Moments of Quadratic Hecke $ L $-Functions of Imaginary Quadratic

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The result of Patterson is more general. In fact, for any algebraic number field F , let µn(F ) be the set of n-th roots of unity in F with n > 2. Suppose that the cardinality of µn(F ) is n. For any n-th order Hecke character χ and any injective character ε : µn(F ) C×, one can define a Gauss sum for ε(χ) in a manner similar to the Gauss sum defined in Section 2.1. Patterson’s result→ then gives that these Gauss sums have a lot of cancellation on average. On the other hand, one does not expect such a result to hold when ε is not injective. For example, when F = Q(i), let ε be the 2 character: x x from µ4(K) to C×. Then for any quartic symbol χ, ε(χ) becomes a quadratic symbol defined in Section 2.1 and7→ the associated Gauss sum becomes a quadratic Gauss sum. To fathom the behavior of these quadratic Gauss sums on average, one can examine their rational analogues: the Gauss sums for quadratic Dirichlet characters. In the work of M. Jutila [19] and K. Soundararajan [24] on the first moment of L(1/2,χ) for quadratic Dirichlet characters χ, it is shown that both sums coming from the approximate functional equation contribute to the main term, unlike the case of quartic symbols discussed above.

Motivated by these observations, it is our goal in this paper to study the ﬁrst and second moments of Hecke L- functions with quadratic characters in Q(i) and Q(ω) at the central point. In this case Patterson’s result is no longer in display and our approach is similar to that used in [24]. In particular, we shall see that both sums coming from the approximate functional equation contribute to the main term, just as the case for the quadratic Dirichlet L-functions.

Let Φ be a smooth Schwarz class function compactly supported in (1, 2) and we assume that 0 Φ(t) 1 for all t. We have ≤ ≤ Theorem 1.1. Let K = Q(i) or Q(ω). For y and any ε> 0, we have → ∞ 2 ∗ 1 N(c) π (3+θ)/4 (1.2) L ,χc Φ = AK BK Φ(0)ˆ y log y + CK Φ(0)ˆ y + O y , 2 y ζK (2) c K ∈OX where ζK (s) is the Dedekind zeta function of K, θ = 131/416, 1 1 (1.3) AQ = 1 , AQ = 1 , (i) − (N(π)+1)N(π) (ω) − (N(π)+1)N(π) π prime in Q(i) π prime in Q(ω) (π,Y2)=1 (π,Y6)=1

2 2+ √2 3+ √3 B = , B = , Φ(0)ˆ = Φ(x)dx, Q(i) 3072 Q(ω) 19440 Z1

CK is a constant and ∗ indicates that the sum runs over square-free elements of Z[i] congruent to 1 (mod 16) if K = Q(i) and square-free elements of Z[ω] congruent to 1 (mod 36) if K = Q(ω). P We note here that θ arises from an application of the currently best known formula in the Gauss circle problem [16]. Therefore, any improvement in the study of that problem will also lead to an improvement in the O-term in (1.2).

As for the second moment, we prove Theorem 1.2. Using the same notation as Theorem 1.1, for y and any ε> 0, we have → ∞ 1 2 N(c) ∗ L ,χ Φ y1+ε. 2 c y ≪ε c K ∈OX

S. Chowla [4] was the ﬁrst to conjecture that a Dirichlet L-function is never zero at s =1/2 and it is believed that an L-function cannot vanish at its central point unless there is a compelling reason (an elliptic curve with a postive algebraic rank or the root number being 1) that there should be a zero there. In this vein, we give the following non-vanishing results for the L-functions under− our consideration. Corollary 1.3. For y and any ε> 0, we have → ∞ 1 1 ε # c Z[i]: c 1 (mod 16), N(c) y, L ,χ =0 y − ∈ ≡ ≤ 2 c 6 ≫ε and 1 1 ε # c Z[ω]: c 1 (mod 36), N(c) y, L ,χ =0 y − . ∈ ≡ ≤ 2 c 6 ≫ε MOMENTS HECKE L-FUNCTIONS WITH QUADRATIC CHARACTERS 3

Proof. The lower bounds follow from Theorems 1.1 and 1.2, via standard arguments using Cauchy’s inequality (see [21]).

2. Preliminaries In this section, we write down the preliminary results required in the proof of our main theorems.

2.1. Quadratic symbol and quadratic Gauss sum in Q(i). The symbol ( n· )4 is the quartic residue symbol in the ring Z[i]. For a prime π Z[i] with N(π) = 2, the quartic character is defined for a Z[i], (a, π) = 1by a (N(π) 1)/4 ∈ a 6 a ∈ π 4 a − (mod π), with π 4 1, i . When π a, we define π 4 = 0. Then the quartic character can ≡ ∈ {± ± } | 2 be extended to any composite n with (N(n), 2) = 1 multiplicatively. We further define ( n· )= n· 4 to be the quadratic residue symbol for all n Z[i] with (N(n), 2) = 1. ∈ Note that in Z[i], every ideal coprime to 2 has a unique generator congruent to 1 modulo (1 + i)3. Such a generator is called primary. Observe that a non-unit n = a + bi in Z[i] is congruent to 1 mod(1+ i)3 if and only if a 1 (mod 4),b 0 (mod 4) or a 3 (mod 4),b 2 (mod 4) by [17, Lemma 6, p. 121]. ≡ ≡ ≡ ≡ Recall that (see [17, Theorem 2, p. 123]) the quartic reciprocity law states that for two primary primes m,n Z[i], ∈ m n ((N(n) 1)/4)((N(m) 1)/4) = ( 1) − − . n 4 m 4 − Thus the following quadratic reciprocity law holds for two primary primes m,n Z[i]: ∈ m n = . n m For a non-unit n Z[i], the quadratic Gauss sumg(n) is defined by ∈ x x z z g(n)= e , where e(z) = exp 2πi . n n 2i − 2i x modX n From the supplement theorem to the quartice reciprocity lawe (see for example, Lemma 8.2.1 and Theorem 8.2.4 in [2]), we have for n = a + bi primary,

i (1 a)/2 1+ i (a b 1 b2)/4 = i − and = i − − − . n n 4 4 It follows that for any c 1 (mod 16), we have ≡ i 1+ i (2.1) = =1. c c 4 4 This shows that χc = ( c· )4 is trivial on units, hence for any c square-free and congruent to 1 (mod 16), χc can be regarded as a primitive character of the ray class group h(c). We recall here that for any c, the ray class group h(c) is deﬁned to be I(c)/P(c), where I(c) = I : ( , (c)) = 1 and P(c) = (a) P : a 1 (mod c) with I and P denoting the group of fractional ideals in{AK ∈and theA subgroup} of principal ideals,{ ∈ respectively.≡ }

We shall determine the exact value of g(c) where c 1 (mod 16) is square-free in Z[i]. We have g(1) = 1 by ≡ 3 definition. If c = 1, we can write c = π1 πk with πi 1 (mod (1 + i) ) being distinct primes. It follows from quadratic reciprocity6 that ··· ≡ k g(c)= g(πi). i=1 Y Thus, it suffices to compute g(π) for a prime π 1 (mod (1 + i)3). This evaluation is available in [22, Proposition 2.2] (be aware that the definition of the Gauss sum≡ in [22] is different from the one here) and we have 1 g(π)= − N(π)1/2. π 4 We conclude readily from the above discussions and (2.1) that for c 1 (mod 16) and square-free, ≡ 1 (2.2) g(c)= − N(c)1/2 = N(c)1/2. c 4 4 PENG GAO AND LIANGYI ZHAO

2.2. Quadratic symbol, Kronecker symbol and quadratic Gauss sum in Q(ω). It is well-known that K = Q(ω) has class number 1, and the symbol ( n· ) is the quadratic residue symbol in the ring of integers K = Z[ω]. For a prime a (NO(π) 1)/2 π Z[ω], π = 2, the quadratic character is defined for a Z[ω], (a, π) = 1 by a − (mod π), with ∈ 6 ∈ π ≡ a 1 . When π a, we define a = 0. Then the quadratic character can be extended to any composite n with π ∈ {± } | π (N(n), 2) = 1 multiplicatively. We further define = 1 when n is a unit in Z[ω]. n· In Z[ω], every ideal co-prime to 3 has a unique generator congruent to 1 (mod 3). Such a generator is called primary. Observe that a non-unit n = a + bω in Z[ω] is congruent to 1 (mod 3) if and only if a 1 (mod 3), and b 0 (mod 3) (see the discussions before [17, Proposition 9.3.5]). ≡ ≡

We shall say that any n Z[ω] is cubic E-primary if n3 = a + bω with a,b Z such that 6 b and a + b 1 (mod 4). Any cubic E-primary number∈ is thus co-prime to 2. ∈ | ≡

It follows from [20, Lemma 7.9] that any n = a + bω Z[ω] is cubic E-primary if and only if ∈ (2.3) a + b 1 (mod 4), if 2 b, ≡ | b 1 (mod 4), if 2 a, ≡ | a 3 (mod 4), if 2 ∤ ab. ≡ Furthermore, the following quadratic reciprocity law holds for two cubic E-primary, co-prime numbers n,m Z[ω]: ∈ n m ((N(n) 1)/2)((N(m) 1)/2) (2.4) = ( 1) − − . m n − r Let c Z[ω], we say that c is E-primary if we can write it as c = ( (1 ω)) c′ with r 0, r Z, (c′, 6) = 1, c′ is ∈ − − ≥ ∈ cubic E-primary and either c′ or c′ is primary. Note that our deﬁnition of E-primary here is the same as that deﬁned in [20, Section 7.3] when (c, 6) =− 1. One checks easily that every ideal co-prime to 2 in Z[ω] has a unique E-primary generator.

One also has the following supplementary laws for n = a + bω, (n, 6) = 1 being E-primary (see [20, Theorem 7.10]),

1 (N(n) 1)/2 1 ω a 2 2 (2.5) − = ( 1) − , − = and = , n − n 3 Z n N(n) Z where · Z denotes the Jacobi symbol in Z. One checks that the last expression above holds in fact for all E-primary numbers.· When c Z[ω] which is square-free and congruent to 1 (mod 36), it follows from (2.5) that ∈ 1 1 ω 2 (2.6) − = − = =1. c c c This shows that χc = ( c· ) is trivial on units and it can be regarded as a primitive character of the ray class group h(c).

( 8c) For any element c Z[ω], (c, 2) = 1, we can define a quadratic Dirichlet character χ − (mod 8c) such that for any ∈ n (Z[ω]/(8cZ[ω]))∗, ∈ ( 8c) 8c χ − (n)= − . n ( 8c) One deduces from (2.5) and the quadratic reciprocity that χ − (n) =1 when n 1 (mod 8c). It follows from this ( 8c) ( 8c) ≡ that χ − (n) is well-defined. As χ − (n) is clearly multiplicative and of order 2 and is trivial on units, it can be ( 8c) regarded as a quadratic Hecke character (mod 8c) of trivial infinite type. We denote χ − for this Hecke character as ( 8c) well and we call it the Kronecker symbol. Furthermore, when c is square-free, χ − is non-principal and primitive. To ( 8c) see this, we write c = u ̟ ̟ with u a unit and ̟ being E-primary primes. Suppose χ − is induced by some c · 1 ··· k c j χ modulo c′ with ̟j ∤ c′, then by the Chinese Remainder Theorem, there exists an n such that n 1 (mod 8c/̟j) ̟j ( 8c) ( 8c) ≡ and n = 1. It follows that χ(n) = 1 but χ − (n) = 1, a contradiction. Thus, χ − can only be possibly induced by some χ6 modulo 4c. By the Chinese Remainder Theorem,6 there exists an n such that n 1 (mod c) and n 1+4ω ≡( 8c) 2 ≡ (mod 8). As this n 1 (mod 4), it follows that n 1 (mod 4c), hence χ(n) = 1 but χ − (n) = = 1 = 1 ≡ ≡ n − 6 (note that u = 1 when u is a unit in Z[ω]) and this implies that χ( 8c) is primitive. This also shows that χ( 8c) is n − − non-principal. MOMENTS HECKE L-FUNCTIONS WITH QUADRATIC CHARACTERS 5

For a non-unit n Z[ω], (n, 2) = 1, the quadratic Gauss sum g(n) is defined by ∈ x x z z g(n)= e , where e(z)= e . n n √ 3 − √ 3 x modX n − − It follows from the definition that g(1) = 1.e The following well-knowne relation (see [6]) now holds for all n: N(n) if n is square-free, g(n) = | | (p0 otherwise. The following properties of g(n) can be easily derived from the definition: n n (2.7) g(n n )= 2 1 g(n )g(n ), (n ,n )=1. 1 2 n n 1 2 1 2 1 2 In what follows we compute the value of g(c) where c 1 (mod 36) is square-free in Z[ω]. We first evaluate the Gauss sum at each E-primary prime ̟. We have the following≡ Lemma 2.3. Let ̟ be an E-primary prime in Z[ω]. Then N(̟)1/2 if N(̟) 1 (mod 4), g(̟)= ≡ iN(̟)1/2 if N(̟) 1 (mod 4). (− ≡− Proof. Note that (1 ω) is an E-primary prime lying above the rational prime 3 and direct computation shows that − − g( (1 ω)) = √3i. Now we consider the case when N(̟)= p is a prime 1 (mod 3) in Z. It follows that ̟ is also a prime− − in Z[ω]− such that (̟, ̟)=1. Here we usen ¯ to denote the complex≡ conjugate of n, for any n C. We consider ∈ the Gauss sum associated to the quadratic Dirichlet character χp = · : p Z x x x x τ(χ )= e = e . p p p ̟ N(̟) 1 x p Z 1 x N(̟) ≤X≤ ≤ X≤ Now write x = y̟ + y̟, where y varies over a set of representatives in Z[ω] (mod ̟), then it is easy to see that as y varies (mod ̟), x varies (mod N(̟)) in Z. We deduce that y̟ y y ̟ y y y τ(χ )= e + = e + . p ̟ ̟ ̟ ̟ ̟ ̟ ̟ y modX ̟ y modX ̟ Observe that x 1 x x √ 3 x x x g(̟)= e = − e + ̟ √ 3 ̟ − ̟ ̟ ̟ ̟ ̟ x modX ̟ − x modX ̟ √ 3̟ ω(1 ω)̟ = − τ(χ )= − τ(χ ). ̟ p ̟ p Next, we shall show that ω(1 ω)̟ ω(1 ω)̟3 − = − . ̟ ̟3 For ̟ being E-primary, our discussion above implies that we can write ̟3 = a + bω with 6 b,a + b 1 (mod 4) and it follows that ̟3 = a + bω2. Hence | ≡ ω(1 ω)̟3 (1 ω)(b + aω) (1 ω)(b + aω a bω) (1 ω)2(b a) b a − = − = − − − = − − = − . ̟3 a + bω a + bω a + bω a + bω Note that when a + bω satisfies 6 b,a + b 1 (mod 4), a b is also E-primary. Thus, it follows from (2.4) that | ≡ − b a 1 a b 1 ((N(a b) 1)/2)((N(a+bω) 1)/2) a + bω − = − − = − ( 1) − − − . a + bω a + bω a + bω a + bω − a b − Note that N(a b) = (a b)2 1 (mod 4). We then conclude that − − ≡ b a 1 a + bω 1 a + bω + (a b)ω 1 a 1+ ω − = − = − − = − . a + bω a + bω a b a + bω a b a + bω a b a b − − − − Using the relation 1 + ω + ω2 = 0, we see that b a 1 a ω2 1 a − = − − = − − . a + bω a + bω a b a b a + bω a b − − − 6 PENG GAO AND LIANGYI ZHAO

We note that for two co-prime a,b Z (see [20, p. 219]), we have ∈ a =1. b We then conclude from the above discussions that we have

1 1 (N(̟) 1)/2 g(̟)= − τ(χ )= − τ(χ ) = ( 1) − τ(χ ). a + bω p ̟ p − p 1/2 1/2 1/2 1/2 As it follows from [5, Chap. 2] that τ(χp)= p = N(̟) when p 1 (mod 4) and τ(χp)= ip = iN(̟) when p 3 (mod 4), this completes the proof for the case when N(̟) is a≡ rational prime 1 (mod 3). ≡ ≡ Next, let p 2 (mod 3),p = 2 be a prime in Z, then p is also a prime in Z[ω], we now compute g(p). As in [5, Chap. ≡ 6 2] (note that in this case, we still have x mod p e˜(x/p) = 0), we have P x2 g(p)= e˜ . p x modX p We now write x = a + bω with a,b (mod p) in Z to see that p p x2 2ab b2 g(p)= e˜ = e − = p = N(p)1/2. p p a=1 x modX p Xb=1 X As N(p)= p2 1 (mod 4), this completes the proof of the lemma. ≡ Now for a fixed square-free c = 1,c 1 (mod 36), c is both primary and E-primary. By writing c as products of 6 ≡ primary primes and adjusting by a possible factor of 1, we can write c = ̟1 ̟k or ̟1 ̟k with ̟i being 3 3 3 −3 3 3 ··· − ··· distinct E-primary primes. Then c = ̟1 ̟k or ̟1 ̟k. We write c = a + bω and note that a,b satisfy (2.3) ··· − 3 ··· 3 by definition. The same consideration for (̟1 ̟k) enables us to conclude that c = ̟1 ̟k as c and c differ only by a possible factor of 1. As N(c) 1 (mod··· 4), we conclude that there must be an even··· number of ̟ in the − ≡ j decomposition of c such that N(̟j ) 1 (mod 4). We may assume that ̟1, ,̟2k0 are such primes. It follows from (2.7) and (2.4) and Lemma 2.3 that≡ − ···

1/2 k k g(c)= g(̟ ̟ ) g(̟ )= g(̟ ̟ )N ̟ . 1 ··· 2k0 j 1 ··· 2k0  j  j=2Yk0+1 j=2Yk0+1   Using Lemma 2.3 and induction on k0 shows that 1/2 2k0 g(̟ ̟ )= N ̟ . 1 ··· 2k0  j  j=1 Y   We then conclude that (2.8) g(c)= N(c)1/2.

2.4. The approximate functional equation. Let χ be a primitive Hecke character (mod m) of trivial inﬁnite type. Let G(s) be any even function which is holomorphic and bounded in the strip 4 < (s) < 4 satisfying G(0) = 1. For t R, by evaluating the integral − ℜ ∈ 1 (s+1/2+it) 1 1 s ds (2π)− Γ s + + it L s + + it,χ G(s)x 2πi 2 2 s (2)Z in two ways, we derive the following expression for L(1/2+ it,χ): 1 χ( ) 2πN( ) L + it,χ = A V A 2 N( )1/2+it t x 0= K A (2.9) 6 A⊂OX W (χ) (2π)2 it Γ(1/2 it) χ( ) 2πN( )x + 1/2 − A1/2 it V t A , N(m) Dk N(m) Γ(1/2+ it) N( ) − − DK N(m) 0= K | | 6 A⊂OX A | | MOMENTS HECKE L-FUNCTIONS WITH QUADRATIC CHARACTERS 7 where W (χ) is as in (1.1) and 1 Γ(s +1/2+ it) ξ s (2.10) V (ξ)= G(s) − ds. t 2πi Γ(1/2+ it) s (2)Z

s2 We write V for V0 and note that for a suitable G(s) (for example G(s) = e− ), we have for any c > 0 (see [18, Proposition 5.4]): c ξ − (2.11) V (ξ) 1+ . t ≪ 1+ t | | On the other hand, when G(s) = 1, we have (see [24, Lemma 2.1]) for the j-th derivative of V (ξ), 1/2 ǫ (j) ξ (2.12) V (ξ)=1+ O(ξ − ) for 0 <ξ< 1 and V (ξ)= O(e− ) for ξ > 0, j 0. ≥ 1/2 If χc is a quadratic Hecke character in Q(i), we derive readily from (2.9) by setting x = ( DK N(c)) the following expression | | 1 χ ( ) πN( ) (2.13) L ,χ =2 c A V A . 2 c N( )1/2 N(c)1/2 0= K 6 A⊂OX A Note that in this situation, W (χ)= g(c) in (2.2).

A similar consideration will give that for a quadratic character χc in Q(ω), 1 χ ( ) 2πN( ) (2.14) L ,χ =2 c A V A . 2 c N( )1/2 (3N(c))1/2 0= K 6 A⊂OX A The above discussions also apply to c = 1, provided that we deﬁne g(1) = 1 and interpret χ1 as the principal character (mod 1) so that L(s,χ1) becomes K, a convention we shall follow in the sequel.

2.5. The large sieve with quadratic symbols. The large sieve inequality for quadratic Hecke characters will be an important ingredient of this paper. The study of the large sieve inequality for characters of a ﬁxed order is of independent interest. We refer the reader to [1, 3, 9, 13–15].

Lemma 2.6. [22, Theorem 1] Suppose that K = Q(i) or Q(ω). Let M, N be positive integers, and let (an)n N be an arbitrary sequence of complex numbers, where n runs over Z[i]. Then we have ∈ 2 n ∗ ∗ a (M + N)(MN)ε a 2, n m ≪ε | n| m K n K N(n) N N(mX∈O) M N(X∈On) N X≤ ≤ ≤ for any ε> 0, where the asterisks indicate that m and n run over square-free elements of and ( · ) is the quadratic K m residue symbol. O

3. Proof of Theorem 1.1 The proofs for Q(i) and Q(ω) are similar. We will give the full details of the proof for Q(i) and a sketch of the proof for Q(ω).

3.1. The main term of the ﬁrst moment. We have, using (2.13) with G(s) = 1, that 1 N(c) χ ( ) πN( ) N(c) ∗ L ,χ Φ =2 ∗ c A V A Φ . 2 c y N( )1/2 N(c)1/2 y c 1 mod 16 c 1 mod 16 0= OK ≡ X ≡ X 6 A⊂X A Since any integral non-zero ideal in Z[i] has a unique generator (1 + i)ra, with r Z, r 0,a Z[i],a 1 (mod (1 + i)3), it follows from the quadraticA reciprocity law and (2.6) that χ ( ) = χ (c)∈ (recall≥ our convention∈ that≡ c A a χ1 is the principal character (mod 1)). This allows us to recast the last expression above as 1 π2rN(a) y1/2 N(c) M =2 M(r, a), where M(r, a)= ∗ χ (c)V Φ . 2r/2N(a)1/2 a y1/2 N(c)1/2 y r 0 c 1 mod 16 ≥ ≡ a 1 modX (1+i)3 X ≡ 8 PENG GAO AND LIANGYI ZHAO

Now we use M¨obius inversion (writing µ[i] for the M¨obius function on Z[i]) to detect the condition that c is square-free, getting π2rN(a) y1/2 N(cl2) M(r, a)= µ (l)χ (l2)M(l,r,a), with M(l,r,a)= χ (c)V Φ . [i] a a y1/2 N(cl2)1/2 y l 1 mod (1+i)3 c 1 mod 16 ≡ X ≡ X By Mellin inversion, we have

r 1/2 2 s ∞ r π2 N(a) y N(cl ) 1 y π2 N(a) s 1 V Φ = f˜(s) ds, where f˜(s)= V Φ(x)x − dx. y1/2 N(cl2)1/2 y 2πi N(cl2) (xy)1/2 (2)Z Z0

Integration by parts and using (2.12) shows f˜(s) is a function satisfying the bound for all (s) > 0, and E > 0, ℜ r E E 2 N(a) − (3.1) f˜(s) (1 + s )− 1+ . ≪ | | y1/2 With this notation, we have 1 y s χ (c) M(l,r,a)= f˜(s) a ds. 2πi N(l2) N(c)s c 1 mod 16 (2)Z ≡ X We now use the ray class characters to detect the condition that c 1 mod 16, getting ≡ s 1 1 ˜ y M(l,r,a)= f(s) 2 L(s,ψχa)ds, #h(16) 2πi N(l ) ψ modX 16 (2)Z where ψ runs over all ray class characters (mod 16), #h(16) = 32 and ψ( )χ ( ) L(s,ψχ )= A a A . a N( )s =0 A6X A We estimate M by shifting the contour to the half line. When ψχa is principal, the Hecke L-function has a pole at s = 1. Weset M0 to be the contribution to M of these residues, and M1 to be the remainder. We shall determine M0 ﬁrst.

Note that ψχa is principal if and only if both ψ and χa are principal. Hence a must be a square. We denote ψ0 for the principal ray class character (mod 16). Then we have

s L(s, ψ χ 2 )= ζQ (s) 1 N(π)− . 0 a (i) − π 2a Y| Let c0 = π/4, the residue of ζQ(i)(s) at s = 1. Then we have

2 2y 1 µ[i](l)χa2 (l ) ˜ 2 M0 = r/2 f(1)Ress=1L(s, ψ0χa ) 2 #h(16) 2 N(a) N(l ) r 0 l 1 mod (1+i)3 ≥ a 1 modX (1+i)3 ≡ X ≡ 2c0y 1 1 ˜ 1 2 − = r/2 f(1) 1 N(π)− 1 N(π)− #h(16)ζQ(i)(2) 2 N(a) − − r 0 π 2a π (2a) ≥ a 1 modX (1+i)3 Y| Y| ≡ 2c0y 1 1 ˜ 1 − = r/2 f(1) 1+ N(π)− . #h(16)ζQ(i)(2) 2 N(a) r 0 π 2a ≥ a 1 modX (1+i)3 Y| ≡ Note that

1 1 µ[i](d) (1 + N(π)− )− = , σ(d) π 2a (d) (2a) Y| X| where σ(d) denotes the sum of the norms of the integral ideal divisors of (d). MOMENTS HECKE L-FUNCTIONS WITH QUADRATIC CHARACTERS 9

Applying this, we see that

1 1 1 µ[i](d) 1 1+ N(π)− − = . N(a) σ(d) N(a) N(a) x π 2a (d) a 1 mod (1+i)3 X≤ 3 Y| X ≡ X a 1 mod (1+i) N(d) 2x d a, N(a) x ≡ ≤ (2,d) | ≤ Note the following result from the Gauss circle problem, π 1= x + O(xθ). 8 N(a) x ≤ a 1 modX (1+i)3 ≡ Here one can take θ to be 131/416 (see [16]).

Applying this and partial summation, we get

1 π θ 1 = log x + C + O(x − ), N(a) 8 0 N(a) x ≤ a 1 modX (1+i)3 ≡ where C0 is a constant.

It follows that

1 1 1 π θ 1 1+ N(π)− − = A log x + C + O(x − ), N(a) 12 Q(i) 1 N(a) x π 2a ≤ | a 1 modX (1+i)3 Y ≡ where AQ(i), C1 are constants with AQ(i) deﬁned in (1.3).

We then deduce that 2 r 2 1 1 1 1 1 1 π2 N(a) f˜(1) 1+ N(π)− − = Φ(x) (1 + N(π)− )− V dx. N(a) N(a) (xy)1/2 a 1 mod (1+i)3 π 2a Z a 1 mod (1+i)3 π 2a ≡ X Y| 1 ≡ X Y| Applying partial summation and (2.12), we get r 2 1 1 1 π2 N(a) (1 + N(π)− )− V N(a) (xy)1/2 a 1 mod (1+i)3 π 2a ≡ X Y| 1 θ 1/4 1/2 r/2 − π (xy) π 2 1/2 r/2 AQ log + C + O (xy) > π2 , 12 (i) π1/22r/2 2 (xy)1/4  !  =   1  1/2 r/2 −  π 2 1/2 r/2 O 1/4 (xy) π2 ,  (xy) ! ≤   with some constant C2.

We then conclude that by a straightforward calculation (we may assume that y is large), (2 + √2)c πA 0 Q(i) ˆ ˆ (3+θ)/4 (3.2) M0 = Φ(0)y log y + CQ(i)Φ(0)y + O(y ), 24#h(16)ζQ(i)(2) where CQ(i) is the same constant CQ(i) appearing in (1.2).

3.2. The remainder terms of the ﬁrst moment. To treat M1, we bound everything by absolute values and use (3.1) to get that for any E > 0,

r E ∞ 1/2 1 1/2 2 N(a) − 1 E (3.3) M y N(a)− 1+ L + it,ψχ (1 + t )− dt. 1 ≪ N(l) y1/2 2 a | | N(l) √y ψ mod 16 r 0 Z X≪ X X≥ 3 a 1 mod (1+i) −∞ ≡ 10 PENG GAO AND LIANGYI ZHAO

Now it follows easily from the Cauchy-Schwarz inequality and (4.1) that

1/2 1 1/2+ǫ N(a)− L + it,ψχ (N(1 + t )) . 2 a ≪ | | a 1 mod (1+i)3 ≡ X N(a) N ≤ Applying this in (3.3) and note that we can restrict the sum over r, a to be 2rN(a) y1/2+ǫ, we immediately deduce that ≤ M y3/4+ǫ. 1 ≪ Combining the results for M0 and M1, we obtain the result of Theorem 1.1 for Q(i).

3.3. The proof for Q(ω). As stated before, the proof for Q(ω) is very similar. Starting with the approximate functional equation, (2.14), with G(s) = 1, we have 1 N(c) χ ( ) 2πN( ) N(c) M = ∗ L ,χ Φ =2 ∗ c A V A Φ . 2 c y N( )1/2 (3N(c))1/2 y c 1 mod 36 c 1 mod 36 0= OK ≡ X ≡ X 6 A⊂X A Since any integral non-zero ideal in Z[ω] has a unique generator 2r1 (1 ω)r2 a, with r , r Z, r , r 0,a Z[ω], A − 1 2 ∈ 1 2 ≥ ∈ (a, 2)=1,a 1 (mod 3), it follows from (2.6) and the deﬁnition of χ(a) that χ ( )= χ (a). ≡ c A c The above discussions allow us to recast M as 1 M =2 M(r, a), 2r1 3r2/2N(a)1/2 r1,r2 0 (a,X2)=1≥ a 1 mod 3 ≡ where π22r1+13r2 1/2N(a) y1/2 N(c) M(r, a)= ∗ χ(a)(c)V − Φ . y1/2 N(c)1/2 y c 1 mod 36 ≡ X Now the proof goes along the same arguments as those in the proof for Q(i). The main term, the analogue of (3.2), that emerges from this computation is √ (1 + 3)πAQ(ω) (3+θ)/4 Ress=1ζQ(ω)(s)Φ(0)y log y + CQ(ω)Φ(0)y + O(y ), 20#h(36)ζQ(ω)(2) where CQ(ω) is the same constant CQ(ω) appearing in (1.2),b b

√3 Res ζ (s)= π, #h = 108. s=1 Q(ω) 9 (36) This completes the proof of the theorem.

4. Proof of Theorem 1.2 The proofs for Q(i) and Q(ω) are very similar and we only give the details for Q(i) here.

3 Let a 1 (mod (1 + i) ) and ψ be a ray class character (mod 16). Let χa be the Hecke character (mod 16a) of trivial inﬁnite≡ type. For any ideal (c) co-prime to (1 + i), with c being the unique generator of (c) satisfying c 1 3 a ≡ (mod (1 + i) ), χa((c)) is deﬁned as χa((c)) = c (see [18, Example 2, p. 62]). Now we show that 1 2 (4.1) L + it,ψχ (N(1 + t ))1+ǫ . 2 a ≪ | | a 1 mod (1+i)3 ≡ X N(a) N ≤ As the proof is similar to the proof of [1, Theorem 1.3] and [3, Corollary 1.4], we only sketch the arguments. Write 2 3 a = a1a2 with a1,a2 1 (mod (1 + i) ) and a1 square-free. Then χa equals χa1 multiplied by a principal character ≡ s2 whose conductor divides a2. We may further assume that ψχa1 is primitive. Then (2.9) is valid with G(s)= e− . By MOMENTS HECKE L-FUNCTIONS WITH QUADRATIC CHARACTERS 11

1/2 inserting (2.9) into the left-hand side expression in (4.1) with x = 2(N(ma1 )) , where ma1 is the conductor of ψχa1 , and applying the Cauchy-Schwarz inequality, we see that it suﬃces to bound 2 ψχ ( ) 2πN( ) ∗ a1 A V A , 1/2+it t 1/2 3 3 N( ) N(ma1 ) a2 1 mod (1+i) a1 1 mod (1+i) 0= K A ≡ X2 ≡ X 2 6 A⊂OX N(a2) N N(a1) N/N(a2) ≤ ≤ where ∗ denotes summation over square-free elements of Z[i].

In viewP of (2.11), we may truncate the sum over above to N( ) (N 1/2(1 + t ))1+ǫ. By shifting the contour in (2.10) to the ǫ line and write s = ǫ + iw, it suﬃces toA bound A ≤ | | 2

∗ ψχa1 ( ) 1/2+ǫA+it+iw . 3 3 N( ) a 1 mod (1+i) a 1 mod (1+i) 0= K 2 1 6 A⊂O A ≡ X2 ≡ X 2 N( ) (NX1/2(1+ t ))1+ǫ N(a2) N N(a1) N/N(a2) ≤ ≤ A ≤ | | r 2 In the inner sum above, writing = (1+ i) 1 2 with 1, 2 co-prime to 1 + i, 1 square-free and using the Cauchy-Schwarz inequality, it is enoughA to estimateA A A A A 1 1 r(1+2ǫ)/2 1+2ǫ 3 2 N( 2) a2 1 mod (1+i) r 0 0= 2 K A ≡ X2 r 1/2X≥ 1/2 1+ǫ 2 6 A1X/2⊂O 1+ǫ r N(a ) N 2 (N (1+ t ) ) N( 2) (N (1+ t )) /2 2 ≤ ≤ | | A ≤ | | 2

ψχ ( ) ∗ ∗ a1 1 1/2+Aǫ+it+iw . × 3 N( 1) a1 1 mod (1+i) 0= 1 K A ≡ X 2 1/26 AX⊂O1+ǫ 2 r N(a ) N/N(a ) N( 1) (N (1+ t )) /(N( 2)2 ) 1 ≤ 2 A ≤ | | A

We now apply the large sieve inequality and arrive at

ψχ ( ) N (N 1/2(1 + t ))1+ǫ ∗ ∗ a1 1 ǫ 1/2+Aǫ+it+iw N 2 + 2 | |r . N( 1) ≪ N(a2) (N( )2 ) 3 0= 2 a1 1 mod (1+i) 1 K A A ≡ X 2 1/26 AX⊂O1+ǫ 2 r N(a ) N/N(a ) N( 1) (N (1+ t )) /(N( 2)2 ) 1 ≤ 2 A ≤ | | A

As the sums over a2, r and 2 all converge, we conclude that A 2

∗ ψχa1 ( ) 1+ǫ 1/2+ǫA+it+iw (N(1 + t )) . 3 3 N( ) ≪ | | a 1 mod (1+i) a 1 mod (1+i) 0= K 2 1 6 A⊂O A ≡ X2 ≡ X 2 N( ) (NX1/2(1+ t ))1+ǫ N(a2) N N(a1) N/N(a2) ≤ ≤ A ≤ | |

This establishes (4.1) from which Theorem 1.2 follows readily.

Acknowledgments. P. G. is supported in part by NSFC grant 11871082 and L. Z. by the FRG grant PS43707. Parts of this work were done when P. G. visited the University of New South Wales (UNSW) in June 2017. He wishes to thank UNSW for the invitation, ﬁnancial support and warm hospitality during his pleasant stay.

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