New Zero-Density Results for Automorphic L -Functions of GL ( n )

: Let L ( s , π ) be an automorphic L -function of GL ( n ) , where π is an automorphic representation of group GL ( n ) over rational number ﬁeld Q . In this paper, we study the zero-density estimates for L ( s , π ) . Deﬁne N π ( σ , T 1 , T 2 ) = (cid:93) { ρ = β + i γ : L ( ρ , π ) = 0, σ < β < 1, T 1 ≤ γ ≤ T 2 }, where 0 ≤ σ < 1 and T 1 < T 2 . We ﬁrst establish an upper bound for N π ( σ , T ,2 T ) when σ is close to 1. Then we restrict the imaginary part γ into a narrow strip [ T , T + T α ] with 0 < α ≤ 1 and prove some new zero-density results on N π ( σ , T , T + T α ) under speciﬁc conditions, which improves previous results when σ near 34 and 1, respectively. The proofs rely on the zero detecting method and the Halász-Montgomery method.


Introduction
Let π be an automorphic representation of group GL(n) over rational number field Q. Then the automorphic (finite-part) L-function related to π can be defined as L(s, π) = ∞ ∑ n=1 A π (n) n s ( s > 1).
Here we take the Riemann zeta function ζ(s) as an example of the automorphic Lfunction L(s, π). To specify π by cusp forms and Maass forms, we refer to the references [16][17][18] and references therein. From the works of Ingham [19] and Huxley [20] we know that About the results on the density hypothesis, in 1977, Jutila [21] proved that In 2000, Bourgain [22] improved Jutila's result to When σ is close to 1, Ivić (see [23], Theorem 11.3) proved that for 9 10 ≤ σ ≤ 1, In this paper, motivated by Ivić's work, we first establish an upper bound of N π (σ, T, 2T), when σ is close to 1 in the following theorem.
Remark 1. Since L(s, π) is a general automorphic L-function in Theorem 1, the mean value estimate now is worse than the case of the Rieman zeta function, which results in the T ε (see the argument around (22) for details).
Now we restrict the imaginary part γ into a narrow strip [T, T + T α ] and suppose for certain 0 < α ≤ 1 and θ ≥ α, where l ≥ 1 is an integer and T ≥ 3. Ye and Zhang [24] established some bounds for N π (σ, T, T + T α ) with 1 2 ≤ σ < 1. Later, Dong, Liu and Zhang [25] obtained a sharper bound for N π (σ, T, T + T α ) when σ is close to 1 and show a range of σ for which the density hypothesis holds. We shall keep on studying zerodensity estimates for L(s, π) in this strip and improve previous results when σ near 3 4 and 1, respectively. Theorem 2. Let L(s, π) be an automorphic L-function satisfying (3). Then for 3 4 ≤ σ ≤ 10 13 and and for 10 13 Theorem 3. Let L(s, π) be an automorphic L-function satisfying (3). Then for 3 4 ≤ σ ≤ 10 13 and and for 10 13 Notation 1. Throughout this paper, the letter ε represents a sufficiently small positive number, whose value may change from statement to statement. Constants, both explicit and implicit, in Vinogradov symbols may depend on ε and π.

Some Lemmas
Proof. We can get this lemma by a standard winding number argument on (3) as in Davenport [26] and Rudnick and Sarnak [15]. Then we have the inequality where a 2 = (a, a).

Lemma 3 ([23]
, (4.60)). Suppose that Y, h > 0 and k ≥ 1 is an integer, we have Lemma 4. For a fixed θ satisfying 1 2 < θ < 1, define where T ≤ t r ≤ T + T α (1 ≤ r ≤ R) and v is defined by Suppose that |t r − t s | ≥ 3 log 4 T for r = s ≤ R, then we have Proof. We can get this lemma by following the argument of ( [23], Lemma 11.6). The main difference is that the interval of t r now is T ≤ t r ≤ T + T α .

Proof of Theorem 1
We first consider the number of zeros ρ = β + iγ of L(s, π) in the rectangle where σ ≥ 1 2 , T ≥ 3, and 0 < α ≤ 1. We define µ π (n) by Consequently µ π (n) is a multiplicative function and Assume Then we have for s > 1, where In terms of (7), for σ > 1 and s > 0, we obtain From the inverse Mellin transform of Γ(ω), the right-hand side of the above formula can be written as Hence we have, for σ > 1 and s > 0, Now we move the line of integration of (8) to ω = u − β < 0 for some suitable 1 2 ≤ u < 1. Then integral on the right hand of (8) becomes where L(s, π)M X (s, π) is the residue of the integrand at ω = 0. Setting ω = u − β + iv and taking s equal to a non-trivial zero ρ = β + iγ, we have Since For the integral in (11) we have due to the fact that they are absolutely convergent. By Stirling's formula again, we can further remove the Γ function in (11) and get that a non-trivial zero ρ = β + iγ counted in N π (σ, T, T + T α ) satisfies either (10) or We divide the elongated rectangle (6) into successive rectangles of length 3 log 4 T noting Lemma 1. These rectangles start at and the last one may have a shorter length. Call these smaller rectangles as I 1 , I 2 , . . .. The number of zeros are denoted by A ij in all odd-numbered rectangles if j = 1 or in all even-numbered rectangles if j = 2, which satisfy (10) We can get a sequence of zeros ρ counted in A ij for r = 1, . . . , R ij , if we choose a zero from each rectangle which contains at least one zero. Here, R ij is the number of rectangles that contains at least one zero counted in A ij . By Lemma 1, we note that each rectangle contains at most 3m 2π log 5 T(1 + o(1)) zeros. Consequently, we have Now we begin to estimate R 1j . By a dyadic subdivision on the sum in (10), we know where there are O(log(Y log 2 Y)) terms. Then there exists Raising (16) to kth power we get with c π (n) = ∑ n=n 1 n 2 ···n k Mr <n i ≤2Mr b π (n 1 ) · · · b π (n k )e − n 1 +n 2 +···+n k Y and k is a natural number depending on M r such that M k r = N, (2M r ) k = P ≤ T c , from where k 1 and P N. We split the sum in (17) into subsums of length N and choose k so that M k r ≤ Y r log 2r Y < M k+1 r , where r is a fixed integer and k ≥ r ≥ 1 is satisfied. Then (17) can be written as for some D 1 and Y r 2 By partial summation and (18), we have We relabel c π (n) to satisfy c π (n) = 0 for n > u. Then simplifying (20), we can get From now on, we let α = 1 in (6). Now we apply Lemma 2 to (21) and take , we then have Moving the line of integration in (24) to w = u < 1, we encounter a simple pole at In view of |Γ(s)| = (2π) Since the γ (1j) r s are at least 3 log 4 T apart, the first sum on the right side of (26) is O(R 1j ).
For the second sum on the right side of (26), we fix each s and let τ r = γ Then we have |τ r | 3T for r = 1, 2, . . . , R 1j and |τ r 1 − τ r 2 | ≥ 3 log 4 T for r 1 = r 2 . Hence we get Inserting (27) into (22) we obtain Then we have for σ ≥ u+1 2 , if X 2σ−1−u M 1 (u, 3T)T ε . Now we turn to estimate R 2j . We may suppose first that σ < 1 − C log T in view of the zero-free region. Note that in this case the zeros ρ From the mean value theorem for integration, we can see that there exists t Thus, from (30) we get We can apply Lemma 2 as in the previous case, but now the choice of ξ and ϕ r are different. We shall take ξ = {ξ n } ∞ n=1 with ξ n = µ π (n)(e − n 2N − e − n N ) − 1 2 n −u for N < n ≤ 2N and zero otherwise, ϕ r = {ϕ r,n } ∞ n=1 with ϕ r,n = (e − n 2N − e − n N ) Therefore X and Y can be chosen as We see that the bound for R 2j is smaller than the one for R 1j . Recalling (14) we have Then in view of (31), we see that it is appropriate to take k = 5. Hence u = 5σ − 4, σ ≥ u+1 2 is satisfied and u ≥ 1 2 for σ ≥ 9 10 . Thus, Theorem 1 follows from (31) with u = 5σ − 4.

Proof of Theorem 2
Throughout the proof of Theorem 2, we restrict the range of zeros to (6). From the estimates of R 2 1j in (22) , we obtain now Recall the functional equation Moving the line of integration in H(it) to w = 1 4 and applying the Cauchy-Schwarz inequality, we get where the S( 3 4 ) is from Lemma 4. Substituting (33) into (32), we can get where we used Lemma 4 to S( 3 4 ). For R 0 points lying in an interval of length In (19), we take r = 2 to get Y It follows from ( [24], (5.21)) that Consequently, we have ). (36) , which is equivalent to Y = T 3α 2 4α(1−σ)−20+32σ . Thus, from (36) we get Comparing the two terms in (37), we can get T ε N 2−2σ for 10 13 < σ < 1.
Then we have ).

Proof of Theorem 3
The proofs of Theorems 2 and 3 are similar, and the main difference is the range of N. Here for completeness, we state some critical details. We let r = 1 in (19) and get Y 1 2 log Y N Y log 2 Y. Unlike the previous (34), we now have For R 0 points lying in an interval of length Consequently, recalling (35), we have ).
Then we have ).