Equidistribution of Eisenstein Series in the Level Aspect

Shin-ya Koyama^{∗}

Abstract

We prove an equidistribution property of the Eisenstein series for congruence subgroups as the level goes to infinity. This is an analogy of the phenomenon called quantum ergodicity.

1 Introduction

Equidistribution of Eisenstein series was discovered by Luo and Sarnak [LS] in 1995. In the remarkable paper they showed for the Eisenstein series E(z, s) for the modular surface X = P SL(2, ℤ)\H^{2} with H^{2} the upper half plane that

lim_{t→∞} \frac{∫_{A} |E(z, \frac{1}{2} + it)|^{2}dV (z)}{∫_{B} |E(z, \frac{1}{2} + it)|^{2}dV (z)} = \frac{vol(A)}{vol(B)}  (1.1)

with A, B compact Jordan measurable subsets of X and dV = \frac{1}{vol(X)} \frac{dxdy}{y^{2}} the normalized volume measure with vol(X) = ∫_{X} \frac{dxdy}{y^{2}} = \frac{π}{3}.

The phenomenon (1.1) is an analogue of so-called quantum ergodicity, which is described as follows. Let u_{j} be eigenfunctions of the Laplacian and all Hecke operators with the normalization ∫_{X} u_{i}(z)\overline{u_{j}(z)}dV (z) = δ_{i,j}. Quantum ergodicity means the property that the probability measures dμ_{j} = |u_{j}(z)|^{2}dV (z) converge weak-∗ to dV (z).

For any X = Γ\P SL_{2}(ℝ) with Γ any congruence lattice, Lindenstrass [L] proved quantum ergodicity for compact cases. He also proved for non-compact cases, that is when Γ is a congruence subgroup, that any limit point of the sequence μ_{j} is a multiple of dV.

The result (1.1) is regarded as a version of quantum ergodicity for continuous spectra. They actually proved

∫_{A} |E(z, \frac{1}{2} + it)|^{2} dV (z) ∼ \frac{48}{π} \frac{vol(A)}{vol(X)} log t

∗Department of Mathematics, Ewha Womans University, Daehyun-dong 11-1, Sedaemoon-ku, 120-750, Seoul, South Korea.

as t → ∞. In the proof they essentially used a subconvexity estimate for the automorphic L-function for a Maass cusp form u:

L(\frac{1}{2} + it, u) = O(t^{\frac{1}{3} +ε}) (t → ∞),

due to Meurman [M]. Such equidistribution was generalized to three-dimensional hyperbolic spaces by the author [K] by use of the subconvexity estimate of Petridis and Sarnak [PS].

According to various aspects of estimates of L-functions, it is possible to consider analogues of such equidistribution. For example, Rudnick and Sarnak [RS] posed a conjecture of quantum ergodicity in the weight aspect ([KMV] Conjecture 1.4), and Sarnak [S] obtained a subconvexity bound for the Rankin-Selberg convolution L-function along this direction.

An analogue of this conjecture in the level aspect is formulated for holomorphic cusp forms by Kowalski, Michel and Vanderkam in [KMV] as follows. Here we put π_{q}: X_{q} → X_{1} to be the canonical projection with X_{q} = Γ_{0}(q)\H^{2}.

Conjecture 1.1 ([KMV] Conjecture 1.5) For k ≥ 2 even and fixed, let {f_{j}}_{j≥1} be any sequence of primitive holomorphic forms of weight k with increasing levels q_{j}. As j → ∞ the sequence of probability measure π_{q_{j},∗}(μ_{f_{j}}), j ≥ 1 converge weakly to the Poincaré measure dV (z) on X_{1}, where μ_{f_{j}} = \frac{3}{π} \frac{|f(z)|^{2}}{(f,f)} y^{k} \frac{dxdy}{y^{2}} and (f, f) is the Petersson inner product.

In this paper we deal with a relevant problem for the Eisenstein series E_{q,κ}(z, s). It is defined for Γ_{0}(q) at a cusp κ as

E_{q,κ}(z, s) := ∑_{γ∈Γ_{q,κ}\Γ_{0}(q)}(Imσ_{κ}^{−1}γz)^{s},

where z ∈ H^{2}, σ_{κ}∞ = κ and Γ_{q,κ} is the stabilizer of the cusp κ in Γ_{0}(q). It is absolutely convergent in Re(s) > 1 and has a meromorphic continuation to s ∈ ℂ. For each prime number q there exist only two inequivalent cusps κ = 0, i∞.

A result corresponding to Conjecture 1.1 is summarized as follows, which is a consequence from our main result (Theorem 1.3).

Theorem 1.2 Let q be a prime number or q = 1. For a congruence surface X_{q} = Γ_{0}(q)\H^{2}, we denote by π_{q} : X_{q} → X_{1} the natural projection from X_{q} onto X_{1}. For a subset A = A_{1} ⊂ X_{1}, we put as

A_{q} := π_{q}^{−1}(A).

For an arbitrary sequence of cusps {κ(q) | κ(q) = 0, i∞ for each prime q} and any compact Jordan measurable subsets A, B ⊂ X_{1}, we have

lim_{q→∞} \frac{∫_{A_{q}} |E_{q,κ(q)}(z, \frac{1}{2} + it)|^{2}dV_{q}(z)}{∫_{B_{q}} |E_{q,κ(q)}(z, \frac{1}{2} + it)|^{2}dV_{q}(z)} = lim_{q→∞} \frac{vol(A_{q})}{vol(B_{q})} = \frac{vol(A)}{vol(B)},  (1.2)

where dV_{q} = \frac{1}{vol(X_{q})} \frac{dxdy}{y^{2}} is the normalized volume measure on X_{q} with vol(X_{q}) = ∫_{X_{q}} dV (z).

This statement is an analog of Conjecture 1.1 in the sense that it concerns the average behavior of the Eisenstein series on the domain A_{q}. The goal of this paper is to establish the following more precise result.

Theorem 1.3 Let q be a prime number or q = 1. For a congruence surface X_{q} = Γ_{0}(q)\H^{2}, we denote by π_{q} : X_{q} → X_{1} the natural projection from X_{q} onto X_{1}. For a connected compact Jordan measurable subset A = A_{1} ⊂ X_{1}, we put {A_{q}^{0} | q : prime} to be an arbitrary sequence of connected components of π_{q}^{−1}(A). For an arbitrary sequence of cusps {κ(q) | κ(q) = 0, i∞ for each prime q} and any connected compact Jordan measurable subsets A, B ⊂ X_{1}, we have

lim_{q→∞} \frac{∫_{A_{q}^{0}} |E_{q,κ(q)}(z, \frac{1}{2} + it)|^{2}dV_{q}(z)}{∫_{B_{q}^{0}} |E_{q,κ(q)}(z, \frac{1}{2} + it)|^{2}dV_{q}(z)} = lim_{q→∞} \frac{vol(A_{q}^{0})}{vol(B_{q}^{0})} = \frac{vol(A)}{vol(B)},  (1.3)

where dV_{q} = \frac{1}{volX_{q}} \frac{dxdy}{y^{2}} is the normalized volume measure on X_{q} with vol(X_{q}) = ∫_{X_{q}} dV (z).

Actually we prove an asymptotic

∫_{A_{q}^{0}} |E_{q,κ(q)}(z, \frac{1}{2} + it)|^{2} dV_{q}(z) ∼ \frac{24vol(A^{0})}{vol(X)} log q ∼ \frac{24vol(A)}{vol(X)} \frac{log q}{q} (q → ∞).

The proofs basically follow the method of [LS]. We compute the integral

∫_{X_{q}} f(z)|E_{q,κ(q)}(z, \frac{1}{2} + it)|^{2}dV_{q}(z)

for f(z) ∈ L^{2}(X_{q}) for the proof of Theorem 1.3. If we carry out the calculation under the restriction to f ∈ L_{old}^{2}(X_{q}), we obtain Theorem 1.2. Here L_{old}^{2}(X_{q}) is the space of oldforms. By the spectral decomposition and the standard approximation argument used in [LS], the problem is reduced to the cases when f(z) is a cusp form u(z) or an incomplete Eisenstein series F_{q,κ(q)}(z|h) defined later. Indeed we will prove Proposition 1.4 below for cusp forms, and Proposition 1.5 for incomplete Eisenstein series:

Proposition 1.4 Let u_{j,q} be a normalized Hecke eigen Maass cusp form for Γ_{0}(q) with Δu_{j,q} = λ_{j,q}u_{j,q} = (\frac{1}{4} + r_{j}^{2})u_{j,q} (λ_{j,q} ≥ 0) with the Laplacian Δ with the normalization ∫_{X_{q}} u_{i,q}(z)\overline{u_{j,q}(z)}dV_{q}(z) = δ_{i,j}. We also assume that u_{j,q} is diagonalized with respect to the antiholomorphic involution ι : H^{2} ∋ x + iy ↦ −x + iy ∈ H^{2}. Then we have

∫_{X_{q}} u_{j,q}(z)|E_{q,κ(q)}(z, \frac{1}{2} + it)|^{2}dV_{q}(z) = o(q^{−1}) (q → ∞).

More precisely, for any sequence in

{(j, q) | j ∈ ℤ, j > 0, q : prime}

with q → ∞, it holds that

∫_{X_{q}} u_{j,q}(z)|E_{q,κ(q)}(z, \frac{1}{2} + it)|^{2}dV_{q}(z) = \begin{cases}O (q^{− \frac{3}{2}} ) & (u_{j,q} ∈ L_{old}^{2}(X_{q}) for all (j, q)) \\ O (q^{−1− \frac{2}{23041} +ε}) & (u_{j,q} ∈ L^{2}(X_{q})). \end{cases}

The exponent −\frac{2}{23041} comes from the bound of Duke-Friedlander-Iwaniec [DFI].

Proposition 1.5 Let F_{q} ∈ C^{∞}(X_{q}) be of the form F_{q,κ(q)}(z|h). Then as q → ∞,

∫_{X_{q}} F_{q}(z) |E_{q,κ(q)}(z, \frac{1}{2} + it)|^{2} dV_{q}(z) = 24 (∫_{X_{q}} F_{q}(z)dV_{q}) \frac{log q}{q} + O (\frac{log q}{q^{2}} ) .

2 Contribution from Cusp Forms

In this section we carry out the calculation of

∫_{X_{q}} u_{j}(z)|E_{q,κ(q)}(z, \frac{1}{2} + it)|^{2}dV_{q}(z),

where u_{j} = u_{j,q} is a Hecke eigen Maass cusp form for Γ_{0}(q) with Δu_{j} = λ_{j}u_{j} = (\frac{1}{4} + r_{j}^{2})u_{j} (λ_{j} ≥ 0) for the Laplacian Δ with the normalization ∫_{X_{q}} u_{i}(z)\overline{u_{j}(z)}dV_{q}(z) = δ_{i,j}. We also assume that u_{j} is diagonalized with respect to the antiholomorphic involution ι : H^{2} ∋ x + iy ↦ −x + iy ∈ H^{2}.

We fix t ∈ ℝ and put

J_{j}(q) := ∫_{X_{q}} u_{j}(z)E_{q,κ(q)}(z, \frac{1}{2} + it)E_{q,κ(q)}(z, \frac{1}{2} − it)dV_{q}(z)

and

I_{j}(q, s) := ∫_{X_{q}} u_{j}(z)E_{q,κ(q)}(z, \frac{1}{2} + it)E_{q,κ(q)}(z, s)dV_{q}(z).

We note J_{j}(q) = I_{j}(q, \frac{1}{2} − it). By unfolding we have

I_{j}(q, s) = ∫_{Γ_{0}(q)\H^{2}} u_{j}(z)E_{q,κ(q)}(z, \frac{1}{2} + it) ∑_{γ∈Γ_{q,κ}\Γ_{0}(q)}(Imσ_{κ}^{−1}γz)^{s}dV_{q}(z)

= ∫_{Γ_{q,κ}\H^{2}} u_{j}(z)E_{q,κ(q)}(z, \frac{1}{2} + it)(Imσ_{κ}^{−1}z)^{s}dV_{q}(z).

Replacing σ_{κ}^{−1}z by z,

I_{j}(q, s) = ∫_{Γ_{q,∞}\H^{2}} u_{j}(σ_{κ}z)E_{q,κ(q)}(σ_{κ}z, \frac{1}{2} + it)(Imz)^{s}dV_{q}(z)
= ∫_{0}^{∞} ∫_{0}^{1} u_{j}(σ_{κ}z)E_{q,κ(q)}(σ_{κ}z, \frac{1}{2} + it)y^{s} \frac{dxdy}{vol(X_{q})y^{2}} .  (2.1)

The Fourier expansion of E_{q,κ(q)}(z, s) at cusp κ is known ([I, Theorem 3.4]):

E_{q,κ}(σ_{κ}z, \frac{1}{2} + it) = y^{\frac{1}{2} +it} + φ_{q,κ,κ}(\frac{1}{2} + it)y^{\frac{1}{2} −it} + ∑_{n≠0} φ_{q,κ,κ}(n, \frac{1}{2} + it)W_{\frac{1}{2} +it}(nz)  (2.2)

with W_{\frac{1}{2} +it}(nz) = 2\sqrt{|n|y}K_{it}(2π|n|y)e(nx), where

φ_{q,κ,κ′}(s) := \frac{π^{\frac{1}{2}} Γ(s − \frac{1}{2})}{Γ(s)} ∑_{c} \frac{S_{q,κ,κ′}(0, 0, c)}{c^{2s}},  (2.3)

φ_{q,κ,κ′}(n, s) := \frac{π^{s}n^{s−1}}{Γ(s)} ∑_{c} \frac{S_{q,κ,κ′}(0, n, c)}{c^{2s}}  (2.4)

with

S_{q,κ,κ′}(0, n, c) := ∑_{(\begin{matrix}a & ∗ \\ c & d \end{matrix})∈B\σ_{κ}^{−1}Γ_{0}(q)σ_{κ′}/B} e (\frac{na}{c})  (2.5)

and

B = \{(\begin{matrix}1 & b \\ 0 & 1 \end{matrix}) : b ∈ ℤ\} .

Lemma 2.1 The value of S_{q,κ,κ′}(0, n, c) vanishes unless c ∈ ℤ for κ = κ′, and c ∈ \sqrt{q}ℤ for κ ≠ κ′. Putting γ := c/\sqrt{q}, we have for n ≠ 0

S_{q,κ,κ}(0, n, c) = \begin{cases}\frac{μ( \frac{c}{(c,n)} )φ(c)}{φ( \frac{c}{(c,n)} )} & (q | c) \\ 0 & (q |̸ c) \end{cases}

and for κ ≠ κ′

S_{q,κ,κ′}(0, n, c) = \begin{cases}0 & (q | γ) \\ \frac{μ( \frac{γ}{(γ,n)} )φ(γ)}{φ( \frac{γ}{(γ,n)} )} & (q |̸ γ) \end{cases}

with μ and φ the Möbius and the Euler functions.

For n = 0 it holds that

S_{q,κ,κ}(0, 0, c) = \begin{cases}φ(c) & (q | c) \\ 0 & (q |̸ c) \end{cases}

and for κ ≠ κ′

S_{q,κ,κ′}(0, 0, c) = \begin{cases}0 & (q | γ) \\ φ(γ) & (q |̸ γ). \end{cases}

Proof. We first compute the case κ = κ′. When q ∤ c, there does not exist a matrix ( \begin{matrix}a & ∗ \\ c & d \end{matrix}) ∈ Γ_{0}(q), and so S_{q,κ,∞}(0, n, c) = 0. Hereafter we compute for the case q | c. The set of representatives of B\σ_{κ}^{−1}Γ_{0}(q)σ_{κ}/B is obtained by [I, Theorem 2.7]:

\{(\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix})\} ∪ \{(\begin{matrix}∗ & ∗ \\ c & d \end{matrix}) ∈ Γ_{0}(q)| c > 0, 1 ≤ d < c, (c, d) = 1\} .  (2.6)

By the formula in [IK, (3.3)], we compute that S_{q,∞,∞}(0, n, c) is the Ramanujan sum:

S_{q,∞,∞}(0, n, c) = ∑_{{1≤d<c \atop (c,d)=1}} e (\frac{na}{c}) = ∑_{δ|(c,n)} μ (\frac{c}{δ}) δ = \frac{μ(\frac{c}{(c,n)})φ(c)}{φ(\frac{c}{(c,n)})}.

Next we treat the case κ ≠ κ′. The set of representatives of B\σ_{a}^{−1}Γ_{0}(q)σ_{κ′}/B is obtained from [I, Theorem 2.7]:

\{(\begin{matrix}α\sqrt{q} & β\sqrt{q} \\ γ\sqrt{q} & δ\sqrt{q} \end{matrix}) ∈ σ_{0}^{−1}Γ_{0}(q)| αδq ≡ 1 (mod γ)\} .  (2.7)

We deduce that

αδq ≡ 1 (mod γ) ⇔ q |̸γ and αδ ≡ q^{−1} (mod γ)
⇔ q |̸γ and δ ∈ (ℤ/γℤ)^{∗},

to get the representatives as

\{(\begin{matrix}∗ & ∗ \\ γ\sqrt{q} & δ\sqrt{q} \end{matrix}) ∈ σ_{0}^{−1}Γ_{0}(q)| γ > 0, q |̸γ, 1 ≤ δ < γ, (γ, δ) = 1\} .

Thus we again have the Ramanujan sum.

When n = 0, the results follow from (2.6) and (2.7).

Lemma 2.2 If we put

a_{n}(c) := \frac{μ(\frac{c}{(c,n)})φ(c)}{φ(\frac{c}{(c,n)})},

we have

L_{n}(s) := ∑_{c=1}^{∞} \frac{a_{n}(c)}{c^{s}} = \frac{σ_{1−s}(n)}{ζ(s)}.

Proof. By a_{n}(1) = 1 and multiplicativity of μ and φ, we find a_{n}(c) is multiplicative as a function in c. So we have an Euler product

L_{n}(s) = ∏_{p: prime} (1 + \frac{a_{n}(p)}{p^{s}} + \frac{a_{n}(p^{2})}{p^{2s}} + ⋯) .

Here a(p^{j}) (j = 1, 2, 3, ⋯) is calculated as follows. Put α = α_{p} to be the exponent such that p^{α_{p}}∥n. Then

a_{n}(p^{j}) = \frac{μ(\frac{p^{j}}{(p^{j},n)})φ(p^{j})}{φ(\frac{p^{j}}{(p^{j},n)})}

= \begin{cases}μ(p^{j}) & (p |̸n, i.e. α = 0) \\ \frac{μ(1)φ(p^{j})}{φ(1)} & (p^{α}∥n, j ≤ α) \\ \frac{μ(p^{j−α})φ(p^{j})}{φ(p^{j−α})} & (p^{α}∥n, 1 ≤ α < j) \end{cases}

= \begin{cases}−1 & (p |̸n, j = 1) \\ 0 & (p |̸n, j ≥ 2) \\ p^{j−1}(p − 1) & (p^{α}∥n, j ≤ α) \\ −p^{α} & (p^{α}∥n, j = α + 1) \\ 0 & (p^{α}∥n, j ≥ α + 2). \end{cases}

For prime divisors of n, the Euler factors are taken as follows:

∏_{p|n} (1 + \frac{p − 1}{p^{s}} + \frac{p(p − 1)}{p^{2s}} + \frac{p^{2}(p − 1)}{p^{3s}} + ⋯ + \frac{p^{α−1}(p − 1)}{p^{αs}} − \frac{p^{α}}{p^{(α+1)s}} )

= ∏_{p|n} (1 + \frac{1}{p^{s−1}} + \frac{1}{p^{2(s−1)}} + ⋯ + \frac{1}{p^{α(s−1)}} − (\frac{1}{p^{s}} + \frac{1}{p^{2s−1}} + \frac{1}{p^{3s−2}} + ⋯ + \frac{1}{p^{(α+1)s−α}} ))

= ∏_{p|n} (1 + \frac{1}{p^{s−1}} + \frac{1}{p^{2(s−1)}} + ⋯ + \frac{1}{p^{α(s−1)}} ) (1 − \frac{1}{p^{s}} )

= σ_{1−s}(n) ∏_{p|n} (1 − \frac{1}{p^{s}} ) .

Therefore

L_{n}(s) = σ_{1−s}(n) ∏_{p primes} (1 − \frac{1}{p^{s}} ) = \frac{σ_{1−s}(n)}{ζ(s)}.

Lemma 2.3 Let α = α_{p} satisfy that q^{α}∥n. Then it holds that

∑_{c=1}^{∞} \frac{S_{q,κ,κ′}(0, n, c)}{c^{2s}} = \frac{1}{ζ(2s)} × \begin{cases}(σ_{1−2s}(n) − \frac{σ_{1−2s}(\frac{n}{q^{α}})}{1 − q^{−2s}}) & (κ = κ′) \\ \frac{σ_{1−2s}(\frac{n}{q^{α}})}{q^{s}(1 − q^{−2s})} & (κ ≠ κ′). \end{cases}

Namely, we have

φ_{q,κ,κ′}(n, s) = \frac{n^{s−1}}{ζ̂(2s)}η_{q,κ,κ′}(n, s),

where

η_{q,κ,κ′}(n, s) = \begin{cases}σ_{1−2s}(n) − \frac{σ_{1−2s}(\frac{n}{q^{α}})}{1 − q^{−2s}} & (κ = κ′) \\ \frac{σ_{1−2s}(\frac{n}{q^{α}})}{q^{s}(1 − q^{−2s})} & (κ ≠ κ′) \end{cases}

with ζ̂(s) = π^{−s/2}Γ(s/2)ζ(s).

Proof. We first deal with the case κ ≠ κ′. By the previous lemma putting c = γ\sqrt{q} we have

∑_{c=1}^{∞} \frac{S_{q,κ,κ′}(0, n, c)}{c^{2s}} = \frac{1}{q^{s}} ∑_{q|̸γ} \frac{a_{n}(γ)}{γ^{2s}} = \frac{L_{n}(2s)}{q^{s} × (the q-Euler factor)}.

As calculated in the proof of the previous lemma,

(the q-Euler factor) = (1 − q^{−2s}) σ_{1−2s}(q^{α}).  (2.8)

Hence we have the result for κ ≠ κ′.

When κ = κ′ we compute

∑_{c=1}^{∞} \frac{S_{q,κ,κ}(0, n, c)}{c^{2s}} = ∑_{q|c} \frac{a_{n}(c)}{c^{2s}}

= ∑_{c=1}^{∞} \frac{a_{n}(c)}{c^{2s}} − ∑_{q|̸c} \frac{a_{n}(c)}{c^{2s}}

= L_{n}(2s) (1 − \frac{1}{(the q-Euler factor)}) .

We again appeal to (2.8) and reach the conclusion.

The Fourier expansion of u_{j} at the cusp i∞ is given by

u_{j}(z) = \sqrt{y} ∑_{n≠0} ρ_{j}(n)K_{ir_{j}}(2πny)e(nx)  (2.9)

with ρ_{j}(n) the Fourier coefficient and λ_{j} = \frac{1}{4} + r_{j}^{2}.

When u_{j} is a newform, put ρ̃_{j}(n) = ρ_{j}(n)/ρ_{j}(1), which is multiplicative. In this case the automorphic L-function is defined by

L(s, u_{j}) = ∑_{n=1}^{∞} \frac{ρ̃_{j}(n)}{n^{s}}

for Re(s) > 1. It is meromorphic in s ∈ ℂ and is expressed as an Euler product

L(s, u_{j}) = ∏_{p: prime} (1 − ρ̃_{j}(p)p^{−s} + χ_{0}(p)p^{−2s})^{−1} ,

where χ_{0}(p) = \begin{cases}0 & (p = q) \\ 1 & (p ≠ q). \end{cases}

On the other hand, when u_{j} is an oldform, either u_{j}(z) is a cusp form for SL(2, ℤ), or there exists a cusp form v_{j} for SL(2, ℤ) such that u_{j}(z) = v_{j}(qz). In the former case, ρ̃_{j}(n) is similarly defined and L(s, u_{j}) has an Euler product as above. In the latter case, putting

v_{j}(z) = \sqrt{y} ∑_{n≠0} τ_{j}(n)K_{ir_{j}}(2πny)e(nx),  (2.10)

we have

u_{j}(z) = v_{j}(qz) = \sqrt{qy} ∑_{n≠0} τ_{j}(n)K_{ir_{j}}(2πqny)e(qnx),

and thus

ρ_{j}(n) = \begin{cases}τ_{j}(n/q) & (q|n) \\ 0 & (otherwise) \end{cases} .

Hence for an oldform u_{j},

∑_{n=1}^{∞} \frac{ρ_{j}(n)}{n^{s}} = ∑_{n=1}^{∞} \frac{ρ_{j}(qn)}{(qn)^{s}} = q^{−s} ∑_{n=1}^{∞} \frac{τ_{j}(n)}{n^{s}} = q^{−s}τ_{j}(1) ∑_{n=1}^{∞} \frac{τ̃_{j}(n)}{n^{s}} = q^{−s}τ_{j}(1)L(s, v_{j})
= q^{−s}τ_{j}(1) ∏_{p: prime} (1 − τ̃_{j}(p)p^{−s} + χ_{0}(p)p^{−2s})^{−1} .

Lemma 2.4 (1) When u_{j} is either a newform for Γ_{0}(q) or a cusp form for SL(2, ℤ), we have

∑_{n=1}^{∞} \frac{ρ̃_{j}(n)σ_{ν}(n)}{n^{s}} = \frac{L(s, u_{j})L(s − ν, u_{j})}{ζ(2s − ν)(1 − q^{−2s+ν})},  (2.11)

∑_{n=1}^{∞} \frac{ρ̃_{j}(n)σ_{ν}(\frac{n}{q^{α}})}{n^{s}} = \frac{L(s, u_{j})L(s − ν, u_{j})}{ζ(2s − ν)} × \frac{1 − ρ̃_{j}(q)q^{−(s−ν)}}{1 − q^{−2s+ν}} .  (2.12)

(2) When u_{j} is an oldform for Γ_{0}(q) and expressed by u_{j}(z) = v_{j}(qz) with v_{j} a cusp form for SL(2, ℤ), we have

∑_{n=1}^{∞} \frac{ρ_{j}(n)σ_{ν}(n)}{n^{s}} = \frac{τ_{j}(1)L(s, v_{j})L(s − ν, v_{j})}{q^{s}ζ(2s − ν)(1 − q^{ν})} \frac{1 − τ̃_{j}(q)q^{−s+ν} − q^{2ν}(1 − τ̃_{j}(q)q^{−s})}{1 − q^{−2s+ν}},

(2.13)

and

∑_{n=1}^{∞} \frac{ρ_{j}(n)σ_{ν}(\frac{n}{q^{α}})}{n^{s}} = \frac{τ_{j}(1)σ_{ν}(q)(1 − τ̃_{j}(q)q^{−(s−ν)})}{q^{s}(1 − q^{−2s+ν})} \frac{L(s, v_{j})L(s − ν, v_{j})}{ζ(2s − ν)}.  (2.14)

Proof. The first identity (2.11) is shown by

∑_{n=1}^{∞} \frac{ρ̃_{j}(n)σ_{ν}(n)}{n^{s}} = ∏_{p: prime} ∑_{k=0}^{∞} \frac{ρ̃_{j}(p^{k})σ_{ν}(p^{k})}{p^{ks}}
= ∏_{p: prime} ∑_{k=0}^{∞} \frac{ρ̃_{j}(p^{k})(1 + p^{ν} + ⋯ + p^{kν})}{p^{ks}}
= ∏_{p: prime} ∑_{k=0}^{∞} \frac{ρ̃_{j}(p^{k})(1 − p^{ν(k+1)})}{p^{ks}(1 − p^{ν})}
= ∏_{p: prime} \frac{1}{1 − p^{ν}} (∑_{k=0}^{∞} \frac{ρ̃_{j}(p^{k})}{p^{ks}} − p^{ν} ∑_{k=0}^{∞} \frac{ρ̃_{j}(p^{k})}{p^{k(s−ν)}} )
= ∏_{p: prime} \frac{1}{1 − p^{ν}} (\frac{1}{1 − ρ̃_{j}(p)p^{−s} + χ_{0}(p)p^{−2s}} − \frac{p^{ν}}{1 − ρ̃_{j}(p)p^{−(s−ν)} + χ_{0}(p)p^{−2(s−ν)}} )
= ∏_{p: prime} \frac{1 − p^{−2s+ν}χ_{0}(p)}{(1 − ρ̃_{j}(p)p^{−s} + χ_{0}(p)p^{−2s})(1 − ρ̃_{j}(p)p^{−(s−ν)} + χ_{0}(p)p^{−2(s−ν)})}
= \frac{L(s, u_{j})L(s − ν, u_{j})}{ζ(2s − ν)(1 − q^{−2s+ν})}.

The second identity (2.12) is proved by (2.11) as

∑_{n=1}^{∞} \frac{ρ̃_{j}(n)σ_{ν}(\frac{n}{q^{α}})}{n^{s}} = ( ∏_{{p: prime \atop p≠q}} ∑_{k=0}^{∞} \frac{ρ̃_{j}(p^{k})σ_{ν}(p^{k})}{p^{ks}} ) (∑_{k=0}^{∞} \frac{ρ̃_{j}(q^{k})}{q^{ks}})
= \frac{L(s, u_{j})L(s − ν, u_{j})}{ζ(2s − ν)} × \frac{1 − ρ̃_{j}(q)q^{−(s−ν)}}{1 − q^{−2s+ν}} .

The third identity (2.13) is shown as follows:

∑_{n=1}^{∞} \frac{ρ_{j}(n)σ_{ν}(n)}{n^{s}} = ∑_{n=1}^{∞} \frac{ρ_{j}(qn)σ_{ν}(qn)}{(qn)^{s}}

= q^{−s} ∑_{n=1}^{∞} \frac{τ_{j}(n)σ_{ν}(qn)}{n^{s}}

= q^{−s}τ_{j}(1)σ_{ν}(q) ∑_{n=1}^{∞} \frac{τ̃_{j}(n)σ̃_{ν}(qn)}{n^{s}}

= q^{−s}τ_{j}(1)σ_{ν}(q) ∏_{p: prime} ∑_{k=0}^{∞} \frac{τ̃_{j}(p^{k})σ̃_{ν}(qp^{k})}{p^{ks}},

where σ̃_{ν}(qn) = σ_{ν}(qn)/σ_{ν}(q) which is multiplicative in n. Here

σ̃_{ν}(qp^{k}) = \begin{cases}\frac{1 + q^{ν} + q^{2ν} + ⋯ + q^{(k+1)ν}}{1 + q^{ν}} & (p = q) \\ 1 + p^{ν} + p^{2ν} + ⋯ + p^{kν} & (p ≠ q) \end{cases}

= \begin{cases}\frac{1 − q^{(k+2)ν}}{(1 + q^{ν})(1 − q^{ν})} & (p = q) \\ \frac{1 − p^{(k+1)ν}}{1 − p^{ν}} & (p ≠ q) \end{cases} .

Thus

∑_{n=1}^{∞} \frac{ρ_{j}(n)σ_{ν}(n)}{n^{s}}

= q^{−s}τ_{j}(1)σ_{ν}(q) ( ∏_{{p: prime \atop p≠q}} ∑_{k=0}^{∞} \frac{τ̃_{j}(p^{k})}{p^{ks}} \frac{1 − p^{(k+1)ν}}{1 − p^{ν}} ) ∑_{k=0}^{∞} \frac{τ̃_{j}(q^{k})}{q^{ks}} \frac{1 − q^{(k+2)ν}}{(1 + q^{ν})(1 − q^{ν})}

= \frac{τ_{j}(1)}{q^{s}} ( ∏_{p: prime} \frac{1}{1 − p^{ν}} ) ( ∏_{{p: prime \atop p≠q}} ∑_{k=0}^{∞} \frac{τ̃_{j}(p^{k})}{p^{ks}}(1 − p^{(k+1)ν})) ∑_{k=0}^{∞} \frac{τ̃_{j}(q^{k})}{q^{ks}}(1 − q^{(k+2)ν}).

Since for p ≠ q,

∑_{k=0}^{∞} \frac{τ̃_{j}(p^{k})}{p^{ks}}(1 − p^{(k+1)ν}) = ∑_{k=0}^{∞} \frac{τ̃_{j}(p^{k})}{p^{ks}} − ∑_{k=0}^{∞} \frac{τ̃_{j}(p^{k})}{p^{k(s−ν)−ν}}
= \frac{1}{1 − τ̃_{j}(p)p^{−s} + p^{−2s}} − \frac{p^{ν}}{1 − τ̃_{j}(p)p^{−s+ν} + p^{−2s+2ν}}
= \frac{(1 − p^{−2s+ν})(1 − p^{ν})}{(1 − τ̃_{j}(p)p^{−s} + p^{−2s})(1 − τ̃_{j}(p)p^{−s+ν} + p^{−2s+2ν})}

and for p = q

∑_{k=0}^{∞} \frac{τ̃_{j}(q^{k})}{q^{ks}}(1 − q^{(k+2)ν}) = ∑_{k=0}^{∞} \frac{τ̃_{j}(q^{k})}{q^{ks}} − ∑_{k=0}^{∞} \frac{τ̃_{j}(q^{k})}{q^{k(s−ν)−2ν}}

= \frac{1}{1 − τ̃_{j}(q)q^{s}} − \frac{q^{2ν}}{1 − τ̃_{j}(q)q^{−s+ν}} ,

it holds that

∑_{n=1}^{∞} \frac{ρ_{j}(n)σ_{ν}(n)}{n^{s}}

= \frac{τ_{j}(1)ζ(−ν)}{q^{s}} ( ∏_{{p: prime \atop p≠q}} \frac{(1 − p^{−2s+ν})(1 − p^{ν})}{(1 − τ̃_{j}(p)p^{−s} + p^{−2s})(1 − τ̃_{j}(p)p^{−s+ν} + p^{−2s+2ν})})

× (\frac{1}{1 − τ̃_{j}(q)q^{s}} − \frac{q^{2ν}}{1 − τ̃_{j}(q)q^{−s+ν}} )

= \frac{τ_{j}(1)L(s, v_{j})L(s − ν, v_{j})}{q^{s}ζ(2s − ν)}

×\frac{(1 − τ̃_{j}(q)q^{−s})(1 − τ̃_{j}(q)q^{−s+ν})}{(1 − q^{−2s+ν})(1 − q^{ν})} (\frac{1}{1 − τ̃_{j}(q)q^{−s}} − \frac{q^{2ν}}{1 − τ̃_{j}(q)q^{−s+ν}} )

= \frac{τ_{j}(1)L(s, v_{j})L(s − ν, v_{j})}{q^{s}ζ(2s − ν)(1 − q^{ν})} \frac{1 − τ̃_{j}(q)q^{−s+ν} − q^{2ν}(1 − τ̃_{j}(q)q^{−s})}{1 − q^{−2s+ν}}.

Finally (2.14) is shown from (2.12) as follows.

∑_{n=1}^{∞} \frac{ρ_{j}(n)σ_{ν}(\frac{n}{q^{α}})}{n^{s}} = ∑_{n=1}^{∞} \frac{ρ_{j}(qn)σ_{ν}(\frac{qn}{q^{α}})}{(qn)^{s}}

= σ_{ν}(q) ∑_{n=1}^{∞} \frac{τ_{j}(n)σ_{ν}(\frac{n}{q^{α}})}{(qn)^{s}}

= \frac{τ_{j}(1)σ_{ν}(q)}{q^{s}} ∑_{n=1}^{∞} \frac{τ̃_{j}(n)σ_{ν}(\frac{n}{q^{α}})}{n^{s}}

= \frac{τ_{j}(1)σ_{ν}(q)(1 − τ̃_{j}(q)q^{−(s−ν)})}{q^{s}(1 − q^{−2s+ν})} \frac{L(s, v_{j})L(s − ν, v_{j})}{ζ(2s − ν)}.

Lemma 2.5 Assume Re(s) = 1/2. Then we estimate as q → ∞ that

I_{j}(q, s) = \begin{cases}0  (if u_{j} is an odd form) \\ \frac{ρ_{j}(1)L(s − it, u_{j})L(s + it, u_{j})}{vol(X_{q})q^{\frac{1}{2}} ζ̂(1 + 2it)ζ(2s)} ∫_{0}^{∞} K_{it}(2πy)K_{ir_{j}}(2πy)y^{s} \frac{dy}{y} (1 + O (q^{− \frac{1}{4}} )) \\ (if u_{j} is an even newform) \\ \frac{1}{vol(X_{q})}O (q^{− \frac{1}{2}} )  (if u_{j} is an even oldform). \end{cases}

Proof. Applying (2.9) and (2.2) into (2.1), we find that

I_{j}(q, s) = \frac{2}{vol(X_{q})} ∑_{n≠0} \frac{φ_{q,κ,κ}(n, \frac{1}{2} + it)ρ_{j}(n)}{|n|^{s− \frac{1}{2}}} ∫_{0}^{∞} K_{it}(2πy)K_{ir_{j}}(2πy)y^{s} \frac{dy}{y} ,

where by the previous lemma

φ_{q,κ,κ}(n, \frac{1}{2} + it) = \frac{n^{− \frac{1}{2} +it}η_{q,κ,κ}(n, \frac{1}{2} + it)}{ζ̂(1 + 2it)}.

When u_{j} is odd, then ρ_{j}(n) = −ρ_{j}(−n) gives I_{j}(q, s) = 0. Hereafter we assume u_{j} is even. The sum over n is

∑_{n=1}^{∞} \frac{φ_{q,κ,κ}(n, \frac{1}{2} + it)ρ_{j}(n)}{n^{s− \frac{1}{2}}} = \frac{1}{ζ̂(1 + 2it)} ∑_{n=1}^{∞} \frac{η_{q,κ,κ}(n, \frac{1}{2} + it)ρ_{j}(n)}{n^{s−it}}.  (2.15)

Since

η_{q,κ,κ}(n, \frac{1}{2} + it) = σ_{−2it}(n) − \frac{σ_{−2it}(\frac{n}{q^{α}})}{1 − q^{−1−2it}}

by Lemma 2.3, we compute from Lemma 2.4 for u_{j} being either a newform for Γ_{0}(q) or a cusp form for SL(2, ℤ) that

(2.15) = \frac{1}{ζ̂(1 + 2it)} (∑_{n=1}^{∞} \frac{ρ_{j}(n)σ_{−2it}(n)}{n^{s−it}} − \frac{1}{1 − q^{−1−2it}} ∑_{n=1}^{∞} \frac{ρ_{j}(n)σ_{−2it}(\frac{n}{q^{α}})}{n^{s−it}})

= ρ_{j}(1)\frac{L(s − it, u_{j})L(s + it, u_{j})}{ζ̂(1 + 2it)ζ(2s)} (1 − \frac{1 − ρ̃_{j}(q)q^{−(s+it)}}{(1 − q^{−1−2it})(1 − q^{−2s})})
= ρ_{j}(1)\frac{L(s − it, u_{j})L(s + it, u_{j})}{ζ̂(1 + 2it)ζ(2s)}ρ̃_{j}(q)q^{−(s+it)} (1 + O (q^{−1})) ,

In particular, when u_{j} is a cusp form for SL(2, ℤ), the value L(s − it, u_{j})L(s + it, u_{j}) is bounded as q → ∞, and thus

(2.15) = O (q^{− \frac{1}{2}} ) ,

when Re(s) = 1/2.

On the other hand, when u_{j} is an oldform expressed by u_{j}(z) = v_{j}(qz) with v_{j} a cusp form of SL(2, ℤ), we compute

(2.15) = \frac{1}{ζ̂(1 + 2it)} (∑_{n=1}^{∞} \frac{ρ_{j}(n)σ_{−2it}(n)}{n^{s−it}} − \frac{1}{1 − q^{−1−2it}} ∑_{n=1}^{∞} \frac{ρ_{j}(n)σ_{−2it}(\frac{n}{q^{α}})}{n^{s−it}})
= \frac{τ_{j}(1)L(s − it, v_{j})L(s + it, v_{j})}{q^{s−it}ζ̂(1 + 2it)ζ(2s)(1 − q^{−2s})}

(\frac{1 − τ̃_{j}(q)q^{−s−it} − q^{−4it}(1 − τ̃_{j}(q)q^{−s+it})}{1 − q^{−2it}} − \frac{σ_{−2it}(q)(1 − τ̃_{j}(q)q^{−(s+it)})}{1 − q^{−1−2it}})

= \frac{τ_{j}(1)L(s − it, v_{j})L(s + it, v_{j})}{q^{s−it}ζ̂(1 + 2it)ζ(2s)(1 − q^{−2s})}

(1 + q^{−2it} − τ̃_{j}(q)q^{−s−it} − σ_{−2it}(q)(1 − τ̃_{j}(q)q^{−(s+it)}) (1 + O(q^{−1})) )

= \frac{τ_{j}(1)L(s − it, v_{j})L(s + it, v_{j})}{q^{2(s+it)}ζ̂(1 + 2it)ζ(2s)(1 − q^{−2s})}τ̃_{j}(q) (1 + O(q^{−1}))
= O (q^{− \frac{3}{4}} ) .

where the last identity holds for Re(s) = 1/2, and where we used the fact that the value L(s − it, v_{j})L(s + it, v_{j}) does not grow as q → ∞, and that τ̃_{j}(q) = O(q^{\frac{1}{4}} ) as q → ∞.

Proposition 2.6 Assume Re(s) = 1/2. Then we have

lim_{q→∞} vol(X_{q})I_{j}(q, s) = 0.

More precisely, as q → ∞

vol(X_{q})I_{j}(q, s) = \begin{cases}O (q^{− \frac{1}{2}} ) & (u_{j} ∈ L_{old}^{2}(X_{q})) \\ O (q^{− \frac{2}{23041} +ε}) & (u_{j} ∈ L^{2}(X_{q})). \end{cases}

Proof. When u_{j} ∈ L_{old}^{2}(X_{q}), the result follows from the previous lemma.

When u_{j} ∈ L^{2}(X_{q}), the subconvexity estimate due to Duke-Friedlander-Iwaniec [DFI] gives

L(s, u_{j}) = O(q^{\frac{1}{4} −δ})

with δ = \frac{1}{23041}.

The Hoffstein-Lockhart estimate [HL]

ρ_{j}(1) = O(q^{ε})

leads to

vol(X_{q})I_{j}(q, s) = O(q^{−2δ+ε}).

Since vol(X_{q}) = \frac{π}{3}(q + 1) for q > 1, Proposition 1.4 follows.

3 Contribution from Incomplete Eisenstein Series

Let h(y) ∈ C^{∞}(ℝ^{+}) be a rapidly decreasing function at 0 and ∞, that is h(y) = O_{N}(y^{N}) as y → 0 or ∞ and N ∈ ℤ. Let H(s) be its Mellin transform

H(s) = ∫_{0}^{∞} h(y)y^{−s} \frac{dy}{y} .

We see H(s) is entire in s and is of Schwartz class in t = Im(s) for fixed σ = Re(s). The inversion formula gives

h(y) = \frac{1}{2πi} ∫_{(σ)} H(s)y^{s}ds

for any σ ∈ ℝ. In this section we compute the integral

∫_{X_{q}} F_{q,κ(q)}(z|h) |E_{q,κ(q)}(z, \frac{1}{2} + it)|^{2} dV_{q}(z),

where F_{q,κ(q)}(z|h) is the incomplete Eisenstein series at cusp κ:

F_{q,κ(q)}(z|h) = ∑_{γ∈Γ_{q,κ(q)}\Γ_{0}(q)} h(Im(σ_{κ}^{−1}γz))

= \frac{1}{2πi} ∫_{(2)} H(s)E_{q,κ(q)}(z, s)ds.

Proposition 3.1 Let F_{q} ∈ C^{∞}(X_{q}) be of the form F_{q,κ(q)}(z|h). Then as q → ∞,

∫_{X_{q}} F_{q}(z) |E_{q,κ(q)}(z, \frac{1}{2} + it)|^{2} dV_{q}(z) = 24 (∫_{X_{q}} F_{q}(z)dV_{q}) \frac{log q}{q} + O (\frac{log q}{q^{2}} ) .

Proof. It holds that F_{q,κ(q)}(z|h)(z) is rapidly decreasing in the cusps. Hence

∫_{X_{q}} F_{q,κ(q)}(z|h) |E_{q,κ(q)}(z, \frac{1}{2} + it)|^{2} dV_{q}(z)

= \frac{1}{2πi} ∫_{X_{q}} ∫_{(2)} H(s)E_{q,κ(q)}(z, s)ds |E_{q,κ(q)}(z, \frac{1}{2} + it)|^{2} dV_{q}(z)
= \frac{1}{2πi} ∫_{X_{q}} ∫_{(2)} H(s) ∑_{γ∈Γ_{q,κ}\Γ_{0}(q)}(Imσ_{κ}^{−1}γz)^{s}ds |E_{q,κ(q)}(z, \frac{1}{2} + it)|^{2} dV_{q}(z)
= \frac{1}{2πi} ∫_{Γ_{q,κ}\H^{2}} ∫_{(2)} H(s)(Imσ_{κ}^{−1}z)^{s}ds |E_{q,κ(q)}(z, \frac{1}{2} + it)|^{2} dV_{q}(z)
= \frac{1}{2πi} ∫_{0}^{∞} ∫_{(2)} H(s)y^{s}ds ∫_{0}^{1} |E_{q,κ(q)}(σ_{κ}z, \frac{1}{2} + it)|^{2} \frac{dxdy}{vol(X_{q})y^{2}} ,

where in the last identity we replaced σ_{κ}^{−1}z with z. It equals by the Fourier expansion

\frac{vol(X_{q})^{−1}}{2πi} ∫_{0}^{∞} ∫_{(2)} H(s)y^{s}ds (|y^{\frac{1}{2} +it} + φ_{q,κ,κ}(\frac{1}{2} + it)y^{\frac{1}{2} −it}|^{2} + ∑_{n=1}^{∞} |φ_{q,κ,κ}(n, \frac{1}{2} + it)W_{\frac{1}{2} +it}(nz)|^{2}) \frac{dy}{y^{2}} .

We put this as vol(X_{q})^{−1}(I_{1} + I_{2}), where

I_{1} = \frac{1}{2πi} ∫_{0}^{∞} ∫_{(2)} H(s)y^{s}ds |y^{\frac{1}{2} +it} + φ_{q,κ,κ}(\frac{1}{2} + it)y^{\frac{1}{2} −it}|^{2} \frac{dy}{y^{2}}
= \frac{1}{2πi} ∫_{0}^{∞} ∫_{(2)} H(s)y^{s}ds\frac{1 + |φ_{q,κ,κ}(\frac{1}{2} + it)|^{2} + 2Re(φ_{q,κ,κ}(\frac{1}{2} + it)y^{2it})}{y}dy

and

I_{2} = \frac{1}{2πi} ∫_{0}^{∞} ∫_{(2)} H(s)y^{s}ds ∑_{n=1}^{∞} |φ_{q,κ,κ}(n, \frac{1}{2} + it)W_{\frac{1}{2} +it}(nz)|^{2} \frac{dy}{y^{2}} .

We first compute I_{1}. By the definition (2.3) and Lemma 2.1, we have

φ_{q,κ,κ}(w) = π^{1/2} \frac{Γ(w − \frac{1}{2})}{Γ(w)} ∑_{c=1}^{∞} \frac{φ(qc)}{(qc)^{2w}}
= π^{1/2} \frac{Γ(w − \frac{1}{2})}{Γ(w)} \frac{ζ(2w − 1)}{ζ(2w)} \frac{q^{2w} − q}{q^{2w} − 1}
= \frac{ζ̂(2w − 1)}{ζ̂(2w)} \frac{q^{2w} − q}{q^{2w} − 1}.

When w = \frac{1}{2} + it, we estimate that

φ_{q,κ,κ}(\frac{1}{2} + it) = \frac{ζ̂(2it)}{ζ̂(1 + 2it)} \frac{2| sin(t log q)|}{q} (1 + O(q^{−1})) = O(q^{−1}) (κ = 0, ∞).

Then we have

|φ_{q,κ,κ}(\frac{1}{2} + it)|^{2} = O(q^{−2})

and that

Re (φ_{q,κ,κ}(\frac{1}{2} + it)y^{2it}) = O(q^{−1})

as q → ∞. Therefore we prove that

I_{1} = \frac{1}{2πi} ∫_{0}^{∞} ∫_{(2)} H(s)y^{s}ds\frac{dy}{y} + O(q^{−1})
= ∫_{0}^{∞} \frac{h(y)}{y}dy + O(q^{−1})

as q → ∞.

Next we deal with I_{2}. By the definition (2.4) we have

|φ_{q,κ,κ}(n, \frac{1}{2} + it)W_{\frac{1}{2} +it}(nz)|^{2} = \frac{4y |K_{it}(2πny)η_{q,κ,κ}(n, \frac{1}{2} + it)|^{2}}{|ζ̂(1 + 2it)|^{2}} ,

where η_{q,κ,κ}(n, \frac{1}{2} + it) is given by Lemma 2.3. Then

I_{2} = \frac{2}{πi |ζ̂(1 + 2it)|^{2}} ∫_{(2)} H(s) ∑_{n=1}^{∞} \frac{|η_{q,κ,κ}(n, \frac{1}{2} + it)|^{2}}{n^{s}} ∫_{0}^{∞} |K_{it}(2πny)|^{2} y^{s}ds\frac{dy}{y} .

We decompose the integer n as n = q^{α}m to get

∑_{n=1}^{∞} \frac{|η_{q,κ,∞}(n, \frac{1}{2} + it)|^{2}}{n^{s}} = ∑_{α=0}^{∞} ∑_{q|̸m} \frac{|σ_{−2it}(m)|^{2}}{(q^{α}m)^{s}} |σ_{−2it}(q^{α}) − \frac{1}{1 − q^{−1−2it}} |^{2}

= ∑_{q|̸m} \frac{|σ_{−2it}(m)|^{2}}{m^{s}} ∑_{α=0}^{∞} \frac{1}{q^{αs}} |\frac{1 − q^{−2(α+1)it}}{1 − q^{−2it}} − \frac{1}{1 − q^{−1−2it}} |^{2}

= \frac{ζ(s)^{2}ζ(s + 2it)ζ(s − 2it)(1 − q^{−s+2it})(1 + q^{−s−2it})}{ζ(2s)(1 + q^{−s})|1 − q^{−2it}|^{2}|1 − q^{−1−2it}|^{2}}

× ∑_{α=0}^{∞} \frac{|(1 − q^{−2(α+1)it})(1 − q^{−1−2it}) − 1 + q^{−2it}|^{2}}{q^{αs}} .  (3.1)

Here putting θ = tα log q and θ′ = t log q,

|(1 − q^{−2(α+1)it})(1 − q^{−1−2it}) − 1 + q^{−2it}|^{2}
= |q^{−2it} − q^{−1−2it} − q^{−2(α+1)it}(1 − q^{−1−2it})|^{2}
= |q^{−2it} − q^{−2(α+1)it} − q^{−1}(q^{−2it} − q^{−2(α+2)it})|^{2}
= |q^{(α+1)it} − q^{−(α−1)it} − q^{−1}(q^{(α+1)it} − q^{−(α+1)it})|^{2}
= |cos(θ + θ′) + i sin(θ + θ′) − (cos(θ − θ′) − i sin(θ − θ′)) − 2iq^{−1} sin(θ + θ′)|^{2}
= (cos(θ + θ′) − cos(θ − θ′))^{2} + (sin(θ + θ′) + sin(θ − θ′) − 2q^{−1} sin(θ + θ′))^{2}
= (−2 sin θ sin θ′)^{2} + (2 sin θ cos θ′ − 2q^{−1} sin(θ + θ′))^{2}
= 4 sin^{2} θ − 8q^{−1} sin θ cos θ′ sin(θ + θ′) + 4q^{−2} sin^{2}(θ + θ′).

The contribution of the first term 4 sin^{2} θ to the sum over α in (3.1) is

4 ∑_{α=0}^{∞} \frac{sin^{2} θ}{q^{αs}} = − ∑_{α=0}^{∞} \frac{q^{2itα} + q^{−2itα} − 2}{q^{αs}}

= − ∑_{α=0}^{∞} (q^{α(2it−s)} + q^{α(−2it−s)} − 2q^{−αs})
= −\frac{1}{1 − q^{2it−s}} − \frac{1}{1 − q^{−2it−s}} + \frac{2}{1 − q^{−s}}
= −\frac{q^{−s}(1 + q^{−s})(1 − q^{2it})^{2}}{q^{2it}(1 − q^{2it−s})(1 − q^{−2it−s})(1 − q^{−s})}
= −\frac{q^{−s}(1 + q^{−s})|1 − q^{−2it}|^{2}}{(1 − q^{2it−s})(1 − q^{−2it−s})(1 − q^{−s})}.

The second and third terms are similarly treated, so that they are estimated as O(q^{−1}) as q → ∞. Then from (3.1) we have

∑_{n=1}^{∞} \frac{|η_{q,κ,∞}(n, \frac{1}{2} + it)|^{2}}{n^{s}}

= \frac{ζ(s)^{2}ζ(s + 2it)ζ(s − 2it)(1 + q^{−s−2it})}{ζ(2s)|1 − q^{−1−2it}|^{2}} (\frac{−q^{−s}}{(1 − q^{−2it−s})(1 − q^{−s})} + O(q^{−1})) .

Consequently,

I_{2} = \frac{2}{πi |ζ̂(1 + 2it)|^{2}} ∫_{(2)} H(s)\frac{ζ(s)^{2}ζ(s + 2it)ζ(s − 2it)(1 + q^{−s−2it})}{ζ(2s)|1 − q^{−1−2it}|^{2}}

\frac{−q^{−s}}{(1 − q^{−2it−s})(1 − q^{−s})} \frac{Γ(s/2)^{2}Γ(\frac{s}{2} − it)Γ(\frac{s}{2} + it)}{Γ(s)}ds + O(q^{−1}).

If we put it by

I_{2} = \frac{2}{πi |ζ̂(1 + 2it)|^{2}} ∫_{(2)} B(s)ds + O(q^{−1}),

then we compute by shifting the contour to Re(s) = 1/2 that

I_{2} = \frac{2}{πi |ζ̂(1 + 2it)|^{2}} (2πiRes_{s=1}B(s) + ∫_{(1/2)} B(s)ds) + O(q^{−1}).

This is justified by the Stirling formula and that H(σ + it) is rapidly decreasing in t. The integral ∫_{(1/2)} B(s)ds converges for any q, and the resulting value is O(1) as q → ∞.

Therefore we have

I_{2} = \frac{4}{|ζ̂(1 + 2it)|^{2}} Res_{s=1}B(s) + O(1)

as q → ∞.

It remains to estimate the residue term. We put B(s) = ζ(s)^{2}G(s), where G(s) is a holomorphic function at s = 1 defined by

G(s) = \frac{H(s)ζ(s + 2it)ζ(s − 2it)(1 + q^{−s−2it})}{ζ(2s)|1 − q^{−1−2it}|^{2}} \frac{−q^{−s}}{(1 − q^{−2it−s})(1 − q^{−s})} \frac{Γ(s/2)^{2}Γ(\frac{s}{2} − it)Γ(\frac{s}{2} + it)}{Γ(s)}.

Then since we know as s → 1 that

ζ(s) = \frac{1}{s − 1} + γ + O(s − 1)

with γ the Euler constant, we have

Res_{s=1}B(s) = G(1) (2γ + \frac{G′}{G} (1)) .

We compute

\frac{G′}{G} (1) = − log q + O(1)

as q → ∞.

We conclude that

I_{2} = \frac{4G(1)}{|ζ̂(1 + 2it)|^{2}} (log q + O(1)) .

As

G(1) = H(1)\frac{π^{2}|ζ̂(1 + 2it)|^{2}(1 + q^{−1−2it})}{ζ(2)|1 − q^{−1−2it}|^{2}} \frac{−q^{−1}}{(1 − q^{−2it−1})(1 − q^{−1})}
= −\frac{6H(1)|ζ̂(1 + 2it)|^{2}}{q} + O(q^{−2}),

we have

I_{2} = 24H(1)\frac{log q}{q} + O (\frac{log q}{q^{2}} ) .

We find that

H(1) = ∫_{0}^{∞} h(y)\frac{dy}{y^{2}}
= ∫_{0}^{∞} ∫_{0}^{1} h(y)\frac{dxdy}{y^{2}}
= ∫_{X_{q}} F_{q,κ(q)}(z|h)dV
= vol(X_{q}) ∫_{X_{q}} F_{q,κ(q)}(z|h)dV_{q},

which shows

I_{2} = 24 (∫_{X_{q}} F_{q,κ(q)}(z|h)dV_{q}) log q + O (\frac{log q}{q} ) .

This completes the proof of the proposition.

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