THE $1$-LEVEL DENSITY FOR ZEROS OF HECKE
L-FUNCTIONS OF IMAGINARY QUADRATIC
NUMBER FIELDS OF CLASS NUMBER $1$
Kristian Holm111email<EMAIL_ADDRESS>
Abstract. Let $\mathbb{K}=\mathbb{Q}(\sqrt{-d})$ be an imaginary quadratic
number field of class number $1$ and $\mathcal{O}_{\mathbb{K}}$ its ring of
integers. We study a family of Hecke $L$-functions associated to angular
characters on the non-zero ideals of $\mathcal{O}_{\mathbb{K}}$. Using the
powerful Ratios Conjecture (RC) due to Conrey, Farmer, and Zirnbauer, we
compute a conditional asymptotic for the average $1$-level density of the
zeros of this family, including terms of lower order than the main term in the
Katz–Sarnak Density Conjecture coming from random matrix theory. We also prove
an unconditional result about the $1$-level density, which agrees with the RC
prediction when our test functions have Fourier transforms with support in
$(-1,1)$.
Keywords: $1$-level density, Hecke $L$-functions, the Ratios Conjecture,
imaginary quadratic number fields
Mathematics Subject Classification: 11R42, 11M50 (Primary), 11R11, 11M41
(Secondary)

##  Introduction

Many problems in modern number theory await progress due to the difficulty of
obtaining exact information about zeros of $L$-functions. Perhaps not
unrelated to this difficulty, the study of the large scale statistics of such
zeros has also become a topic of much interest, the underlying philosophy
being that a collection of objects is often more regular and well-behaved than
the objects themselves. This line of research began with the work of
Montgomery in the 1970’s who famously conjectured [16] that the pair
correlation of normalized zeros of the Riemann zeta function, quantifying the
”probability” of $\zeta$ having two zeros within some prescribed distance of
each other, is the same as the pair correlation for eigenvalues of random
Hermitian matrices. Today, there is a large body of conjectures describing the
links between $L$-functions and random matrices, or formulating properties
that should hold for $L$-functions by analogy with random matrices. In this
paper, we will focus on two of these: The Katz–Sarnak Density Conjecture and
the $L$-functions Ratios Conjecture.

The Katz–Sarnak conjecture is a statement about another statistic of the zeros
of (a family of) $L$-functions, namely the $1$-level density, which was first
studied by Özlük and Snyder in [18]. For our purposes, it can be defined as
follows: If $\mathcal{F}=\mathopen{}\mathclose{{}\left\\{L_{k}:k\geqslant
1}\right\\}$ is a family of $L$-functions indexed by some parameter $k$, and
$\mathcal{F}(K)=\mathopen{}\mathclose{{}\left\\{L_{k}:1\leqslant k\leqslant
K}\right\\}$, let

$\displaystyle
Z_{k}:=\mathopen{}\mathclose{{}\left\\{z:0\leqslant\mathrm{Re}(z)\leqslant
1,\,L_{k}(z)=0}\right\\}$

be the set of zeros of $L_{k}$ in the critical strip. If $\rho\in Z_{k}$, let
$\gamma(\rho)=-i\mathopen{}\mathclose{{}\left(\rho-1/2}\right)$. Thus, under
the Riemann Hypothesis for the family $\mathcal{F}(K)$, $\gamma(\rho)$ is the
imaginary part of the zero $\rho$ of $L_{k}$. Furthermore, let
$f:\mathbb{R}\rightarrow\mathbb{R}$ be an even Schwarz function with the
property that its Fourier transform $\widehat{f}$ has compact support. Then
the $1$-level density of the zeros of the family $\mathcal{F}(K)$ is the
number

$\displaystyle
D\mathopen{}\mathclose{{}\left(\mathcal{F}(K);f}\right):=\frac{1}{K}{\sum}_{k=1}^{K}\,{\sum}_{\rho\in
Z_{k}}f\mathopen{}\mathclose{{}\left(\frac{\gamma(\rho)\log K}{\pi}}\right).$

Here, the scaling of $\gamma(\rho)$ ensures that the average spacing between
the zeros is approximately $1$.

Under the Grand Riemann Hypothesis, the $1$-level density measures the average
density of the normalized zeros of the family $\mathcal{F}(K)$ in a weak
sense. That is, for zeros on the critical line $\mathrm{Re}(z)=1/2$, and
especially such zeros close to the real line, $D(\mathcal{F}(K),\cdot)$ is a
functional that sees their (scaled) distribution through the lens of a
suitable test function. When taking more and more $L$-functions of the family
and their zeros into account, one thus obtains a sequence of distributions.
For the purposes of studying this sequence, and in particular its limit, such
weak characterizations in fact give a complete picture of the distribution of
the zeros of the family. (At least, this holds if one knows the weak
characterizations of the distribution for a sufficiently large class of test
functions, thanks to classic results in probability theory such as the
Portmanteau theorem.)

Regarding the limiting distribution of the zeros of a family of $L$-functions,
Katz and Sarnak conjectured [14, 15] that when $K$ tends to infinity, the
functional $D(\mathcal{F}(K),\cdot)$ converges weakly to some integral kernel
that arises in the dimensional limit of the $1$-level density of eigenangles
of random unitary matrices, chosen uniformly at random with respect to Haar
measure either from the full unitary group $\mathrm{U}(N)$ or from one of the
subgroups $\mathrm{USp}(N)$ (when $N$ is even), $\mathrm{O}(N)$, or
$\mathrm{SO}(N)$. Specifically, if the family $\mathcal{F}$ has so-called
unitary symplectic symmetry type, the Katz–Sarnak Density Conjecture states
that

$\displaystyle\lim_{K\rightarrow\infty}D\mathopen{}\mathclose{{}\left(\mathcal{F}(K);f}\right)={\int}_{\mathbb{R}}f(x)\mathopen{}\mathclose{{}\left(1-\frac{\sin(2\pi
x)}{2\pi x}}\right)\hskip 2.84544pt\mathrm{d}x$ (1.1)

for any even Schwarz function $f$ whose Fourier transform has compact support.
We emphasize that if the family $\mathcal{F}(K)$ has a different symmetry
type, the integral kernel which is conjectured to appear in the limit has a
different form.

Next, the very powerful $L$-functions Ratios Conjecture due to Conrey, Farmer,
and Zirnbauer [5] asserts that averages of quotients of $L$-functions
evaluated at certain parameters satisfy asymptotics that parallel those of
quotients of characteristic polynomials of matrices. (See Section 3 for a more
detailed statement.) Many authors have used the Ratios Conjecture to study
statistical aspects of the zeros of $L$-functions, or various other aspects of
such functions. For example, Conrey and Snaith [6] studied the pair
correlation of the zeros of the Riemann zeta function. In the same paper, they
also studied the $1$-level density for zeros of quadratic Dirichlet
$L$-functions. Later, the Ratios Conjecture was used to study zeros of
$L$-functions of a more general class of characters, namely Hecke characters
of a number field. The Hecke $L$-functions considered in [20] are those
associated to angular characters of the Gaussian integers, and Waxman here
used the Ratios Conjecture to compute the $1$-level density and identify
lower-order terms (compared with the Katz–Sarnak heuristic) in this
asymptotic. The goal of this paper is to follow [20] and do such a study for a
general imaginary quadratic number field of class number $1$. Thus, we
consider a family $\mathcal{F}(K)$ of $L$-functions associated to angular
Hecke characters of such fields, which we will describe now.

By the Baker–Heegner–Stark Theorem, a complete list of imaginary quadratic
number fields with class number $1$ is given by
$\mathbb{K}:=\mathbb{K}_{d}:=\mathbb{Q}(\sqrt{-d})$, where $d$ is one of the
Heegner numbers,

$\displaystyle d=1,2,3,7,11,19,43,67,163.$

Since the case $d=1$ has already been treated in [20], we will let $d$ denote
any of the eight remaining numbers on the list above. (This restriction will
also make certain computations simpler, as the arguments involve several
functions defined conditionally on the value of $d$.) Moreover, we will also
let $N\geqslant 1$ denote any fixed positive multiple of
$\mathopen{}\mathclose{{}\left|\mathcal{O}_{\mathbb{K}}^{\times}}\right|$,
where
$\mathopen{}\mathclose{{}\left|\mathcal{O}_{\mathbb{K}}^{\times}}\right|<\infty$
is the order of the group of units in the ring $\mathcal{O}_{\mathbb{K}}$. Our
family of $L$-functions is then given by
$\mathcal{F}=\mathopen{}\mathclose{{}\left\\{L_{k}(s):k\geqslant 1}\right\\}$,
where

$\displaystyle L_{k}(s):={\sum}_{I\subset\mathcal{O}_{\mathbb{K}}\atop I\neq
0}\frac{\psi_{k}(I)}{\mathbb{N}(I)^{s}},\quad\quad\psi_{k}(\langle\alpha\rangle)=\mathopen{}\mathclose{{}\left(\frac{\alpha}{\overline{\alpha}}}\right)^{Nk},$

when $\mathrm{Re}(s)>1$.

We note that such $L$-functions have been studied for arithmetic purposes on
several occasions in the past. To give a few examples, Harman and Lewis
considered the functions $L(s,\Xi_{k})$, $k\geqslant 1$, with the Hecke
character $\Xi_{k}$ given by
$\Xi_{k}(\alpha)=\mathopen{}\mathclose{{}\left(\alpha/\,\overline{\alpha}}\right)^{2k}$
for $\alpha\in\mathbb{Z}\mathopen{}\mathclose{{}\left[i}\right]$, and proved
[10, Thm. 1] the existence of infinitely many rational primes $p$ that have a
Gaussian prime factor with a small argument (depending on the size of $p$).
Later, in [19] Rudnick and Waxman considered the same family of $L$-functions
and counted Gaussian primes in more general sectors of the complex plane, in a
sense quantifying Hecke’s classical theorem about the equidistribution of the
angles of Gaussian primes on the circle ([11], [12]). In particular, the
authors in [19] studied the variance of such smooth counts of Gaussian primes
and conjectured an asymptotic ([19, Conjecture 1.2]) for this statistic based
on a random matrix model and an analogue with similar counts over function
fields. The asymptotic behaviour of this variance was, in fact, investigated
quite recently from a different point of view in [3], with a particular point
of interest being the nature of the lower order terms in the asymptotic. An
important aspect of the work in [3] relied on a study of the Hecke
$L$-functions $L(s,\Xi_{k})$ discussed above in combination with the Ratios
Conjecture.

We now state our main results. In the formulations of these (and throughout
the paper), $D$ denotes the discriminant of our number field $\mathbb{K}$, and
$\chi(n)=(-d/n)$ denotes the Dirichlet character coming from the Kronecker
symbol (see Section 2.3). Moreover, $\gamma$ denotes the Euler–Mascheroni
constant.

######  Theorem 1.1.

Suppose that $f:\mathbb{R}\rightarrow\mathbb{R}$ is an even Schwarz function
with
$\mathrm{supp}\,\widehat{f}\subset\mathopen{}\mathclose{{}\left(-1,1}\right)$.
Then we have

$\displaystyle
D\mathopen{}\mathclose{{}\left(\mathcal{F}(K);f}\right)={\int}_{\mathbb{R}}f(x)\mathopen{}\mathclose{{}\left(1-\frac{\sin(2\pi
x)}{2\pi x}}\right)\hskip 2.84544pt\mathrm{d}x+\frac{\ell_{0}\hat{f}(0)}{\log
K}+O\mathopen{}\mathclose{{}\left(\frac{1}{(\log K)^{2}}}\right),$

where

$\displaystyle\ell_{0}$
$\displaystyle=-{\int}_{1}^{\infty}t^{-2}\mathopen{}\mathclose{{}\left(-t+{\sum}_{n\leqslant
t}\Lambda(n)}\right)\hskip
2.84544pt\mathrm{d}t-\frac{L^{\prime}(1,\chi)}{L(1,\chi)}-2{\sum}_{p\geqslant
3\atop(-d/p)=-1}\frac{\log p}{p^{2}-1}$
$\displaystyle\quad\quad+\log\sqrt{|D|}-\log 2\pi+\log N-2-\frac{\sqrt{d}\log
d}{d-1}-\frac{2\log 2}{3}\cdot\mathbbm{1}(d\neq 2,7),$

and $\Lambda(n)$ denotes the von Mangoldt function.

By assuming the Grand Riemann Hypothesis (GRH) and the Ratios Conjecture
([5]), we also prove the following result.

######  Theorem 1.2.

Suppose that $f:\mathbb{R}\rightarrow\mathbb{R}$ is an even Schwarz function
whose Fourier transform has compact support. Assume the GRH and the Ratios
Conjecture. Then

$\displaystyle\begin{split}D\mathopen{}\mathclose{{}\left(\mathcal{F}(K);f}\right)={\int}_{\mathbb{R}}f(x)\mathopen{}\mathclose{{}\left(1-\frac{\sin(2\pi
x)}{2\pi x}}\right)\hskip 2.84544pt\mathrm{d}x+\frac{\ell_{0}}{\log
K}\mathopen{}\mathclose{{}\left(\hat{f}(0)-\hat{f}(1)}\right)+O\mathopen{}\mathclose{{}\left(\frac{1}{(\log
K)^{2}}}\right),\end{split}$

where $\ell_{0}$ is as in the statement of Theorem 1.1.

Remarks.

1) Although we give an explicit value for $L^{\prime}(1,\chi)/L(1,\chi)$ in
Lemma 4.5, this expression is rather intricate, and we therefore decided to
keep the notation $L^{\prime}(1,\chi)/L(1,\chi)$ in the statements of the
theorems.

2) We note that both Theorem 1.1 and Theorem 1.2 verify the Katz–Sarnak
Density Conjecture, but to different extents: While Theorem 1.1 requires $f$
to have a Fourier transform with very small support, Theorem 1.2 holds without
any such assumption.

3) The appearance of
$\mathopen{}\mathclose{{}\left|\mathcal{O}_{\mathbb{K}}^{\times}}\right|$ in
the exponents of the characters $\psi_{k}$ is very natural, since $\psi_{k}$
must satisfy a condition related to the units in order to define a Hecke
character on the ideals of $\mathcal{O}_{\mathbb{K}}$ (cf. Section 2.1).
Compared to the setup in [20], we are considering a more general family of
characters since we allow $N$ to be any multiple of the order of the unit
group. The reason why we are able to handle this more general case is that we
formulate and prove a generalization (Theorem 6.1) of the result [19, Lemma
2.1] relating the arguments and norms of certain elements of
$\mathcal{O}_{\mathbb{K}}$.

Our approach is based on [20]. We first prove Theorem 1.2, and this is
accomplished in Section 3, where we also describe the Ratios Conjecture in
detail for our family $\mathcal{F}$, and in Section 4. In Section 5, we use
the explicit formula for our family $\mathcal{F}$ to give an unconditional
asymptotic for the $1$-level density. Section 6 is then a comparison between
this and the conditional asymptotic, which leads to a proof of Theorem 1.1. As
mentioned, this comparison is facilitated by Theorem 6.1, which we also state
and prove in Section 6.

Acknowledgements. We are most grateful to Anders Södergren for suggesting the
problem considered in this article, and for many helpful discussions.
Moreover, we are indebted to Victor Ahlquist for pointing out the symmetry
between the coefficients of $\widehat{f}(0)$ and $\widehat{f}(1)$ in the
statement of Theorem 1.2. Finally, we would like to thank Ezra Waxman, Daniel
Fiorilli, Julia Brandes, and Michael Björklund for valuable discussions and
comments in relation to this project.

##  Preliminaries

We will now introduce Hecke characters on imaginary quadratic number fields,
describe our concrete family of $L$-functions in more detail, and mention
various standard results that we will need later.

###  Hecke Characters in Imaginary Quadratic Number Fields

An equivalent formulation of $\mathbb{K}$ having class number $1$ is that its
ring of integers $\mathcal{O}_{\mathbb{K}}$ is a principal ideal domain.
Explicitly, we have

$\displaystyle\mathcal{O}_{\mathbb{K}}=\begin{cases}\mathbb{Z}[\sqrt{-d}]&\text{if
}d\equiv 1,2\text{ (mod $4$)},\\\ \mathbb{Z}[(1+\sqrt{-d})/2]&\text{if
}d\equiv 3\text{ (mod $4$)}.\end{cases}$ (2.1)

By using the fact that any unit in $\mathcal{O}_{\mathbb{K}}$ must have norm
$1$, one may easily prove that

$\displaystyle\mathcal{O}_{\mathbb{K}}^{\times}\simeq\begin{cases}\mathbb{Z}/2\mathbb{Z}&\text{if
}d=2\text{ or }d\geqslant 5,\\\ \mathbb{Z}/6\mathbb{Z}&\text{if
}d=3.\end{cases}$

Since we will later make use of the lattice structure of
$\mathcal{O}_{\mathbb{K}}$, we also describe these rings in the following way.

######  Lemma 2.1.

Under the identification $\mathbb{C}\simeq\mathbb{R}^{2}$, we have

$\displaystyle\mathcal{O}_{\mathbb{K}}=2^{1/4}\mathopen{}\mathclose{{}\left(\begin{matrix}2^{-1/4}&0\\\
0&2^{1/4}\end{matrix}}\right)\mathbb{Z}^{2},$

when $d=2$; or, when $d\geqslant 3$,

$\displaystyle\mathcal{O}_{\mathbb{K}}=d^{1/4}2^{-1/2}\mathopen{}\mathclose{{}\left(\begin{matrix}2^{-1/2}d^{-1/4}&0\\\
0&2^{1/2}d^{1/4}\end{matrix}}\right)\mathopen{}\mathclose{{}\left(\begin{matrix}1&0\\\
1/2&1\end{matrix}}\right)\mathbb{Z}^{2}.$

###### Proof.

When $d=2$, this is simply a matter of expanding (2.1). When $d\geqslant 3$,
(2.1) shows that any $\alpha\in\mathcal{O}_{\mathbb{K}}$ can be written as

$\displaystyle-a+b/2+b\sqrt{-d}/2=s/2+(s+2a)\sqrt{-d}/2$

with $s=-2a+b$. The decomposition of $\mathcal{O}_{\mathbb{K}}$ in this case
now follows once we express $\mathcal{O}_{\mathbb{K}}$ using the variables $s$
and $a$. ∎

When $\mathbb{K}$ has class number $1$, a Hecke character $\psi$ on
$\mathbb{K}$ corresponds to a unique pair $(\chi,\chi_{\infty})$ consisting of
a generalized Dirichlet character $\chi$ (modulo some ideal
$\mathfrak{m}\subset\mathcal{O}_{\mathbb{K}}$) and a unitary character
$\chi_{\infty}$ on $\mathbb{C}^{\times}$. Conversely, given a pair of such
characters, their (pointwise) product is a Hecke character provided that
$\chi\cdot\chi_{\infty}$ is constant on $\mathcal{O}_{\mathbb{K}}^{\times}$,
cf. [13, eq. (3.80)]. Thus, in order for us to specify a Hecke character, it
is enough to specify two characters

$\displaystyle\chi:\big{(}\mathcal{O}_{\mathbb{K}}/\mathfrak{m}\big{)}^{\times}\rightarrow\mathbb{C}^{\times},\quad\quad\chi_{\infty}:\mathbb{C}^{\times}\rightarrow
S^{1}$

satisfying $\chi(u)\chi_{\infty}(u)=1$ for all
$u\in\mathcal{O}_{\mathbb{K}}^{\times}$.

We now describe our concrete family of Hecke characters. Let $N$ be any
positive integer multiple of
$\mathopen{}\mathclose{{}\left|\mathcal{O}_{\mathbb{K}}^{\times}}\right|$.
Since $\mathcal{O}_{K}$ is a principal ideal domain, for any $k\geqslant 1$ we
can define the unitary character $\chi_{\infty,k}$ by

$\displaystyle\chi_{\infty,k}(I)=\chi_{\infty,k}(\alpha)=\mathopen{}\mathclose{{}\left(\alpha/\bar{\alpha}}\right)^{Nk}$

whenever $I=\langle\alpha\rangle$. This is well-defined since any two
generators of $I$ will differ by a factor in $\mathcal{O}_{K}^{\times}$ where
$\chi_{\infty,k}$ is identically equal to $1$. To make this into a Hecke
character, we also need to specify a Dirichlet character that is compatible
with $\chi_{\infty,k}$ in the above sense. However, we can simply take $\chi$
to be the trivial generalized Dirichlet character of modulus
$\mathfrak{m}=\mathcal{O}_{\mathbb{K}}$. In this way we obtain the family of
Hecke characters given by

$\displaystyle\psi_{k}(\langle\alpha\rangle)=\psi_{k}(\alpha):=\chi(\alpha)\cdot\chi_{\infty,k}(\alpha)=\chi_{\infty,k}(\alpha)=\mathopen{}\mathclose{{}\left(\frac{\alpha}{\overline{\alpha}}}\right)^{Nk}$

for
$\alpha\in\mathcal{O}_{\mathbb{K}}\setminus\mathopen{}\mathclose{{}\left\\{0}\right\\}$.
Since conjugation is an automorphism of $\mathbb{C}$, we note the relation
$\overline{\psi_{k}}=\psi_{-k}$, which will be useful later on.

In the literature it is common to write such unitary characters as
$\chi_{\infty}(\alpha)=\mathopen{}\mathclose{{}\left(\alpha/|\alpha|}\right)^{\ell}$
for a suitable integer $\ell$ called the frequency. In the case of our
character $\psi_{k}$, we see that
$\psi_{k}(\alpha)=\mathopen{}\mathclose{{}\left(\alpha/|\alpha|}\right)^{2Nk}$,
so that $\psi_{k}$ has frequency $2Nk$.

We note that $\psi_{k}$ can also be described explicitly as a function of the
argument of $\alpha$, which will be convenient at certain points in the paper.
Namely, if we write $\alpha=re^{i\theta_{\alpha}}$, we have

$\displaystyle\psi_{k}(\alpha)=\mathopen{}\mathclose{{}\left(\frac{re^{i\theta_{\alpha}}}{re^{-i\theta_{\alpha}}}}\right)^{Nk}=e^{2iNk\theta_{\alpha}}.$

We also wish to speak of the ”argument of the ideal $\langle\alpha\rangle$.” A
priori, this is not well-defined since $\langle
u\alpha\rangle=\langle\alpha\rangle$ for any unit
$u\in\mathcal{O}_{\mathbb{K}}^{\times}$. However, since any unit has argument
equal to a multiple of $2\pi/|\mathcal{O}_{\mathbb{K}}^{\times}|$, the effect
of multiplying $\alpha$ with a unit $u$ is to change $\theta_{\alpha}$ by such
a multiple. For this reason, by choosing $u$ appropriately, we can always
ensure that the argument of $u\alpha$ lies in
$\mathopen{}\mathclose{{}\left[0,2\pi/|\mathcal{O}_{\mathbb{K}}^{\times}|}\right)$.
Accordingly, the angle $\theta_{\langle\alpha\rangle}$ of the ideal
$\langle\alpha\rangle$ is well-defined when taken in the interval
$\mathopen{}\mathclose{{}\left[0,2\pi/|\mathcal{O}_{\mathbb{K}}^{\times}|}\right)$.

###  Hecke $L$-Functions and Their Zeros

To each of the characters $\psi_{k}$ ($k\geqslant 1$) we can associate a Hecke
$L$-function given initially by the series and corresponding Euler product

$\displaystyle
L_{k}(s)=L_{k}(s,\psi_{k}):={\sum}_{I\subset\mathcal{O}_{\mathbb{K}}\atop
I\neq
0}\frac{\psi_{k}(I)}{\mathbb{N}(I)^{s}}={\prod}_{\mathfrak{p}}\frac{1}{1-\psi_{k}(\mathfrak{p})/\mathbb{N}(\mathfrak{p})^{s}},\quad\quad\mathrm{Re}(s)>1.$

Let us immediately note that $L_{k}=L_{-k}$. Indeed, if $N(a,b)$ denotes the
norm (see Section 2.3) of an element

$\displaystyle j(a,b)=\begin{cases}a+i\sqrt{2}b&\text{ if }d=2,\\\
a+b(1+i\sqrt{d})/2&\text{ if }d\geqslant 3,\end{cases}$

we note that the map
$A_{d}=\mathopen{}\mathclose{{}\left(\begin{smallmatrix}1&0\\\
0&-1\end{smallmatrix}}\right)+\mathbbm{1}(d\geqslant
3)\cdot\mathopen{}\mathclose{{}\left(\begin{smallmatrix}0&1\\\
0&0\end{smallmatrix}}\right)$ preserves the norm $N(a,b)$ and satisfies
$j\mathopen{}\mathclose{{}\left(A_{d}(a,b)^{\intercal}}\right)=\overline{j(a,b)}$.
In particular, $A_{d}$ defines a bijection on the set
$\mathopen{}\mathclose{{}\left\\{(a,b)\in\mathbb{Z}^{2}:N(a,b)\neq
0}\right\\}$. Therefore, for $\mathrm{Re}(s)>1$, the trivial identity [20, eq.
(2.1)] gives

$\displaystyle L_{-k}(s)$
$\displaystyle=\frac{1}{\mathopen{}\mathclose{{}\left|\mathcal{O}_{\mathbb{K}}^{\times}}\right|}{\sum}_{N(a,b)\neq
0}\mathopen{}\mathclose{{}\left(\frac{j(a,b)}{|j(a,b)|}}\right)^{-2Nk}N(a,b)^{-s}$
$\displaystyle=\frac{1}{\mathopen{}\mathclose{{}\left|\mathcal{O}_{\mathbb{K}}^{\times}}\right|}{\sum}_{N(a,b)\neq
0}\mathopen{}\mathclose{{}\left(\frac{|j(a,b)|}{j(a,b)}}\right)^{2Nk}N(a,b)^{-s}$
$\displaystyle=\frac{1}{\mathopen{}\mathclose{{}\left|\mathcal{O}_{\mathbb{K}}^{\times}}\right|}{\sum}_{N(a,b)\neq
0}\mathopen{}\mathclose{{}\left(\frac{\overline{j(a,b)}}{|j(a,b)|}}\right)^{2Nk}N(a,b)^{-s}$
$\displaystyle=\frac{1}{\mathopen{}\mathclose{{}\left|\mathcal{O}_{\mathbb{K}}^{\times}}\right|}{\sum}_{N(A_{d}(a,b)^{\intercal})\neq
0}\mathopen{}\mathclose{{}\left(\frac{j\mathopen{}\mathclose{{}\left(A_{d}(a,b)^{\intercal}}\right)}{|j\mathopen{}\mathclose{{}\left(A_{d}(a,b)^{\intercal}}\right)|}}\right)^{2Nk}N\mathopen{}\mathclose{{}\left(A_{d}(a,b)^{\intercal}}\right)^{-s}$
$\displaystyle=\frac{1}{\mathopen{}\mathclose{{}\left|\mathcal{O}_{\mathbb{K}}^{\times}}\right|}{\sum}_{N(a,b)\neq
0}\mathopen{}\mathclose{{}\left(\frac{j(a,b)}{|j(a,b)|}}\right)^{2Nk}N(a,b)^{-s}=L_{k}(s).$

By a theorem of Hecke [13, Theorem 3.8], if $k\neq 0$ (so that $\psi_{k}$ is
not the trivial character), $L_{k}$ admits an analytic continuation (which we
will also denote by $L_{k}$) to the entire complex plane, and it satisfies the
functional equation

$\displaystyle\Lambda\mathopen{}\mathclose{{}\left(s,\psi_{k}}\right)=\frac{\tau\mathopen{}\mathclose{{}\left(\psi_{k}}\right)}{i^{\ell}\sqrt{\mathbb{N}(\mathfrak{m})}}\Lambda\mathopen{}\mathclose{{}\left(1-s,\overline{\psi_{k}}}\right),$

where $\mathfrak{m}=\mathcal{O}_{\mathbb{K}}$ is the modulus of $\psi_{k}$,
$\ell=2Nk$ is the frequency, and
$\Lambda\mathopen{}\mathclose{{}\left(s,\psi_{k}}\right)$ denotes the
completed $L$-function

$\displaystyle\Lambda\mathopen{}\mathclose{{}\left(s,\psi_{k}}\right)=\Lambda_{k}(s):=L_{k}(s)\frac{\big{(}|D|\mathbb{N}(\mathfrak{m})\big{)}^{s/2}}{(2\pi)^{s}}\Gamma\big{(}s+|\ell|/2\big{)},$

and where $\tau(\psi_{k})$ denotes the Gauss sum

$\displaystyle\tau\mathopen{}\mathclose{{}\left(\psi_{k}}\right)=\psi_{k}(\gamma)\psi_{k}(\mathfrak{c})^{-1}{\sum}_{\alpha\in\mathfrak{c}/\mathfrak{c}\mathfrak{m}}\exp\mathopen{}\mathclose{{}\left(2\pi
i\mathrm{Tr}\mathopen{}\mathclose{{}\left(\alpha/\gamma}\right)}\right),$

cf. [13, eq. (3.86)]. Here $\gamma\in\mathcal{O}_{K}$ and
$\mathfrak{c}\subset\mathcal{O}_{K}$ are arbitrary except for the requirements
that $\mathfrak{c}$ should be an ideal, and that $\gamma$ and $\mathfrak{c}$
should satisfy $(\mathfrak{c},\mathfrak{m})=1$ and
$\mathfrak{c}\mathfrak{d}\mathfrak{m}=\langle\gamma\rangle$, where
$\mathfrak{d}$ is the different of $K$. In our case, we have
$\tau(\psi_{k})=1$: Since we have $\mathfrak{m}=\mathcal{O}_{K}$ and
$\mathfrak{d}=\langle\sqrt{D}\rangle$, these conditions are satisfied with
$\mathfrak{c}=\mathcal{O}_{K}$ and $\gamma=\sqrt{D}$. In combination with the
relation $\overline{\psi_{k}}=\psi_{-k}$ and the fact that the frequency
$\ell$ of $\psi_{k}$ is $2Nk\equiv 0$ (mod $4$), this means that the root
number of $L_{k}$ is $1$, and the functional equation assumes the simpler form

$\displaystyle\Lambda_{k}(s)=\Lambda_{-k}(1-s)=\Lambda_{k}(1-s),$ (2.2)

where also the completed $L$-function can be described in the simpler form

$\displaystyle\Lambda_{k}(s)=L_{k}(s)\frac{|D|^{s/2}}{(2\pi)^{s}}\Gamma\big{(}s+Nk\big{)}.$
(2.3)

Of course, it is also possible to recast the identity (2.2) as a statement
about $L_{k}$ that does not explicitly involve $\Lambda_{k}$. Doing so, we
find that

$\displaystyle
L_{k}\mathopen{}\mathclose{{}\left(s}\right)=L_{k}(1-s)X_{k}(s),$ (2.4)

where

$\displaystyle X_{k}(s):$
$\displaystyle=\frac{\Gamma(1-s+Nk)}{\Gamma(s+Nk)}|D|^{1/2-s}(2\pi)^{2s-1}.$
(2.5)

If $K\geqslant 1$ is an integer, then as we mentioned in the introduction, we
will use the notation

$\displaystyle\mathcal{F}(K):=\mathopen{}\mathclose{{}\left\\{L_{k}:1\leqslant
k\leqslant K}\right\\}$

to denote our family of $L$-functions. We wish to normalize the zeros of this
family so that they have mean spacing $1$. This of course warrants an
understanding of the asymptotic number $N_{k}(T)$ of zeros of $L_{k}$ in the
critical strip $0\leqslant\mathrm{Re}(z)\leqslant 1$ up to a given height $T$
as $k\rightarrow\infty$. Estimates of the count $N_{k}(T)$ for general
$L$-functions are abundant in the literature, but these usually provide an
asymptotic as $T\rightarrow\infty$. We therefore prove the following result,
which gives an asymptotic expression for $N_{k}(T)$ when it is not the height,
but rather the size of our family that tends to infinity.

######  Proposition 2.2.

Let $k\geqslant 1$, and assume the Riemann Hypothesis for $L_{k}$. For $T>0$,
let

$\displaystyle
N_{k}(T):=\\#\Big{\\{}z\in\mathbb{C}:L_{k}(z)=0,\,0\leqslant\mathrm{Re}(z)\leqslant
1,\,-T\leqslant\mathrm{Im}(z)\leqslant T\Big{\\}}$

be the number of zeros of $L_{k}$ in the critical strip up to absolute height
$T$. Then as $k\rightarrow\infty$,

$\displaystyle N_{k}(T)\sim\frac{2T\log k}{\pi}.$

###### Proof.

Let us write $X:=1/2+Nk$. By [2, Thm. 5] and [2, Eq. (4.1)], the main term of
$N_{k}(T)$ comes from the integral

$\displaystyle\frac{1}{\pi}{\int}_{-T}^{T}\mathrm{Re}\mathopen{}\mathclose{{}\left(\frac{\Gamma^{\prime}(X+it)}{\Gamma(X+it)}}\right)\hskip
2.84544pt\mathrm{d}t=\frac{1}{\pi}\mathrm{Re}\mathopen{}\mathclose{{}\left({\int}_{-T}^{T}\frac{\Gamma^{\prime}(X+it)}{\Gamma(X+it)}\hskip
2.84544pt\mathrm{d}t}\right).$

Indeed, in the notation of that paper, we have

$\displaystyle
L(z,\pi_{\infty})=\frac{|D|^{z/2}}{(2\pi)^{z}}\Gamma\big{(}z+Nk\big{)};$

and since the logarithmic derivative of the factor $|D|^{z/2}(2\pi)^{-z}$ is
constant, the integral of the logarithmic derivative of
$\mathopen{}\mathclose{{}\left|D}\right|^{(X+it)/2}(2\pi)^{-X-it}$ over the
line $-T\leqslant t\leqslant T$ is at most a constant times $T$.

Since $-i\log\Gamma(X+it)$ is a primitive function for the logarithmic
derivative in the integrand above, and since our domain of integration lies in
the right half-plane, we can use Stirling’s formula $\log\Gamma(z)=z\log
z-z-(\log
z)/2+O\mathopen{}\mathclose{{}\left(\max\mathopen{}\mathclose{{}\left\\{1,1/z}\right\\}}\right)$
to obtain

$\displaystyle{\int}_{-T}^{T}\frac{\Gamma^{\prime}(X+it)}{\Gamma(X-it)}\hskip
2.84544pt\mathrm{d}t$ $\displaystyle=-i\cdot
X\log\frac{X+iT}{X-iT}+T\log\mathopen{}\mathclose{{}\left(X^{2}+T^{2}}\right)$
$\displaystyle\quad\quad-2T+i\cdot\frac{1}{2}\log\frac{X+iT}{X-iT}+O\mathopen{}\mathclose{{}\left(1}\right).$

By taking the real part and using that
$T\log\mathopen{}\mathclose{{}\left(X^{2}+T^{2}}\right)=2T\log k+O_{T}(1)$, we
therefore have

$\displaystyle\frac{1}{\pi}{\int}_{-T}^{T}\mathrm{Re}\mathopen{}\mathclose{{}\left(\frac{\Gamma^{\prime}(X+it)}{\Gamma(X+it)}}\right)\hskip
2.84544pt\mathrm{d}t=\frac{2T\log k}{\pi}+O_{T}(1)$

as $k\rightarrow\infty$. This concludes the proof. ∎

###  Splitting Behaviour of Rational Primes

We now record some facts about how a prime $p\in\mathbb{Z}$ behaves in the
extension $\mathbb{K}=\mathbb{Q}(\sqrt{-d})$, where we recall that $d$ is one
of the integers $2,3,7,11,19,43,$ $67,163$. For a rational prime $p$, we say
that $p$ is ramified if $p=q^{2}u$ for a prime $q\in\mathcal{O}_{\mathbb{K}}$
and $u\in\mathcal{O}_{\mathbb{K}}^{\times}$, split if $p=q\overline{q}u$ for a
prime $q\in\mathcal{O}_{\mathbb{K}}$ and
$u\in\mathcal{O}_{\mathbb{K}}^{\times}$, or inert if $p$ is a prime element in
$\mathcal{O}_{\mathbb{K}}$. Any rational prime belongs to exactly one of these
categories.

By using the norm map $\mathcal{O}_{\mathbb{K}}\rightarrow\mathbb{Z}_{+}$, we
immediately see that if $p$ is ramified or split, then some prime element
$q\in\mathcal{O}_{\mathbb{K}}$ has norm $p$. Thus, we can reduce certain
aspects of the study of the splitting behaviour in $\mathbb{K}$ of rational
primes to studying which primes are represented over $\mathbb{Z}$ by the
quadratic forms defined by the norm. We now describe these forms in more
detail.

If $d\equiv 3$ (mod $4$), any integer in $\mathbb{K}$ has the form
$a+b(1+\sqrt{-d})/2$, and its norm is

$\displaystyle\mathbb{N}\mathopen{}\mathclose{{}\left(a+\frac{b(1+\sqrt{-d})}{2}}\right)=\mathopen{}\mathclose{{}\left(a+\frac{b}{2}}\right)^{2}+\frac{db^{2}}{4}=a^{2}+ab+\frac{d+1}{4}b^{2}.$
(2.6)

Note that this form has integer coefficients precisely because of the
congruence condition on $d$. On the other hand, if $d\equiv 1,2$ (mod $4$),
then any integer in $\mathbb{K}$ has the form $a+b\sqrt{-d}$, and its norm is

$\displaystyle\mathbb{N}\mathopen{}\mathclose{{}\left(a+b\sqrt{-d}}\right)=a^{2}+db^{2}.$
(2.7)

It is well-known that a rational prime $p$ is ramified in $\mathbb{K}$ if and
only if $p\mid D$, the discriminant of $\mathbb{K}$. Since

$\displaystyle|D|=\begin{cases}d&\text{ if }d\equiv 3\text{ (mod $4$)},\\\
4d&\text{ if }d\equiv 1,2\text{ (mod $4$)},\end{cases}$

we see that $p$ is ramified in $\mathbb{K}$ if and only if $p=d$.

By [17, Prop I.8.5] we know that if $p\neq 2$, then $p$ splits in $\mathbb{K}$
if and only if the Legendre symbol $(-d/p)$ equals $1$. On the other hand, if
$p=2$ we can use (2.6) to see that $p$ only splits in $\mathbb{K}$ when $d=7$.
(This also implies that whenever $d\neq 2,7$, the prime $2$ is inert in
$\mathcal{O}_{\mathbb{K}}$.) Indeed, if $d\equiv 3$ (mod $4$) and $2$ splits
in $\mathbb{K}$, there are rational integers $a$ and $b$ such that
$q=a+b\mathopen{}\mathclose{{}\left(1+\sqrt{-d}}\right)/2$ and

$\displaystyle\mathbb{N}(q)-2=a^{2}+ab+\frac{d+1}{4}b^{2}-2=0,$

which implies (by solving the quadratic equation in $a$) that $8-db^{2}$ has
to be a square number. Given that $d\equiv 3$ (mod $4$), this is only possible
if $d=7$ and $b=\pm 1$. This shows that the element $q=(1+\sqrt{-7})/2$ has
norm $2$. It is clear that $2=q\overline{q}$ and that $q$ and $\overline{q}$
are not equivalent up to multiplication by units.

In the remainder of the paper, we will also need to work with the ”Legendre
symbol modulo composite numbers.” We therefore introduce the Kronecker symbol,
which will also be convenient for the purpose of expressing the splitting
behavior of a prime. If $n=p_{1}^{e_{1}}\cdots p_{k}^{e_{k}}$ is (the prime
factorization of) a positive integer and $a\in\mathbb{Z}$, the Kronecker
symbol $(a/n)$ is defined as

$\displaystyle\mathopen{}\mathclose{{}\left(\frac{a}{n}}\right)={\prod}_{i=1}^{k}\mathopen{}\mathclose{{}\left(\frac{a}{p_{i}}}\right)^{e_{i}},$

where the symbol $(a/p_{i})$ appearing on the right-hand side is the Legendre
symbol if $p_{i}$ is an odd prime, or otherwise given by

$\displaystyle\mathopen{}\mathclose{{}\left(\frac{a}{2}}\right)=\begin{cases}0&\text{
if $a$ is even,}\\\ 1&\text{ if $a\equiv\pm 1$ (mod $8$),}\\\ -1&\text{ if
$a\equiv\pm 3$ (mod $8$).}\end{cases}$

We note that if $a\not\equiv 3$ (mod $4$), then the map $n\mapsto(a/n)$ is a
quadratic Dirichlet character, cf. [7, §5]. In particular, for $d$ equal to
any of the eight non-trivial Heegner numbers that we are considering,
$\chi(n):=(-d/n)$ is a Dirichlet character.

We can now sum up the above discussion about the splitting behavior of
rational primes in the ring $\mathcal{O}_{\mathbb{K}}$ in the following lemma.

######  Lemma 2.3.

Let $p$ be a rational prime. In the number field $\mathbb{K}$,

$\displaystyle p\text{ is }\begin{cases}\text{ramified}&\text{if
}\chi(p)=0\text{ (that is, $p=d$)},\\\ \text{split}&\text{if }\chi(p)=1,\\\
\text{inert}&\text{if }\chi(p)=-1.\end{cases}$

We end this section with the following elementary lemma whose proof we include
for the sake of completeness.

######  Lemma 2.4.

Let $\mathfrak{p}\subset\mathcal{O}_{\mathbb{K}}$ be a prime ideal. Then
$\mathfrak{p}$ lies over a rational prime $p$ (that is,
$\mathfrak{p}\cap\mathbb{Z}=p\mathbb{Z}$) if and only if $\mathfrak{p}\mid
p\mathcal{O}_{\mathbb{K}}$. Hence, the prime ideals with norm equal to a power
of $p$ are precisely the prime ideals dividing $p\mathcal{O}_{\mathbb{K}}$.

###### Proof.

Suppose that $\mathfrak{p}$ lies over $p$. Then clearly $p\in\mathfrak{p}$,
and by the ideal property of $\mathfrak{p}$ we therefore have
$p\mathcal{O}_{\mathbb{K}}\subset\mathfrak{p}$, which means $\mathfrak{p}\mid
p\mathcal{O}_{\mathbb{K}}$.

Conversely, if $\mathfrak{p}\mid p\mathcal{O}_{\mathbb{K}}$, then
$\mathfrak{p}\supset p\mathcal{O}_{\mathbb{K}}$, and hence
$\mathfrak{p}\cap\mathbb{Z}\supset
p\mathcal{O}_{\mathbb{K}}\cap\mathbb{Z}\supset p\mathbb{Z}$. Since
$p\mathbb{Z}$ is a maximal ideal in $\mathbb{Z}$ and
$\mathfrak{p}\cap\mathbb{Z}\subset\mathbb{Z}$ must be a proper ideal, it
follows that $p\mathbb{Z}=\mathfrak{p}\cap\mathbb{Z}$. Therefore
$\mathfrak{p}$ lies over $p$. ∎

##  Implications of the Ratios Conjecture

In this section, we describe the $L$-functions Ratios Conjecture which is due
to Conrey, Farmer, and Zirnbauer ([5]), generalizing a conjecture of Farmer
about the Riemann zeta function (see [8]). We show how the conjecture implies
strong estimates for the $1$-level density of the zeros of the family
$\mathcal{F}(K)$ as $K\rightarrow\infty$.

The Ratios Conjecture states that a sum of ratios of (products of)
$L$-functions evaluated at certain parameters should obey a specific
asymptotic estimate. We now describe the recipe from [5] for conjecturing such
an asymptotic. However, for the sake of simplicity, we do not describe the
most general case possible. In the case of two $L$-functions in the numerator
and denominator, one considers

$\displaystyle
Q(s,\bm{\alpha},\bm{\gamma};\chi)=\frac{L(s+\alpha_{1},\chi)L(s+\alpha_{2},\chi)}{L(s+\gamma_{1},\chi)L(s+\gamma_{2},\chi)},$

where $L(s,\chi)$ is the $L$-function associated to a character $\chi$ and
satisfying the functional equation

$\displaystyle L(s,\chi)=X(s,\chi)L(1-s,\overline{\chi}).$

The recipe is as follows:

  * •

Approximate functional equation for $L$
Replace each $L$-function in the numerator of
$Q(s,\bm{\alpha},\bm{\gamma};\chi)$ with the two main terms from its
approximate functional equation, completely disregarding the remainder term.

  * •

Infinite series for $1/L$
Replace each reciprocal $L$-function with its expression as an infinite series
involving a suitable Möbius function.

  * •

Extend ranges and regroup factors
Extending the ranges of all series to infinity and multiplying out the
resulting expression, write each of the four resulting terms as

$\displaystyle\big{(}\text{product of root numbers
$\varepsilon_{\chi}$}\big{)}\times\big{(}\text{product of
$X(\cdot,\chi)$-factors}\big{)}\times\big{(}\text{sum over
$n_{1},n_{2},n_{3},n_{4}$}\big{)},$

where $n_{1},n_{2},n_{3},n_{4}$ are the indexing variables from the two
approximate functional equations and the two infinite series giving the
reciprocal $L$-functions.

  * •

Average factors over the family
Replace each product of root numbers, each product of $X(\cdot,\chi)$-factors,
and each summand in the last factor with their respective averages over the
family $\mathcal{F}=\mathopen{}\mathclose{{}\left\\{\chi}\right\\}$. Denote
the resulting expression by $M(s,\bm{\alpha},\bm{\gamma})$.

  * •

Statement of the conjecture
The conjecture now states that for any $\varepsilon>0$,

$\displaystyle{\sum}_{\chi\in\mathcal{F}}Q(s,\bm{\alpha},\bm{\gamma};\chi)w\mathopen{}\mathclose{{}\left(q(\chi)}\right)$
$\displaystyle=\mathopen{}\mathclose{{}\left(1+O\mathopen{}\mathclose{{}\left(e^{(-1/2+\varepsilon)q(\chi)}}\right)}\right){\sum}_{\chi\in\mathcal{F}}M(s,\bm{\alpha},\bm{\gamma})w\mathopen{}\mathclose{{}\left(q(\chi)}\right),$

where $q(\chi):=\mathopen{}\mathclose{{}\left|X^{\prime}(1/2,\chi)}\right|$
denotes the log conductor of $\chi$, and $w$ is a suitable weight function.

A later addition to the Ratios Conjecture [6, Sect. 2] is that such an
asymptotic is expected to hold provided that $\alpha$ and $\gamma$ satisfy the
constraints

$\displaystyle-\frac{1}{4}<\mathrm{Re}(\alpha)<\frac{1}{4},\quad\frac{1}{\log
K}\ll\mathrm{Re}(\gamma)<\frac{1}{4},\quad\mathrm{Im}(\alpha),\,\mathrm{Im}(\gamma)\ll_{\varepsilon}K^{1-\varepsilon}.$
(3.1)

###  Ingredients for the Recipe

To follow the recipe outlined above in the case of the expression

$\displaystyle
R_{K}\mathopen{}\mathclose{{}\left(\alpha,\gamma}\right)=\frac{1}{K}{\sum}_{k=1}^{K}\frac{L_{k}(1/2+\alpha)}{L_{k}(1/2+\gamma)},$
(3.2)

we first need to describe the approximate functional equation for $L_{k}$ and
obtain an expression for the reciprocal $L_{k}^{-1}$ as an infinite series.

We begin by describing the approximate functional equation, which involves
writing $L_{k}$ in a different way that is more reminiscent of classical
Dirichlet $L$-functions. To this end, we observe that

$\displaystyle L_{k}(s)={\sum}_{I\subset\mathcal{O}\atop I\neq
0}\frac{\psi_{k}(I)}{\mathbb{N}(I)^{s}}={\sum}_{n\geqslant
1}\mathopen{}\mathclose{{}\left({\sum}_{\mathbb{N}(I)=n}\psi_{k}(I)}\right)n^{-s}={\sum}_{n\geqslant
1}\frac{A_{k}(n)}{n^{s}}$

for $\mathrm{Re}(s)>1$, with

$\displaystyle A_{k}(n):={\sum}_{\mathbb{N}(I)=n}\psi_{k}(I),\quad\quad
n\geqslant 1.$

We note that $A_{k}$ is real-valued. Indeed, this follows from the definition
of $\psi_{k}$ and the fact that for any $\alpha\in\mathcal{O}_{\mathbb{K}}$,

$\displaystyle\mathbb{N}\mathopen{}\mathclose{{}\left(\langle\alpha\rangle}\right)=\alpha\overline{\alpha}=\mathbb{N}\mathopen{}\mathclose{{}\left(\langle\overline{\alpha}\rangle}\right).$

Hence, any ideal that contributes to the sum defining $A_{k}(n)$ is
accompanied by its conjugate ideal, provided that these are different. (If
they are not, then their contribution to $A_{k}$ is clearly real.) Complex
conjugation of $A_{k}$ therefore amounts to permuting the terms in the sum,
which of course leaves $A_{k}$ unchanged.

Using the series defining $L_{k}$ and the functional equation (2.4), we can
now describe the approximate functional equation of $L_{k}$ as

$\displaystyle L\mathopen{}\mathclose{{}\left(s,\psi_{k}}\right)$
$\displaystyle\approx{\sum}_{n<x}\frac{A_{k}(n)}{n^{s}}+X_{k}(s){\sum}_{n<y}\frac{\overline{A_{k}(n)}}{n^{1-s}}$
$\displaystyle={\sum}_{n<x}\frac{A_{k}(n)}{n^{s}}+X_{k}(s){\sum}_{n<y}\frac{A_{k}(n)}{n^{1-s}},$
(3.3)

where $x$ and $y$ are positive real parameters.

As mentioned above, the recipe of the Ratios Conjecture also requires us to
obtain a formula for the reciprocal function $L_{k}^{-1}$. Taking the
reciprocal of the Euler product and using the fact that $\psi_{k}$ and the
ideal norm are completely multiplicative, we see that

$\displaystyle\frac{1}{L_{k}(s)}$
$\displaystyle={\prod}_{\mathfrak{p}}\mathopen{}\mathclose{{}\left(1-\frac{\psi_{k}(I)}{\mathbb{N}(I)^{s}}}\right)$
$\displaystyle=1-{\sum}_{n=\mathfrak{p}_{1}}\frac{\psi_{k}\mathopen{}\mathclose{{}\left(\mathfrak{p}_{1}}\right)}{\mathbb{N}\mathopen{}\mathclose{{}\left(\mathfrak{p}_{1}}\right)^{s}}+{\sum}_{n=\mathfrak{p}_{1}\mathfrak{p}_{2}}\frac{\psi_{k}\mathopen{}\mathclose{{}\left(\mathfrak{p}_{1}\mathfrak{p}_{2}}\right)}{\mathbb{N}\mathopen{}\mathclose{{}\left(\mathfrak{p}_{1}\mathfrak{p}_{2}}\right)^{s}}-{\sum}_{n=\mathfrak{p}_{1}\mathfrak{p}_{2}\mathfrak{p}_{3}}\frac{\psi_{k}\mathopen{}\mathclose{{}\left(\mathfrak{p}_{1}\mathfrak{p}_{2}\mathfrak{p}_{3}}\right)}{\mathbb{N}\mathopen{}\mathclose{{}\left(\mathfrak{p}_{1}\mathfrak{p}_{2}\mathfrak{p}_{3}}\right)^{s}}+\cdots$
$\displaystyle={\sum}_{I\subset\mathcal{O}\atop I\neq
0}\frac{\mu(I)\psi_{k}(I)}{\mathbb{N}(I)^{s}},$

where

$\displaystyle\mu(I)=\begin{cases}(-1)^{n}&\text{if }I\text{ is the product of
$n$ distinct primes},\\\ 0&\text{otherwise},\end{cases}$

is the natural analogue of the Möbius function. With

$\displaystyle\mu_{k}(n):={\sum}_{\mathbb{N}(I)=n}\mu(I)\psi_{k}(I),$

we therefore obtain the formula

$\displaystyle\frac{1}{L_{k}(s)}={\sum}_{n\geqslant
1}\frac{\mu_{k}(n)}{n^{s}},\quad\quad\mathrm{Re}(s)>1.$ (3.4)

######  Lemma 3.1.

For any $k\geqslant 1$, the functions $\mu_{k}$ and $A_{k}$ are
multiplicative.

###### Proof.

It suffices to show that if $p\neq q$ are rational primes and $a,b\geqslant 1$
are rational integers, then

$\displaystyle\mu_{k}\mathopen{}\mathclose{{}\left(p^{a}q^{b}}\right)=\mu_{k}\mathopen{}\mathclose{{}\left(p^{a}}\right)\mu_{k}\mathopen{}\mathclose{{}\left(q^{b}}\right),\quad\quad
A_{k}\mathopen{}\mathclose{{}\left(p^{a}q^{b}}\right)=A_{k}\mathopen{}\mathclose{{}\left(p^{a}}\right)A_{k}\mathopen{}\mathclose{{}\left(q^{b}}\right).$

We prove the claim for $\mu_{k}$ as the proof of the claim for $A_{k}$ is
analogous.

Since $\psi_{k}$ is completely multiplicative, we have

$\displaystyle\mu_{k}\mathopen{}\mathclose{{}\left(p^{a}}\right)\mu_{k}\mathopen{}\mathclose{{}\left(q^{b}}\right)={\sum}_{\mathbb{N}\mathopen{}\mathclose{{}\left(I_{1}}\right)=p^{a}\atop\mathbb{N}\mathopen{}\mathclose{{}\left(I_{2}}\right)=q^{b}}\mu\mathopen{}\mathclose{{}\left(I_{1}I_{2}}\right)\psi_{k}\mathopen{}\mathclose{{}\left(I_{1}I_{2}}\right).$

It remains to show that all ideals $I$ of norm $p^{a}q^{b}$ have the form
$I_{1}I_{2}$ for ideals $I_{1}$ and $I_{2}$ of norm $p^{a}$ and $q^{b}$,
respectively. However, this follows immediately from the unique factorization
of ideals in $\mathcal{O}_{\mathbb{K}}$ in combination with the properties of
the norm map. ∎

###  Following the Recipe

In accordance with the recipe outlined above, we now use (3.1) and (3.4) and
compute

$\displaystyle\frac{L_{k}(1/2+\alpha)}{L_{k}(1/2+\gamma)}\approx{\sum}_{m,n\geqslant
1}\frac{\mu_{k}(m)A_{k}(n)}{m^{1/2+\gamma}n^{1/2+\alpha}}+X_{k}(1/2+\alpha){\sum}_{m,n\geqslant
1}\frac{\mu_{k}(m)A_{k}(n)}{m^{1/2+\gamma}n^{1/2-\alpha}},$ (3.5)

where we extended the ranges of summation to infinity in accordance with the
recipe provided above. We now average the individual factors. We begin with
the following lemma.

######  Lemma 3.2.

As $K\rightarrow\infty$, we have

$\displaystyle\begin{split}\frac{1}{K}{\sum}_{k=1}^{K}X_{k}(1/2+\alpha)&=\frac{1}{1-2\alpha}\mathopen{}\mathclose{{}\left(\frac{2\pi}{KN\sqrt{|D|}}}\right)^{2\alpha}+O_{\alpha}\mathopen{}\mathclose{{}\left(\frac{1}{K}}\right)+O_{\alpha}\mathopen{}\mathclose{{}\left(\frac{1}{K^{1+2\alpha}}}\right).\end{split}$
(3.6)

###### Proof.

We can argue exactly as in the proof of [20, Lemma 3.1]. The only thing we
need to take into account is that we have $\textbf{k}=1/2+Nk$ in the notation
of that paper. ∎

It remains for us to compute the average of the summands in (3.5). Here it is
fruitful to rewrite the sums appearing there as products in order to take
advantage of the multiplicative nature of the function $\mu_{k}$. To that end,
with the help of Lemma 3.1 we note that

$\displaystyle{\sum}_{m,n\geqslant
1}\frac{\mu_{k}(m)A_{k}(n)}{m^{1/2+\gamma}n^{1/2+\alpha}}={\prod}_{p\atop\text{
prime}}{\sum}_{m,n\geqslant
0}\frac{\mu_{k}\mathopen{}\mathclose{{}\left(p^{m}}\right)A_{k}\mathopen{}\mathclose{{}\left(p^{n}}\right)}{p^{m(1/2+\gamma)}p^{n(1/2+\alpha)}}.$
(3.7)

At this point, we will describe the values of $\mu_{k}$ and $A_{k}$ on prime
powers more precisely.

######  Lemma 3.3.

Let $p$ be a rational prime. Then

$\displaystyle\mu_{k}\mathopen{}\mathclose{{}\left(p^{m}}\right)=\begin{cases}1&\text{if
}m=0,\\\ -A_{k}(p)&\text{if }m=1,\\\ -1&\text{if $m=2$ and $p$ is inert,}\\\
1&\text{if $m=2$ and $p$ splits,}\\\ 0&\text{otherwise}.\end{cases}$

###### Proof.

The first two claims follow immediately from the definition of $\mu_{k}$. As
for the third claim, we have
$\mu_{k}\mathopen{}\mathclose{{}\left(p^{2}}\right)=-1$ since only the prime
ideal $\langle p\rangle$ has norm $p^{2}$ by Lemma 2.4.

Turning to the fourth claim, let us suppose that $\langle p\rangle$ has prime
factors $\mathfrak{p}_{1}$ and $\mathfrak{p}_{2}$. In this case, there is only
one squarefree ideal of norm $p^{2}$, namely $\langle p\rangle$, since only
the prime ideals dividing $\langle p\rangle$ have norms equal to a power of
$p$ by Lemma 2.4.

We now prove the fifth and final claim. We first examine the case where $m=2$
and $p$ is ramified in $\mathcal{O}_{\mathbb{K}}$ with $\langle
p\rangle=\mathfrak{q}^{2}$ for some prime ideal $\mathfrak{q}$. This
immediately implies that $\mu_{k}(p^{2})=0$ since the only ideal of norm
$p^{2}$ is $\mathfrak{q}^{2}$ by Lemma 2.4. For the final case, we assume
$m\geqslant 3$. To prove the claim, it is enough to show that the norm of a
product of distinct prime ideals lying over $p$ is at most $p^{2}$. However,
this is clear from Lemma 2.4 once we consider the three possible splitting
behaviours of $p$. ∎

We also need the following description of the function $A_{k}$ on prime
powers.

######  Lemma 3.4.

$A_{k}$ assumes the following values on prime powers (where we understand that
the case $n=0$ takes precedence over the remaining cases):

$\displaystyle A_{k}(p^{n})=\begin{cases}1&\text{if }n=0,\\\\[5.0pt]
(q^{n}/\,\overline{q}^{n})^{Nk}&\text{if $\langle p\rangle=\langle
q\rangle^{2}$,}\\\\[5.0pt] 1&\text{if }p\text{ is inert and }n\text{ is
even},\\\\[5.0pt] 0&\text{if }p\text{ is inert and }n\text{ is
odd},\\\\[5.0pt] \sum_{j=-n/2}^{n/2}\psi_{k}(q)^{2j}&\text{if $\langle
p\rangle=\langle q\rangle\langle\overline{q}\rangle$ and $n$ is
even},\\\\[5.0pt] \sum_{j=-(n+1)/2}^{(n-1)/2}\psi_{k}(q)^{2j+1}&\text{if
$\langle p\rangle=\langle q\rangle\langle\overline{q}\rangle$ and $n$ is
odd}.\end{cases}$

###### Proof.

Since only $\langle 1\rangle=\mathcal{O}_{\mathbb{K}}$ has norm $1$, the first
claim follows immediately from the definition. In the following we suppose
that $n\geqslant 1$.

Suppose that $p$ is ramified in $\mathcal{O}_{\mathbb{K}}$ with $\langle
p\rangle=\mathfrak{q}^{2}$. Then Lemma 2.4 immediately implies that the only
ideal of norm $p^{n}$ is $\mathfrak{q}^{n}$, which proves the claim.

Suppose that $p$ is inert. Then there is no prime ideal of norm $p$. Moreover,
the ideal $\langle p\rangle$ is prime and has norm $p^{2}$, and it is the only
such prime ideal. It follows that if $I$ is any ideal of norm $p^{n}$, then
$n$ must be even, and in this case $I=\langle p\rangle^{n/2}=\langle
p^{n/2}\rangle$. This yields the third and fourth claims.

Turning to the final claim, we assume that $p$ splits as $q\overline{q}$ in
$\mathcal{O}_{\mathbb{K}}$. Then clearly $\langle q\rangle$ and
$\langle\overline{q}\rangle$ are the only prime ideals of norm $p$. Moreover,
as we assumed that $p$ is not ramified, it is clear that $\langle
q\rangle\neq\langle\overline{q}\rangle$. Similarly as before, we also see that
no prime ideals of norm $p^{2}$ can exist. It follows that if $I$ has norm
$p^{n}$, then $I=\langle q\rangle^{j}\langle\overline{q}\rangle^{n-j}=\langle
q^{j}\overline{q}^{n-j}\rangle$ for some $j=0,\ldots,n$. Thus,

$\displaystyle
A_{k}(p^{n})={\sum}_{j=0}^{n}\mathopen{}\mathclose{{}\left(\frac{q^{j}\overline{q}^{n-j}}{\overline{q}^{j}q^{n-j}}}\right)^{Nk}={\sum}_{j=0}^{n}\mathopen{}\mathclose{{}\left(\frac{q^{2j-n}}{\overline{q}^{2j-n}}}\right)^{Nk},$

and therefore

$\displaystyle
A_{k}(p^{n})=\begin{cases}{\sum}_{j=-n/2}^{n/2}\psi_{k}(q)^{2j}&\text{if $n$
is even},\\\\[15.00002pt]
{\sum}_{j=-(n+1)/2}^{(n-1)/2}\psi_{k}(q)^{2j+1}&\text{if $n$ is
odd}.\end{cases}$

This concludes the proof. ∎

In accordance with the Ratios Conjecture, we now compute the (asymptotic)
averages of the function $\mu_{k}(p^{m})A_{k}(p^{n})$ as $k=1,\ldots,K$. We
will denote this by $\delta_{p}(m,n)$ so that

$\displaystyle\delta_{p}(m,n)=\lim_{K\rightarrow\infty}\frac{1}{K}{\sum}_{k=1}^{K}\mu_{k}(p^{m})A_{k}(p^{n}).$

The existence of this limit will be clear from the consideration of the
special cases of $p$ (split, inert, or ramified). In anticipation of this, we
furthermore write

$\displaystyle\delta_{p}(m,n)=\begin{cases}\delta_{\mathrm{in}}(m,n)&\text{if
$p$ is inert}\\\ \delta_{\mathrm{sp}}(m,n)&\text{if $p$ splits}\\\
\delta_{\mathrm{ram}}(m,n)&\text{if $p$ is ramified}.\end{cases}$

######  Lemma 3.5.

We have

$\displaystyle\delta_{\mathrm{in}}(m,n)$
$\displaystyle=\begin{cases}0&\text{if }m=1\text{ or }m\geqslant 3\text{ or
}n\text{ is odd,}\\\ 1&\text{if }m=0\text{ (and $n$ is even),}\\\ -1&\text{if
}m=2\text{ (and $n$ is even),}\end{cases}$
$\displaystyle\delta_{\mathrm{sp}}(m,n)$
$\displaystyle=\begin{cases}1&\text{if $m=0$ and $n$ is even,}\\\ -2&\text{if
$m=1$ and $n$ is odd,}\\\ 1&\text{if $m=2$ and $n$ is even,}\\\
0&\text{otherwise,}\\\ \end{cases}$ $\displaystyle\delta_{\mathrm{ram}}(m,n)$
$\displaystyle=\begin{cases}1&\text{if $m=0$,}\\\ -1&\text{if $m=1$,}\\\
0&\text{if $m\geqslant 2$.}\end{cases}$

###### Proof.

This follows immediately from Lemma 3.3 and Lemma 3.4. In the case of a
ramified prime $p$, we also use the fact that
$A_{k}(p^{n})=\mathopen{}\mathclose{{}\left(q^{n}/\,\overline{q}^{n}}\right)^{Nk}=1$,
since in this case $q/\,\overline{q}\in\mathcal{O}_{\mathbb{K}}^{\times}$. ∎

Using Lemma 3.5, we can describe the limiting average of the right-hand side
of (3.7) as follows. If the prime $p$ is inert, we write

$\displaystyle G_{\mathrm{in}}(p;\alpha,\gamma):={\sum}_{m,n\geqslant
0}\frac{\delta_{\mathrm{in}}(m,n)}{p^{m(1/2+\gamma)+n(1/2+\alpha)}},$

and we define $G_{\mathrm{sp}}(p;\alpha,\gamma)$ and
$G_{\mathrm{ram}}(p;\alpha,\gamma)$ analogously. By using Lemma 3.5, we then
deduce that

$\displaystyle\allowdisplaybreaks G_{\mathrm{in}}(p;\alpha,\gamma)$
$\displaystyle={\sum}_{n\geqslant
0}p^{-n(1+2\alpha)}-p^{-(1+2\gamma)}{\sum}_{n\geqslant
0}p^{-n(1+2\alpha)}=\mathopen{}\mathclose{{}\left(1-p^{-(1+2\gamma)}}\right)\mathopen{}\mathclose{{}\left(1-p^{-(1+2\alpha)}}\right)^{-1},$
$\displaystyle G_{\mathrm{sp}}(p;\alpha,\gamma)$
$\displaystyle=\mathopen{}\mathclose{{}\left(1-2p^{-(1+\alpha+\gamma)}+p^{-(1+2\gamma)}}\right)\mathopen{}\mathclose{{}\left(1-p^{-(1+2\alpha)}}\right)^{-1},$
$\displaystyle G_{\mathrm{ram}}(p;\alpha,\gamma)$
$\displaystyle={\sum}_{n\geqslant
0}\mathopen{}\mathclose{{}\left(p^{-n(1/2+\alpha)}-p^{-(1/2+\gamma)}p^{-n(1/2+\alpha)}}\right)=\mathopen{}\mathclose{{}\left(1-p^{-(1/2+\gamma)}}\right)\mathopen{}\mathclose{{}\left(1-p^{-(1/2+\alpha)}}\right)^{-1}.$

Moreover, we let

$\displaystyle F_{2}(\alpha,\gamma):=\begin{cases}1,&\text{if }d=2,\\\
\mathopen{}\mathclose{{}\left(1-2^{-(\alpha+\gamma)}+2^{-(1+2\gamma)}}\right)\mathopen{}\mathclose{{}\left(1-2^{-(1+2\alpha)}}\right)^{-1}&\text{if
}d=7,\\\
\mathopen{}\mathclose{{}\left(1-2^{-(1+2\gamma)}}\right)\mathopen{}\mathclose{{}\left(1-2^{-(1+2\alpha)}}\right)^{-1}&\text{otherwise}.\end{cases}$

We now see that the product of $G_{*}(p;\alpha,\gamma)$ over all rational
primes equals

$\displaystyle\begin{split}G(\alpha,\gamma)&=F_{2}(\alpha,\gamma)\,G_{\mathrm{ram}}\mathopen{}\mathclose{{}\left(d;\alpha,\gamma}\right){\prod}_{p\geqslant
3\atop(-d/p)=1}G_{\mathrm{sp}}(p;\alpha,\gamma){\prod}_{p\geqslant
3\atop(-d/p)=-1}G_{\mathrm{in}}(p;\alpha,\gamma)\\\
&=\mathopen{}\mathclose{{}\left(1-d^{-(1/2+\gamma)}}\right)\mathopen{}\mathclose{{}\left(1-d^{-(1/2+\alpha)}}\right)^{-1}F_{2}(\alpha,\gamma){\prod}_{p\geqslant
3\atop(-d/p)=1}\mathopen{}\mathclose{{}\left(1-2p^{-(1+\alpha+\gamma)}+p^{-(1+2\gamma)}}\right)\\\
&\quad\quad\times{\prod}_{p\geqslant
3\atop(-d/p)=-1}\mathopen{}\mathclose{{}\left(1-p^{-(1+2\gamma)}}\right){\prod}_{p\geqslant
3\atop p\neq
d}\mathopen{}\mathclose{{}\left(1-p^{-(1+2\alpha)}}\right)^{-1}\\\
&=\mathopen{}\mathclose{{}\left(1-\mathbbm{1}(d\geqslant 3)\cdot
d^{-(1+2\alpha)}}\right)\mathopen{}\mathclose{{}\left(1-d^{-(1/2+\gamma)}}\right)\mathopen{}\mathclose{{}\left(1-d^{-(1/2+\alpha)}}\right)^{-1}F_{2}(\alpha,\gamma)\\\\[13.00005pt]
&\quad\quad\times{\prod}_{p\geqslant
3\atop(-d/p)=1}\mathopen{}\mathclose{{}\left(1-2p^{-(1+\alpha+\gamma)}+p^{-(1+2\gamma)}}\right){\prod}_{p\geqslant
3\atop(-d/p)=-1}\mathopen{}\mathclose{{}\left(1-p^{-(1+2\gamma)}}\right)\\\\[5.0pt]
&\quad\quad\times\hskip 10.00002pt{\prod}_{p\geqslant
3}\mathopen{}\mathclose{{}\left(1-p^{-(1+2\alpha)}}\right)^{-1}\\\\[5.0pt]
&=\mathopen{}\mathclose{{}\left(1+d^{-(1/2+\alpha)}}\right)\mathopen{}\mathclose{{}\left(1-d^{-(1/2+\gamma)}}\right)\tilde{F}_{2}(\alpha,\gamma){\prod}_{p\geqslant
3\atop(-d/p)=1}\mathopen{}\mathclose{{}\left(1-2p^{-(1+\alpha+\gamma)}+p^{-(1+2\gamma)}}\right)\\\
&\quad\quad\times{\prod}_{p\geqslant
3\atop(-d/p)=-1}\mathopen{}\mathclose{{}\left(1-p^{-(1+2\gamma)}}\right)\zeta(1+2\alpha),\end{split}$
(3.8)

when $\mathrm{Re}(\alpha)>0$. Here $\tilde{F}_{2}$ is the function obtained by
possibly removing the factor
$\mathopen{}\mathclose{{}\left(1-2^{-(1+2\alpha)}}\right)^{-1}$ from $F_{2}$,
i.e.

$\displaystyle\tilde{F}_{2}(\alpha,\gamma):=\begin{cases}1&\text{if }d=2,\\\
1-2^{-(\alpha+\gamma)}+2^{-(1+2\gamma)}&\text{if }d=7,\\\
1-2^{-(1+2\gamma)}&\text{otherwise}.\end{cases}$

It will be convenient to introduce convergence factors in the two Euler
products and thus bundle together all singularities in a number of zeta
functions and Dirichlet $L$-functions. Doing so, and writing $\chi(p)=(-d/p)$,
we find that our expression (3.8) equals

$\displaystyle\frac{\zeta(1+2\alpha)L(1+2\gamma,\chi)}{\zeta(1+\alpha+\gamma)L(1+\alpha+\gamma,\chi)}\mathopen{}\mathclose{{}\left(1+d^{-(1/2+\alpha)}}\right)\mathopen{}\mathclose{{}\left(1-d^{-(1/2+\gamma)}}\right)\mathopen{}\mathclose{{}\left(1-d^{-(1+\alpha+\gamma)}}\right)^{-1}H_{2}(\alpha,\gamma)$
$\displaystyle\quad\quad\times{\prod}_{p\geqslant
3\atop(-d/p)=1}\frac{\mathopen{}\mathclose{{}\left(1-2p^{-(1+\alpha+\gamma)}+p^{-(1+2\gamma)}}\right)\mathopen{}\mathclose{{}\left(1-p^{-(1+2\gamma)}}\right)}{\mathopen{}\mathclose{{}\left(1-p^{-(1+\alpha+\gamma)}}\right)^{2}}$
$\displaystyle\quad\quad\times{\prod}_{p\geqslant
3\atop(-d/p)=-1}\frac{\mathopen{}\mathclose{{}\left(1-p^{-(1+2\gamma)}}\right)\mathopen{}\mathclose{{}\left(1+p^{-(1+2\gamma)}}\right)}{\mathopen{}\mathclose{{}\left(1-p^{-(1+\alpha+\gamma)}}\right)\mathopen{}\mathclose{{}\left(1+p^{-(1+\alpha+\gamma)}}\right)},$

where the function $H_{2}$, defined by

$\displaystyle H_{2}(\alpha,\gamma)=\begin{cases}1&\text{if }d=2,\\\
\mathopen{}\mathclose{{}\left(1-2^{-(1+2\gamma)}}\right)\mathopen{}\mathclose{{}\left(1-2^{-(1+\alpha+\gamma)}}\right)^{-2}\mathopen{}\mathclose{{}\left(1-2^{-(\alpha+\gamma)}+2^{-(1+2\gamma)}}\right)&\text{if
}d=7,\\\
\mathopen{}\mathclose{{}\left(1-2^{-2(1+2\gamma)}}\right)\mathopen{}\mathclose{{}\left(1-2^{-2(1+\alpha+\gamma)}}\right)^{-1}&\text{otherwise},\end{cases}$

keeps track of the contributions from the prime $p=2$ when $d\neq 2$. Indeed;
for example, in case of the $L$-function $L(1+2\gamma,\chi)$, we see that all
of its factors are cancelled out by reciprocal factors in the two Euler
products, except for factors corresponding to the primes $p=2$ and $p=d$. If
$d=2$, then no such factor is missing as $\chi(2)=0$, and if $d\geqslant 3$,
only the factor corresponding to $p=2$ is missing, namely the factor
$1-\chi(2)2^{-(1+2\gamma)}$.

If we let

$\displaystyle A_{1}(\alpha,\gamma)\,\,$ $\displaystyle:={\prod}_{p\geqslant
3\atop(-d/p)=1}\frac{\mathopen{}\mathclose{{}\left(1-2p^{-(1+\alpha+\gamma)}+p^{-(1+2\gamma)}}\right)\mathopen{}\mathclose{{}\left(1-p^{-(1+2\gamma)}}\right)}{\mathopen{}\mathclose{{}\left(1-p^{-(1+\alpha+\gamma)}}\right)^{2}},$
$\displaystyle A_{-1}(\alpha,\gamma)\,\,$ $\displaystyle:={\prod}_{p\geqslant
3\atop(-d/p)=-1}\frac{\mathopen{}\mathclose{{}\left(1-p^{-(1+2\gamma)}}\right)\mathopen{}\mathclose{{}\left(1+p^{-(1+2\gamma)}}\right)}{\mathopen{}\mathclose{{}\left(1-p^{-(1+\alpha+\gamma)}}\right)\mathopen{}\mathclose{{}\left(1+p^{-(1+\alpha+\gamma)}}\right)},$
$\displaystyle F_{d}(\alpha,\gamma)\,\,$
$\displaystyle:=\frac{\mathopen{}\mathclose{{}\left(1+d^{-(1/2+\alpha)}}\right)\mathopen{}\mathclose{{}\left(1-d^{-(1/2+\gamma)}}\right)}{1-d^{-(1+\alpha+\gamma)}},$

we therefore have

$\displaystyle G(\alpha,\gamma)$
$\displaystyle=\frac{\zeta(1+2\alpha)L(1+2\gamma,\chi)}{\zeta(1+\alpha+\gamma)L(1+\alpha+\gamma,\chi)}F_{d}(\alpha,\gamma)H_{2}(\alpha,\gamma)A_{1}(\alpha,\gamma)A_{-1}(\alpha,\gamma).$
(3.9)

Combining this with (3.6), we see that the prediction of the Ratios Conjecture
is that with $\alpha$ and $\gamma$ subject to the conditions (3.1), the
average $R_{K}(\alpha,\gamma)$ is equal to

$\displaystyle R_{K}(\alpha,\gamma)$
$\displaystyle=G(\alpha,\gamma)+\frac{1}{1-2\alpha}\mathopen{}\mathclose{{}\left(\frac{2\pi}{KN\sqrt{|D|}}}\right)^{2\alpha}G(-\alpha,\gamma)+O_{\varepsilon}\mathopen{}\mathclose{{}\left(K^{-1/2+\varepsilon}}\right).$
(3.10)

###  The Logarithmic Derivative

If we know that the effective estimate (3.10) is invariant under
differentiation (in the sense that the derivatives of the main terms are
equal, at least up to an error of roughly the same order of magnitude), we can
use (3.10) to describe an asymptotic for the average of the logarithmic
derivative of $L_{k}$ over the family $\mathcal{F}(K)$. Since the logarithmic
derivative $L_{k}^{\prime}/L_{k}$ is intimately connected with the zeros and
poles of $L_{k}$ by Cauchy’s residue theorem, such an asymptotic estimate will
allow us to study the asymptotics of the one-level density of the family
$\mathcal{F}(K)$.

To see that (3.10) is, in fact, invariant under differentiation in the sense
above, let $\Omega\subset\mathbb{C}$ be any open set where the function

$\displaystyle\alpha\mapsto
f(\alpha):=R_{K}(\alpha,\gamma)-G(\alpha,\gamma)-\frac{1}{1-2\alpha}\mathopen{}\mathclose{{}\left(\frac{2\pi}{KN\sqrt{|D|}}}\right)^{2\alpha}G(-\alpha,\gamma)$

is holomorphic. For $\alpha_{0}\in\Omega$, choose $\delta>0$ so that $\Omega$
contains the circle $C$ centered at $\alpha_{0}$ with radius $\delta$. Then by
Cauchy’s integral formula,

$\displaystyle|f^{\prime}(\alpha_{0})|$
$\displaystyle\leqslant\frac{1}{2\pi}{\int}_{C}\mathopen{}\mathclose{{}\left|\frac{f(z)}{\mathopen{}\mathclose{{}\left(z-\alpha_{0}}\right)^{2}}}\right|\hskip
2.84544pt|\mathrm{d}z|\leqslant\frac{1}{2\pi}\frac{1}{\delta^{2}}\cdot
2\pi\delta\cdot\sup_{z\in C}|f(z)|=\frac{1}{\delta}\cdot
O_{\varepsilon}\mathopen{}\mathclose{{}\left(K^{-1/2+\varepsilon}}\right),$

and the claim follows by linearity of differentiation.

Since only the numerators in the sum defining $R_{K}(\alpha,\gamma)$ depend on
$\alpha$, (3.10) and the estimate obtained by differentiating (3.10) therefore
imply that

$\displaystyle\begin{split}\frac{1}{K}{\sum}_{k=1}^{K}\frac{L_{k}^{\prime}(1/2+r)}{L_{k}(1/2+r)}&=\frac{\partial}{\partial\alpha}R_{K}(\alpha,\gamma)\bigg{\rvert}_{\alpha=\gamma=r}\\\
&\approx\frac{\partial}{\partial\alpha}G(\alpha,\gamma)\bigg{\rvert}_{\alpha=\gamma=r}+\frac{\partial}{\partial\alpha}\frac{1}{1-2\alpha}\mathopen{}\mathclose{{}\left(\frac{2\pi}{KN\sqrt{|D|}}}\right)^{2\alpha}G(-\alpha,\gamma)\bigg{\rvert}_{\alpha=\gamma=r}\end{split}$
(3.11)

for any $r\in\mathbb{C}$ satisfying the conditions

$\displaystyle\frac{1}{\log
K}\ll\mathrm{Re}(r)<1/4,\quad\quad\mathrm{Im}(r)\ll_{\varepsilon}K^{1-\varepsilon}.$
(3.12)

By going through a computation identical to that in the proof of [20, Lemma
3.4] and observing that $F_{d}(r,r)=H_{2}(r,r)=A_{1}(r,r)=A_{-1}(r,r)=1$, we
now see that

$\displaystyle\frac{\partial}{\partial\alpha}G(\alpha,\gamma)\bigg{\rvert}_{\alpha=\gamma=r}$
$\displaystyle=\frac{\zeta^{\prime}(1+2r)}{\zeta(1+2r)}-\frac{L^{\prime}(1+2r,\chi)}{L(1+2r,\chi)}-\frac{d^{r+1/2}\log
d}{d^{2r+1}-1}$ $\displaystyle\quad+H_{2}^{\prime}(r)-2{\sum}_{p\geqslant
3\atop(-d/p)=-1}\frac{\log p}{p^{4r+2}-1},$

where

$\displaystyle
H_{2}^{\prime}(r):=\frac{\partial}{\partial\alpha}H_{2}(\alpha,\gamma)\bigg{\rvert}_{\alpha=\gamma=r}=\begin{cases}0&\text{if
}d=2,\,7\\\ -2\log
2\mathopen{}\mathclose{{}\left(2^{2(2r+1)}-1}\right)^{-1}&\text{otherwise}.\end{cases}$

It remains to compute the other partial derivative in (3.11). However, in the
resulting sum only the term coming from differentiating the quotient of zeta-
and $L$-functions survives on account of the pole of $\zeta(s)$ at $s=1$.
Thus, with

$\displaystyle F_{d}(-r,r)$ $\displaystyle=1+\frac{d^{1/2+r}-d^{1/2-r}}{d-1},$
$\displaystyle H_{2}(-r,r)$ $\displaystyle=\begin{cases}1&\text{if }d=2,\\\
2^{1-2r}\mathopen{}\mathclose{{}\left(1-2^{-1-2r}}\right)&\text{if }d=7,\\\
\tfrac{4}{3}\mathopen{}\mathclose{{}\left(1-2^{-2(2r+1)}}\right)&\text{otherwise},\end{cases}$
$\displaystyle A_{1}(-r,r)$ $\displaystyle={\prod}_{p\geqslant
3\atop(-d/p)=1}\frac{\mathopen{}\mathclose{{}\left(1-2p^{-1}+p^{-(1+2r)}}\right)\mathopen{}\mathclose{{}\left(1-p^{-(1+2r)}}\right)}{\mathopen{}\mathclose{{}\left(1-p^{-1}}\right)^{2}},$
$\displaystyle A_{-1}(-r,r)$ $\displaystyle={\prod}_{p\geqslant
3\atop(-d/p)=-1}\frac{1-p^{-2(1+2r)}}{1-p^{-2}},$

we conclude that

$\displaystyle J(r):$
$\displaystyle=\frac{\partial}{\partial\alpha}\frac{1}{1-2\alpha}\mathopen{}\mathclose{{}\left(\frac{2\pi}{KN\sqrt{|D|}}}\right)^{2\alpha}G(-\alpha,\gamma)\bigg{\rvert}_{\alpha=\gamma=r}$
$\displaystyle=-\frac{\zeta(1-2r)L(1+2r,\chi)}{L(1,\chi)}\frac{1}{1-2r}\mathopen{}\mathclose{{}\left(\frac{2\pi}{KN\sqrt{|D|}}}\right)^{2r}F_{d}(-r,r)H_{2}(-r,r)A_{1}(-r,r)A_{-1}(-r,r).$

We sum up this discussion in the following proposition.

######  Proposition 3.6.

The Ratios Conjecture implies that for any $r\in\mathbb{C}$ satisfying (3.12),
we have

$\displaystyle\frac{1}{K}{\sum}_{k=1}^{K}\frac{L_{k}^{\prime}(1/2+r)}{L_{k}(1/2+r)}$
$\displaystyle=\frac{\zeta^{\prime}(1+2r)}{\zeta(1+2r)}-\frac{L^{\prime}(1+2r,\chi)}{L(1+2r,\chi)}-\frac{d^{r+1/2}\log
d}{d^{2r+1}-1}+H_{2}^{\prime}(r)$ $\displaystyle\quad\quad-2{\sum}_{p\geqslant
3\atop(-d/p)=-1}\frac{\log
p}{p^{4r+2}-1}+J(r)+O_{\varepsilon}\mathopen{}\mathclose{{}\left(K^{-1/2+\varepsilon}}\right).$

We now let $f:\mathbb{R}\rightarrow\mathbb{R}$ be an even Schwarz function
with $\mathrm{supp}\,\widehat{f}$ compact. By the argument given in the
beginning of [20, Sect. 4], the above result allows us to express the
$1$-level density $D\mathopen{}\mathclose{{}\left(\mathcal{F}(K);f}\right)$
(conditionally on the Ratios Conjecture) as

$\displaystyle\frac{1}{2\pi
i}{\int}_{(c)}\frac{1}{K}\mathopen{}\mathclose{{}\left({\sum}_{k=1}^{K}\mathopen{}\mathclose{{}\left(2\cdot\frac{L_{k}^{\prime}(1/2+r)}{L_{k}(1/2+r)}-\frac{X_{k}^{\prime}(1/2+r)}{X_{k}(1/2+r)}}\right)}\right)f\mathopen{}\mathclose{{}\left(\frac{ir\log
K}{\pi}}\right)\hskip 2.84544pt\mathrm{d}r$
$\displaystyle\quad=S_{X}+S_{\zeta}+S_{L}+S_{A^{\prime}}+S_{J}+S_{d}+S_{H}+O_{\varepsilon}\mathopen{}\mathclose{{}\left(K^{-1/2+\varepsilon}}\right),$
(3.13)

where $c$ is any real number satisfying $1/\log K<c<1/4$, and

$\displaystyle S_{X}$ $\displaystyle:=-\frac{1}{2K\log
K}{\int}_{(C)}{\sum}_{k=1}^{K}\frac{X_{k}^{\prime}\mathopen{}\mathclose{{}\left(1/2+\pi
i\tau/\log K}\right)}{X_{k}\mathopen{}\mathclose{{}\left(1/2+\pi i\tau/\log
K}\right)}f(\tau)\hskip 2.84544pt\mathrm{d}\tau,$ (3.14) $\displaystyle
S_{\zeta}$ $\displaystyle:=\frac{1}{\log
K}{\int}_{(C)}\frac{\zeta^{\prime}\mathopen{}\mathclose{{}\left(1+2\pi
i\tau/\log K}\right)}{\zeta\mathopen{}\mathclose{{}\left(1+2\pi i\tau/\log
K}\right)}f(\tau)\hskip 2.84544pt\mathrm{d}\tau,$ (3.15) $\displaystyle S_{L}$
$\displaystyle:=-\frac{1}{\log
K}{\int}_{(C)}\frac{L^{\prime}\mathopen{}\mathclose{{}\left(1+2\pi i\tau/\log
K,\chi}\right)}{L\mathopen{}\mathclose{{}\left(1+2\pi i\tau/\log
K,\chi}\right)}f(\tau)\hskip 2.84544pt\mathrm{d}\tau,$ (3.16) $\displaystyle
S_{A^{\prime}}$ $\displaystyle:=-\frac{2}{\log
K}{\int}_{(C)}{\sum}_{p\geqslant 3\atop(-d/p)=-1}\frac{\log p}{p^{4\pi
i\tau/\log K+2}-1}f(\tau)\hskip 2.84544pt\mathrm{d}\tau,$ (3.17)
$\displaystyle S_{J}$ $\displaystyle:=\frac{1}{\log
K}{\int}_{(C)}J\mathopen{}\mathclose{{}\left(\frac{\pi i\tau}{\log
K}}\right)f(\tau)\hskip 2.84544pt\mathrm{d}\tau,$ (3.18) $\displaystyle S_{d}$
$\displaystyle:=-\frac{\log d}{\log K}{\int}_{(C)}\frac{d^{\pi i\tau/\log
K+1/2}}{d^{2\pi i\tau/\log K+1}-1}f(\tau)\hskip 2.84544pt\mathrm{d}\tau,$
(3.19) $\displaystyle S_{H}$ $\displaystyle:=\frac{1}{\log
K}{\int}_{(C)}H_{2}^{\prime}\mathopen{}\mathclose{{}\left(\frac{\pi
i\tau}{\log K}}\right)f(\tau)\hskip 2.84544pt\mathrm{d}\tau,$ (3.20)

where we denoted by $(C)$ the set of all $\tau$ with $\mathrm{Im}(\tau)=-c\log
K/\pi$.

In order to determine explicitly the prediction of the one-level density
$D\mathopen{}\mathclose{{}\left(\mathcal{F}(K);f}\right)$ offered by the
Ratios Conjecture, our next goal will be to provide estimates of each of these
integrals in terms of $f$ and the relevant parameters of our family
$\mathcal{F}(K)$. This is the point of the next section.

##  Computations of the Integrals (3.14)-(3.20)

In the following we will allow all implicit constants to depend on the test
function $f$. We start by recalling and adapting some results from [20].

######  Lemma 4.1.

Let $B$ be a positive integer. As $K\rightarrow\infty$, we have

$\displaystyle
S_{\zeta}=-\frac{f(0)}{2}-{\sum}_{j=1}^{B}\frac{c_{j}\hat{f}^{(j-1)}(0)}{\mathopen{}\mathclose{{}\left(\log
K}\right)^{j}}+O_{B}\mathopen{}\mathclose{{}\left(\frac{1}{\mathopen{}\mathclose{{}\left(\log
K}\right)^{B+1}}}\right),$

where $c_{1},c_{2},c_{3},\ldots$ are numbers defined in [20, eq. (5.4),
(5.5)]. In particular,

$\displaystyle c_{1}=1+{\int}_{1}^{\infty}\frac{\psi(t)-t}{t^{2}}\hskip
2.84544pt\mathrm{d}t=-\gamma,$

where $\gamma$ denotes the Euler–Mascheroni constant and
$\psi(t)=\sum_{n\leqslant t}\Lambda(n)$ is the second Chebyshev function.

###### Proof.

The asymptotic expression for $S_{\zeta}$ is simply [20, Lemma 5.2], so we
only need to prove that

$\displaystyle{\int}_{1}^{\infty}\frac{\psi(t)-t}{t^{2}}\hskip
2.84544pt\mathrm{d}t=-\gamma-1.$ (4.1)

We begin by rewriting the integral as

$\displaystyle{\int}_{1}^{\infty}\frac{\psi(t)-t}{t^{2}}\hskip
2.84544pt\mathrm{d}t=\lim_{T\rightarrow\infty}{\int}_{1}^{T}\frac{\psi(t)-t}{t^{2}}\hskip
2.84544pt\mathrm{d}t=\lim_{T\rightarrow\infty}{\int}_{1}^{T}\frac{\psi(t)}{t^{2}}\hskip
2.84544pt\mathrm{d}t-\log T.$ (4.2)

To evaluate the integral of $\psi(t)/t^{2}$, we let
$\mathopen{}\mathclose{{}\left\\{q_{k}:k\geqslant 1}\right\\}$ denote the
sequence of prime powers in increasing order, and we write
$p_{k}:=\exp(\Lambda(q_{k}))$ for the unique prime dividing $q_{k}$. It will
also be convenient to write $\Pi_{k}:=p_{1}\cdot p_{2}\cdots p_{k}$. Then,
using that $\psi(t)$ is constantly equal to $\log\Pi_{k}=\psi(q_{k})$ on the
interval $\mathopen{}\mathclose{{}\left[q_{k},q_{k+1}}\right)$, we compute
that for any large prime power $q_{M}$,

$\displaystyle{\int}_{1}^{q_{M}}\frac{\psi(t)}{t^{2}}\hskip
2.84544pt\mathrm{d}t$
$\displaystyle={\sum}_{k=1}^{M}\log\Pi_{k}{\int}_{q_{k}}^{q_{k+1}}\frac{1}{t^{2}}\hskip
2.84544pt\mathrm{d}t={\sum}_{k=1}^{M}\log\Pi_{k}\mathopen{}\mathclose{{}\left(\frac{1}{q_{k}}-\frac{1}{q_{k+1}}}\right)$
$\displaystyle=\frac{\log\Pi_{1}}{q_{1}}-\frac{\log\Pi_{M}}{q_{M+1}}+{\sum}_{k=2}^{M-1}\frac{1}{q_{k}}\Big{(}\log\Pi_{k}-\log\Pi_{k-1}\Big{)}$
$\displaystyle=-\frac{\log\Pi_{M}}{q_{M+1}}+{\sum}_{k=1}^{M-1}\frac{\log
p_{k}}{q_{k}}=-\frac{\psi(q_{M})}{q_{M+1}}+{\sum}_{n=1}^{q_{M-1}}\frac{\Lambda(n)}{n}.$

Inserting this into (4.2) and replacing $T$ by $q_{M-1}$, we obtain from the
prime number theorem that

$\displaystyle{\int}_{1}^{\infty}\frac{\psi(t)-t}{t^{2}}\hskip
2.84544pt\mathrm{d}t$
$\displaystyle=\lim_{M\rightarrow\infty}-\frac{\psi(q_{M})}{q_{M}}\cdot\frac{q_{M}}{q_{M+1}}+{\sum}_{n=1}^{q_{M-1}}\frac{\Lambda(n)}{n}-\log
q_{M-1}+\log\frac{q_{M-1}}{q_{M}}$
$\displaystyle=-1+\gamma+{\sum}_{n=1}^{\infty}\frac{\Lambda(n)-1}{n}=\lim_{s\rightarrow
1^{+}}\mathopen{}\mathclose{{}\left(-\frac{\zeta^{\prime}(s)}{\zeta(s)}-\zeta(s)}\right)=-1-\gamma,$

where we also used the familiar asymptotic for the harmonic numbers and noted
that $q_{M}/q_{M+1}\rightarrow 1$ as $M\rightarrow\infty$. ∎

Although our Dirichlet $L$-function $L(s,\chi)$ is defined in terms of the
character $\chi(\cdot)=\mathopen{}\mathclose{{}\left(-d/\cdot}\right)$ and not
the non-principal character modulo $4$ as in [20, Lemma 5.4], we note that
this result, including the statement about $S_{A^{\prime}}$, remains valid in
our case with exactly the same proof. We record these two results in the
following two separate lemmas.

######  Lemma 4.2 ([20, Lemma 5.4]).

As $K\rightarrow\infty$, we have

$\displaystyle S_{L}=-\frac{\hat{f}(0)}{\log
K}\frac{L^{\prime}(1,\chi)}{L(1,\chi)}+O\mathopen{}\mathclose{{}\left(\frac{1}{(\log
K)^{2}}}\right).$

######  Lemma 4.3 ([20, Lemma 5.4]).

As $K\rightarrow\infty$, we have

$\displaystyle S_{A^{\prime}}=-\frac{2\hat{f}(0)}{\log K}{\sum}_{p\geqslant
3\atop(-d/p)=-1}\frac{\log
p}{p^{2}-1}+O\mathopen{}\mathclose{{}\left(\frac{1}{(\log K)^{2}}}\right).$

Moreover, we can adapt [20, Lemma 5.1] to our situation and obtain the
following result.

######  Lemma 4.4.

As $K\rightarrow\infty$, we have

$\displaystyle
S_{X}=\hat{f}(0)\mathopen{}\mathclose{{}\left(1+\frac{\log\sqrt{|D|}-\log
2\pi+\log N-1}{\log
K}}\right)+O\mathopen{}\mathclose{{}\left(\frac{1}{K}}\right).$

###### Proof.

We have

$\displaystyle\frac{X_{k}^{\prime}(s)}{X_{k}(s)}=\frac{\mathrm{d}}{\mathrm{d}s}\log\Gamma(1-s+Nk)-\frac{\mathrm{d}}{\mathrm{d}s}\log\Gamma(s+Nk)-\log|D|+2\log
2\pi.$

The contribution of the constants to the integral (3.14) is

$\displaystyle\frac{1}{\log
K}\widehat{f}(0)\mathopen{}\mathclose{{}\left(\log\sqrt{|D|}-\log
2\pi}\right).$

As for the contribution of the gamma functions, we recall that, in the
notation of [20], we have $\textbf{k}=1/2+Nk$. It then follows immediately
from [20, Eq. (5.1)], that this contribution equals

$\displaystyle\frac{1}{K\log
K}\widehat{f}(0){\sum}_{k=1}^{K}\log\textbf{k}+O\mathopen{}\mathclose{{}\left(\frac{1}{K}}\right).$

By arguing as in [20, Eq. (5.2)], we find that

$\displaystyle{\sum}_{k=1}^{K}\log\,\textbf{k}$
$\displaystyle={\sum}_{k=1}^{K}\log\mathopen{}\mathclose{{}\left(1/2+Nk}\right)=K\log
N+\log K!+O(\log K)$ $\displaystyle=K\log N+K\log K-K+O(\log K)$
$\displaystyle=K\log K+K\mathopen{}\mathclose{{}\left(\log N-1}\right)+O(\log
K).$

This proves the claim. ∎

We now turn to the integral (3.18). As Lemma 4.6 below will show, the
asymptotic expression for this integral will involve a special value of the
logarithmic derivative of $L(s,\chi)$. In anticipation of this, we will first
compute this special value.

######  Lemma 4.5.

Let $\eta$ denote Dedekind’s eta function,

$\displaystyle\eta(\tau)=e^{\pi i\tau/12}{\prod}_{n\geqslant
1}\mathopen{}\mathclose{{}\left(1-e^{2\pi
in\tau}}\right),\quad\quad\mathrm{Im}(\tau)>0,$

and let $\gamma$ denote the Euler–Mascheroni constant. Then we have

$\displaystyle\frac{L^{\prime}(1,\chi)}{L(1,\chi)}=\gamma-\log
2-\mathopen{}\mathclose{{}\left(\log|D|}\right)/2-\log\mathrm{Im}(\tau_{0})-4\log\mathopen{}\mathclose{{}\left|\eta(\tau_{0})}\right|$

where $C(d)=2\gamma-\log
2-\mathopen{}\mathclose{{}\left(\log|D|}\right)/2-\log\mathrm{Im}(\tau_{0})-4\log\mathopen{}\mathclose{{}\left|\eta(\tau_{0})}\right|$,
and

$\displaystyle\tau_{0}=\begin{cases}i\sqrt{2}&\text{ if }d=2,\\\
(-1+i\sqrt{d})/2&\text{ if }d\neq 2.\end{cases}$

###### Proof.

If $\zeta_{\mathbb{K}}(s)$ denotes the Dedekind zeta function of $\mathbb{K}$,
[4, Prop. 10.5.5] gives the factorization
$\zeta_{\mathbb{K}}(s)=\zeta(s)L(s,\chi)$. Indeed, this is clear if
$d\geqslant 3$ since then $D=-d$, whereas for $d=2$ and $n\geqslant 1$, the
fact that $(-2/n)$ can only assume the values $0,\pm 1$ immediately shows that

$\displaystyle\mathopen{}\mathclose{{}\left(\frac{-2}{n}}\right)=\mathopen{}\mathclose{{}\left(\frac{-2}{n}}\right)^{3}=\mathopen{}\mathclose{{}\left(\frac{-8}{n}}\right),$

so that $(-d/n)=(D/n)$ even in this case.

On the other hand, since any non-zero ideal
$\mathfrak{m}\subset\mathcal{O}_{\mathbb{K}}$ has exactly
$\mathopen{}\mathclose{{}\left|\mathcal{O}_{\mathbb{K}}^{\times}}\right|$
generators, we also have

$\displaystyle\zeta_{\mathbb{K}}(s)=\frac{1}{\mathopen{}\mathclose{{}\left|\mathcal{O}_{\mathbb{K}}^{\times}}\right|}{\sum}_{N(a,b)\neq
0}\frac{1}{N(a,b)^{s}},\quad\quad\mathrm{Re}(s)>1,$

where $N(a,b)$ denotes the norm of the element $j(a,b)$ as defined in Section
2.2. It now follows from this and [4, Corollary 10.4.8] that

$\displaystyle\zeta(s)L(\chi,s)$
$\displaystyle=\frac{1}{\mathopen{}\mathclose{{}\left|\mathcal{O}_{\mathbb{K}}^{\times}}\right|}{\sum}_{N(a,b)\neq
0}\frac{1}{N(a,b)^{s}}=\frac{2}{\mathopen{}\mathclose{{}\left|\mathcal{O}_{\mathbb{K}}^{\times}}\right|}\frac{\pi}{\sqrt{|D|}}\mathopen{}\mathclose{{}\left(\frac{1}{s-1}+C(d)+O(s-1)}\right).$
(4.3)

Furthermore, by writing $L$ and $\zeta$ as Laurent series around $s=1$, we
find that

$\displaystyle L(s,\chi)$
$\displaystyle=L(1,\chi)+L^{\prime}(1,\chi)(s-1)+O\mathopen{}\mathclose{{}\left((s-1)^{2}}\right),$
$\displaystyle\zeta(s)$ $\displaystyle=\frac{1}{s-1}+\gamma+O(s-1),$

and hence

$\displaystyle\zeta(s)L(s,\chi)=\frac{L(1,\chi)}{s-1}+L^{\prime}(1,\chi)+\gamma
L(1,\chi)+O(s-1).$ (4.4)

Together, the two different expressions (4.3) and (4.4) for
$\zeta(s)L(s,\chi)$ now force an equality of coefficients, namely

$\displaystyle
L(1,\chi)=\frac{2\pi}{\mathopen{}\mathclose{{}\left|\mathcal{O}_{\mathbb{K}}^{\times}}\right|\sqrt{|D|}},\quad\quad
L^{\prime}(1,\chi)=\frac{2\pi\mathopen{}\mathclose{{}\left(C(d)-\gamma}\right)}{\mathopen{}\mathclose{{}\left|\mathcal{O}_{\mathbb{K}}^{\times}}\right|\sqrt{|D|}}.$

By computing the quotient $L^{\prime}(1,\chi)/L(1,\chi)$, we obtain the claim.
∎

######  Lemma 4.6.

As $K\rightarrow\infty$, we have

$\displaystyle
S_{J}=\frac{f(0)}{2}-\frac{1}{2}{\int}_{\mathbb{R}}\hat{f}(\tau)\mathbbm{1}_{\mathopen{}\mathclose{{}\left[-1,1}\right]}(\tau)\hskip
2.84544pt\mathrm{d}\tau+\frac{\hat{f}(1)}{\log
K}\ell_{1}+O_{d}\mathopen{}\mathclose{{}\left(\frac{1}{(\log K)^{2}}}\right),$

where

$\displaystyle\ell_{1}$
$\displaystyle=\frac{L^{\prime}(1,\chi)}{L(1,\chi)}+2{\sum}_{p\geqslant
3\atop(-d/p)=-1}\frac{\log p}{p^{2}-1}+\log\frac{2\pi
e}{N\sqrt{|D|}}+\frac{\sqrt{d}\log d}{d-1}-\gamma-\frac{2a\log 2}{3},$

$a=-\mathbbm{1}(d\neq 2,7)$, and $L^{\prime}(1,\chi)/L(1,\chi)$ is given in
Lemma 4.5.

###### Proof.

We proceed as in the proof of [20, Lemma 5.5], which relies on the methods of
[9]. That is, we will replace the domain of integration $(C)$ with the union
of the compact and non-compact contours $C_{1}\cup C_{\eta}$ and $C_{0}$,
respectively, where

$\displaystyle C_{0}$
$\displaystyle:=\mathopen{}\mathclose{{}\left\\{\tau\in\mathbb{C}:\mathrm{Im}(\tau)=0,\,|\mathrm{Re}(\tau)|\geqslant(\log
K)^{\varepsilon}}\right\\},$ $\displaystyle C_{1}$
$\displaystyle:=\mathopen{}\mathclose{{}\left\\{\tau\in\mathbb{C}:\mathrm{Im}(\tau)=0,\,\eta\leqslant|\mathrm{Re}(\tau)|\leqslant(\log
K)^{\varepsilon}}\right\\},$ $\displaystyle C_{\eta}$
$\displaystyle:=\mathopen{}\mathclose{{}\left\\{\eta
e^{i\theta}:-\pi\leqslant\theta\leqslant 0}\right\\}.$

The utility of such a decomposition is two-fold: On the non-compact part
$C_{0}$ we can bound the integrand using the rapid decay of our test function
$f$. On the other hand, on the compact part we can estimate the individual
factors in the integrand with the first few terms of their Taylor expansions.
Moreover, the fact that $C_{1}\cup C_{\eta}$ becomes symmetric when we let
$\eta\rightarrow 0$ means that we do not have to take into account
contributions from any odd, positive powers of $\tau$ in these expansions, as
the integrals of $\tau f(\tau)$, $\tau^{3}f(\tau)$, $\tau^{5}f(\tau)$,
$\ldots$ over this set vanish. We exploit this fact to get rid of any
occurrences of $\varepsilon$ in the error terms. We now proceed to the
details.

Initially, we recall the following elementary estimates: If $\gamma$ denotes
the Euler–Mascheroni constant, then as $K\rightarrow\infty$,

$\displaystyle\zeta\mathopen{}\mathclose{{}\left(1-\frac{2\pi i\tau}{\log
K}}\right)$ $\displaystyle=-\frac{\log K}{2\pi
i\tau}+\gamma+O\mathopen{}\mathclose{{}\left(\frac{|\tau|}{\log K}}\right),$
(4.5) $\displaystyle\frac{L(1+2\pi i\tau/\log K,\chi)}{L(1,\chi)}$
$\displaystyle=1+\frac{L^{\prime}(1,\chi)}{L(1,\chi)}\frac{2\pi i\tau}{\log
K}+O\mathopen{}\mathclose{{}\left(\frac{|\tau|^{2}}{(\log K)^{2}}}\right),$
(4.6)

and

$\displaystyle\begin{split}&A_{1}\mathopen{}\mathclose{{}\left(-\frac{\pi
i\tau}{\log K},\frac{\pi i\tau}{\log
K}}\right)A_{-1}\mathopen{}\mathclose{{}\left(-\frac{\pi i\tau}{\log
K},\frac{\pi i\tau}{\log K}}\right)\\\
&\quad=1+\mathopen{}\mathclose{{}\left(2{\sum}_{p\geqslant
3\atop(-d/p)=-1}\frac{\log p}{p^{2}-1}}\right)\frac{2\pi i\tau}{\log
K}+O\mathopen{}\mathclose{{}\left(\frac{|\tau|^{2}}{(\log
K)^{2}}}\right).\end{split}$ (4.7)

We now obtain Taylor expansions of the other factors in $J$. First of all,

$\displaystyle\begin{split}&\frac{1}{1-2r}\mathopen{}\mathclose{{}\left(\frac{2\pi}{KN\sqrt{|D|}}}\right)^{2r}\bigg{\rvert}_{r=\tfrac{\pi
i\tau}{\log K}}\\\ &\quad=\exp\mathopen{}\mathclose{{}\left(\frac{2\pi
i\tau}{\log K}\log\frac{2\pi}{N\sqrt{|D|}}-2\pi
i\tau}\right)\mathopen{}\mathclose{{}\left(1+\frac{2\pi i\tau}{\log
K}+O\mathopen{}\mathclose{{}\left(\frac{|\tau|^{2}}{(\log
K)^{2}}}\right)}\right)\\\ &\quad=e^{-2\pi
i\tau}\mathopen{}\mathclose{{}\left(1+\frac{2\pi i\tau}{\log
K}\log\frac{2\pi}{N\sqrt{|D|}}+O\mathopen{}\mathclose{{}\left(\frac{|\tau|^{2}(\log|D|)^{2}}{(\log
K)^{2}}}\right)}\right)\mathopen{}\mathclose{{}\left(1+\frac{2\pi i\tau}{\log
K}+O\mathopen{}\mathclose{{}\left(\frac{|\tau|^{2}}{(\log
K)^{2}}}\right)}\right)\\\ &\quad=e^{-2\pi i\tau}+e^{-2\pi i\tau}\frac{2\pi
i\tau}{\log K}\log\frac{2\pi
e}{N\sqrt{|D|}}+O\mathopen{}\mathclose{{}\left(\frac{|\tau|^{2}(\log|D|)^{2}}{(\log
K)^{2}}}\right),\end{split}$ (4.8)

whenever $K$ is so large that

$\displaystyle\mathopen{}\mathclose{{}\left|\frac{2\pi i\tau}{\log
K}}\right|<1,\quad\quad\mathopen{}\mathclose{{}\left|\frac{2\pi i\tau}{\log
K}\log\frac{2\pi}{N\sqrt{|D|}}}\right|<1.$

(Note that this is certainly satisfied for $\tau\in C_{1}\cup C_{\eta}$.)
Next, we see that

$\displaystyle F_{d}\mathopen{}\mathclose{{}\left(-\frac{\pi i\tau}{\log
K},\frac{\pi i\tau}{\log K}}\right)=1+\frac{\sqrt{d}\log d}{d-1}\frac{2\pi
i\tau}{\log K}+O_{d}\mathopen{}\mathclose{{}\left(\frac{|\tau|^{2}}{(\log
K)^{2}}}\right).$ (4.9)

Finally, we also record the bound

$\displaystyle H_{2}\mathopen{}\mathclose{{}\left(-\frac{\pi i\tau}{\log
K},\frac{\pi i\tau}{\log K}}\right)=1-\frac{2a\log 2}{3}\frac{2\pi i\tau}{\log
K}+O\mathopen{}\mathclose{{}\left(\frac{|\tau|^{2}}{(\log K)^{2}}}\right).$
(4.10)

By taking the product of all the Taylor expansions (4.5)-(4.10) and
disregarding all those resulting terms which have order at least $\tau/\log
K$, we obtain that

$\displaystyle J\mathopen{}\mathclose{{}\left(\frac{\pi i\tau}{\log
K}}\right)$
$\displaystyle\quad=\mathopen{}\mathclose{{}\left(\frac{1}{x}-\gamma+O\mathopen{}\mathclose{{}\left(|x|}\right)}\right)\Biggl{(}e^{-2\pi
i\tau}+x\Biggl{(}e^{-2\pi
i\tau}\frac{L^{\prime}(1,\chi)}{L(1,\chi)}{\color[rgb]{1,1,1}\Biggr{)}}+2e^{-2\pi
i\tau}{\sum}_{p\geqslant 3\atop(-d/p)=-1}\frac{\log p}{p^{2}-1}$
$\displaystyle\quad\quad\hskip 8.00003pt+\,e^{-2\pi i\tau}\log\frac{2\pi
e}{N\sqrt{|D|}}\Biggl{.}\Biggl{.}+\,e^{-2\pi i\tau}\frac{\sqrt{d}\log
d}{d-1}-e^{-2\pi i\tau}\frac{2a\log
2}{3}\Biggr{)}+O_{d}\mathopen{}\mathclose{{}\left(|x|^{2}}\right)\Biggr{)}$
$\displaystyle\quad=\frac{e^{-2\pi i\tau}}{x}+\,e^{-2\pi
i\tau}\Biggl{(}-\gamma+\frac{L^{\prime}(1,\chi)}{L(1,\chi)}+2{\sum}_{p\geqslant
3\atop(-d/p)=-1}\frac{\log p}{p^{2}-1}{\color[rgb]{1,1,1}\Biggr{)}}$
$\displaystyle\quad\quad{\color[rgb]{1,1,1}\Biggl{)}}+\log\frac{2\pi
e}{N\sqrt{|D|}}+\frac{\sqrt{d}\log d}{d-1}-\frac{2a\log
2}{3}\Biggr{)}+O_{d}\mathopen{}\mathclose{{}\left(|x|}\right),$

where we wrote $x=2\pi i\tau/\log K$ for simplicity. Thus, we obtain

$\displaystyle\begin{split}S_{J}&={\int}_{C_{1}\cup
C_{\eta}}f(\tau)\frac{e^{-2\pi i\tau}}{2\pi i\tau}\hskip
2.84544pt\mathrm{d}\tau\\\\[8.00003pt] &\quad+\frac{1}{\log
K}\Biggl{(}-\gamma+\frac{L^{\prime}(1,\chi)}{L(1,\chi)}+2{\sum}_{p\geqslant
3\atop(-d/p)=-1}\frac{\log p}{p^{2}-1}+\log\frac{2\pi
e}{N\sqrt{|D|}}+\frac{\sqrt{d}\log d}{d-1}-\frac{2a\log 2}{3}\Biggr{)}\\\
&\quad\times{\int}_{C_{1}\cup C_{\eta}}f(\tau)e^{-2\pi i\tau}\hskip
2.84544pt\mathrm{d}\tau+\frac{1}{\log
K}{\int}_{C_{0}}J\mathopen{}\mathclose{{}\left(\frac{\pi i\tau}{\log
K}}\right)f(\tau)\hskip
2.84544pt\mathrm{d}\tau+O_{d}\mathopen{}\mathclose{{}\left(\frac{1}{(\log
K)^{2}}}\right),\end{split}$ (4.11)

where we made use of the evenness of $f$ as described earlier. By arguing
exactly as in [20, Lemma 5.5], we relate the integrals above to special values
of the Fourier transform $\hat{f}$, namely

$\displaystyle{\int}_{C_{1}\cup C_{\eta}}f(\tau)e^{-2\pi i\tau}\hskip
2.84544pt\mathrm{d}\tau$
$\displaystyle=\hat{f}(1)+O\mathopen{}\mathclose{{}\left(\frac{1}{(\log
K)^{3}}}\right),$ (4.12) $\displaystyle{\int}_{C_{1}\cup
C_{\eta}}f(\tau)\frac{e^{-2\pi i\tau}}{2\pi i\tau}\hskip
2.84544pt\mathrm{d}\tau$
$\displaystyle=\frac{f(0)}{2}-\frac{1}{2}{\int}_{-1}^{1}\hat{f}(\tau)\hskip
2.84544pt\mathrm{d}\tau+O\mathopen{}\mathclose{{}\left(\frac{1}{(\log
K)^{3}}}\right).$ (4.13)

Since the rapid decay of $f$ on $\mathbb{R}$ shows that the integral over
$C_{0}$ in (4.11) is at most a constant times $(\log K)^{-2}$ (for example),
it now follows from (4.12) and (4.13) that

$\displaystyle S_{J}$
$\displaystyle=\frac{f(0)}{2}-\frac{1}{2}{\int}_{-1}^{1}\hat{f}(\tau)\hskip
2.84544pt\mathrm{d}\tau+\frac{\hat{f}(1)}{\log
K}\Biggl{(}\frac{L^{\prime}(1,\chi)}{L(1,\chi)}+2{\sum}_{p\geqslant
3\atop(-d/p)=-1}\frac{\log p}{p^{2}-1}{\color[rgb]{1,1,1}\Biggr{)}}$
$\displaystyle\quad\quad{\color[rgb]{1,1,1}\Biggl{)}}+\log\frac{2\pi
e}{N\sqrt{|D|}}+\frac{\sqrt{d}\log d}{d-1}-\gamma-\frac{2a\log
2}{3}\Biggr{)}+O_{d}\mathopen{}\mathclose{{}\left(\frac{1}{(\log
K)^{2}}}\right),$

which completes the proof. ∎

We now turn to the final two integrals (3.19) and (3.20).

######  Lemma 4.7.

As $K\rightarrow\infty$, we have the estimate

$\displaystyle S_{d}=-\frac{\log d}{\log
K}\frac{\sqrt{d}}{d-1}\hat{f}(0)+O_{d}\mathopen{}\mathclose{{}\left(\frac{1}{(\log
K)^{2}}}\right).$

###### Proof.

The integrand only has poles when $\mathrm{Im}(\tau)=(\log K)/2\pi>0$, so
analogously to the proof of Lemma 4.6, we use Cauchy’s residue theorem and the
rapid decay of $f$ to move the contour $(C)$ to the real line without changing
the value of the integral. As before, we partition this set into a compact and
a non-compact part in order to, respectively, use the (even-indexed terms of
the) Taylor expansion of the integrand and bound the integral using the decay
of the test function. Concretely, we write $\mathbb{R}=C_{0}\cup C_{1}$ with

$\displaystyle C_{0}$
$\displaystyle:=\mathopen{}\mathclose{{}\left\\{\tau\in\mathbb{C}:\mathrm{Im}(\tau)=0,\,|\mathrm{Re}(\tau)|>(\log
K)^{\varepsilon}}\right\\},$ $\displaystyle C_{1}$
$\displaystyle:=\mathopen{}\mathclose{{}\left\\{\tau\in\mathbb{C}:\mathrm{Im}(\tau)=0,\,|\mathrm{Re}(\tau)|\leqslant(\log
K)^{\varepsilon}}\right\\}.$

As in the proof of Lemma 4.6, we note that

$\displaystyle{\int}_{C_{0}}\frac{d^{\pi i\tau/\log K+1/2}}{d^{2\pi i\tau/\log
K+1}-1}f(\tau)\hskip 2.84544pt\mathrm{d}\tau\ll_{d}\frac{1}{(\log K)^{2}},$

whereas for the integral over $C_{1}$, we use the Taylor expansion

$\displaystyle\frac{d^{r+1/2}}{d^{2r+1}-1}=\frac{\sqrt{d}}{d-1}+r\cdot\frac{\mathrm{d}}{\mathrm{d}r}\mathopen{}\mathclose{{}\left(\frac{d^{r+1/2}}{d^{2r+1}-1}}\right)\Biggr{|}_{r=0}+O_{d}\mathopen{}\mathclose{{}\left(r^{2}}\right)\quad\quad(r\rightarrow
0)$

and the fact that $\tau f(\tau)$ is odd to obtain

$\displaystyle{\int}_{C_{1}}\frac{d^{\pi i\tau/\log K+1/2}}{d^{2\pi i\tau/\log
K+1}-1}f(\tau)\hskip 2.84544pt\mathrm{d}\tau$
$\displaystyle={\int}_{C_{1}}\mathopen{}\mathclose{{}\left(\frac{\sqrt{d}}{d-1}+O_{d}\mathopen{}\mathclose{{}\left(\frac{\tau^{2}}{\mathopen{}\mathclose{{}\left(\log
K}\right)^{2}}}\right)}\right)f(\tau)\hskip 2.84544pt\mathrm{d}\tau$
$\displaystyle=\mathopen{}\mathclose{{}\left(\frac{\sqrt{d}}{d-1}+O_{d}\mathopen{}\mathclose{{}\left(\frac{1}{\mathopen{}\mathclose{{}\left(\log
K}\right)^{2-2\varepsilon}}}\right)}\right){\int}_{C_{1}}f(\tau)\hskip
2.84544pt\mathrm{d}\tau$
$\displaystyle=\mathopen{}\mathclose{{}\left(\frac{\sqrt{d}}{d-1}+O_{d}\mathopen{}\mathclose{{}\left(\frac{1}{\mathopen{}\mathclose{{}\left(\log
K}\right)^{2-2\varepsilon}}}\right)}\right)\mathopen{}\mathclose{{}\left({\int}_{\mathbb{R}}f(\tau)\hskip
2.84544pt\mathrm{d}\tau+O\mathopen{}\mathclose{{}\left(\frac{1}{(\log
K)^{2}}}\right)}\right),$

where we used the rapid decay of $f$ in the last step. The claim now follows.
∎

Finally, we have the following asymptotic estimate. We assume that $d\neq 2,7$
since otherwise $S_{H}=0$.

######  Lemma 4.8.

Suppose that $d\neq 2,\,7$. As $K\rightarrow\infty$, we have

$\displaystyle S_{H}=\frac{-2\log 2}{3\log
K}\hat{f}(0)+O\mathopen{}\mathclose{{}\left(\frac{1}{(\log K)^{2}}}\right).$

###### Proof.

The method of proof is identical to that of the previous lemma. Once again, we
note that $H_{2}^{\prime}(\pi i\tau/\log K)$ only has poles if
$\mathrm{Im}(\tau)=\log K/(2\pi\log 2)>0$, so that we are justified in moving
the contour to the real line. As before, the integral over the non-compact
part $C_{0}$ of our partition $C_{0}\cup C_{1}$ of $\mathbb{R}$ simply
contributes to the error term, while the integral over the compact part is

$\displaystyle{\int}_{C_{1}}\mathopen{}\mathclose{{}\left(\frac{-2\log
2}{3}+\frac{\pi i\tau}{\log
K}\cdot\frac{\mathrm{d}}{\mathrm{d}r}H_{2}^{\prime}(r)\Bigg{\rvert}_{r=0}+O\mathopen{}\mathclose{{}\left(\frac{|\tau|^{2}}{\mathopen{}\mathclose{{}\left(\log
K}\right)^{2}}}\right)}\right)f(\tau)\hskip 2.84544pt\mathrm{d}\tau,$

which follows from the Taylor expansion

$\displaystyle H_{2}^{\prime}(r)=\frac{-2\log
2}{3}+r\cdot\frac{\mathrm{d}}{\mathrm{d}r}H_{2}^{\prime}(r)\Bigg{\rvert}_{r=0}+O\mathopen{}\mathclose{{}\left(r^{2}}\right).$

We now proceed exactly as in the proof of the previous lemma. ∎

By combining the results from Lemma 4.1 to Lemma 4.8 with (3.3), we have
completed the proof of Theorem 1.2.

##  An Unconditional Asymptotic for the One-Level Density

In this section, we use the following formula for logarithmic derivatives of
the $L$-functions $L_{k}$ in order to give an unconditional expression for the
one-level density $D\mathopen{}\mathclose{{}\left(\mathcal{F}(K);f}\right)$.

######  Lemma 5.1.

For $k\geqslant 1$ and $\mathrm{Re}(s)>1$, we have

$\displaystyle\frac{L_{k}^{\prime}(s)}{L_{k}(s)}=-{\sum}_{n\geqslant
1}\frac{c_{k}(n)}{n^{s}},\quad\quad
c_{k}(n)=\Lambda(n){\sum}_{\mathbb{N}(\mathfrak{p}^{m})=n}\Big{(}1+\mathbbm{1}\big{(}\mathfrak{p}=\langle
p\rangle\big{)}\Big{)}\psi_{k}\mathopen{}\mathclose{{}\left(\mathfrak{p}^{m}}\right),$

where we understand that the indicator function specifies whether or not
$\mathfrak{p}$ lies over an inert rational prime.

###### Proof.

By taking the logarithmic derivative of the (absolutely convergent) Euler
product that defines $L_{k}(s)$ in the half-plane $\mathrm{Re}(s)>1$, we get

$\displaystyle\frac{L_{k}^{\prime}(s)}{L_{k}(s)}=-{\sum}_{\mathfrak{p}}\log\mathbb{N}(\mathfrak{p})\frac{\psi_{k}(\mathfrak{p})/\mathbb{N}(\mathfrak{p})^{s}}{1-\psi_{k}(\mathfrak{p})/\mathbb{N}(\mathfrak{p})^{s}}=-{\sum}_{\mathfrak{p}}\log\mathbb{N}(\mathfrak{p}){\sum}_{m\geqslant
1}\frac{\psi_{k}\mathopen{}\mathclose{{}\left(\mathfrak{p}^{m}}\right)}{\mathbb{N}\mathopen{}\mathclose{{}\left(\mathfrak{p}^{m}}\right)^{s}},$

where we also used that $\psi_{k}$ and the norm are completely multiplicative.

To finish the computation we note that, if $\mathfrak{p}$ lies over a ramified
or split prime $p$, then for any $m\geqslant 1$,

$\displaystyle\log\mathbb{N}(\mathfrak{p})=\log
p=\Lambda\mathopen{}\mathclose{{}\left(\mathbb{N}(\mathfrak{p})}\right)=\Lambda\mathopen{}\mathclose{{}\left(\mathbb{N}(\mathfrak{p}^{m})}\right),$

whereas if $\mathfrak{p}=\langle p\rangle$ lies over an inert prime,

$\displaystyle\log\mathbb{N}(\mathfrak{p})=2\log
p=2\Lambda\mathopen{}\mathclose{{}\left(\mathbb{N}(\mathfrak{p})}\right)=2\Lambda\mathopen{}\mathclose{{}\left(\mathbb{N}(\mathfrak{p}^{m})}\right)$

for any $m\geqslant 1$. In light of this and the previous computation, we can
rewrite the logarithmic derivative as

$\displaystyle\frac{L_{k}^{\prime}(s)}{L_{k}(s)}$
$\displaystyle=-{\sum}_{\mathfrak{p}}\Big{(}1+\mathbbm{1}\big{(}\mathfrak{p}=\langle
p\rangle\big{)}\Big{)}{\sum}_{m\geqslant
1}\frac{\psi_{k}\mathopen{}\mathclose{{}\left(\mathfrak{p}^{m}}\right)}{\mathbb{N}\mathopen{}\mathclose{{}\left(\mathfrak{p}^{m}}\right)^{s}}\Lambda\mathopen{}\mathclose{{}\left(\mathbb{N}\mathopen{}\mathclose{{}\left(\mathfrak{p}^{m}}\right)}\right).$

We now obtain the claim by grouping together all $\mathfrak{p}^{m}$ with norm
$n$, for all $n\geqslant 1$. ∎

Towards a computation of the $1$-level density, we note that just as in [20,
Section 6], we have

$\displaystyle D(\mathcal{F}(K);f)=\frac{1}{2\pi
i}{\int}_{(c^{\prime})}\frac{1}{K}\mathopen{}\mathclose{{}\left({\sum}_{k=1}^{K}2\frac{L_{k}^{\prime}(1/2+r)}{L_{k}(1/2+r)}-\frac{X_{k}^{\prime}(1/2+r)}{X_{k}(1/2+r)}}\right)f\mathopen{}\mathclose{{}\left(\frac{ir\log
K}{\pi}}\right)\hskip 2.84544pt\mathrm{d}r,$

where now $c^{\prime}\geqslant 1/2$. Using Lemma 5.1 and arguing as in [13,
Thm. 5.12], we now obtain

$\displaystyle D\mathopen{}\mathclose{{}\left(\mathcal{F}(K);f}\right)$
$\displaystyle=S_{X}-\frac{1}{\pi
i}\frac{1}{K}{\int}_{(c^{\prime})}\,{\sum}_{k=1}^{K}\,{\sum}_{n\geqslant
1}\frac{c_{k}(n)}{n^{1/2+r}}f\mathopen{}\mathclose{{}\left(\frac{ir\log
K}{\pi}}\right)\hskip 2.84544pt\mathrm{d}r$ $\displaystyle=S_{X}-\frac{1}{\pi
i}\frac{1}{K}{\sum}_{k=1}^{K}\,{\sum}_{n\geqslant
1}\frac{c_{k}(n)}{\sqrt{n}}{\int}_{(c^{\prime})}e^{-r\log
n}f\mathopen{}\mathclose{{}\left(\frac{ir\log K}{\pi}}\right)\hskip
2.84544pt\mathrm{d}r$ $\displaystyle=S_{X}-\frac{1}{K\log
K}{\sum}_{k=1}^{K}\,{\sum}_{n\geqslant
1}\frac{c_{k}(n)}{\sqrt{n}}{\int}_{\mathopen{}\mathclose{{}\left\\{\mathrm{Im}(r)=c^{\prime}\log
K/\pi}\right\\}}e^{2\pi ir\log n/2\log K}f(r)\hskip 2.84544pt\mathrm{d}r$
$\displaystyle=S_{X}-\frac{1}{K\log K}{\sum}_{k=1}^{K}{\sum}_{n\geqslant
1}\frac{c_{k}(n)}{\sqrt{n}}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{\log
n}{2\log K}}\right),$

where we replaced
$\mathopen{}\mathclose{{}\left\\{\mathrm{Im}(r)=c^{\prime}\log
K/\pi}\right\\}$ with $\mathbb{R}$ due to the rapid decay of $f$ on horizontal
strips.

The occurrence of the von Mangoldt function in $c_{k}(n)$ means that the only
indices $n$ contributing to the sum above are the prime powers. If we replace
$n$ with $p^{n}$ (to save notation), we can then describe the resulting values
of $c_{k}(p^{n})$ depending on the splitting behavior of $p$. Thus, with the
help of Lemma 2.4, we find that

$\displaystyle
c_{k}(p^{n})=\begin{cases}\mathopen{}\mathclose{{}\left(\psi_{k}(\mathfrak{q}_{1}^{n})+\psi_{k}(\mathfrak{q}_{2}^{n})}\right)\log
p&\text{if }\langle
p\rangle=\mathfrak{q}_{1}\mathfrak{q}_{2},\,\mathfrak{q}_{1}\neq\mathfrak{q}_{2},\\\
2\log p&\text{if }p\text{ is inert and }n\text{ is even},\\\ 0&\text{if
}p\text{ is inert and }n\text{ is odd},\\\ \log p&\text{if }\langle
p\rangle=\mathfrak{q}^{2}.\end{cases}$

Considering these special values of $c_{k}(p^{n})$, we now define

$\displaystyle S_{\mathrm{inert}}$ $\displaystyle=-\frac{2}{\log
K}{\sum}_{\text{$p$ inert}}\hskip 2.84544pt{\sum}_{n\geqslant 1}\frac{\log
p}{p^{n}}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{n\log p}{\log
K}}\right),$ $\displaystyle S_{\mathrm{split}}$ $\displaystyle=-\frac{1}{K\log
K}{\sum}_{\text{$p$ split}}\hskip 2.84544pt{\sum}_{n\geqslant 1}\frac{\log
p}{p^{n/2}}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{n\log p}{2\log
K}}\right){\sum}_{k=1}^{K}\mathopen{}\mathclose{{}\left(\psi_{k}(\mathfrak{q}_{1}^{n})+\psi_{k}(\mathfrak{q}_{2}^{n})}\right),$
$\displaystyle S_{\mathrm{ram}}$ $\displaystyle=-\frac{1}{\log
K}{\sum}_{n\geqslant 1}\frac{\log
d}{d^{n/2}}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{n\log d}{2\log
K}}\right).$

Thus,

$\displaystyle
D\mathopen{}\mathclose{{}\left(\mathcal{F}(K);f}\right)=S_{X}+S_{\mathrm{inert}}+S_{\mathrm{split}}+S_{\mathrm{ram}}.$
(5.1)

We now want to compare the terms $S_{\mathrm{inert}}$ and $S_{\mathrm{ram}}$
with the terms (3.14)-(3.20) appearing in the expression of the one-level
density conditional on the Ratios Conjecture. To facilitate this comparison,
we begin by rewriting these terms as follows.

######  Lemma 5.2.

We have that

$\displaystyle S_{\zeta}$ $\displaystyle=-\frac{1}{\log K}{\sum}_{n\geqslant
1}\frac{\Lambda(n)}{n}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{\log
n}{\log K}}\right),$ (5.2) $\displaystyle S_{A^{\prime}}$
$\displaystyle=-\frac{2}{\log K}{\sum}_{p\geqslant
3\atop(-d/p)=-1}{\sum}_{n\geqslant 1}\frac{\log
p}{p^{2n}}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{2n\log p}{\log
K}}\right),$ (5.3) $\displaystyle S_{L}$ $\displaystyle=\frac{1}{\log
K}{\sum}_{n\geqslant
1}\frac{\Lambda(n)\chi(n)}{n}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{\log
n}{\log K}}\right),$ (5.4) $\displaystyle S_{d}$ $\displaystyle=-\frac{\log
d}{\sqrt{d}\log K}{\sum}_{n\geqslant
0}\frac{1}{d^{n}}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{(1/2+n)\log
d}{\log K}}\right),$ (5.5) $\displaystyle S_{H}$
$\displaystyle=\frac{a}{2}\frac{\log 2}{\log K}{\sum}_{n\geqslant
0}\frac{1}{4^{n}}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{(2n+2)\log
2}{\log K}}\right).$ (5.6)

###### Proof.

The proofs of the equalities (5.2) and (5.3) follow immediately from the
corresponding proofs in [20, Lemma 4.2], once we substitute our character
$\chi$ for the character $\chi_{1}$ in that paper.

As for the equality (5.4), we note the standard formula

$\displaystyle\frac{L^{\prime}(s,\chi)}{L(s,\chi)}=-{\sum}_{n\geqslant
1}\frac{\Lambda(n)\chi(n)}{n^{s}},\quad\quad\mathrm{Re}(s)>1.$

Substituting this infinite sum for the logarithmic derivative appearing in the
definition of $S_{L}$, we obtain the claim by changing variables and moving
the contour of integration to the real line, which is justified due to the
rapid decay of $f$ in combination with the lack of poles of the integrand in
the part of the complex plane enclosed by these contours.

Turning to $S_{d}$, we note that when $\tau$ has imaginary part $-c\log K/\pi$
where $c>0$, the number $-2\pi i\tau/\log K-1$ has real part $-2c-1<0$, and so
we can write

$\displaystyle\frac{d^{\pi i\tau/\log K+1/2}}{d^{2\pi i\tau/\log K+1}-1}$
$\displaystyle=d^{-\pi i\tau/\log K-1/2}{\sum}_{n\geqslant
0}\mathopen{}\mathclose{{}\left(d^{-2\pi i\tau/\log K-1}}\right)^{n}.$

Inserting this into the expression (3.19), we get that

$\displaystyle S_{d}$ $\displaystyle=-\frac{\log d}{\sqrt{d}\log
K}{\int}_{(C)}\exp\mathopen{}\mathclose{{}\left(-\pi i\tau\frac{\log d}{\log
K}}\right){\sum}_{n\geqslant 0}\exp\mathopen{}\mathclose{{}\left(-n\log
d\mathopen{}\mathclose{{}\left(\frac{2\pi i\tau}{\log
K}+1}\right)}\right)f(\tau)\hskip 2.84544pt\mathrm{d}\tau$
$\displaystyle=-\frac{\log d}{\sqrt{d}\log K}{\sum}_{n\geqslant
0}d^{-n}{\int}_{(C)}f(\tau)\exp\mathopen{}\mathclose{{}\left(-2\pi
i\tau(1/2+n)\frac{\log d}{\log K}}\right)\hskip 2.84544pt\mathrm{d}\tau,$

which equals the claimed expression for $S_{d}$ as we can move the contour
$(C)$ to the real line for the usual reasons.

Finally, we turn our attention to $S_{H}$. In the same way as before, we
rewrite the function

$\displaystyle H_{2}^{\prime}(r)=\begin{cases}0&\text{if }d=2,7,\\\ -2\log
2\mathopen{}\mathclose{{}\left(2^{2(2r+1)}-1}\right)^{-1}&\text{otherwise},\end{cases}$

as a geometric series. Letting $r=\pi i\tau/\log K$, we then find that

$\displaystyle H_{2}^{\prime}\mathopen{}\mathclose{{}\left(\frac{\pi
i\tau}{\log K}}\right)=\frac{a\log 2}{2}{\sum}_{n\geqslant
0}4^{-n}\exp\mathopen{}\mathclose{{}\left(-2\pi i\tau\frac{(2n+2)\log 2}{\log
K}}\right),$

where $a=-\mathbbm{1}(d\neq 2,7)$. The equality (5.6) now follows once we
substitute this expression for $H_{2}^{\prime}$ in (3.20) and shift the
contour of integration. ∎

It now follows from (5.2) and (5.4) that

$\displaystyle S_{\zeta}+S_{L}=\frac{1}{\log K}{\sum}_{n\geqslant
1}\frac{\Lambda(n)(\chi(n)-1)}{n}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{\log
n}{\log K}}\right).$

Since those $d$ we are interested in satisfy $d\not\equiv 1$ (mod $4$), we
know that $\chi(n)$ is a quadratic Dirichlet character of modulus $4d=d^{3}$
(in case $d=2$) or modulus $d$ (otherwise). This fact means that for all prime
powers $n$ appearing in the above sum, $\chi(n)=0$ if and only if $n$ is a
power of $d$. Moreover, since $\chi(n)-1=0$ whenever $n$ is the power of a
split prime or an even power of an inert prime, the computation above shows
that

$\displaystyle S_{\zeta}+S_{L}$ $\displaystyle=-\frac{\log d}{\log
K}{\sum}_{n\geqslant
1}d^{-n}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{n\log d}{\log
K}}\right)-\frac{2}{\log K}{\sum}_{n\geqslant 0}\hskip
2.84544pt{\sum}_{\text{$p$ inert}}\frac{\log
p}{p^{2n+1}}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{(2n+1)\log p}{\log
K}}\right).$ (5.7)

Regarding the first infinite sum, we note with the help of Lemma 5.2 that

$\displaystyle\begin{split}-\frac{\log d}{\log K}{\sum}_{n\geqslant
1}d^{-n}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{n\log d}{\log
K}}\right)&=S_{\mathrm{ram}}+\frac{\log d}{\log K}{\sum}_{n\geqslant
0}d^{-n-1/2}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{(2n+1)\log d}{2\log
K}}\right)\\\ &=S_{\mathrm{ram}}-S_{d}.\end{split}$ (5.8)

Similarly, in the case of the second infinite sum, we see that

$\displaystyle\begin{split}&-\frac{2}{\log K}{\sum}_{n\geqslant 0}\hskip
2.84544pt{\sum}_{\text{$p$ inert}}\frac{\log
p}{p^{2n+1}}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{(2n+1)\log p}{\log
K}}\right)\\\ &\quad=S_{\mathrm{inert}}+\frac{2}{\log K}{\sum}_{\text{$p$
inert}}\hskip 2.84544pt{\sum}_{n\geqslant 1}\frac{\log
p}{p^{2n}}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{2n\log p}{\log
K}}\right)\\\ &\quad=S_{\mathrm{inert}}+\frac{2}{\log K}{\sum}_{p\geqslant
3\atop\text{$p$ inert}}{\sum}_{n\geqslant 1}\frac{\log
p}{p^{2n}}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{2n\log p}{\log
K}}\right)\\\ &\quad\quad+\mathbbm{1}\mathopen{}\mathclose{{}\left(\text{$2$
inert}\,}\right)\frac{2\log 2}{\log K}\,{\sum}_{n\geqslant
1}\,2^{-2n}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{2n\log 2}{\log
K}}\right)\\\\[5.0pt]
&\quad=S_{\mathrm{inert}}-S_{A^{\prime}}+\mathbbm{1}\mathopen{}\mathclose{{}\left(2\text{
inert}}\right)\frac{2\log 2}{\log K}{\sum}_{n\geqslant
1}2^{-2n}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{2n\log 2}{\log
K}}\right),\end{split}$ (5.9)

again using Lemma 5.2. Since $2$ is inert in $\mathbb{K}$ if and only if
$d\neq 2,\,7$, we see that when $d\neq 2,\,7$, the last term above is

$\displaystyle\frac{2\log 2}{\log K}{\sum}_{n\geqslant
1}2^{-2n}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{2n\log 2}{\log
K}}\right)$ $\displaystyle=\frac{4}{2}\frac{\log 2}{\log K}{\sum}_{n\geqslant
0}2^{-2n-2}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{(2n+2)\log 2}{\log
K}}\right)$ $\displaystyle=\frac{1}{2}\frac{\log 2}{\log K}{\sum}_{n\geqslant
0}2^{-2n}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{(2n+2)\log 2}{\log
K}}\right)$ $\displaystyle=-S_{H}.$

Since $S_{H}$ furthermore vanishes if $d=2,\,7$, combining (5.7), (5.8), and
(5.9), we therefore obtain that

$\displaystyle
S_{\mathrm{inert}}+S_{\mathrm{ram}}=S_{\zeta}+S_{L}+S_{A^{\prime}}+S_{d}+S_{H}.$
(5.10)

It now follows from (5.1) and (3.3) that our unconditional expression for the
$1$-level density $D\mathopen{}\mathclose{{}\left(\mathcal{F}(K);f}\right)$
agrees with the expression conditional on the Ratios Conjecture, and hence
with the Katz–Sarnak prediction (1.1), if

$\displaystyle S_{\mathrm{split}}\approx S_{J}.$

##  Comparison with the Katz–Sarnak Density Conjecture

The goal of this section is to unify our explicit computation with the
prediction of the Katz–Sarnak Density Conjecture. As we described earlier,
this goal amounts to verifying that the term $S_{\mathrm{split}}$ coming from
the split rational primes is equal to the term $S_{J}$ predicted by the Ratios
Conjecture, at least up to some small error.

We now generalize the result [19, Lemma 2.1] which provides a useful relation
between the angle $\theta_{I}$ and the norm $\mathbb{N}(I)$ of a non-zero
ideal $I\subset\mathcal{O}_{\mathbb{K}}$.

######  Theorem 6.1.

Let $\Lambda=g\mathbb{Z}^{2}\subset\mathbb{R}^{2}$ be a unimodular lattice
with

$\displaystyle g=\mathopen{}\mathclose{{}\left(\begin{matrix}a&0\\\
0&a^{-1}\end{matrix}}\right)\mathopen{}\mathclose{{}\left(\begin{matrix}1&0\\\
n&1\end{matrix}}\right)\in\mathrm{SL}_{2}(\mathbb{R}).$

Let $\ell$ be a line through the origin such that the angle between the
positive $x$-axis and $\ell$ is
$\theta\in\mathopen{}\mathclose{{}\left[0,\pi}\right)$. If $\theta=\pi/2$ (in
which case we let $q:=1$), or if $-n+a^{2}\tan\theta$ is an algebraic number
of degree $q\geqslant 1$, then there exists $C=C(\Lambda,\ell)>0$ such that
for every $\textbf{v}\in\Lambda\setminus\ell$,

$\displaystyle\mathopen{}\mathclose{{}\left|\alpha(\textbf{v})}\right|\geqslant\frac{C}{\|\textbf{v}\|^{q}},$

where $\alpha(\textbf{v})$ denotes the angle between v and $\ell$.

###### Proof.

Let us first note that if $\ell$ is the $y$-axis, then the claim follows
easily: Indeed, the set of first coordinates of lattice points in $\Lambda$ is
discrete, so for any $\textbf{v}\in\Lambda$, if the angle $\alpha(\textbf{v})$
between v and the $y$-axis is non-zero, but small, we have

$\displaystyle
2|\alpha(\textbf{v})|\geqslant|\tan\alpha(\textbf{v})|=\frac{|as|}{\mathopen{}\mathclose{{}\left|(sn+t)/a}\right|}\gg_{a}\frac{1}{\mathopen{}\mathclose{{}\left|(sn+t)/a}\right|}\geqslant\frac{1}{\sqrt{(sn+t)^{2}/a^{2}+a^{2}s^{2}}}=\frac{1}{\|\textbf{v}\|}.$

We can therefore assume that $\theta\neq\pi/2$, so that $\cos\theta\neq 0$.

Similarly, if $\ell$ is the $x$-axis, Liouville’s Theorem [1, Thm. 1.1] and
the assumption that $-n$ is algebraic of degree $q$ imply that
$|sn+t|\gg|s|^{1-q}$ uniformly in
$(s,t)\in\mathbb{Z}^{2}\setminus\mathopen{}\mathclose{{}\left\\{{\bf
0}}\right\\}$, and hence

$\displaystyle 2|\alpha(\textbf{v})|$
$\displaystyle\geqslant|\tan\alpha(\textbf{v})|=\frac{|sn+t|}{a^{2}|s|}\gg_{a}\frac{1}{|a^{q}s^{q}|}\geqslant\frac{1}{\sqrt{a^{2q}s^{2q}+(sn+t)^{2q}/a^{2q}}}=\frac{1}{\|\textbf{v}\|^{q}},$

if $\alpha(\textbf{v})$ is non-zero and small. (Note that, in particular, this
implies that $s\neq 0$ and $sn+t\neq 0$.) We can therefore also assume that
$\theta\neq 0$, so that $\sin\theta\neq 0$.

Let us now rotate $\ell$ and $\Lambda$ clockwise by the angle $\theta$, which
transforms $\ell$ into the $x$-axis and $\Lambda$ into the lattice

$\displaystyle\Lambda^{\prime}=\mathopen{}\mathclose{{}\left\\{\mathopen{}\mathclose{{}\left(\begin{matrix}x\cdot\cos\theta+y\cdot\sin\theta\\\
-x\cdot\sin\theta+y\cdot\cos\theta\end{matrix}}\right):x=as,\,y=(sn+t)/a,\,s,t\in\mathbb{Z}}\right\\}.$

Since the substance of the claim pertains to the situation where
$\alpha(\textbf{v})\neq 0$ is very small, we assume that
$\textbf{v}\in\Lambda^{\prime}$ is any non-zero lattice point with
$\alpha(\textbf{v})$ non-zero, but small. Under this assumption, we have the
estimate

$\displaystyle\begin{split}2\mathopen{}\mathclose{{}\left|\alpha(\textbf{v})}\right|\geqslant\mathopen{}\mathclose{{}\left|\tan\alpha(\textbf{v})}\right|=\frac{|-x\cdot\sin\theta+y\cdot\cos\theta|}{|x\cdot\cos\theta+y\cdot\sin\theta|}=\frac{|y-x\cdot\tan\theta|}{|x+y\cdot\tan\theta|}\gg_{a}\frac{|t-s\cdot\mathopen{}\mathclose{{}\left(a^{2}\tan\theta-n}\right)|}{|x+y\cdot\tan\theta|}.\end{split}$
(6.1)

If $s=0$, we have $x=0$ and $y=t/a\neq 0$, and and the right-hand side of
(6.1) is bounded from below (independently of v) by $|a/\tan\theta|>0$. We can
therefore suppose that $s\neq 0$. In this case, the assumption about
$a^{2}\tan\theta-n$ implies, by Liouville’s Theorem, that there exists
$C=C(\Lambda,\ell)>0$ such that

$\displaystyle\mathopen{}\mathclose{{}\left|t-s\cdot(a^{2}\tan\theta-n)}\right|\geqslant\frac{C}{|s|^{q-1}}.$
(6.2)

Since we have
$|s|^{q-1}\ll_{a}\mathopen{}\mathclose{{}\left(a^{2}s^{2}+(sn+t)^{2}/a^{2}}\right)^{(q-1)/2}=\|\textbf{v}\|^{q-1}$
and

$\displaystyle|x+y\cdot\tan\theta|\leqslant\sqrt{|x+y\cdot\tan\theta|^{2}+|y-x\cdot\tan\theta|^{2}}=\frac{\|\textbf{v}\|}{|\cos\theta|},$

we obtain the claim from (6.1). ∎

Remark. In anticipation of the lemma below, we use Theorem 6.1 to define

$\displaystyle Q:=\max_{0\leqslant m\leqslant 2N-1}\Big{\\{}q\geqslant
1:{\small\begin{array}[]{l}-n(\mathcal{O}_{\mathbb{K}})+a(\mathcal{O}_{\mathbb{K}})^{2}\tan\mathopen{}\mathclose{{}\left(\pi
m/(2N)}\right)\\\ \text{ is algebraic of degree $q$}\end{array}}\Big{\\}},$

where $n(\mathcal{O}_{\mathbb{K}})$ and $a(\mathcal{O}_{\mathbb{K}})$ denote
the parameters appearing in the Iwasawa decomposition of the lattice
$\mathcal{O}_{\mathbb{K}}$ from Lemma 2.1. That is, $Q$ is the largest of the
degrees of all the algebraic numbers
$-n(\mathcal{O}_{\mathbb{K}})+a(\mathcal{O}_{\mathbb{K}})^{2}\tan\mathopen{}\mathclose{{}\left(\pi
m/(2N)}\right)$, where $m=0,\ldots,2N-1$. Note that these numbers are indeed
algebraic: This follows from Lemma 2.1 and from the fact that $\tan(\pi m/2N)$
is algebraic, as $\tan(\pi m)=0$ can be written as a quotient of two
polynomials in $\tan(\pi m/2N)$ with integer coefficients.

In particular, there exists a constant $0<c_{0}<1/4N$, which only depends on
$\mathcal{O}_{\mathbb{K}}$ and $N$, such that for $m=0,\ldots,2N-1$, we have
$|\alpha(\textbf{v})|\geqslant c_{0}/\|\textbf{v}\|^{Q}$, where
$\alpha(\textbf{v})$ denotes the angle between $\ell_{m}$ and
$\textbf{v}\in\mathcal{O}_{\mathbb{K}}\setminus\ell_{m}$.

We can now repeat the argument of Waxman in [20, Section 6] to prove that, at
least when $\alpha:=\sup\mathrm{supp}\,\widehat{f}<1$, the unconditional
asymptotic for the one-level density obtained above is in agreement with the
prediction of the Katz–Sarnak Density Conjecture. Observe that when
$\alpha<1$, this is the case precisely if $S_{\mathrm{split}}$ is very small,
cf. Lemma 4.6.

######  Lemma 6.2.

Suppose that $\alpha<1$. Then
$S_{\mathrm{split}}\ll_{\widehat{f},\varepsilon}K^{\alpha-1+\varepsilon}$.

###### Proof.

Note that the character sum appearing in the definition of
$S_{\mathrm{split}}$ satisfies

$\displaystyle\mathopen{}\mathclose{{}\left|{\sum}_{k=1}^{K}\psi_{k}(I)}\right|=\mathopen{}\mathclose{{}\left|{\sum}_{k=1}^{K}e^{2iNk\theta_{I}}}\right|\leqslant\frac{2}{\mathopen{}\mathclose{{}\left|e^{2iN\theta_{I}}-1}\right|}$
(6.3)

whenever $\theta_{I}$ is not a multiple of $\pi/N$. Also, note that
$\theta_{I}$ can’t even be a multiple of $\pi/2N$, for in that case, if
$I=\langle\beta\rangle$, we have $(\beta/\overline{\beta})^{N}=\pm 1$, and
hence

$\displaystyle\langle\beta\rangle^{N}=\langle\beta^{N}\rangle=\langle\overline{\beta}^{N}\rangle=\langle\overline{\beta}\rangle^{N},$

which forces $\beta\equiv\overline{\beta}$ (mod
$\mathcal{O}_{\mathbb{K}}^{\times}$) due to the unique factorization of ideals
in $\mathcal{O}_{\mathbb{K}}$. It therefore follows that $\langle\beta\rangle$
can’t lie over a split prime, which is a contradiction. Now, if

$\displaystyle\frac{n\pi}{2N}<\theta_{I}<\frac{(n+1)\pi}{2N}$ (6.4)

for some $n=0,\ldots,2N-1$, we have

$\displaystyle\frac{2}{\mathopen{}\mathclose{{}\left|e^{2iN\theta_{I}}-1}\right|}\ll\frac{1}{(n+1)\pi/(2N)-\theta_{I}}+\frac{1}{\theta_{I}-n\pi/(2N)}.$

Since we can always take
$\theta_{I}\in\mathopen{}\mathclose{{}\left(0,\pi}\right)$ because $\pm
1\in\mathcal{O}_{\mathbb{K}}^{\times}$ for all possible values of $d$, we can
define $n\mathopen{}\mathclose{{}\left(\theta_{I}}\right)$ to be the unique
$n\in\mathopen{}\mathclose{{}\left\\{0,\ldots,2N-1}\right\\}$ such that (6.4)
is satisfied. In combination with (6.3), we therefore obtain

$\displaystyle\mathopen{}\mathclose{{}\left|{\sum}_{k=1}^{K}\psi_{k}(I)}\right|\ll\frac{1}{(n(\theta_{I})+1)\pi/(2N)-\theta_{I}}+\frac{1}{\theta_{I}-n(\theta_{I})\pi/(2N)}.$
(6.5)

Now, from the definition of $S_{\mathrm{split}}$ and (6.5) it is clear that

$\displaystyle\begin{split}S_{\mathrm{split}}&\ll\frac{1}{K\log
K}{\sum}_{n=0}^{2N-1}{\sum}_{I\subset\mathcal{O}_{\mathbb{K}}\atop
n(\theta_{I})=n}\hskip
2.84544pt\frac{\Lambda\mathopen{}\mathclose{{}\left(N(I)}\right)}{\sqrt{N(I)}}\cdot\mathopen{}\mathclose{{}\left|\widehat{f}\mathopen{}\mathclose{{}\left(\frac{\log\sqrt{N(I)}}{\log
K}}\right)}\right|\\\
&\quad\times\mathopen{}\mathclose{{}\left(\frac{1}{(n+1)\pi/(2N)-\theta_{I}}+\frac{1}{\theta_{I}-n\pi/(2N)}}\right)\end{split}$
(6.6)

since all the powers $p^{m}$ ($m\geqslant 1$) of a split prime will appear as
norms of suitable ideals $I\subset\mathcal{O}_{\mathbb{K}}$ with
$\theta_{I}\neq 0$.

Since $\mathcal{O}_{\mathbb{K}}$ is a two-dimensional lattice in $\mathbb{C}$,
the basic idea is that the right-hand side of (6.6) can be estimated rather
sharply by replacing it with an integral over certain parts of the ambient
complex plane, where a change to polar coordinates will simplify the integrand
greatly. To specify these domains of integration, for each
$I\subset\mathcal{O}_{\mathbb{K}}$ we let

$\displaystyle
S_{I}\subset\mathopen{}\mathclose{{}\left\\{re^{i\theta}\in\mathbb{C}:n(\theta_{I})\pi/2N+c_{0}<\theta<(n(\theta_{I})+1)\pi/2N-c_{0}}\right\\}$

be a small annulus sector with distance $\gg 1$ to the origin and containing
the generator $\sqrt{\mathbb{N}(I)}e^{i\theta_{I}}$ of the ideal $I$ in the
direction given by $\theta_{I}$. Furthermore, we choose these sectors to have
the same area and so that $S_{I}\cap S_{J}=\varnothing$ if $I\neq J$. Now, we
find that

$\displaystyle\frac{1}{K\log
K}{\sum}_{n=0}^{2N-1}{\sum}_{I\subset\mathcal{O}_{\mathbb{K}}\atop
n(\theta_{I})=n}\hskip
2.84544pt\frac{\Lambda\mathopen{}\mathclose{{}\left(N(I)}\right)}{\sqrt{N(I)}}\mathopen{}\mathclose{{}\left|\widehat{f}\mathopen{}\mathclose{{}\left(\frac{\log\sqrt{N(I)}}{\log
K}}\right)}\right|\mathopen{}\mathclose{{}\left(\frac{1}{(n+1)\pi/(2N)-\theta_{I}}+\frac{1}{\theta_{I}-n\pi/(2N)}}\right)$
$\displaystyle\quad=\frac{1}{K\log
K}{\sum}_{n=0}^{2N-1}{\sum}_{I\subset\mathcal{O}_{\mathbb{K}}\atop
n(\theta_{I})=n}{\int}_{S_{I}}\log
r\mathopen{}\mathclose{{}\left|\widehat{f}\mathopen{}\mathclose{{}\left(\frac{\log
r}{\log
K}}\right)}\right|\mathopen{}\mathclose{{}\left(\frac{1}{(n+1)\pi/(2N)-\theta_{I}}+\frac{1}{\theta_{I}-n\pi/(2N)}}\right)\hskip
2.84544pt\mathrm{d}r\,\mathrm{d}\theta$ $\displaystyle\quad\ll\frac{1}{K\log
K}{\sum}_{n=0}^{2N-1}{\int}_{r=1}^{\infty}\log
r\mathopen{}\mathclose{{}\left|\widehat{f}\mathopen{}\mathclose{{}\left(\frac{\log
r}{\log
K}}\right)}\right|{\int}_{n\pi/2N+c_{0}/r^{Q}}^{(n+1)\pi/2N-c_{0}/r^{Q}}\frac{1}{(n+1)\pi/2N-\theta}\hskip
2.84544pt\mathrm{d}\theta\,\mathrm{d}r$ $\displaystyle\quad\ll\frac{1}{K\log
K}{\int}_{r=1}^{\infty}\log
r\mathopen{}\mathclose{{}\left|\widehat{f}\mathopen{}\mathclose{{}\left(\frac{\log
r}{\log
K}}\right)}\right|{\int}_{c_{0}/r^{Q}}^{\pi/2N-c_{0}/r^{Q}}\frac{1}{\theta}\hskip
2.84544pt\mathrm{d}\theta\,\mathrm{d}r,$

where we used Lemma 6.1 and the remark following it to bound the angular
parameter $\theta$ in terms of $r$. Continuing, we obtain from (6.6) and this
estimate that

$\displaystyle S_{\mathrm{split}}\ll_{Q}\frac{1}{K\log
K}{\int}_{r=1}^{\infty}\big{(}\log
r\big{)}^{2}\mathopen{}\mathclose{{}\left|\widehat{f}\mathopen{}\mathclose{{}\left(\frac{\log
r}{\log K}}\right)}\right|\,\mathrm{d}r\ll_{\widehat{f}}\frac{1}{K\log
K}{\int}_{r=0}^{K^{\alpha}}\big{(}\log r\big{)}^{2}\,\mathrm{d}r\ll
K^{\alpha-1}\log K.$

This concludes the proof. ∎

By the remark after (5.10), we have therefore proved Theorem 1.1.

It would be satisfactory to give a reason why $S_{\mathrm{split}}$ and $S_{J}$
would have anything to do with each other, so that the approximate equality
given by Lemma 6.2 is not just ”accidental.” It seems difficult to give any
such reason, considering the particularly involved arithmetic quality of the
term $S_{J}$, and we have not succeeded in this matter. However, we can make
the following observation, which may or may not be relevant in order to make
further progress: The character sum appearing in $S_{\mathrm{split}}$ is
expressible in terms of the Dirichlet kernel

$\displaystyle D_{K}(x)=\frac{\sin(Kx+x/2)}{\sin
x/2}=1+2\,{\sum}_{k=1}^{K}\cos(kx).$

More precisely, we have the following result.

######  Proposition 6.3.

Suppose $\mathfrak{q}_{1}$ and $\mathfrak{q}_{2}$ are different prime ideals
in $\mathcal{O}_{\mathbb{K}}$ lying over a rational prime $p$, and let
$n\geqslant 1$. Then we have

$\displaystyle{\sum}_{k=1}^{K}\Big{(}\psi_{k}(\mathfrak{q}_{1}^{n})+\psi_{k}(\mathfrak{q}_{2}^{n})\Big{)}=-1+D_{K}\big{(}2Nn\theta_{\mathfrak{q}_{1}}\big{)},$

where
$\theta_{\mathfrak{q}_{1}}\in\mathopen{}\mathclose{{}\left(0,\pi}\right)$
denotes the argument of a generator of $\mathfrak{q}_{1}$ in the upper half-
plane.

###### Proof.

Suppose that $z=a+ib$ is any generator of $\mathfrak{q}_{1}$. Since
$\bigcap_{d}\mathcal{O}_{\mathbb{K}}^{\times}=\mathbb{Z}/2\mathbb{Z}$, we can
assume that $b>0$. Moreover, since the generator of the ideal
$\mathfrak{q}_{1}\mathfrak{q}_{2}$ is a rational prime, the conjugate
$\overline{z}=a-ib$ must be a generator of $\mathfrak{q}_{2}$. Therefore,

$\displaystyle\psi_{k}(\mathfrak{q}_{1}^{n})+\psi_{k}(\mathfrak{q}_{2}^{n})$
$\displaystyle=\mathopen{}\mathclose{{}\left(\frac{z}{\overline{z}}}\right)^{Nnk}+\mathopen{}\mathclose{{}\left(\frac{\overline{z}}{z}}\right)^{Nnk}$
$\displaystyle=\exp\big{(}Nnk\log\big{(}z/\,\overline{z}\big{)}\big{)}+\exp\big{(}\\!-\\!\\!Nnk\log\big{(}z/\,\overline{z}\big{)}\big{)}$
$\displaystyle=2\cos\big{(}Nink\log\big{(}z/\,\overline{z}\big{)}\big{)}$
$\displaystyle=2\cos\mathopen{}\mathclose{{}\left(Nink\log\frac{a/b+i}{a/b-i}}\right)$
$\displaystyle=2\cos\mathopen{}\mathclose{{}\left(2Nnk\cdot\frac{1}{2i}\log\frac{a/b-i}{a/b+i}}\right)$
$\displaystyle=2\cos\big{(}2Nnk\cdot\arctan(b/a)\big{)},$

where we used the identity

$\displaystyle\arctan\frac{1}{x}=-\frac{1}{2i}\log\frac{x-i}{x+i},$

valid for all $x\neq 0$. Moreover, since the angle $\theta_{\mathfrak{q}_{1}}$
between $1$ and $a+ib$ is given by

$\displaystyle\theta_{\mathfrak{q}_{1}}=\begin{cases}\arctan(b/a)&\text{if
}a>0,\\\ \arctan(b/a)+\pi&\text{if }a<0,\end{cases}$

the computation above and the periodicity of cosine show that

$\displaystyle\psi_{k}(\mathfrak{q}_{1}^{n})+\psi_{k}(\mathfrak{q}_{2}^{n})=2\cos\big{(}2Nnk\theta_{\mathfrak{q}_{1}}\big{)}.$

It now follows that

$\displaystyle{\sum}_{k=1}^{K}\mathopen{}\mathclose{{}\left(\psi_{k}(\mathfrak{q}_{1}^{n})+\psi_{k}(\mathfrak{q}_{2}^{n})}\right)=2{\sum}_{k=1}^{K}\cos\big{(}2Nnk\theta_{\mathfrak{q}_{1}}\big{)}=-1+D_{K}\big{(}2Nn\theta_{\mathfrak{q}_{1}}\big{)},$

which completes the proof. ∎

Making use of this observation, we can then rewrite $S_{\mathrm{split}}$ as

$\displaystyle S_{\mathrm{split}}$ $\displaystyle=\frac{1}{K\log
K}{\sum}_{(-d/p)=1}\hskip 2.84544pt{\sum}_{n\geqslant 1}\frac{\log
p}{p^{n/2}}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{n\log p}{2\log
K}}\right)$ $\displaystyle\quad\quad-\frac{1}{\log K}{\sum}_{(-d/p)=1}\hskip
2.84544pt{\sum}_{n\geqslant 1}\frac{\log
p}{p^{n/2}}\widehat{f}\mathopen{}\mathclose{{}\left(\frac{n\log p}{2\log
K}}\right)\frac{D_{K}\big{(}2Nn\theta_{p}\big{)}}{K},$

where $\theta_{p}\in(0,\pi)$ denotes the argument of one of the generators of
either of the prime ideals lying over $p$. Thus, the primary benefit of Lemma
6.3 is that it allows one to express the contribution of the split primes as a
sum of two terms, among which the first one can be understood with partial
summation, and the other one involves a well-studied function in a normalized
form (due to the factor $1/K$). At any rate, the presence of the angles
$\theta_{p}$ in the last contribution seems to be quite a difficult obstacle.
A better understanding of these angles is likely a crucial step if one wishes
to understand the relation $S_{\mathrm{split}}\approx S_{J}$ on a deeper
level.

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