ω(`, q) q (log T )` +(`−1) + . When we sum over m, we see that 2 +(`−1)2

K`,` =

X (λ2f (m))∗`

res r`,` (α, β, i, j, s, u)

m

m≤M`

s=u=0

+ O(Li−1−`(`−1) Ω(`, j)(log Y )`(`−1)+` + Lj−1−`(`−1) Ω(`, i)(log Y )`(`−1)+` + Ω(`, i + j)(log Y )2(`(`−1)+`) ) 2 +(`−1)2

=

X (λ2f (m))∗`

res r`,` (α, β, i, j, s, u)

m

m≤M`

s=u=0

2 +(`−1)2 +

+ O((log T )`

(Li−1−`(`−1) + Lj−1−`(`−1) + 1)),

Recall that we have i, j ≥ `2 − ` + 1. Therefore, K`,` =

X (λ2f (m))∗` m≤M`

m

2 +(`−1)2

res r`,` (α, β, i, j, s, u) + O(log T i+j−1+ε ).

s=u=0


20

Let us now move on to the main term. We first notice that `2 +(`−1)2

{ζ (N ) (2(1 + s + u))} Aα,β (s, u, 0) L`−1 (f ⊗ f, 1 + α + s)L`−1 (f ⊗ f, 1 + β + u) L`(`−1) (f ⊗ f, 1 + 2s)L`(`−1) (f ⊗ f, 1 + 2u) L` (f ⊗ f, 1 + β + s)L` (f ⊗ f, 1 + α + u) `2 +(`−1)2

=

{ζ (N ) (2)}

(2s)`(`−1) (2u)`(`−1) 2 +(`−1)2 +1

(ress=1 L(f ⊗ f, s))`

(α + u)` (β + s)` (α + s)`−1 (β + u)`−1

+ O(1/L2`(`−1)+3 ),

since A0,0 (0, 0, 0) = 1. We now get the product of two neatly separated integrals `2 +(`−1)2 ζ (N ) (2) 1 2`(`−1) res r`,` (α, β, i, j, s, u) = 2 s=u=0 ress=1 L(f ⊗ f, s) ress=1 L(f ⊗ f, s) s I ` ds 1 M` (β + s) × `−1 si+1−`(`−1) 2πi D(0,L−1 ) m (α + s) u I 1 M` du (α + u)` × . `−1 j+1−`(`−1) 2πi D(0,L−1 ) m u (β + u) Let us remark that the second integral is the same as the first integral except that i has to be replaced by j and α has to be replaced by β. Consequently, it is enough to compute any of these two integrals. The first integral is computed below. Lemma 4.2. One has that I 1 M` s (β + s)` 1 d` ds 1 M` `−1+i−`(`−1) = x + log 2πi D(0,L−1 ) m (` − 2)! (i − `(` − 1))! dx` m (α + s)`−1 si+1−`(`−1) −αu Z 1 i−`(`−1) x(β−αu) M` `−2 × u (1 − u) e du , m 0 x=0 for i ≥ `(` − 1) + 1. Proof. The first remark is that d` (β+s)x (β + s) = ` e dx x=0 `

for all integer values of `. Next, set 1 Υ1 (α, β, `) = 2πi

I D(0,L−1 )

M` m

s

(β + s)` (α + s)

`−1

ds si+1−`(`−1)

so that d` Υ1 (α, β, `) = ` eβx Υ11 (x)|x=0 dx

where

1 Υ11 (x) = 2πi

s 1 ds x M` e . `−1 i+1−`(`−1) m s (α + s) D(0,L−1 )

I

Now taking a power series of the exponential inside Υ11 yields I X 1 M` r 1 sr−i−1+`(`−1) Υ11 (x) = x + log ds. r! m 2πi D(0,L−1 ) (α + s)`−1 r≥0 The poles of the integrand are s = −α and when r − i − 1 + `(` − 1) ≤ −1, thus the easiest approach is the one put forward in [8], namely that of computing the residue at infinity. By making the change of variables s 7→ 1/s we get I 2 X 1 M` r 1 si−r−` +2`−2 Υ11 (x) = x + log ds. r! m 2πi D(0,L−1 ) (1 + αs)`−1 r≥0


21

Moreover, we take a power series of (1 + αs)1−` by the use of the binomial theorem with fractional powers X 1 − ` 1 1−` (αs)k . = (1 + αs) = `−1 k (1 + αs) k≥0 Here 1−` (1 − `)(1 − ` − 1)(1 − ` − 2) · · · (1 − ` − k + 1) = . k k! When we insert this into Υ11 we have I X 1 M` r X 1 − ` k 1 2 α x + log sk+i−r−` +2`−2 ds. Υ11 (x) = k r! m 2πi D(0,L−1 ) r≥0

k≥0

This integral picks out r = k + i − `2 + 2` − 1, thus 2 αk M` k M` i−` +2`−1 X 1 − ` x + log . Υ11 (x) = x + log k m (k + i − `2 + 2` − 1)! m k≥0

To end this calculation we invoke the confluent hypergeometric function of the first kind 1 F1 , see e.g. [1]. This allows us to write 2 M` i−` +2`−1 1 Υ11 (x) = x + log 2 m (i − ` + 2` − 1)! M` 2 × 1 F1 ` − 1, i + 2` − ` , −a x + log m i−`2 +2`−1 M` 1 Γ(i + 2` − `2 ) = x + log 2 m (i − ` + 2` − 1)! Γ(i + ` − `2 + 1)Γ(` − 1) Z 1 M` 2 × e−a(x+log m )u u`−2 (1 − u)i+2`−` −` du 0

2 −au Z 1 2 M` i−` +2`−1 1 −aux M` = x + log e u`−2 (1 − u)i+`−` du, m (i + ` − `2 )!(` − 2)! 0 m provided ` > 1. Moreover, we remark that Z 1 2 1 (4.11) lim e−au u`−2 (1 − u)i+`−` du = 1 `→1 (` − 2)! 0 provided i > −1. Putting these results together we see that 2 d` βx 1 M` i−` +2`−1 Υ1 (α, β, `) = ` e x + log m (i + ` − `2 )!(` − 2)! dx −au Z 1 i+`−`2 −aux M` `−2 × e u (1 − u) du , m 0 x=0 as it was to be shown. This ends the proof. We can now insert this result in the residue at s = u = 0 to obtain `2 +(`−1)2 1 ζ (N ) (2) 2`(`−1) K`,` = 2 ress=1 L(f ⊗ f, s) ress=1 L(f ⊗ f, s) 1 1 1 × 2 (i − `(` − 1))! (j − `(` − 1))! ((` − 2)!)


22

2 2 X (λ2f (n))∗` +(`−1) M` `−1+i−`(`−1) M` `−1+j−`(`−1) d2` x + log y + log × ` ` m m m dx dy m≤M` −αu−βv Z 1Z 1 i−`(`−1) j−`(`−1) x(β−αu) y(α−βv) M` `−2 `−2 × u v (1 − u) (1 − v) e e dudv m 0 0 x=y=0

+ O(Li+j−2+ε ). 0 Let us perform the sums over i and j in the expression for I`,`,0 . For the first sum we have

X a`,i i! 1 M` `−1+i−`(`−1) x + log (1 − u)i−`(`−1) i (i − `(` − 1))! m log M` i M` `−1 −`(`−1) x + log = (log M` ) m i−`(`−1) ` X (x + log M m ) × a`,i i(i − 1)(i − 2) · · · (i − `(` − 1) + 1) (1 − u) log M` i `−1 ` (x + log M M` (`(`−1)) −`(`−1) m ) P` . = (log M` ) x + log (1 − u) m log M` Similarly, for the second sum we have X a`,j j! 1 M` `−1+j−`(`−1) y + log (1 − u)j−`(`−1) j (j − `(` − 1))! m log M ` j ` (y + log M M` `−1 (`(`−1)) −`(`−1) m ) P` . = (log M` ) y + log (1 − u) m log M` 0 becomes Therefore, the expression for I`,`,0 0 (α, β) I`,`,0

`2 +(`−1)2 22`(`−1) w(0) b ζ (N ) (2) = α+β ress=1 L(f ⊗ f, s) Z 1Z 1 1 d2` αy+βx × e u`−2 v `−2 e−αux−βvy 2 dx` dy ` ((` − 2)!) 0 0 2 +(`−1)2

×

X (λ2f (m))∗`

` (x + log M m )

`−1

` (y + log M m )

`−1

M` m

−αu−βv

log2`(`−1) M` ` ` (x + log M (y + log M (`(`−1)) (`(`−1)) m ) m ) × P` (1 − u) P` (1 − v) dudv , log M` log M` x=y=0 m

m≤M`

(4.12)

+ O(T L−1+ε ),

where we have used the Laurent expansion L(f ⊗ g, 1 + α + β) =

ress=1 L(f ⊗ g, s) + O(1). α+β

We shall write the main in a more convenient way as `2 +(`−1)2 22`(`−1) w(0) b ζ (N ) (2) 0 I`,`,0 (α, β) = α+β ress=1 L(f ⊗ f, s)


d2` 1 × ((` − 2)!)2 dx` dy `

1Z 1

Z 0

∗`2 +(`−1)2

×

X (λ2f (m))

(`(`−1)) P`

x(β−αu)+y(α−βv)

u`−2 v `−2 M`

(x +

m

m≤M`

×

0

23

`−1 log(M` /m) `−1 ` /m) (y + log(M M` −αu−βv log M` ) log M` ) m log2`(`−1) M` log2 M`

` ` log M log M (`(`−1)) m m P` (1 − v) y + dudv (1 − u) x + log M` log M` x=y=0

+ O(T 1−ε ). By the Euler-Maclaurin result of Lemma 3.5, with k = `2 + (` − 1)2 , s = −αu − βv, x = z = M` , (`(`−1)) (`(`−1)) F (r) = (x + r)`−1 P` ((1 − u)(x + r)) as well as H(r) = (y + r)`−1 P` ((1 − v)(y + r)), we obtain 2 2 X (λ2f (m))∗` +(`−1) log M` /n `−1 log M` /n `−1 x+ y+ log M` log M` m1−αu−βv m≤M` log M` /n log M` /n (`(`−1)) (`(`−1)) × P` P` (1 − u) x + (1 − v) y + log M` log M` 2 `2 +(`−1)2 2 ress=1 L(f ⊗ f, s) (log M` )` +(`−1) = ζ (N ) (2) (`2 + (` − 1)2 − 1)!M`−αu−βv Z 1 2 2 × (1 − r)` +(`−1) −1 (x + r)`−1 (y + r)`−1 ×

0 (`(`−1)) P` ((1

(`(`−1))

− u)(x + r))P`

r(−αu−βv)

((1 − v)(y + r))M`

2 +(2`−1)2 −1

dr + O(L`

).

Consequently, we are left with 0 (α, β) = I`,`

22`(`−1) w(0) b 1 1 2 2 (α + β) log M` ((` − 2)!) (` + (` − 1)2 − 1)! Z 1Z 1Z 1 d2` β(x−v(y+r))+α(y−u(x+r)) × ` ` M` dx dy 0 0 0 × u`−2 v `−2 (1 − r)` ×

(`(`−1)) P` ((1

2 +(`−1)2 −1

− u)(x +

(x + r)`−1 (y + r)`−1

(`(`−1)) r))P` ((1

− v)(y + r))drdudv

+ O(T L−1+ε ).

x=y=0 0 I`,` (α, β) and

00 (α, β), where As we discussed earlier, to form the full I`,` (α, β) we need to add I`,` 00 0 I`,` (α, β) is formed by taking I`,` (α, β), then we switch α and −β, and finally we multiply by T −α−β . To accomplish this, we first let β(x−v(y+r))+α(y−u(x+r))

U (α, β) =

M`

−α(x−v(y+r))−β(y−u(x+r))

− T −α−β M` α+β

.

This implies that I`,` (α, β) =

22`(`−1) w(0) b 1 1 (α + β) log M` ((` − 2)!)2 (`2 + (` − 1)2 − 1)! Z 1Z 1Z 1 d2` β(x−v(y+r))+α(y−u(x+r)) × ` ` M` dx dy 0 0 0 × U (α, β)u`−2 v `−2 (1 − r)`

2 +(`−1)2 −1

(x + r)`−1 (y + r)`−1 + O(T L−1+ε ).


24

However, we can also write U (α, β) =

β(x−v(y+r))+α(y−u(x+r)) 1 M`

x+y−v(y+r)−u(x+r) −α−β

− (T M`

)

.

α+β

Finally, the identity 1 − z −α−β = log z α+β combined with the fact that M` = T ν` yields c`,` (α, β) =

Z

1

z −t(α+β) dt,

0

22`(`−1) 1 d2` ((` − 2)!)2 (`2 + (` − 1)2 − 1)! dx` dy ` Z 1Z 1Z 1Z 1 2 2 × (1 − r)` +(`−1) −1 u`−2 v `−2 0

0

0

0

× T ν` (β(x−v(y+r))+α(y−u(x+r))) (T 1+ν` (x+y−v(y+r)−u(x+r)) )−t(α+β) 1 × + (x + y − v(y + r) − u(x + r)) (x + r)`−1 (y + r)`−1 ν` (`(`−1)) (`(`−1)) × P` ((1 − u)(x + r))P` ((1 − v)(y + r))dtdrdudv

.

x=y=0

This proves Lemma 2.1. Theorem 2.3 follows by using −1 d −1 d Q c`,` (α, β) c`,` = Q log T dα log T dβ α=β=−R/L =

d2` 1 22`(`−1) ((` − 2)!)2 (`2 + (` − 1)2 − 1)! dx` dy ` Z 1Z 1Z 1Z 1 2 1 2 × + (x + y − v(y + r) − u(x + r)) (1 − r)` +(`−1) −1 ν` 0 0 0 0 × e−ν` R[x+y−v(y+r)−u(x+r)] e2Rt[1+ν` (x+y−v(y+r)−u(x+r))] × Q(ν` (−x + v(y + r)) + t(1 + ν` (x + y − v(y + r) − u(x + r)))) × Q(ν` (−y + u(x + r)) + t(1 + ν` (x + y − v(y + r) − u(x + r)))) × (x + r)`−1 (y + r)`−1 u`−2 v `−2 ×

(`(`−1)) P` ((1

− u)(x +

(`(`−1)) r))P` ((1

− v)(y + r))dtdrdudv

. x=y=0

This ends the computation of the I`,` term. 4.2. The mean value integral I`,`+1 (α, β). We shall follow a similar strategy to that of the case I`,`+1 , except that now we will have the factor χf (1/2 + it) inside the integral J2,f below. This fact will account for the presence of the arithmetic term σα,−β (f, l) in the p-adic sum. We start by plugging in the definitions of ψ` and ψ`+1 into the mean value integral I`,`+1 so that Z ∞ I`,`+1 (α, β) = w(t)L(f, 12 + α + it)L(f, 12 + β − it)ψ` ψ`+1 (σ0 + it)dt Z−∞ ∞ ` 1 1 1 1 = w(t)χ`−1 f ( 2 − it)χf ( 2 + it)L(f, 2 + α + it)L(f, 2 + β − it) −∞

×

X h1 k1 ≤M`

µf,` (h1 )λf∗`−1 (k1 ) 1/2−it 1/2+it k1

h1

P` [h1 k1 ]

X h2 k2 ≤M`+1

µf,`+1 (h2 )λ∗` f (k2 ) 1/2+it 1/2−it k2

h2

P`+1 [h2 k2 ]dt


X

=

X

µf,` (h1 )µf,`+1 (h2 )λ∗`−1 (k1 )λ∗` f (k2 ) f (h1 h2 k1 k2 )1/2

h1 k1 ≤M` h2 k2 ≤M`+1

25

P` [h1 k1 ]P`+1 [h2 k2 ]J1,f ,

where Z

J1,f

∞

h2 k1 w(t) = h1 k2 −∞

−it

χf ( 12 + it)L(f, 12 + α + it)L(f, 12 + β − it)dt,

since χf ( 12 + it) = χf ( 12 − it)−1 for all values of t. At this point we employ the functional equation of L(f, 1/2 + β − it) as well as the Stirling approximation −β t χf (1/2 + β − it)χf (1/2 + it) = (1 + O(t−1 )), 2π to write h2 k1 −it t −β w(t) = L(f, 12 + α + it)L(f, 12 − β + it)dt + O(T ε ). h1 k2 2π −∞ Z

J1,f

∞

Now that we have opposite signs in front of α and β we apply Lemma 3.4 to get Z ∞ X σα,−β (f, l) −l/T 3 ∞ h2 k1 l −it t −β e w(t) dt + O(T ε ). J1,f = 1/2 h1 k2 2π l −∞ l=1 When we plug this back into I`,`+1 we see that X

I`,`+1 (α, β) =

∞ X µf,` (h1 )µf,`+1 (h2 )λ∗`−1 (k1 )λ∗` f (k2 )σα,−β (f, l) f

X

h1 k1 ≤M` h2 k2 ≤M`+1 l=1

× P` [h1 k1 ]P`+1 [h2 k2 ]c w0

(h1 h2 k1 k2 l) h2 k1 l 1 log , 2π h1 k2

1/2

e−l/T

3

t −β where w0 (t) := w(t)( 2π ) .

4.3. Bounding the off-diagonal terms. Let C`,`+1 denote the contribution to I`,`+1 from the off-diagonal terms, so that C`,`+1 (α, β) =

µf,` (h1 )µf,`+1 (h2 )λ∗`−1 (k1 )λ∗` f (k2 )σα,−β (f, l) f

X

(h1 h2 k1 k2 l)

h1 k1 ≤M` h2 k2 ≤M`+1 l≥1 h1 k2 6=h2 k1 l

×w c0

h2 k1 l 1 log 2π h1 k2

1/2

3

e−l/T P` [h1 k1 ]P`+1 [h2 k2 ]

.

ν`+1 , we have to estimate the above term. Since we define Given that M` = T ν` and M`+1 R ∞= T t −β w0 (x) = w(x)( 2π ) , we have −∞ w0 (x) dx T . Furthermore, it was shown in [8] that 1 T (4.13) log x B w0 T 2π (1 + L log x)B

for any B ≥ 0. Let us split C`,`+1 into (4.14)

0 00 C`,`+1 = C`,`+1 + C`,`+1

0 with C`,`+1 =

X 1≤l≤T 4

00 and C`,`+1 =

X l≥T 4

.


26

For the second term, we get the bound 00 C`,`+1

X

|µf,` (h1 )µf,`+1 (h2 )λ∗`−1 (k1 )λ∗` f (k2 )σα,−β (l)| f

X

(h1 h2 k1 k2 l)1/2

l≥T 4 h1 k1 ≤M` h2 k2 ≤M`+1 h1 k2 6=h2 k1 l

X

` T

X

l≥T 4 h1 k1 ≤M` h2 k2 ≤M`+1 h1 k2 6=h2 k1 l

T

X

l

e

−l/T 3

Z

∞

w0 (x) dx −∞

(h1 h2 k1 k2 l)ε −l/T 3 e (h1 h2 k1 k2 l)1/2

−1/2+ε −l/T 3

X

e

h1 k1 ≤M`

l≥T 4

1 (h1 k1 )1/2−ε

X h1 k1 ≤M`+1

1 (h2 k2 )1/2−ε

T 4 e−T T 1/2(ν` +ν`+1 )+3ε T −2017 , 0 where we have used (4.3). We now come to C`,`+1 . We choose ν` and ν`+1 so that ν` + ν`+1 < 1 h2 k1 l and thus we have for h1 k2 6= 1 that 1 1 h2 k1 l −1+ε (4.15) . 1 − h1 k2 ≥ h1 k2 ≥ M` M`+1 ≥ T

Therefore w0

1 h2 k1 l log 2π h1 k2

B B

T = (1 + log( hh21kk12l ))B

(1 +

T L

(1 +

T −1+ε B ) LT

T

T L

T log(1 + ( hh21kk12l − 1)))B

T 1−εB .

Using this, we obtain 0 C`,`+1

X

|µf,` (h1 )µf,`+1 (h2 )λ∗`−1 (k1 )λ∗` f (k2 )σα,−β (l)| f

X

1≤l≤T 4 h1 k1 ≤M` h2 k2 ≤M`+1 h1 k2 6=h2 k1 l

X

` T 1−εB

X

l≥T 4 h1 k1 ≤M` h2 k2 ≤M`+1 h1 k2 6=h2 k1 l

(h1 h2 k1 k2 l)1/2

−l/T 3

e

1 h2 k1 l w b0 2π log h1 k2

(h1 h2 k1 k2 l)ε −l/T 3 e T 4−εB T 1/2(ν` +ν`+1 )+3ε T −2017 , (h1 h2 k1 k2 l)1/2

again by the use of (4.3). This shows that for ν` + ν`+1 < 1 the off-diagonal terms get absorbed in the error term and do not contribute to our final results. 4.4. The diagonal terms h1 k2 = h2 k1 l and their reduction to a contour integral. By using the Mellin identities deg XP` a`,i X a`,i i! 1 Z M` s ds i P` [h1 k1 ] = (log M` /h1 k1 ) = , logi M` logi M` 2πi (1) h1 k1 si+1 i=0

i

and deg P`+1

P`+1 [h2 k2 ] =

X j=0

X a`+1,j j! 1 a`+1,j j (log M /h k ) = 2 2 `+1 logj M`+1 logj M`+1 2πi j

Z (1)

M`+1 h2 k2

u

du , uj+1


as well as the Cahen-Mellin integral Z 1 −y e = Γ(z)y −z dz, 2πi (c)

c > 0,

27

Re(y) > 0,

we arrive at I`,`+1 (α, β) = w c0 (0)

XX i

j

a`,i i!a`+1,j j! logi M` logj M`+1

1 2πi

3 Z

Z

(1)

Z

(1)

u T 3z Γ(z)M`s M`+1

(1)

X


h2 k1 l=h1 k2

1/2+s 1/2+u 1/2+s 1/2+u 1/2+z h1 h2 k1 k2 l

×

We must now evaluate the arithmetic sum

P

dz

ds du + O(T 1−ε ). si+1 uj+1

by p-adic methods.

h2 k1 l=h1 k2

Lemma 4.3. Let Υα,β be the set of vectors u, s, z ∈ C3 satisfying Re(s) > −1/4, Re(u) > −1/4, Re(z) + Re(u) > −1/2 − Re(α), Re(z) + Re(u) > −1/2 + Re(β), Re(s) + Re(z) > −1/2 − Re(α), Re(s) + Re(z) > −1/2 + Re(β). Then one has X (k1 )λ∗` µf,` (h1 )µf,`+1 (h2 )λ∗`−1 f (k2 )σα,−β (f, l) f 1/2+s 1/2+u 1/2+s 1/2+u 1/2+z h2 k1 k2 l

h1

h2 k1 l=h1 k2

2

=

L2` (f ⊗ f, 1 + s + u)L` (f ⊗ f, 1 + α + u + z)L` (f ⊗ f, 1 − β + u + z) L`(`−1) (f ⊗ f, 1 + 2s)L`(`+1) (f ⊗ f, 1 + 2u)L` (f ⊗ f, 1 + α + s + z)L` (f ⊗ f, 1 − β + s + z) × Bα,β (s, u, z),

where Bα,β (s, u, z) is given by an absolutely convergent Euler product on Υα,β . Proof. Let us set S`,`+1 =

(k1 )λ∗` µf,` (h1 )µf,`+1 (h2 )λ∗`−1 f (k2 )σα,−β (f, l) f

X

1/2+s 1/2+u 1/2+s 1/2+u 1/2+z h2 k1 k2 l

The definition of σα,−β (f, l) = S`,`+1 =

P

ab=l

X

.

h1

h2 k1 l=h1 k2

λf (a)λf (b)a−α bβ allows us to write

µf,` (h1 )µf,`+1 (h2 )λ∗`−1 (k1 )λ∗` f (k2 )λf (a)λf (b) f

h2 k1 ab=h1 k2

1/2+s 1/2+u 1/2+s 1/2+u 1/2+α+z 1/2−β+z h2 k1 k2 a b

.

h1

We now translate this into an Euler product over primes so that S`,`+1 =

Y

`4 `5 `6 µf,` (p`1 )µf,`+1 (p`2 )λ∗`−1 (p`3 )λ∗` f (p )λf (p )λf (p ) f

X

p `2 +`3 +`5 +`6 =`1 +`4

(p`1 )

1/2+s

(p`2 )

1/2+u

1/2+s

(p`3 )

1/2+u

(p`4 )

1/2+α+z

(p`5 )

1/2−β+z

(p`6 )

,

where we have used the substitutions h1 = p`1 , h2 = p`2 , k1 = p`3 , k2 = p`4 and a = p`5 , b = p`6 . Note that we can consider only first order terms in p since they are enough to determine the expression in the lemma. Using the facts that µf,` (p) = −`λf (p) and λ∗` f (p) = `λf (p) we have Y (` + 1)` λf (p)2 `(` − 1) λf (p)2 `λf (p)2 `λf (p)2 (` + 1)` λf (p)2 S`,`+1 = 1+ − − − − p1+s+u p1+2s p1+α+s+z p1+2u p1−β+s+z p


28

(` − 1)` λf (p)2 `λf (p)2 `λf (p)2 −2+ε(s,u,z,α,β) + + 1+α+u+z + 1−β+u+z + O(p ) p1+s+u p p 2

L2` (f ⊗ f, 1 + s + u)L` (f ⊗ f, 1 + α + u + z)L` (f ⊗ f, 1 − β + u + z) = `(`−1) L (f ⊗ f, 1 + 2s)L`(`+1) (f ⊗ f, 1 + 2u)L` (f ⊗ f, 1 + α + s + z)L` (f ⊗ f, 1 − β + s + z) × Bα,β (s, u, z), where ε(s, u, z, α, β) ∈ Υα,β and X Y Bα,β (s, u, z) = 1+ p

r,l

, (s,u,z,α,β)

bp,l,` (p) pr+Yr,l,`

with |bp,l,` (p)| `2 and Yr,l,` (u, s, z, α, β) are linear forms in s, u, z, α, β and the sum over r, l is absolutely convergent in Υα,β . This means that we are left with XX I`,`+1 (α, β) = w c0 (0) i

×

j


1 2πi

3 Z

Z

(1)

(1)

Z

u T 3z Γ(z)M`s M`+1

(1)

1 L`(`+1) (f ⊗ f, 1 + 2u)L` (f ⊗ f, 1 + α + s + z)` (f ⊗ f, 1 − β + s + z)

L` (f ⊗ f, 1 + α + u + z)` (f ⊗ f, 1 − β + u + z) L`(`+1) (f ⊗ f, 1 + 2u) ds du 2 × L2` (f ⊗ f, 1 + s + u)Bα,β (s, u, z)dz i+1 j+1 + O(T 1−ε ). s u

×

The next step is to move the s- and u- contours of integration to Re(s) = Re(u) = δ, and then move the z-contour to −2δ/3, where δ > 0 is some fixed constant such that the arithmetical factor converges absolutely. This has the effect of crossing a simple pole at z = 0. Also, on the new paths the integral can be bounded in a straightforward way by using absolute values. Thus, the contribution to I`,`+1 is Z Z Z ∞ X X a`,i i!a`+1,j j! 1 3 Z u T 3z Γ(z)M`s M`+1 w(t) i j 2πi log M` log M`+1 Re(s)=δ Re(u)=δ Re(z)=−2δ/3 −∞ i j 2

L2` (f ⊗ f, 1 + s + u)L` (f ⊗ f, 1 + α + u + z)L` (f ⊗ f, 1 − β + u + z) × `(`−1) L (f ⊗ f, 1 + 2s)L`(`+1) (f ⊗ f, 1 + 2u)L` (f ⊗ f, 1 + α + s + z)L` (f ⊗ f, 1 − β + s + z) Z ∞ ds du M` M`+1 δ × Bα,β (s, u, z)dz i+1 j+1 dt |w(t)|dt T 1−ε , 2 s u T −∞ since ν` + ν`+1 < 2. This implies that I`,`+1 (α, β) = w c0 (0)

XX i

×

j


1 2πi

2 Z (δ)

Z

u res T 3z Γ(z)M`s M`+1

(δ) z=0

1 L`(`+1) (f

⊗ f, 1 +

2u)L` (f

⊗ f, 1 + α + s + z)` (f ⊗ f, 1 − β + s + z)

L` (f ⊗ f, 1 + α + u + z)` (f ⊗ f, 1 − β + u + z) L`(`+1) (f ⊗ f, 1 + 2u) ds du 2 × L2` (f ⊗ f, 1 + s + u)Bα,β (s, u, z) i+1 j+1 + O(T 1−ε ) s u

×


=w c0 (0)

XX i

j

1 2πi

29

a`,i i!a`+1,j j! K`,`+1 (α, β, i, j) + O(T 1−ε ), logi M` logj M`+1

where K`,`+1 (α, β, i, j) =

2 Z

Z

(δ)

2

u M`s M`+1 L2` (f ⊗ f, 1 + s + u)Bα,β (s, u, 0)

(δ)

1

×

L`(`+1) (f ⊗ f, 1 + 2u)L` (f ⊗ f, 1 + α + s)` (f ⊗ f, 1 − β + s)

L` (f ⊗ f, 1 + α + u)` (f ⊗ f, 1 − β + u) ds du × . si+1 uj+1 L`(`+1) (f ⊗ f, 1 + 2u) Before we proceed we note that Bα,β (s, s, s) =

=

=

X


h2 k1 l=h1 k2

(h1 h2 k1 k2 l)1/2+s

X

µf,` (h1 )µf,`+1 (h2 )λ∗`−1 (k1 )λ∗` f (k2 )σα,−β (f, l) f (h1 k2 )1+2s X µf,` (h1 )λ∗` f (k2 )

h2 k1 l=h1 k2 ∞ X X j=1

h1 k2 =j

(h1 k2 )1+2s

(k1 ) σα,−β (f, l)µf,`+1 (h2 )λ∗`−1 f

= 1,

h2 k1 l=j

because the first bracket is 0 if j 6= 1 and 1 if j = 1 and since X (k1 ) = 1, σα,−β (f, l)µf,`+1 (h2 )λ∗`−1 f h2 k1 l=j

when j = 1 by the definition of σα,−β (f, l). This means that Bα,β (s, s, s) = 1 for all values of s. Let us recall that the Rankin-Selberg convolution L-function is given ∞ X λf (n)λg (n) L(f ⊗ g, s) = L(χ, 2s) . ns n=1

By moving the u-integral to the far right (since ν`+1 ≤ ν` ) we obtain 2 X (λ2f (n))∗2` 1 2 Z Z 2`2 K`,`+1 (α, β, i, j) = {ζ (N ) (2(1 + s + u))} n 2πi (δ) (δ) n≤M`+1

×

1 L`(`+1) (f

⊗ f, 1 +

2u)L` (f

⊗ f, 1 + α + s)` (f ⊗ f, 1 − β + s)

L` (f ⊗ f, 1 + α + u)` (f ⊗ f, 1 − β + u) L`(`+1) (f ⊗ f, 1 + 2u) s M` M`+1 u ds du × Bα,β (s, u, 0) i+1 j+1 . n n s u The double integral K`,`+1 can be computed by similar methods to those employed in the calculation of K`,` . We define the integrand to be ×

r`,`+1 (α, β, i, j, s, u) =

{ζ (N ) (2(1 + s + u))}

2`2

L`(`+1) (f ⊗ f, 1 + 2u)L` (f ⊗ f, 1 + α + s)` (f ⊗ f, 1 − β + s) L` (f ⊗ f, 1 + α + u)` (f ⊗ f, 1 − β + u) M` s M`+1 u 1 1 × Bα,β (s, u, 0). i+1 j+1 `(`+1) n n s u L (f ⊗ f, 1 + 2u)


30

Let us follow the strategy of I`,` and that of Lemma 5.7 of [8] by using the zero-free region of L(f ⊗ f, s), see [21]. Since L(f ⊗ f, s) does not vanish, we replace the double integrals of Re(u) = Re(v) = δ by the contour of integration γ on page 18. We get by the Cauchy residue theorem Z Z 1 2 r`,`+1 (α, β, i, j, s, u)dsdu 2πi (δ) (δ) Z Z Z 1 1 2 = res r`,`+1 (α, β, i, j, s, u)du + r`,`+1 (α, β, i, j, s, u)dsdu s=0 2πi Re(u)=δ 2πi s∈γ Re(u)=δ Z 1 r`,`+1 (α, β, i, j, s, u)du = res r`,`+1 (α, β, i, j, s, u) + res s=u=0 s=0 2πi u∈γ Z Z Z 1 2 1 r`,`+1 (α, β, i, j, s, u)ds + r`,`+1 (α, β, i, j, s, u)dsdu. + res u=0 2πi s∈γ 2πi s∈γ u∈γ R 1 Again, the first estimation will again be that of ress=0 2πi s∈γ r`,`+1 (α, β, i, j, s, u)ds. We start by writing the residue as a contour integral over a small circle of radius 1/L centered at 0, i.e. Z 1 r`,`+1 (α, β, i, j, s, u)du res s=0 2πi u∈γ Z 1 2 M`+1 u L` (f ⊗ f, 1 + α + u)L` (f ⊗ f, 1 − β + u) = 2πi n L`(`+1) (f ⊗ f, 1 + 2u) u∈γ I 2`2 {ζ (N ) (2(1 + s + u))} Bα,β (s, u, 0) M` s ds du × . `(`−1) ` ` n L (f ⊗ f, 1 + 2s)L (f ⊗ f, 1 + α + s)L (f ⊗ f, 1 − β + s) si+1 uj+1 D(0,L−1 ) Next we use the fact that ζ (N ) (2(1 + s + u))Bα,β (s, u, 0) 1 in this contour of integration and L`(`−1) (f

⊗ f, 1 +

2s)L` (f

1 (2s)`(`−1) (α + s)` (−β + s)` ⊗ f, 1 + α + s)L` (f ⊗ f, 1 − β + s) L−`(`+1) ,

since s 1/L, to write Z 1 res r`,`+1 (α, β, i, j, s, u)du s=0 2πi u∈γ Z M`+1 Re(u) L` (f ⊗ f, 1 + α + u)L` (f ⊗ f, 1 − β + u) du Li−`(`+1) |u|j+1 . n L`(`+1) (f ⊗ f, 1 + 2u) u∈γ The novelty is that in addition to the bound (4.9), we shall also use [21, Chapter 5] (4.16)

0

L(f ⊗ f, σ + iτ ) N,ε |τ |4(α +ε) ,

α0 = max{ 12 (1 − σ), 0},

for all ε > 0 and where N is the level of the L-function. This enables us to obtain Z M`+1 Re(u) L` (f ⊗ f, 1 + α + u)L` (f ⊗ f, 1 − β + u) du |u|j+1 n L`(`+1) (f ⊗ f, 1 + 2u) u∈γ Z Z 0 log`(`+1) |τ | dσ `(`+1) dτ + (log Y ) j+1−ε j+1−4c`/ log Y −ε |τ | |τ |≥Y −c/ log Y |σ + iY | M`+1 −c/ log Y + (log Y )`(`+1) n Z Z dτ × + |τ − ic/ log Y |j+1−4c`/ log Y −ε c/ log Y ≤|τ |≤Y 0≤|τ |≤c/ log Y


31

−c/ log Y (log Y )`(`+1) j+`(`+1)+ε M`+1 + (log Y ) . Yj n This means that −c/ log Y Z 1 1 j M`+1 i−`(`+1) `(`+1)+ε res r`,`+1 (α, β, i, j, s, u)du L (log Y ) . + (log Y ) s=0 2πi u∈γ Yj n By using a similar technique, one gets −c/ log Y Z 1 1 i M` j−`(`−1) `(`+1)+ε r`,`+1 (α, β, i, j, s, u)ds L (log Y ) + (log Y ) . res u=0 2πi s∈γ Yi n Now we bound the double integral over s ∈ γ and u ∈ γ, i.e. Z Z 1 2 r`,`+1 (α, β, i, j, s, u)dsdu 2πi s∈γ u∈γ Z M` Re(s) ds 1 `(`−1) (f ⊗ f, 1 + 2s)L` (f ⊗ f, 1 + α + s)L` (f ⊗ f, 1 − β + s) si+1 n L s∈γ Z M`+1 Re(u) L` (f ⊗ f, 1 + α + u)L` (f ⊗ f, 1 − β + u) du × n uj+1 L`(`+1) (f ⊗ f, 1 + 2u) u∈γ `(`+1) −c/ log Y log Y j+`(`+1) M`+1 + (log Y ) Yj n `(`+1) −c/ log Y log Y i+`(`+1) M` × + (log Y ) Yi n −c/ log Y 1 i+j M` 2`(`+1) (log Y ) + (log Y ) , Y i+j n since max(M` , M`+1 ) = M` . Let us recall that 2 −c/ log Y X (λ2f (n))∗2` 1 q M` . Ω(q) = + (log Y ) n Yq n n≤M`+1

We use Lemma 3.5 regarding the convolution λ2f (n)∗k , to write Ω(q) =

∗2`2

(λ2f (n))

X

n

n≤M`+1

1 q Y

−c/ log Y 1 q M` + (log Y ) Yq n 2

(λ2f (n))∗2`

X n≤M`+1

+ (log Y )

n

q

X n≤M`+1

(λ2f (n))∗2`

2

n

M` n

−c/ log Y

2

log2` (M`+1 ) 2 + (log Y )q (log M`+1 )2` . q Y The choice of Y has to be such that Y = o(T ), specifically we take Y = log T = L. Thus we get 2 Ω(q) L2` + . Using that i ≥ (` + 1)2 − (` + 1) + 1 and j ≥ `2 − ` + 1 and putting all pieces together, we obtain ∗2`2

K`,`+1 (α, β, i, j) =

X n≤M`+1

(λ2f (n)) n

res r`,`+1 (α, β, i, j, s, u)

s=u=0

+ O(Li−`(`+1) Ω(j) log Y + Lj−`(`−1) Ω(i) log Y + Ω(i + j)log2 Y )


32

∗2`2

(λ2f (n))

X

=

n≤M`+1

res r`,`+1 (α, β, i, j, s, u) + O(Li+j−1 ).

n

s=u=0

In order to get the main term of the lemma we need to compute the residue at s = u = 0 of r`,`+1 . This is accomplished by expressing the residue as two contour integrals over small circles of radii 1/L centered at 0. In other words, I I 1 2 M` s M`+1 u res r`,`+1 (α, β, i, j, s, u) = s=u=0 2πi n n D(0,L−1 ) D(0,L−1 ) 2`2

× ×

{ζ (N ) (2(1 + s + u))}

L`(`+1) (f ⊗ f, 1 + 2u)L` (f ⊗ f, 1 + α + s)` (f ⊗ f, 1 − β + s) ds du L` (f ⊗ f, 1 + α + u)` (f ⊗ f, 1 − β + u) Bα,β (s, u, 0) i+1 j+1 . `(`+1) s u L (f ⊗ f, 1 + 2u)

Now we must separate the complex variables s and u to decouple these two integrals. To do this, we recall that s u 1/L and hence ζ (N ) (2(1 + s + u))2 = ζ (N ) (2)2 + O(1/L) Bα,β (s, u, 0) = B0,0 (0, 0, 0) + O(1/L) 1 α+s = (1 + O(1/L)) L(f ⊗ f, 1 + α + s) ress=1 L(f ⊗ f, s) resu=1 L(f ⊗ f, u) L(f ⊗ f, 1 + α + u) = (1 + O(1/L)), α+u and we recall that we had shown that B0,0 (0, 0, 0) = 1. Thus 2`2

{ζ (N ) (2(1 + s + u))} L` (f ⊗ f, 1 + α + u)L` (f ⊗ f, 1 − β + u)Bα,β (s, u, 0) L`(`−1) (f ⊗ f, 1 + 2s)L`(`+1) (f ⊗ f, 1 + 2u)L` (f ⊗ f, 1 + α + s)L` (f ⊗ f, 1 − β + s) =

{ζ (N ) (2)}

2`2

(α + s)` (−β + s)` 2`2

(ress=1 L(f ⊗ f, s))

`

(α + u) (−β + u)

`

2 −1

(2s)`(`−1) (2u)`(`+1) + O(L−2`

).

Indeed, we now get the product of two cleanly separated integrals 2`2

res r`,`+1 (α, β, i, j, s, u) =

s=u=0

{ζ (N ) (2)}

2

(ress=1 L(f ⊗ f, s))2` `(`−1) I 2 M` s ds ` ` (α + s) (−β + s) i+1−`(`−1) × 2πi D(0,L−1 ) n s `(`+1) I u 2 M`+1 1 du × 2πi D(0,L−1 ) n (α + u)` (−β + u)` uj+1−`(`+1) + O(Li+j−1 ).

We shall compute these integrals by the use of Cauchy’s integral theorem. We will proceed in more generality than strictly needed. First, we have αx−βy I I 1 e M` s dp+q M` x+y s ds p q ds (α + s) (−β + s) k+1 = p q e 2πi D(0,L−1 ) n dx dy 2πi n s sk+1 x=y=0 D(0,L−1 ) 1 dp+q M` k αx−βy = e x + y + log k! dxp dy q n x=y=0


33

Our case of interest naturally follows by taking p = q = ` and k = i − `(` − 1). For the u-integral we proceed in a slightly different way. We will use the equality Z 1 (−1)τ τ ! q −α−u (4.17) rα+u−1 logτ rdr = − P (u, α, log q), (α + u)τ +1 (α + u)τ +1 1/q which is valid for all complex numbers α, u, positive q and τ = 0, 1, 2, · · · . Here P is a polynomial in log q of degree τ − 1. We temporarily set q = M`+1 /n so that I 1 1 du qu m n k+1 2πi D(0,L−1 ) (α + u) (−β + u) u I Z 1 (−1)m−1 1 1 du = qu rα+u−1 logm−1 rdr k+1 + E(q), n (m − 1)! 2πi D(0,L−1 ) (−β + u) 1/q u where E(q) is term arising from the second part of (4.17), i.e. I P (u, α, log q) du (−1)m−1 −α 1 (4.18) q E(q) = − m n k+1 . (m − 1)! 2πi (α + u) (−β + u) u It can be seen, by taking the contour to be arbitrarily large, that this term vanishes. Reversing the order of integration in the main term yields I 1 1 du qu m n k+1 2πi D(0,L−1 ) (α + u) (−β + u) u Z I (−1)m−1 1 α−1 m−1 1 1 du = r log r (qr)u n k+1 dr. (m − 1)! 1/q 2πi D(0,L−1 ) (−β + u) u Applying again (4.17) but with the lower boundary of integration at 1/(qr) and seeing that the second term of (4.17) will also vanish by the same reason as the previous second term (4.18), we then obtain Z I (−1)m−1 1 α−1 1 1 du m−1 r (qr)u rdr m j−k+1 log (m − 1)! 1/q 2πi D(0,L−1 ) (−β + u) u Z 1 Z 1 I (−1)m+n 1 du = rα−1 t−β−1 logm−1 rlogn−1 t (qrt)u k+1 dtdr (m − 1)!(n − 1)! 1/q 1/qr 2πi u Z 1 Z 1 (−1)m+n M2 k α−1 −β−1 m−1 n−1 = r t log rlog t log rt dtdr. k!(m − 1)!(n − 1)! 1/q 1/qr n The last step is to make the change of variables r = q −a and t = q −b so that the above becomes −aα+bβ ZZ logk+n+m q k M2 (1 − a − b) am−1 bn−1 dadb. k!(m − 1)!(n − 1)! a+b≤1 n a,b≥0

Our case of interest in this setting follows by taking m = n = ` and k = j − `(` + 1). The end result of this reasoning is that 2`2

res r`,`+1 (α, β, i, j, s, u) =

s=u=0

{ζ (N ) (2)}

2

(ress=1 L(f ⊗ f, s))2`

2`(`−1) d2` M` i−`(`−1) αx−βy × e x + y + log (i − `(` − 1))! dx` dy ` n x=y=0 ×

2`(`+1) logj−`(`+1)+2` (M`+1 /n) (j − `(` + 1))!

34


ZZ ×

j−`(`+1)

a+b≤1 a,b≥0

(1 − a − b) 2 −1

+ O(Li+j−2`

M`+1 n

−aα+bβ

(ab)`−1 dadb

).

When we go back to K`,`+1 with this new information we get Z Z 1 2 r`,`+1 (α, β, i, j, s, u)dsdu K`,`+1 (α, β, i, j, s, u) = 2πi (δ) (δ) 2`2

=

2`(`−1) d2` 2`(`+1) logj−`(`+1)+2` (M`+1 /n) 2 ` ` (j − `(` + 1))! (ress=1 L(f ⊗ f, s))2` (i − `(` − 1))! dx dy 2 X (λ2f (n))∗2` M` i−`(`−1) αx−βy x + y + log × e n n x=y=0 n≤M`+1 −aα+bβ ZZ j−`(`+1) M`+1 × (1 − a − b) (ab)`−1 dadb a+b≤1 n {ζ (N ) (2)}

a,b≥0

+ O(Li+j−2`

2 −1

).

Recall that in the last expression for I`,`+1 , we have sums over i and over j. The sum over i is X a`,i i! M` i−`(`−1) x + y + log n logi M` (i − `(` − 1))! i X 1 log(M` /n) i−`(`−1) x+y = + a`,i i(i − 1) · · · (i − `(` − 1) + 1) log M` log M` log`(`−1) M` i 1 log(M` /n) x+y (`(`−1)) = P` + , `(`−1) log M` log M` log M` whereas the sum over j is X a`+1,j j! logj−`(`+1)+2` (M`+1 /n) (1 − a − b)j−`(`+1) j (j − `(` − 1))! log M`+1 j log2` (M`+1 /n) X

log(M`+1 /n) = a j(j − 1) · · · (j − `(` − 1) + 1) (1 − a − b) `+1,j log M`+1 log`(`+1) M`+1 j log2` (M`+1 /n) (`(`+1)) log(M`+1 /n) P`+1 (1 − a − b) = . log M`+1 log`(`+1) M`+1

j−`(`+1)

Plugging these results into I`,`+1 we see that 2`2 2 22` w c0 (0) {ζ (N ) (2)} d2` I`,`+1 (α, β) = dx` dy ` log`(`−1) M` log`(`+1) M`+1 ress=1 L(f ⊗ f, s) 2 X (λ2f (n))∗2` x+y log(M` /n) (`(`−1)) αx−βy 2` ×e log (M`+1 /n)P` + |x=y=0 n log M` log M` n≤M`+1 ZZ log(M`+1 /n) M`+1 −aα+bβ (`(`+1)) × P`+1 (1 − a − b) (ab)`−1 dadb a+b≤1 log M`+1 n a,b≥0

+ O(T /L).


35

Making the changes x → x/ log M` and y → y/ log M` puts in this in a slightly more comfortable form 2`2 2 22` T −β w(0) b d2` {ζ (N ) (2)} I`,`+1 (α, β) = dx` dy ` log`(`−1)+2` M` log`(`+1)−2` M`+1 ress=1 L(f ⊗ f, s) 2 X (λ2f (n))∗2` log2` (M`+1 /n) (`(`−1)) log(M` /n) αx−βy P` x+y+ |x=y=0 × M` 2` n log M` log M `+1 n≤M`+1 ZZ log(M`+1 /n) M`+1 −aα+bβ (`(`+1)) P (1 − a − b) (ab)`−1 dadb × a+b≤1 `+1 log M`+1 n a,b≥0

+ O(T /L). The sum over n is computed by the use of Lemma 3.5 with k = 2`2 , s = −αa + bβ, x = M` , (`(`−1)) (`(`+1)) z = M`+1 , F (u) = P` (x + y + u), H(u) = u2` P`+1 ((1 − a − b)u). The result is ∗2`2

(λ2f (n))

X

log(M`+1 /n) log M`+1

2`

n1−aα+bβ log(M` /n) log(M`+1 /n) (`(`−1)) (`(`+1)) × P` x+y+ P`+1 (1 − a − b) log M` log M`+1 2`2 2 ress=1 L(f ⊗ f, s) log2` M`+1 = −aα+bβ {ζ (N ) (2)} (2`2 − 1)!M`+1 Z 1 log M`+1 2`2 −1 (`(`−1)) (1 − u) P` × x + y + 1 − (1 − u) log M` 0

n≤M`+1

(`(`+1))

× u2` P`+1

u(−aα+bβ)

((1 − a − b)u)M`+1

2 −1

du + O(L2`

).

This means that we are left with 2 22` T −β w(0) b log M`+1 `(`+1) d2` I`,`+1 (α, β) = (2`2 − 1)! log M` dx` dy ` ZZ Z 1 log M`+1 αx−βy 2`2 −1 (`(`−1)) × M` (1 − u) P` x + y + 1 − (1 − u) a+b≤1 0 log M` a,b≥0 (`(`+1)) u(−aα+bβ) 2` + O(T /L). ((1 − a − b)u)M`+1 u dudadb × (ab)`−1 P`+1 x=y=0

Setting M` =

T ν`

T ν`+1

and M`+1 = we obtain Lemma 2.2, i.e. 2 22` ν`+1 `(`+1) d2` c`,`+1 (α, β) = (2`2 − 1)! ν` dx` dy ` ZZ Z 1 2 −β −bu au −α × u2` (1 − u)2` −1 (M`−x M`+1 ) (M`y M`+1 T) a+b≤1 a,b≥0

0

ν`+1 (`(`+1)) (`(`−1)) `−1 P`+1 ((1 − a − b)u)(ab) dudadb . × P` x + y + 1 − (1 − u) ν` x=y=0 Therefore, the main term of Theorem 2.4 is given by −1 d −1 d c`,`+1 = Q Q c`,`+1 (α, β)|α=β=−R/L log T dα log T dβ


36

ZZ Z 1 2 2 ν`+1 `(`+1) R d2` 22` u2` (1 − u)2` −1 eR[ν` (y−x)+uν`+1 (a−b)] = e 2 ` ` a+b≤1 0 (2` − 1)! ν` dx dy a,b≥0

× Q(−xν` + auν`+1 )Q(1 + yν` − buν`+1 ) ν`+1 (`(`−1)) (`(`+1)) `−1 × P` x + y + 1 − (1 − u) P`+1 ((1 − a − b)u)(ab) dudadb . ν` x=y=0 This ends the computation of I`,`+1 . 4.5. The mean value integral I`,`+j (α, β) for j ∈ N\{1}. We must now examine the case I`,`+j (α, β) where j = 2, 3, 4, · · · . We will show that I`,`+j (α, β) T L−1+ε and therefore the mean value integral contribution of these terms to κ is zero. As before, we start by inserting ψ` (s) = χ`−1 f (s +

1 2

− σ0 )

X

σ −1/2 1/2−σ0 k1 P` [h1 k1 ], 1−s hs1 k1

µf,` (h1 )λ∗`−1 (k1 )h1 0 f

h1 k1 ≤M`

and (s + ψ`+j (s) = χ`−1+j f

1 2

− σ0 )

X

σ −1/2 1/2−σ0 k2

(k2 )h2 0 µ`+j (h2 )λ∗`−1+j f hs2 k21−s

h2 k2 ≤M`+j

P`+j [h2 k2 ],

into the mean value integral I, i.e. Z ∞ I`,`+j (α, β) = w(t)L(f, 12 + α + it)L(f, 12 + β − it)ψ` ψ`+j (σ0 + it)dt −∞

=

X

X

(k1 ) µf,` (h1 )λ∗`−1 f 1/2 1/2

h1 k1 ≤M` h2 k2 ≤M`+j

h1 k1

P` [h1 k1 ]

(k2 ) µ`+j (h2 )λ∗`−1+j f 1/2 1/2

h2 k2

P`+j [h2 k2 ]J3,f ,

where k1 h2 −it j 1 χf ( 2 + it)dt = w(t)L(f, + α + it)L(f, + β − it) h1 k2 −∞ −β Z ∞ t k1 h2 −it j−1 1 1 1 w(t) = L(f, 2 + α + it)L(f, 2 − β + it) χf ( 2 + it)dt + O(T ε ), 2π h k 1 2 −∞ Z

J3,f

∞

1 2

1 2

by the use of the functional equation of L(f, s). We remark that so far this is the same procedure in the integrand, and in this case j is an integer as in the I`,`+1 case except for the presence of χj−1 f strictly greater than one. One can show using Stirling’s approximation formula that t −it iπ/4 1 χf ( 2 + it) = F (t) + E(t), where F (t) = e , and E(t) t−1 . 2πe Let us handle the error term first. By power moment estimates (see e.g. [4, Corollary 2] or [30]) we have −β Z ∞ t k1 h2 −it j−1 1 1 w(t) L(f, 2 + α + it)L(f, 2 − β + it) E (t)dt 2π h1 k2 −∞ Z 2T 1 j−1 |L(f, 12 + α + it)||L(f, 21 − β + it)|dt T T /4 1/2 Z 2T 1/2 Z 2T 1 2 2 1 1 ≤ j−1 |L(f, 2 + α + it)| dt |L(f, 2 − β + it)| dt T T /4 T /4 1 log T j−1 (T log T )1/2 (T log T )1/2 = j−2 , T T


37

by the Cauchy-Schwarz inequality. Therefore, the estimate on J3,f so far is −β Z ∞ t k1 h2 −it j−1 1 1 1 J3,f = w(t) L(f, 2 + α + it)L(f, 2 − β + it) F ( 2 + it)dt + Oε,f (T ε ). 2π h1 k2 −∞ We now use the definition of σα,−β (f, l) to further write −β Z ∞ X σα,−β (f, l) −l/T 3 ∞ t k1 h2 l −it j−1 1 w(t) e F ( 2 + it)dt + Oε,f (T ε ). J3,f = 2π h k l1/2 1 2 −∞ l=1

The key observation comes from noticing that for all 1 ≤ h1 , k1 ≤ M` and 1 ≤ h2 , k2 ≤ M`+j as well as for any l ≥ 1, one has T j−1 1 t j−1 k1 h2 l T j−1 (4.19) ≥ ≥ T ε0 , = 2πe h1 k2 4 2πeM` M`+j 8πeT ν` +ν`+j provided that ν` + ν`+j < j − 1 − ε.

(4.20)

From the conditions of the test function w we have w(r) (t) (L/T )r . Hence, picking up from (4.19) and using integration by parts, we arrive at −β −it Z ∞ t t j−1 k1 h2 l 1 w(t) (4.21) dt r,ε0 r , 2π 2πe h1 k2 T −∞ for any fixed integer r. Thus, using (4.21) we can further bound J3,f as J3,f r,ε0

∞ 1 X σα,−β (f, l) −l/T 3 e + Oε,f (T ε ) ε,ε0 T ε . Tr l1/2 l=1

The last step is to now plug this back into I`,`+j (α, β) to see that I`,`+j (α, β) ε,ε0 T

X

ε

(k2 )| (k1 )λ∗`−1+j |µf,` (h1 )µ`+j (h2 )λ∗`−1 f f

X

(h1 h2 k1 k2 )1/2

h1 k1 ≤M` h2 k2 ≤M`+j

`,ε,ε0 T 2ε

X

1

X

h1 k1 ≤M` h2 k2 ≤M`+j

(h1 h2 k1 k2 )1/2−ε

|P` [h1 k1 ]P`+j [h2 k2 ]|

,

by the use of (4.3). Recognizing that the sums can be consolidated by employing the divisor functions we obtain X τ (n) X τ (m) 1/2+ε 1/2+ε I`,`+j (α, β) `,ε,ε0 T 2ε `,ε,ε0 T 4ε M` M`+j 1/2−ε 1/2−ε n m n≤M m≤M `

(4.22)

=T

6ε+(ν` +ν`+j )/2

`+j

.

Since ν` + ν`+j < 2, we can choose ε so small that I`,`+j (α, β) T /L. This ends the proof of Lemma 2.3 and Theorem 2.5. 5. Applications Let us define Nf,0 (T ) to be the number of non-trivial zeros up to heigh T > 0 such that β = 1/2. We also set Nf,0 (T ) κf = lim inf . T →∞ Nf (T ) For the case of the Riemann zeta-function, i.e. an L-function of degree one, Selberg [28] established that 0 < κ ≤ 1. Later on Levinson changed this technique considerably and he managed to show


38

that κ ≥ .3474. This was later twice refined by Conrey to κ ≥ .4088, see [10], using the work on Kloosterman sums of Deshouillers and Iwaniec [14, 15]. In 2010, Young [29] simplified Levinson’s proof substantially. In the early 2010’s, Bui, Conrey and Young [8] established that κ ≥ .4105. The current bound is κ ≥ .4107, see [20, 23, 27]. Adapting the work of Levinson [24] and Conrey [10], it is well understood (see [4, Proposition 5] as well as [16, 17, 18, 26]) that if one applies Littlewood’s lemma, and then the arithmetic and geometric mean inequalities, one arrives at Z T 1 1 2 (5.1) |V ψ(σ0 + it)| dt + o(1), κf ≥ 1 − log R T 1 where, we recall, σ0 = 1/2 − R/L with R is a bounded positive of our choice. In 2015, Bernard [4] revised Young’s paper and adapted it to modular forms. For holomorphic primitive cusp forms of even weight, square-free level and trivial character, his results are that κf ≥ .0297 unconditionally and κf ≥ .0693 on the Ramanujan hypothesis. Unfortunately, as mentioned on [4, p. 203], the size of the mollifier, even under the Ramanujan hypothesis, is too small to establish results for simple zeros on the critical line. Here, we will improve those results a little bit by taking L = 2 in (1.16). The reason behind this choice is due to the fact that one may think of ψ1 (s) as the main term of the mollifier and of ψ2 (s), ψ3 (s) and so on as the perturbations to the main piece. While the contribution to κ of these perturbations is strictly greater than zero, their magnitudes decrease very rapidly. 5.1. Numerical evaluations. As a consequence of our previous results we can now establish the following. Theorem 5.1. One has ( .0307506, κ≥ .0693754,

unconditionally, under the Selberg conjecture.

We use Mathematica to numerically evaluate c(PL , Q, R, νL ) with the following choices of pa5 rameters: L = 2 and θ = 7/64 so that ν = 27 and note that ν1 + ν2 = 5/54 + 5/54 = 10/54 < 1. Then, with R = 6.301 we have Q(x) = .499312 + 1.59864(1 − 2x) − 3.03514(1 − 2x)3 + 3.08498(1 − 2x)5 − 1.14779(1 − 2x)7 , P1 (x) = 1.0997x + .186107x2 − 1.45602x3 + 2.34073x4 − 1.16983x5 , P2 (x) = −.000120465x3 − .000319875x4 + .000125939x5 , we have κ ≥ .0307506. Finally if we take θ = 0, then ν = 1/4 and we note that ν1 +ν2 = 1/8+1/8 = 1/4 < 1. Finally, taking R = 5.5 we obtain, Q(x) = .498981 + 1.53751(1 − 2x) − 2.79365(1 − 2x)3 + 2.77541(1 − 2x)5 − 1.01825(1 − 2x)7 , P1 (x) = .941033x − .00864408x2 + .0457966x3 + .0456478x4 − .0238337x5 , P2 (x) = −.0000416836x3 + 4.1767 × 10−6 x4 − .000089891x5 , we have κ ≥ .0693754. As mentioned earlier, from numerical experiments, one would need a size of about θ = 1/5 to get a proportion of simple critical zeros. The following plots will illustrate this phenomenon. If we take only L = 1, and for the sake of simplicity P (x) = x and Q(x) = 1 − x (in fact, having Q(x)


39

be a polynomial of degree one is necessary to obtain simple critical zeros, see [2] and [19] as well as [8, p. 37] and [20, p. 513]), then 2 Z Z 1 1 1 2Rv d Rν1 x e c1,1 = 1 + e P (x + u)Q(v + ν1 x)|x=0 dudv ν1 0 0 dx (5.2)

=1−

3 + 6R − 2R3 (ν1 − 3)ν1 + 2R4 ν12 + R2 (6 + ν12 ) − e2R (3 + R2 ν12 ) , 12R3 ν1

and

1 log c1,1 + o(1). R Let us now graph κ b=κ b(R, ν1 ) = RHS of (5.2) to see the proportion of simple zeros as a function of R and of ν1 . κ≥1−

Figure 5.1. Left-hand side: κ b(R, ν1 ) for ν1 = 1/2 (light blue), ν1 = 1/3 (brown), ν1 = 1/4 (purple), ν1 = 1/5 (red), ν1 = 1/6 (green), ν1 = 1/8 (orange), ν1 = 5/54 1 (dark blue). Right-hand side: surface plot of κ b(R, ν1 ) with 10 ≤ R ≤ 10 and 1 100 ≤ θ ≤ 1. We note that when ν1 = 1/2, this was considered in [29] and the optimal value is R ≈ 1.3. Adding further pieces to ψ(s), i.e. increasing L will not help in securing a stronger bound on ν1 but will increase the value of κ. 6. Acknowledgments The first author wishes to acknowledge partial support from SNF grant PP00P2 138906. References [1] M. Abramowitz and I. A. Stegun (Eds.). Confluent Hypergeometric Functions in Ch 13 of Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, 1972. [2] R. J. Anderson. Simple zeros of the Riemann zeta-function. J. Number Theory, (17):176–182, 1983. [3] R. Balasubramanian, B. Conrey and D. R. Heath-Brown. Asymptotic mean square of the product of the Riemann zeta-function and a Dirichlet polynomial. J. reine angew. Math., (357):161–181, 1985. [4] D. Bernard. Modular case of Levinson’s theorem. Acta Arith., (167.3):201–237, 2015. [5] V. Blomer. Shifted convolution sums and subconvexity bounds for automorphic L-functions. Int. Math. Res. Notices, 3905–3926, 2004. [6] V. Blomer. Rankin-Selbeg L-functions on the critical line. Manuscripta Math., (117):111–133, 2005. [7] H. M. Bui. Critical zeros of the Riemann zeta-function. https://arxiv.org/abs/1410.2433 [8] H. M. Bui, B. Conrey, and M. P. Young. More than 41% of the zeros of the zeta function are on the critical line. Acta Arith., (150.1):35–64, 2011.

40


[9] J. B. Conrey. Zeros of derivatives of the Riemann’s ξ-function on the critical line. J. Number Theory, (16):49–74, 1983. [10] J. B. Conrey. More than two fifths of the zeros of the Riemann zeta function are on the critical line. J. reine angew. Math., (399):1–26, 1989. [11] J. B. Conrey, D. W. Farmer and M. R. Zirnbauer. Autocorrelation of ratios of L-functions. Comm. Number Theory Phys., (2)94:593-636, 2008. [12] J. B. Conrey and N. C. Snaith. Applications of the L-functions ratios conjectures. Proc. London Math. Soc., (3)94:594-646, 2007. ´ (43):273–307, 1974. [13] P. Deligne. La conjecture de Weil. I. Publications Math´ematiques de l’IHES, [14] J. M. Deshouillers and H. Iwaniec. Kloosterman sums and Fourier coefficients of cusp forms. Invent. Math., (70):219–288, 1982. [15] J. M. Deshouillers and H. Iwaniec. Power mean values of the Riemann zeta function II. Acta Arith., (48):305–312, 1984. [16] D. W. Farmer. Mean value of Dirichlet series associated with holomorphic cusp forms. J. Number Theory, (49):209–245, 1994. [17] J. L. Hafner. Zeros on the critical line for Dirichlet series attached to certain cusp forms. Math. Ann., (264):21– 37, 1983. [18] J. L. Hafner. Zeros on the critical line for Maass wave form L-functions. J. reine angew. Math., (377):127–158, 1987. [19] D. R. Heath-Brown. Simple zeros of the Riemann zeta-function on the critical line. Bull. Lond. Math. Soc., (11):17–18, 1979. [20] S. Feng. Zeros of the Riemann zeta function on the critical line. J. Number Theory, (132):511–542, 2012. [21] H. Iwaniec and E. Kowalski. Analytic Number Theory. Amer. Math. Soc., Providence, RI, 2004. [22] I. I. Kim, Functoriality for the exterior square of GL4 and the symmetric fourth of GL2 (with appendices by I. Ramakrishnan, I. I. Kim and P. Sarnak), J. Amer. Math. Soc., (16): 139–183, 2003. [23] P. K¨ uhn, N. Robles and D. Zeindler. On a mollifier of the perturbed Riemann zeta-function. https://arxiv.org/abs/1605.02604 [24] N. Levinson. More than One Third of Zeros of Riemann’s Zeta-Function are on σ = 21 . Adv. Math., (13):383–436, 1974. [25] R. A. Rankin. Contributions to the theory of Ramanujan’s function τ (n) and similar arithmetical functions II. The order of the Fourier coefficients of integral modular forms. Proc. Cambridge Pilos. Soc., (35):351–372, 1939. [26] I. S. Rezvyakova. On the zeros of Hecke L-functions and of their linear combinations on the critical line. Dokl. Akad. Nauk, (431):741–746, 2010 (in Russian). [27] N. Robles, A. Roy, and A. Zaharescu. Twisted second moments of the Riemann zeta-function and applications. J. Math. Anal. Appl., (434):271–314, 2016. [28] A. Selberg. On the zeros of Riemann’s zeta-function. Skr. Norske Vid. Akad. Oslo I, (10):1–59, 1942. [29] M. P. Young. A short proof of Levinson’s theorem. Arch. Math., (95):539–548, 2010. [30] Q. Zhang. Integral mean values of modular L-functions. J. Number Theory, (115):100–122, 2005. ¨ r Mathematik, Universita ¨ t Zu ¨ rich, Winterthurerstrasse 190, CH-8057 Zu ¨ rich, SwitzerInstitut fu land E-mail address: [email protected] Department of Mathematics, University of Illinois, 1409 West Green Street, Urbana, IL 61801, United States E-mail address: [email protected] Department of Mathematics and Statistics, Lancaster University, Fylde College, Bailrigg, Lancaster LA1 4YF, United Kingdom E-mail address: [email protected]

On mean values for Dirichlet's polynomials and L-functions

On mean values for Dirichlet's polynomials and L-functions

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