Abstract We prove an asymptotic for the third moment of symmetric-square L-functions associated to holomorphic Hecke cusp forms of level one and weight between K and 2K, as K→∞. 1. Introduction Many questions in number theory boil down to problems concerning the central values of L-functions, and thus we would like to understand these objects (see [8] for a nice survey). One way to proceed is to consider L-functions in spectrally complete families, over which we can try to evaluate moments of the L-values. In principle, knowledge of all moments can recover the full distribution of the L-values. But even the moments of low order can give us valuable information about L-functions. Indeed, this is the starting point in the amplifier, mollifier and resonance methods [4, 7, 18], which are used to prove upper bounds, non-vanishing and large values of L-functions. Moments of L-functions have been extensively studied for GL(1) and GL(2) L-functions, but there are relatively few results in higher rank that provide asymptotic evaluations of moments. There are well-established conjectures [3] for these asymptotics and it turns out that the shape of the asymptotic depends on the symmetry type of the family in the sense of the so-called Katz–Sarnak philosophy [12]. In this paper, we are interested in the symplectic symmetry type. Many papers offer the following two examples of families that have this symmetry type: the Dirichlet L-functions associated to quadratic characters and the symmetric-square L-functions described below. For the former family, the first three moments are known by the work of Jutila [11] and Soundararajan [17], and Young [21] later improved the error term of the third moment. The goal of this paper is to evaluate for the first time the third moment of the symmetric-square family, so that now it is on par with the quadratic Dirichlet L-functions in terms of moments known. Let Hk denote the set of holomorphic Hecke cusp forms f of level 1 and weight k, with Fourier expansion f(z)=∑n=1∞af(n)nk−12e2πinz for I(z)>0, where the coefficients af(n) are real and af(1)=1. The set Hk has k/12+O(1) elements and forms an orthogonal basis of the space of cusp forms of level 1 and weight k. The symmetric-square L-function associated to f∈Hk is defined as L(s,sym2f)=ζ(2s)∑n=1∞af(n2)ns for R(s)>1, where R(s) and I(s) denote the real and imaginary parts of s. This converges absolutely for R(s)>1 since af(n)≤d(n) by the work of Deligne. As shown by Shimura [16], the series given above analytically extends to an entire function on C and satisfies the functional equation L∞(s)L(s,sym2f)=L∞(1−s)L(1−s,sym2f), where L∞(s)=π−32sΓ(s+12)Γ(s+k−12)Γ(s+k2)=π−32s+1222−s−kΓ(s+12)Γ(s+k−1). Thus (under our normalization) the central point is s=12, and the central value L(12,sym2f) is real. We also see from the functional equation that the conductor of L(s,sym2f) in the weight aspect is k2. Gelbart and Jacquet [5] showed that L(s,sym2f) is attached to an automorphic form on GL(3). As far as moments are concerned, only the first moment of L(12,sym2f) is known when averaging f over the set Hk, as k→∞. An asymptotic for the first moment was proved by Lau [14], which has been improved since then; the current best result is due to Balkanova and Frolenkov [1]. The second moment over the set Hk seems to be completely out of reach of current methods. However, if we enlarge the family to ⋃K≤k≤2Kk≡0mod2Hk, (1.1) where K→∞, then the ratio of log(conductor) to log(family size) is the same for both quadratic Dirichlet L-functions and symmetric-square L-functions, and so that the playing field is level. In [13], the second author proved the following asymptotic for the second twisted moment, which yields the second moment by putting r=1 (in which case the right-hand size has leading term of size log3K). Theorem 1.1. ([13]). Let h be an infinitely differentiable and compactly supported function on the positive reals. For any positive integer d, write d=d1d22with d1square-free. For any (Convention: throughout this paper, ϵwill always denote an arbitrarily small positive constant, but not necessarily the same one from one occurrence to the next. Unless stated otherwise, all error terms may depend on ϵand h.) ϵ>0and r≤K1−ϵwe have 1K∑k≡0mod2h(k−1K)∑f∈Hk†L(12,sym2f)2af(r2)=1r∑d∣r21d1∫0∞h(u)(14log2uKd1d2loguKd2r−112log3uKd1d2+P2(loguKd1d2)+loguKd2rP1(loguKd1d2))du+O(r12K−1+ϵ),where Piis a polynomial of degree i, logix=(logx)i, and ∑f∈Hk†αf=∑f∈Hk2π2(k−1)L(1,sym2f)αffor any complex numbers αfdepending on f. The ‘harmonic’ average ∑† arises naturally in the Petersson trace formula. We should think of the weights in this sum as being of size 1/∣Hk∣ on average, for ∑f∈Hk†1∼1 as k→∞. In a different setting, Blomer [2] obtained the (twisted) first and second moments of symmetric-square L-functions attached to newforms of fixed weight, prime level and real primitive nebentypus, as the level tends to infinity. This family has the same log(conductor) to log(family size) ratio as the family in [13], so the problems are of roughly the same level of difficulty. Nothing is known for the third moment in the level aspect. Now we come to the third moment of L(12,sym2f) over the family (1.1), which is the main result of this paper. We first observe that Theorem 1.1 comes quite close to giving us the third moment. This is because after writing the third moment as 1K∑k≡0mod2h(k−1K)∑f∈Hk†L(12,sym2f)2L(12,sym2f) (1.2) and then using Lemma 2.3 to write L(12,sym2f)=2∑r<k1+ϵaf(r2)r12Vk(r)+O(k−100), we see that a result like Theorem 1.1 would suffice if it allowed us to take r to be as large as K1+ϵ, and if the error term were improved a bit. But the restriction to r<K1−ϵ in the twisted second moment seems to require new ideas to overcome. Our approach to prove an asymptotic for the third moment is to combine the ideas of [13] and those of Sprung [19]. In his Masters thesis, Sprung proved a sharp upper bound of Kϵ for the third moment (1.2) with additional twisting. Consider the off-diagonal part of the third moment in dyadic intervals, which as we will see, roughly has the shape ∑n∼N,m∼M,r∼Rc≤NMR/K2−ϵS(n2,m2r2,c)ce(2nmrc) for N,M,R<K1+ϵ. Sprung’s method was to apply Poisson summation in n,m,r over residue classes mod c. This process leads to a certain exponential sum, for which Sprung proved an upper bound that reflects square-root cancellation. The upper bound leads to a sharp estimate of the third moment, but not an asymptotic. A full evaluation of the exponential sum could lead to an asymptotic, but this seems to be difficult. Our observation is that Sprung’s upper bound suffices for the purposes of an asymptotic when c≥K1−2ϵ. If c<K1−2ϵ and R≥K1−ϵ, then we need only a simplified version of Sprung’s exponential sum, which can be evaluated. This is because Poisson summation over r∼R in residue classes mod c will lead to only the zero frequency sum on the dual side. In the case that R<K1−ϵ, we can use a result like Theorem 1.1. Making this idea work, by putting together all these different cases, is a little delicate. We really need to put together the different pieces, because upper bounds for each piece may be too large; but when everything is put together the large term (seen at the very end of the paper) will vanish. This kind of phenomenon was also present in [2, 13], where parts of the diagonal and off-diagonal were shown to cancel out. Our main result is Theorem 1.2. (Third moment). Let h be an infinitely differentiable and compactly supported function on the positive reals. We have 1K∑k≡0mod2h(k−1K)∑f∈Hk†L(12,sym2f)3=P6(logK)+O(K−1/2+ϵ),for some polynomial P6of degree 6, depending on h. Although we do not explicitly compute the polynomial above, we note that its degree is exactly what is expected for the symplectic family type. It would be interesting to fully work out the cubic moment with small shifts away from the central point, as in [3], and compare with the conjectures given there. 2. Preliminaries 2.1. Petersson’s trace formula Lemma 2.1. ∑f∈Hk†af(n)af(m)=δm,n+2πik∑c=1∞S(n,m,c)cJk−1(4πmnc),where the value of δm,nis 1 if m=nand 0 otherwise, S(n,m,c)is a Kloosterman sum, and Jk−1(x)is the J-Bessel function. This may be found in [10, Proposition 14.5]. 2.2. Hecke multiplicativity Recall the Hecke multiplicative relation af(n2)af(m2)=∑d∣(n2,m2)af(n2m2/d2). (2.1) We will need the following observation. For any integers n,m,r, we have ∑d∣(n2,m2)r2=n2m2/d21=∑d∣(r2,m2)n2=r2m2/d21. (2.2) To justify this, first recall that by the Taylor expansion of the J-Bessel function [6, 8.402], we have the estimate Jk−1(x)≪C(x/k)k uniformly for 0≤x≤C. Now by Lemma 2.1 and (2.1), we have that the left-hand side of (2.2) equals limk→∞∑f∈Hk†∑d∣(n2,m2)af(n2m2/d2)af(r2)=limk→∞∑f∈Hk†af(n2)af(m2)af(r2). Similarly, the right-hand side equals limk→∞∑f∈Hk†∑d∣(r2,m2)af(r2m2/d2)af(n2)=limk→∞∑f∈Hk†af(n2)af(m2)af(r2). 2.3. Sum of Kloosterman sums We will need the following exponential sums. The first part of this Lemma is from [13, Lemma 3.2] and the second part is from [19, Lemma 5]. Lemma 2.2. Suppose that d,l1,l2,l3are integers. We have ∑amodcS(a2,d2,c)e(2adc)={ϕ(c)c−12ifcisasquare,0otherwise, ∑a1,a2,a3modcS(a12,a22a32d,c)e(2a1a2a3dc)e(a1l1+a2l2+a3l3c)≪c2+ϵ(dl1,c)(l12,l2,l3,c)12. 2.4. Approximate functional equation The following is quoted from [13, Lemma 2.2]. Lemma 2.3. We have L(12,sym2f)=2∑n≥1af(n2)n12Vk(n),where for any real ξ>0and A>0, Vk(ξ)=12πi∫(A)L∞(12+y)L∞(12)ζ(1+2y)ξ−ydyyis real valued and satisfies Vk(B)(ξ)≪A,Bξ−B(kξ)A,foranyA>0andintegerB≥0.Vk(ξ)=12(log(k/ξ)+C)+O(ξk),whereC=2γ−3logπ2−log2+Γ′(34)2Γ(34). Thus L(12,sym2f) can be expressed as a Dirichlet series of length about k1+ϵ, up to negligible error. 2.5. Average of J-Bessel functions The following is quoted from [13, Lemma 2.3], but it originates from [9, Lemma 5.8]. Lemma 2.4. We have for t>0, 1K∑k≡0mod22ikh(k−1K)Jk−1(t)=−1tI(e−2πi/8eitℏ(K22t))+O(tK5∫−∞∞v4∣hˆ(v)∣dv), (2.3)where ℏ(v)=∫0∞h(u)2πueiuvduand hˆdenotes the Fourier transform of h. The implied constant is absolute. By integrating by parts several times we get that ℏ(v)≪∣v∣−B for any B≥0, where the implied constant depends on B (such dependence will not be explicitly stated every time in subsequent bounds). The main term of (2.3) is O(K−B) for B as large as we like if t<K2−ϵ. For future use, define for any complex number w the more general function ℏw(v)=∫0∞h(u)2πuuw/2eiuvdu. By integration by parts, we have the bound ℏw(j)(v)≪(1+∣w∣)B∣v∣−B (2.4) for any integers j,B≥0, where the implied constant depends on R(w), and on j and B of course. 2.6. A Mellin transform Let ℏ˜w(s)=∫0∞vs−1ℏ(v)dv denote the Mellin transform of ℏ. This converges absolutely for R(s)>0 by (2.4) and is holomorphic in this half plane. By integration by parts and (2.4), we have ℏ˜w(s)≪(1+∣w∣)B+R(s)+1(1+∣s∣)−B, (2.5) where the implied constant depends on R(w) and B. The following result (taken from [13, Lemma 3.3]) evaluates ℏ˜w(s). In particular, this gives its analytic continuation to C. Lemma 2.5. For 0<R(s)<1, we have ℏ˜w(s)=∫0∞h(u)2πuuw/2−sΓ(s)e(s/4)du,where e(s)=e2πis. 2.7. A weight function For future use, we define for positive integers n,m,r and real v, WK(n,m,r,v)=1(2πi)3∫(A1)∫(A2)∫(A3)G(y)G(x)G(z)ζ(1+2y)yζ(1+2x)xζ(1+2z)z1nxmyrz×∫0∞Γ(uK+y+12)Γ(uK+12)Γ(uK+x+12)Γ(uK+12)Γ(uK+z+12)Γ(uK+12)×h(u)2πueiuvdudxdydz, (2.6) where A1,A2,A3>0 and G(z)=π−32z2−zΓ(z2+34)Γ(34). By Stirling’s approximation of the gamma function, G(z) decays exponentially as ∣I(z)∣→∞. By Stirling’s approximation applied to the gamma functions involving u, we have for n,m,r<K1+ϵ, WK(n,m,r,v)=W(nK,mK,rK,v)+1KW0(nK,mK,rK,v)+O(K−2+ϵ), (2.7) where for ξ1,ξ2,ξ3>0, we define W(ξ1,ξ2,ξ3,v)=1(2πi)3∫(A1)∫(A2)∫(A3)G(y)G(x)G(z)ζ(1+2y)yζ(1+2x)xζ(1+2z)z×ξ1−xξ2−yξ3−zℏx+y+z(v)dxdydz, (2.8) and W0(ξ1,ξ2,ξ3,v) is defined in the same way except that it has an additional factor of P(x,y,z)/u12 for some polynomial P(x,y,z). All this is similar to [13, Section 2.4], except that there the estimate WK=W+O(K−1+ϵ) was used, while here we need the additional term 1KW0 from Stirling’s expansion (in order to arrive at the desired error term in (3.1)). Using (2.4) we have ∂j1∂ξ1j1∂j2∂ξ2j2∂j3∂ξ3j3∂j4∂vj4W(ξ1,ξ2,ξ3,v)≪ξ1−j1−A1ξ2−j2−A2ξ3−j3−A3v−B, (2.9) for x1,x2,x3,v>0, and any real A1,A2,A3>0 and integers j1,j2,j3,j4,B≥0. We have a similar bound for W0. 2.8. Poisson summation We will make vital use of the Poisson summation formula (the proof of which is standard). Lemma 2.6. Given a rapidly decaying smooth function Φ(x)and an arithmetic function Sc(n)with period c, we have ∑−∞<n<∞Φ(nN)Sc(n)=Nc∑−∞<l<∞Φˆ(lNc)∑amodcSc(a)e(alc),where Φˆis the Fourier transform of Φ. 3. Pieces of the third moment Let δ>0 be an arbitrarily small positive constant. Let U(x) be a smooth function with bounded derivatives that is compactly supported on the interval (12,2K2δ), such that U(x)=1 for 1≤x≤K2δ. Let L=K1−δ. Using Lemma 2.3, we have 1K∑k≡0mod2h(k−1K)∑f∈Hk†L(12,sym2f)3=8K∑k≡0mod2h(k−1K)∑f∈Hk†∑n,m,r≥1af(n2)af(m2)af(r2)(nmr)12Vk(n)Vk(m)Vk(r)=S1+S2, where S1=8K∑k≡0mod2h(k−1K)∑f∈Hk†∑n,m,r≥1af(n2)af(m2)af(r2)(nmr)12Vk(n)Vk(m)Vk(r)U(rL) and S2=8K∑k≡0mod2h(k−1K)∑f∈Hk†∑n,m,r≥1af(n2)af(m2)af(r2)(nmr)12Vk(n)Vk(m)Vk(r)(1−U(rL)). We should think of the sum S1 as consisting of the terms with r large, because U(rL) will restrict to 12K1−δ≤r≤2K1+δ, and the terms with r≥2K1+δ are negligible by the decay of Vk(r). We should think of S2 as consisting of the terms with r small, because once we restrict to r≤K1+δ by the decay of Vk(r), the function 1−U(rL) vanishes unless r≤K1−δ. 3.1. The sum S1 Using (2.1), we get S1=8K∑k≡0mod2h(k−1K)∑f∈Hk†∑n,m,r≥1d∣(n2,m2)af(n2m2/d2)af(r2)(nmr)12Vk(n)Vk(m)Vk(r)U(rL). We apply Lemma 2.1 to get S1=D1+OD1, where the diagonal is D1=8K∑k≡0mod2h(k−1K)∑n,m,r≥1r2=n2m2/d2d∣(n2,m2)1(nmr)12Vk(n)Vk(m)Vk(r)U(rL) and the off-diagonal is OD1=8K∑n,m,r≥1r2=n2m2/d2d∣(n2,m2)S(r2,n2m2/d2,c)c(nmr)12∑k≡0mod2h(k−1K)2πikJk−1(4πrnmcd)×Vk(n)Vk(m)Vk(r)U(rL). Writing d=d1d22, where d1 is square-free, we have that d∣n2 if and only if d1d2∣n. Thus we can replace n by nd1d2, and m by md1d2, to get OD1=8K∑n,m,r,d1,d2,c≥1S(r2,n2m2d12,c)μ2(d1)c(nmr)12d1d2∑k≡0mod2h(k−1K)2πikJk−1(4πrnmd1c)×Vk(nd1d2)Vk(md1d2)Vk(r)U(rL). Using Lemma 2.4 now we get OD1=−4π12I(∑n,m,r,d1,d2,c≥1S(r2,n2m2d12,c)μ2(d1)c12nmrd132d2e(2nmrd1/c)e(1/8)×WK(nd1d2,md1d2,r,K2c8πrnmd1)U(rL))+O(K−1/2+ϵ), where WK was defined in (2.6) and the error term comes from the error in Lemma 2.4, by the same argument as [13, Section 3.4]. By the comments following Lemma 2.4, the range of c is essentially c<nmrd1K2−ϵ<K1+ϵ. Now we can use (2.7) to get (we first truncate the sums to n,m,r<K1+ϵ by Lemma 2.3 before applying (2.7), then again expand the sums to n,m,r<∞ using (2.9), all up to negligible error) OD1=−4π12I(∑n,m,r,d1,d2,c≥1S(r2,n2m2d12,c)μ2(d1)c12nmrd132d2e(2nmrd1/c)e(1/8)×W(nd1d2K,md1d2K,rK,K2c8πrnmd1)U(rL))+⋯+O(K−1/2+ϵ), (3.1) where the ellipsis signifies the contribution coming from the lower order term W0. The treatment for this omitted term will be similar unless indicated otherwise. We divide OD1 into two cases: OD1=OD1,1+OD1,2, where OD1,1 is the same as OD1 but restricted to c≤L/Kϵ, and OD1,2 is restricted to c>L/Kϵ. 3.2. The sum OD1,1 We sum over r by Poisson summation (Lemma 2.6) after splitting into residue classes modulo c. We get OD1,1=∑−∞<l<∞−4π12I(∑n,m,d1,d2≥1c≤LK−ϵμ2(d1)e(1/8)c12nmd132d2LW^(l)Lc×∑amodcS(a2,n2m2d12,c)e(2anmd1c)e(alc))+⋯+O(K−1/2+ϵ), where W^(l)=∫−∞∞1xW(nd1d2K,md1d2K,xLK,K2c8πnmd1Lx)U(x)e(xlLc)dx. If l≠0, we integrate by parts j times (integrating the exponential and differentiating the rest) using (2.9) and the fact that U(x) is supported away from zero on the interval (12,2K2δ), to see that W^(l)≪Kϵ∣clL∣j. Since c≤L/Kϵ, the contribution of l≠0 is O(K−100) say, by taking j large enough, and so we have OD1,1=−4π12I(∑n,m,d1,d2≥1c≤LK−ϵμ2(d1)e(1/8)c12nmd132d2LW^(0)Lc∑amodcS(a2,n2m2d12,c)e(2anmd1/c))+⋯+O(K−1/2+ϵ). (3.2) By an identical argument, up to an error of O(K−100) the first line of (3.2) is the same expression one gets by applying Poisson summation in r to the sum −4π12I(∑n,m,r,d1,d2≥1c≤LK−ϵμ2(d1)e(1/8)c12nmd132d2rW(nd1d2K,md1d2K,rK,K2c8πnmd1r)U(rL)×1c∑amodcS(a2,n2m2d12,c)e(2anmd1/c)). (3.3) Thus OD1,1 equals (3.3) up to an error of O(K−12+ϵ); the reason for converting back to a sum over r is that this will later facilitate evaluation by residue calculations. By Lemma 2.2, the complete sum of (3.3) vanishes unless c is a square, so we replace c by c2 after evaluating the complete sum, and get OD1,1=−4π12I(∑n,m,r,d1,d2≥11≤c<LK−ϵϕ(c2)μ2(d1)e(1/8)c2nmrd132d2W(nd1d2K,md1d2K,rK,K2c28πrnmd1)U(rL))+⋯+O(K−1/2+ϵ). 3.3. The sum OD1,2 We sum over r,n,m by Poisson summation in residue classes modulo c. Thus OD1,2=∑−∞<l1,l2,l3<∞−4π12I(∑d1,d2≥1c>LK−ϵd112d2μ2(d1)e(1/8)c12K2LW^(l1,l2,l3)K2Ld12d22c3×∑a1,a2,a3modcS(a12,a22a32d12,c)e(2a1a2a3d1c)e(a1l1+a2l2+a3l3c))+⋯+O(K−1/2+ϵ), (3.4) where W^(l1,l2,l3)=∫−∞∞∫−∞∞∫−∞∞W(x1,x2,x3LK,cd1d228πx1x2x3L)U(x3)x1x2x3×e(−x1l1Kd1d2c+−x2l2Kd1d2c+−x3l3Lc)dx1dx2dx3. (3.5) Note that even though W(x1,x2,x3LK,cd1d228πx1x2x3L) is initially defined only for x1,x2,x3>0, we can set it to be zero if xi≤0 for any i; by the bound (2.9), it still remains a smooth function. Thus the Poisson summation is valid. Now we claim that by integrating by parts in x1,x2,x3 multiple times, we can restrict to −Kϵ<l1,l2,l3<Kϵ, up to an error of O(K−100). The argument needs a bit more work than what we saw in Section 3.2. This is because multiple differentiation of the weight functions in the integrand of (3.5) leads to negative powers of x1,x2,x3, and we must take care to be able to integrate up to zero (this was not an issue in Section 3.2 because there the function U(x) restricted the integral to a positive compact interval). By (2.9), we have ∂j1∂x1j1∂j2∂x2j2∂j3∂x3j3W(x1,x2,x3LK,cd1d228πx1x2x3L)U(x3)x1x2x3≪Kϵ(x1x2x3Lcd1d22)B∏i=13min{1xiji+2,1xiji+10+j1+j2+j3} (3.6) for any integer B≥0, where the minimum occurs by taking Ai=1 or Ai=9+j1+j2+j3 in (2.9). Integrating by parts ji times with respect to xi, for each i, we get W^(l1,l2,l3)≪Kϵ∫0∞∫0∞∫0∞∣d1d2cl1K∣j1∣d1d2cl2K∣j2∣cl3L∣j3×∣∂j1∂x1j1∂j2∂x2j2∂j3∂x3j3W(x1,x2,x3LK,cd1d228πx1x2x3L)U(x3)x1x2x3∣dx1dx2dx3. We can now show that up to negligible error we may assume ∣li∣<Kϵ. We first deal with i=1 and suppose that l1≠0. For i≠1, we take ji=0 if li=0 and ji=2 if li≠0 (in order to facilitate convergence of the sums over l2,l3). Take B=2+j1+j2+j3 in (3.6). Then taking j1 as large as we like, we get that the contribution to (3.4) of ∣l1∣>Kϵ∣cK∣∣Lc∣ is O(K−100). Since L<K this shows that the contribution of ∣l1∣>Kϵ is negligible. We can argue similarly for i=2,3. Now we keep the terms of (3.4) with l1=0, and estimate the rest. By Lemma 2.2, we have that OD1,2=(∑l1=0)+O(∑−Kϵ<l1,l2,l3<Kϵd1,d2<K1+ϵl1≠0Kϵμ2(d1)d132d2∑c>LK−ϵ(c,l1d1)(c,l12,l2l3)c32)+⋯+O(K−1/2+ϵ)=(∑l1=0)+⋯+O(K−1/2+ϵ). This uses the fact that (c,l12,l2l3)≪Kϵ because 0<∣l1∣<Kϵ, and that ∑c>LK−ϵ(c,l1d1)c32≤∑δ∣l1d1∑c>LK−ϵδ∣cδc32≤∑δ∣l1d11δ12∑c>L(Kϵδ)−11c32≪KϵL−1/2. The error can be recast as O(K−1/2+ϵ) because L=K1−δ and δ is arbitrarily small. The sum with l1=0 is converted back to a sum over n,m,r by Poisson summation again, up to an error of O(K−100) say, as we did to the r-sum in Section 3.2. We get OD1,2=−4π12I(∑n,m,r,d1,d2≥1c>LK−ϵμ2(d1)e(1/8)c12nmrd132d2W(nd1d2K,md1d2K,rK,K2c8πrnmd1)U(rL)×1c∑a1modcS(a12,n2m2d12,c)e(2nmd1c))+⋯+O(K−1/2+ϵ). By Lemma 2.2, we evaluate the complete sum and, replacing c with c2 we get, OD1,2=−4π12I(∑n,m,r,d1,d2≥1c>LK−ϵϕ(c2)μ2(d1)e(1/8)c2nmrd132d2W(nd1d2K,md1d2K,rK,K2c28πrnmd1)U(rL))+⋯+O(K−1/2+ϵ). 3.4. Combining OD1,1 and OD1,2 to recover OD1 Adding together the final expressions for OD1,1 and OD1,2, we get a sum over all c: OD1=−4π12I(∑n,m,r,d1,d2,c≥1ϕ(c2)μ2(d1)e(1/8)c2nmrd132d2W(nd1d2K,md1d2K,rK,K2c28πrnmd1)U(rL))+⋯+O(K−1/2+ϵ). 3.5. The sum S2 Using Hecke multiplicativity in a different way than we did for S1, because we want to apply Poisson summation to a different variable this time, we get S2=8K∑k≡0mod2h(k−1K)∑f∈Hk†∑n,m,r≥1d∣(r2,m2)af(r2m2/d2)af(n2)(nmr)12Vk(n)Vk(m)Vk(r)(1−U(rL)). Applying Lemmas 2.1 and 2.4, we get S2=D2+OD2, where D2=8K∑k≡0mod2h(k−1K)∑n,m,r≥1n2=m2r2/d2d∣(r2,m2)1(nmr)12Vk(n)Vk(m)Vk(r)(1−U(rL)) and (after writing d=d1d22 and replacing r by rd1d2 and m by md1d2) the off-diagonal part is OD2=−4π12I(∑n,m,r,d1,d2,c≥1S(n2,r2m2d12,c)μ2(d1)c12nmrd132d2e(2nmrd1/c)e(1/8)×W(nK,md1d2K,rd1d2K,K2c8πrnmd1)(1−U(rd1d2L)))+⋯+O(K−1/2+ϵ). (3.7) 3.6. Combining D1 and D2 Define the total diagonal to be D=D1+D2. By (2.2), we get D=8K∑k≡0mod2h(k−1K)∑n,m,r≥1n2=m2r2/d2d∣(r2,m2)1(nmr)12Vk(n)Vk(m)Vk(r). Now writing d=d1d22 as before, we can replace r by rd1d2 and m by md1d2, to get D=8K∑k≡0mod2h(k−1K)∑n,m,r,d1,d2≥1n=mrd1μ2(d1)(nmr)12d1d2Vk(n)Vk(md1d2)Vk(rd1d2). (3.8) 3.7. The sum OD2 By (2.9) and the remarks given just before Section 3.1, we may restrict (3.7) to r≤K1−δ and md1d2≤K1+ϵ, up to an error of O(K−100). Following the argument given in [13, Lemma 3.2], we sum over n by Poisson summation after splitting into residue classes modulo c. We get OD2=∑−∞<l<∞−4π12I×(∑m,r,d1,d2,c≥1r≤K1−δmd1d2≤K1+ϵμ2(d1)e(1/8)c12mrd132d2KWˇ(l)Kc∑amodcS(a2,r2m2d12,c)e(2armd1/c))+⋯+O(K−1/2+ϵ), (3.9) where Wˇ(l)=∫−∞∞1xW(x,md1d2K,rd1d2K,Kc8πmrd1x)(1−U(rd1d2L))e(xlKc)dx. If l≠0, we integrate by parts j times and use (2.9) to see that Wˇ(l)≪Kϵ∫0Kϵ∣clK∣j∣mrd1xKc∣Bdx∣x∣2+j. Taking B=j+2, we get Wˇ(l)≪Kϵ∫0Kϵ∣1l∣j∣mrd1K2∣j∣mrd1Kc∣2dx. Keeping in mind that r≤K1−δ and md1d2<K1+ϵ, we can choose 0<ϵ<δ and take j as large as we like to get that the contribution of l≠0 is O(K−100). Now we can convert the l=0 contribution back to a sum over n, as we saw in Section 3.2 and [13, Lemma 3.2], and then remove the truncations on r and md1d2. Evaluating the innermost exponential sum in (3.9) using Lemma 2.2, we get OD2=−4π12I(∑n,m,r,d1,d2,c≥1ϕ(c2)μ2(d1)e(1/8)c2nmrd132d2W(nK,md1d2K,rd1d2K,K2c28πrnmd1)×(1−U(rd1d2L)))+⋯+O(K−1/2+ϵ). 3.8. Combining OD1 and OD2 Define the total off-diagonal to be OD=OD1+OD2. We have Lemma 3.1. OD=−4π12I(∑n,m,r,d1,d2,c≥1ϕ(c2)μ2(d1)e(1/8)c2nmrd132d2W(nK,md1d2K,rd1d2K,K2c28πrnmd1))+⋯+O(K−1/2+ϵ). (3.10) Proof Adding together OD1 and OD2, we see that what we need to prove is that OD2′=∑n,m,r,d1,d2,c≥1ϕ(c2)μ2(d1)c2nmrd132d2W(nK,md1d2K,rd1d2K,K2c28πrnmd1)U(rd1d2L) (3.11) equals OD1′=∑n,m,r,d1,d2,c≥1ϕ(c2)μ2(d1)c2nmrd132d2W(nd1d2K,md1d2K,rK,K2c28πrnmd1)U(rL) (3.12) up to an error of O(K−1/2+ϵ). For the secondary term with 1KW0 in place of W, the corresponding expressions will both be O(K−1/2+ϵ). Write U˜ and ℏ˜w for the Mellin transforms of U and ℏw. For bounds, we have (2.5) and, since U is compactly supported on (12,2K2δ) with bounded derivatives, we have U˜(s)≪R(s),BK2δ(1+∣s∣)−B for any integer B≥0. Using (2.8) and Mellin inversion, we have that OD2′=1(2πi)5∫(1)∫(1)∫(1+ϵ)∫(1)∫(1−ϵ)(8π)sG(x)G(y)G(z)ζ(1+2x)xζ(1+2y)yζ(1+2z)z×Kx+y+zLuK2sζ(1+x−s)ζ(1+y−s)ζ(1+z+u−s)ζ(1+y+z+u)×ζ(2s)ζ(2s+1)−1(∑d1≥1μ2(d1)d132+y+z+u−s)ℏ˜x+y+z(s)U˜(u)dsdxdydzdu. Here is a word of explanation: the zeta functions ζ(1+2x),ζ(1+2y),ζ(1+2z) come from the integrand in (2.8), the zeta functions ζ(1+x−s),ζ(1+y−s),ζ(1+z+u−s) come from the sums over n,m,r, respectively, the zeta function ζ(1+y+z+u) comes from the sum over d2, and the zeta functions ζ(2s)ζ(2s+1)−1 come from the sum over c. Moving the line of integration from R(u)=1 to R(u)=−3/2+ϵ, we cross a simple pole of ζ(1+z+u−s) at u=s−z. The integral on the new line is the furthest we can shift left before we would encounter another pole, of the d1-sum. On the new line, the integral is bounded by K1+1+1+2δ+ϵL−3/2K2≪K−1/2+2δ+ϵ. Thus up to this error, which is O(K−12+ϵ) since δ>0 is arbitrarily small, taking the residue at u=s−z, we have that OD2′ equals 1(2πi)4∫(1)∫(1+ϵ)∫(1)∫(1−ϵ)(8π)sG(x)G(y)G(z)ζ(1+2x)xζ(1+2y)yζ(1+2z)zKx+y+zLs−zK2s×ζ(1+x−s)ζ(1+y−s)ζ(1+y+s)ζ(2s)ζ(2s+1)−1×(∑d1≥1μ2(d1)d132+y)ℏ˜x+y+z(s)U˜(s−z)dsdxdydz. Now we move the line of integration from R(s)=1−ϵ to R(s)=100, crossing simple poles at s=x and s=y. On the new line, the integral is bounded by K1+1+1+2δ+ϵL100−1K200≪K−10. From the residues, we get OD2′=−1(2πi)3∫(1)∫(1+ϵ)∫(1)(8π)xG(x)G(y)G(z)1xζ(1+2y)yζ(1+2z)zKy+zLx−zKx×ζ(1+y−x)ζ(1+y+x)ζ(2x)(∑d1≥1μ2(d1)d132+y)ℏ˜x+y+z(x)U˜(x−z)dxdydz−1(2πi)3∫(1)∫(1+ϵ)∫(1)(8π)yG(x)G(y)G(z)ζ(1+2x)xζ(1+2y)yζ(1+2z)zKx+zLy−zKy×ζ(1+x−y)ζ(2y)(∑d1≥1μ2(d1)d132+y)ℏ˜x+y+z(y)U˜(y−z)dxdydz+O(K−1/2+ϵ). (3.13) The second integral of (3.13) falls into the error term; this can be seen by moving the y-integral far to the right (no poles are crossed). For the first integral of (3.13), we move the x-integral far to the right. We cross a simple pole at x=y and the shifted integral falls into the error term. From the residue, we get OD2′=1(2πi)2∫(1)∫(1+ϵ)(8π)yG(y)2G(z)ζ(1+2y)2y2ζ(1+2z)zKzLy−z×ζ(2y)(∑d1≥1μ2(d1)d132+y)ℏ˜2y+z(y)U˜(y−z)dydz+O(K−1/2+ϵ). (3.14) Now we turn to OD1′. We have OD1′=1(2πi)5∫(1)∫(1)∫(1+ϵ)∫(1)∫(1−ϵ)(8π)sG(x)G(y)G(z)ζ(1+2x)xζ(1+2y)yζ(1+2z)z×Kx+y+zLuK2sζ(1+x−s)ζ(1+y−s)ζ(1+z+u−s)ζ(1+x+y)ζ(2s)ζ(2s+1)−1×(∑d1≥1μ2(d1)d132+x+y−s)ℏ˜x+y+z(s)U˜(u)dsdxdydzdu. Moving the line of integration from R(u)=1 to R(u)=−100, we cross a simple pole at u=s−z. The shifted integral is negligible and from the residue we get OD1′=1(2πi)4∫(1)∫(1+ϵ)∫(1)∫(1−ϵ)(8π)sG(x)G(y)G(z)ζ(1+2x)xζ(1+2y)yζ(1+2z)z×Kx+y+zLs−zK2sζ(1+x−s)ζ(1+y−s)ζ(1+x+y)ζ(2s)ζ(2s+1)−1×(∑d1≥1μ2(d1)d132+x+y−s)ℏ˜x+y+z(s)U˜(s−z)dsdxdydz+O(K−10). Now we move the s-integral to R(s)=5/2−ϵ, which is just before the pole of the d1-sum, crossing simple poles at s=x and s=y. We get OD1′=−1(2πi)3∫(1)∫(1+ϵ)∫(1)(8π)xG(x)G(y)G(z)1xζ(1+2y)yζ(1+2z)zKy+zLx−zKx×ζ(1+y−x)ζ(1+x+y)ζ(2x)(∑d1≥1μ2(d1)d132+y)ℏ˜x+y+z(x)U˜(x−z)dxdydz−1(2πi)3∫(1)∫(1+ϵ)∫(1)(8π)yG(x)G(y)G(z)ζ(1+2x)x1yζ(1+2z)zKx+zLy−zKy×ζ(1+x−y)ζ(1+x+y)ζ(2y)(∑d1≥1μ2(d1)d132+x)ℏ˜x+y+z(y)U˜(y−z)dxdydz+O(K−1/2+ϵ). (3.15) The second integral of (3.15) falls into the error term; this can be seen by moving the y-integral far to the right (no poles are crossed). For the first integral of (3.15), we move the x-integral far to the right. We cross a simple pole at x=y and the shifted integral falls into the error term. From the residue, we get OD1′=1(2πi)2∫(1)∫(1+ϵ)(8π)yG(y)2G(z)ζ(1+2y)2y2ζ(1+2z)zKzLy−z×ζ(2y)(∑d1≥1μ2(d1)d132+y)ℏ˜2y+z(y)U˜(y−z)dydz+O(K−1/2+ϵ). (3.16) Comparing with (3.14), we see that OD1′=OD2′+O(K−1/2+ϵ), just as we wanted to prove. Also observe by moving the lines of integration to R(z)=R(y)=1/2+ϵ in (3.14) or (3.16) that OD1′ and OD2′ are bounded by K1/2+ϵ. Thus the corresponding expressions with 1KW0 in place of W are bounded by K−1/2+ϵ.□ 4. The diagonal In this section, we evaluate the contribution of the diagonal (3.8). Lemma 4.1. D=P(logk)+O(k−1/2+ϵ)for some polynomial Pof degree at most 6. Proof By Stirling’s approximation, we have Vk(ξ)=12πi∫(ϵ)G(z)ζ(1+2z)z(kξ)zdz+O(k−1+ϵ). Using this, we get D=8K∑k≡0mod2h(k−1K)∑n,m,r,d1,d2≥1n=mrd1μ2(d1)(nmr)12d1d2Vk(n)Vk(md1d2)Vk(rd1d2)=8K∑k≡0mod2h(k−1K)1(2πi)3∫(3ϵ)∫(2ϵ)∫(ϵ)G(x)G(y)G(z)ζ(1+2x)ζ(1+2y)ζ(1+2z)xyz×kx+y+zζ(1+x+y)ζ(1+x+z)ζ(1+y+z)(∑d1≥1μ2(d1)d132+y+z)dxdydz+O(K−1+ϵ). We will prove the claim by shifting the lines of integration to R(x)=R(y)=R(z)=−1/2+ϵ. This ensures that we stay in the region of absolute convergence of the d1-sum, and all the shifted integrals will be bounded by K−1/2+ϵ. Showing that the main term is at most a degree 6 polynomial in logk is equivalent to showing that the following is at most a degree 6 polynomial in logk up to the same error: 1(2πi)3∫(3ϵ)∫(2ϵ)∫(ϵ)G(x)G(y)G(z)x2y2z2(x+y)(y+z)(x+z)kx+y+zdxdydz. (4.1) There are several cases that arise when we shift integrals and we explain just one of them as the argument for the rest are similar. If we first shift the x-integral of (4.1) to R(x)=−1/2+ϵ, we will encounter a simple pole at x=−y and a double pole at x=0. The residue at the double pole yields the integrals logkG(0)+G′(0)(2πi)2∫(3ϵ)∫(2ϵ)G(y)G(z)y3z3(y+z)ky+zdydz−1(2πi)2∫(3ϵ)∫(2ϵ)G(y)G(z)G(0)y4z4ky+zdydz. Consider the last integral above, and shift the integral to R(y)=R(z)=−1/2+ϵ, crossing poles of order 4 at y=0 and z=0. Each residue is a degree 3 polynomial in logk, which together gives a degree 6 polynomial in logK, and the shifted integral is O(K−1/2+ϵ).□ 5. The off-diagonal In this section, we evaluate the off-diagonal contribution (3.10). Lemma 5.1. We have OD=P(logK)+O(K−1/2+ϵ)for some polynomial Pof degree at most 6 (not the same polynomial from the previous result). Proof We have by Lemma 3.1 that OD−4π12=I(e(−1/8)1(2πi)4∫(34+2ϵ)∫(34+ϵ)∫(34)∫(34−ϵ)(8π)sG(x)G(y)G(z)ζ(1+2x)x×ζ(1+2y)yζ(1+2z)zKx+y+zK2sζ(1+x−s)ζ(1+y−s)ζ(1+z−s)×ζ(1+y+z)ζ(2s)ζ(2s+1)−1(∑d1≥1μ2(d1)d132+y+z−s)ℏ˜x+y+z(s)dsdxdydz). We shift the line of integration from R(s)=1−ϵ to R(s)=2−ϵ, stopping just before the pole of the d1-sum. The integral on the new line is O(K−7/4+ϵ). This shift crosses simple poles at s=x, s=y, and s=z, and contribution of the residues gives OD−4π12=−I1−I2−I3+O(K−3+ϵ), where I1=I(e(−1/8)1(2πi)3∫(34+2ϵ)∫(34+ϵ)∫(34)(8π)xG(x)G(y)G(z)ζ(2x)xζ(1+2y)yζ(1+2z)z×Ky+zKxζ(1+y−x)ζ(1+z−x)ζ(1+y+z)(∑d1≥1μ2(d1)d132+y+z−x)ℏ˜x+y+z(x)dxdydz),I2=I(e(−1/8)1(2πi)3∫(34+2ϵ)∫(34+ϵ)∫(34)(8π)yG(x)G(y)G(z)ζ(1+2x)xζ(2y)yζ(1+2z)z×Kx+zKyζ(1+x−y)ζ(1+z−y)ζ(1+y+z)(∑d1≥1μ2(d1)d132+z)ℏ˜x+y+z(y)dxdydz),I3=I(e(−1/8)1(2πi)3∫(34+2ϵ)∫(34+ϵ)∫(34)(8π)zG(x)G(y)G(z)ζ(1+2x)xζ(1+2y)yζ(2z)z×Kx+yKzζ(1+x−z)ζ(1+y−z)ζ(1+y+z)(∑d1≥1μ2(d1)d132+y)ℏ˜x+y+z(z)dxdydz). The treatment of these integrals is similar, so we just look at one of them. For I1, we shift the x-integral to R(x)=2−ϵ, where the shifted integral is O(K−1/2+ϵ). This shift crosses simple poles at x=y and x=z, giving I1=−I1,1−I1,2+O(K−1/2+ϵ), where I1,1=I(e(−1/8)1(2πi)2∫(34+2ϵ)∫(34+ϵ)(8π)yG(y)2G(z)ζ(2y)ζ(1+2y)y2ζ(1+2z)z×Kzζ(1+z−y)ζ(1+y+z)(∑d1≥1μ2(d1)d132+z)ℏ˜2y+z(y)dydz) and I1,2=I(e(−1/8)1(2πi)2∫(34+2ϵ)∫(34+ϵ)(8π)zG(y)G(z)2ζ(1+2y)yζ(1+2z)ζ(2z)z2Kyζ(1+y−z)ζ(1+y+z)(∑d1≥1μ2(d1)d132+y)ℏ˜y+2z(z)dydz). In I1,2, we shift the y-integral to R(y)=−1/2+ϵ, crossing a double pole at y=0. The residue is a degree 1 polynomial in logK and the shifted integral has size O(K−1/2+ϵ). In I1,1, we shift the z-integral to R(z)=−1/2+ϵ, crossing a simple pole at z=y and a double pole at z=0. The shifted integral has size O(K−1/2+ϵ) and the residue is some constant. The residue at the simple pole of I1,2 is I(e(−1/8)12πi∫(34+ϵ)(8πK)yG(y)3ζ(1+2y)3ζ(2y)y3(∑d1≥1μ2(d1)d132+y)ℏ˜3y(y)dy)=I(12πi∫(34+ϵ)∫0∞(8πK)yG(y)3ζ(1+2y)3ζ(2y)y3(∑d1≥1μ2(d1)d132+y)×h(u)2πuy−12Γ(y)e(y4−18)dudy), on using Lemma 2.5 to evaluate the Mellin transform ℏ˜3y(y). Now we shift the line of integration to R(y)=−1/2+ϵ, where the new integral is O(K−1/2+ϵ). The shift crosses a simple pole at y=1/2 and a pole of order 7 at y=0. The residue from the pole at y=0 is a degree 6 polynomial in logK. The residue from the simple pole is I(16K12G(12)3Γ(12)ζ(2)3(∑d1≥1μ2(d1)d12)∫0∞h(u)u−14du). This would have been too big because of the presence of K12, but fortunately it vanishes because we are taking the imaginary part.□ 6. Conclusion We have so far shown that the main term of the cubic moment is a polynomial in logK of degree at most 6. To show that the degree is exactly 6, we could carefully work out all the residue contributions from above and verify that the leading term of the polynomial is non-zero. But as a shortcut we can complete the proof by using a result of Tang [20] which proves, following ideas of Rudnick and Soundararajan [15], that the cubic moment is ≫log6K. Acknowledgements We thank B. Sprung for sending us a copy of his Masters thesis. The first author would like to acknowledge financial support from IISc. Bangalore, DST (India) and the UGC Centre for advanced studies. We also thank the referee for a careful reading of the manuscript and for many useful suggestions. References 1 O. Balkanova and D. Frolenkov , On the mean value of symmetric square L-functions, preprint, arXiv:1610.06331. 2 V. Blomer , On the central value of symmetric square L-functions , Math. Z. 260 ( 2008 ), 755 – 777 . Google Scholar Crossref Search ADS 3 J. B. Conrey , D. W. Farmer , J. P. Keating , M. O. Rubinstein and N. C. Snaith , Integral moments of L-functions , Proc. London Math. Soc. (3) 91 ( 2005 ), 33 – 104 . Google Scholar Crossref Search ADS 4 W. Duke , J. B. Friedlander and H. Iwaniec , Bounds for automorphic L-functions. II , Invent. Math. 115 ( 1994 ), 219 – 239 . Google Scholar Crossref Search ADS 5 S. Gelbart and H. Jacquet , A relation between automorphic representations of GL(2) and GL(3) , Ann. Sci. École Norm. Sup. 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For permissions, please e-mail: journals.permissions@oup.com This article is published and distributed under the terms of the Oxford University Press, Standard Journals Publication Model (https://academic.oup.com/journals/pages/open_access/funder_policies/chorus/standard_publication_model)
The Quarterly Journal of Mathematics – Oxford University Press
Published: Sep 1, 2018
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