The low-lying zeros of L-functions
associated to non-Galois cubic fields
Master’s thesis in Mathematics
Victor Ahlquist
 
Department of Mathematical Sciences
UNIVERSITY OF GOTHENBURG
Gothenburg, Sweden 2023

Master’s thesis 2023
The low-lying zeros of L-functions
associated to non-Galois cubic fields
Victor Ahlquist
Department of Mathematical Sciences
University of Gothenburg
Gothenburg, Sweden 2023
The low-lying zeros of L-functions associated to non-Galois cubic fields
Victor Ahlquist
© Victor Ahlquist, 2023.
Supervisor: Anders Södergren, Department of Mathematical Sciences
Examiner: Julia Brandes, Department of Mathematical Sciences
Master’s Thesis 2023
Department of Mathematical Sciences
University of Gothenburg
iv
Abstract
We study the low-lying zeros of Artin L-functions associated to non-Galois cubic number
fields through their one- and two-level densities. In particular, we find new precise estimates
for the two-level density with a power-saving error term. We apply the L-functions Ratios
Conjecture to study these densities for a larger class of test functions than unconditional
computations allow. By reviewing a known Ratios Conjecture prediction, due to Cho,
Fiorilli, Lee, and Södergren, for the one-level density, we isolate a phase transition in the
lower-order terms, which reveals a striking symmetry. Our computations show that the
same symmetry exists in the one-level density of several other families, that have previously
been studied in the literature, and this motivates us to formulate a conjecture extending one
part of the Katz–Sarnak prediction for families of symplectic symmetry type. Moreover, we
isolate several phase transitions in the lower-order terms of the two-level density. To the
best of our knowledge, this is the first time such phase transitions have been observed in
any n-level density with n ≥ 2.
Keywords: Mathematics, number theory, cubic fields, L-functions, low-lying zeros, one-
level density, two-level density, phase transition.
v

Acknowledgements
I would like to express my gratitude to my supervisor Anders Södergren. First for intro-
ducing me to the fascinating subject of low-lying zeros of L-functions and sharing your
knowledge. Second, for taking your time to carefully read through every line of this thesis
and pointing out every imprecision.
I would also like to thank my friends and family. Your support gave me the endurance
necessary to finish this project.
Victor Ahlquist, Gothenburg, February 2023
vii

Contents
1 Introduction 1
1.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Results and outline of the report . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 Preliminaries 5
2.1 The Dedekind zeta-function of a non-Galois cubic field . . . . . . . . . . . . . 5
2.2 General L-functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.3 The low-lying zeros of L-functions . . . . . . . . . . . . . . . . . . . . . . . . 8
2.4 The low-lying zeros of L(s, fK) . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3 The one-level density 11
3.1 The explicit formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3.2 Summing over K . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.3 Interpreting Theorem 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.4 An application to counting cubic fields . . . . . . . . . . . . . . . . . . . . . . 18
4 The two-level density 19
4.1 The explicit formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.2 A refined counting function for the number of cubic fields . . . . . . . . . . . 20
4.3 Finding the two-level density . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
4.3.1 Summing C1,KC2,K over K . . . . . . . . . . . . . . . . . . . . . . . . 22
4.3.2 Summing Cℓ,KDi,K over K . . . . . . . . . . . . . . . . . . . . . . . . 22
4.3.3 Summing D1,KD2,K over K . . . . . . . . . . . . . . . . . . . . . . . . 23
4.4 Interpreting Theorem 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
5 Counting cubic fields 27
5.1 Cubic rings and forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
5.2 Counting integral binary cubic forms . . . . . . . . . . . . . . . . . . . . . . . 32
5.2.1 The group GL2(R) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
5.2.2 Reducing to an integral . . . . . . . . . . . . . . . . . . . . . . . . . . 34
5.2.3 Integrating over GL2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
5.3 Sieving for maximality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
5.4 The secondary term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
6 The one-level density through the Ratios Conjecture 41
6.1 The Ratios Conjecture recipe . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
6.2 The one-level density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
6.3 Averages over K . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
6.4 Finding a differentiated estimate . . . . . . . . . . . . . . . . . . . . . . . . . 47
6.5 Estimating R′2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
6.6 Finding the one-level density . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
ix
Contents
7 Interpreting the Ratios Conjecture prediction for the one-level density 53
7.1 Two expressions for the one-level density . . . . . . . . . . . . . . . . . . . . . 53
7.2 A phase transition in the one-level density . . . . . . . . . . . . . . . . . . . . 56
7.2.1 An integral of Gamma functions . . . . . . . . . . . . . . . . . . . . . 57
7.2.2 Estimating prime sums . . . . . . . . . . . . . . . . . . . . . . . . . . 58
7.2.3 Estimating an integral . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
7.2.4 Completing the proof of Theorem 7.3 . . . . . . . . . . . . . . . . . . 61
7.3 A symplectic phenomenon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
7.3.1 Quadratic Dirichlet characters . . . . . . . . . . . . . . . . . . . . . . 62
7.3.2 Hecke characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
7.3.3 A polynomial analogue . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
8 The Ratios Conjecture prediction for the two-level density 65
8.1 Sums over K . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
8.2 Estimating R′1,1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
8.3 Finishing the estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
8.4 Finding the two-level density . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
9 Interpreting the Ratios Conjecture prediction for the two-level density 79
9.1 Two expressions for the two-level density . . . . . . . . . . . . . . . . . . . . . 79
9.2 Phase transitions in the two-level density . . . . . . . . . . . . . . . . . . . . 84
Bibliography 93
A Infinite products, the gamma function and integration I
A.1 Infinite products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I
A.2 The Gamma function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II
A.3 Stieltjes integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II
x
1
Introduction
We begin by introducing the study of L-functions, with a particular focus on their low-lying
zeros. Next, we describe the outline of the report and present our results. The chapter ends
with a brief description of notation that we will be using in the sequel.
1.1 Background
An L-function is a meromorphic function associated to a mathematical object. More pre-
cisely, given complex numbers an related to the object of interest, we may define a corre-
sponding generating function as the series
∑∞ an
, s ∈ C,
=1 n
s
n
as long as the series converges. Series of this form are known as Dirichlet series, and all
L-functions take the form of such a series, at least when the real part of s is greater than 1.
Aside from being given by a Dirichlet series, all L-functions have a meromorphic continuation
to the entire complex plane, along with a functional equation. The functional equation
relates the value of the L-function at a point s to the value of the dual L-function at the
point 1−s. Here, the dual L-function is simply the function obtained by taking the complex
conjugate of the coefficients an in the Dirichlet series above. In particular, the self-dual
L-functions are precisely those with real coefficients an.
One of the most well-known L-functions is the Riemann zeta function ζ, associated to the
integers and initially defined by the Dirichlet series where one takes all an to equal 1. From
classical analytic number theory, we know that the zeros of ζ encode information about the
distribution of the primes. To make this more precise, we define a function π(x), for real
x ≥ 2, counting the number of primes less than x. Then by studying the zeros of ζ, one can
prove the Prime Number Theo∫rem, i.e. thatx 1 ( )
π(x) = log dt+O x exp(−c(log x)
1/2) ,
2 t
for a specific constant c > 0, see e.g. [D3, Ch. 18]. Moreover, the value of the integral above
is approximately equal to x/(log x), where log x denotes the natural logarithm of x.
The celebrated Riemann Hypothesis asserts that all the so-called nontrivial zeros of the
Riemann zeta function lie on the critical line, defined as the set {s ∈ C : Re(s) = 1/2}. The
Riemann Hypothesis has several important implications. In(√particul)ar, it implies that the
error term in the Prime Number Theorem can be taken as O x log x , which is a significant
improvement compared to the error term above.
The success of using the zeros of ζ to study primes indicates that one may study an object
by studying the zeros of its associated L-function. We remark that all L-functions have so-
called trivial zeros, which are very easy to locate. The Generalised Riemann Hypothesis
(GRH) asserts that the nontrivial zeros of any L-function all lie on the critical line. If one
believes in the GRH, then the point s = 1/2 seems to be special, in a sense being the
midpoint of the critical line with respect to the transformation s 7→ 1 − s. Even without
assuming the GRH, the point 1/2 is the only fixed point with respect to s 7→ 1 − s, which
1
1. Introduction
indicates that it may have important properties. Indeed, The Birch and Swinnerton-Dyer
Conjecture claims that this point is of special importance, at least for certain L-functions.
Specifically, the conjecture asserts that if one considers the L-function associated to an
elliptic curve, then the multiplicity of a potential zero at the point s = 1/2 is equal to the
rank of the group of rational points on the elliptic curve.
Instead of directly studying zeros of L-functions at s = 1/2, it may be easier to study
the zeros lying close to this point. Such zeros are known as low-lying zeros. Furthermore,
instead of studying the zeros of a single L-function, one can study the zeros of an entire
collection, or family, of L-functions. There are many examples of families of L-functions,
e.g. L-functions associated to certain elliptic curves, modular forms, or number fields. The
distribution of the low-lying zeros of such families of L-functions can be analysed by studying
a certain average, known as the n-level density, where n ≥ 1 is an integer.
A conjecture of Katz and Sarnak asserts that every natural family is associated to one of
five symmetry types, which determines the main term of the n-level density for any n. The
five different symmetry types are unitary, symplectic, and one of three different orthogonal
types. The terminology comes from random matrix theory, where the same main terms
arise if one studies the eigenvalues close to 1 of large random matrices. We remark that
this conjecture has been partially confirmed in several different families, see e.g. [CK], [HR],
[ILS], [ÖS], [Rb], [Ya], [Yo].
We will be interested in studying the n-level densities of the family F of L-functions
associated to non-Galois cubic number fields, for n = 1, 2. The one-level density for this
family was first studied in [Ya], who showed that the main term is the one given by the
symplectic symmetry type. The main terms of the n-level densities for n ≥ 2 were found in
[CK], and as expected they were also of symplectic type. In [CFLS, Ch. 3], the one-level
density was evaluated precisely, taking into account not only the main term but also several
lower-order terms. We remark that in all of these cases, the n-level density was found by
relating it to certain prime sums. Unfortunately, this method leads to an error term that
can become quite large.
The Ratios Conjecture gives a method for evaluating certain averages of ratios of L-
functions. We describe this conjecture more closely in Chapter 6. By using this conjecture
to compute certain integrals, one can evaluate the n-level density by relating it to new prime
sums that are more well-behaved compared to the prime sums that arise in the unconditional
calculations. In particular, one obtains an error term of substantially better quality. This
was done for the one-level density, once again taking into account many lower-order terms,
in [CFLS, Ch. 4-5]. Aside from its application to computing the n-level density, the Ratios
Conjecture has been successfully applied to finding moments of zeta functions, and it can
also be used to study the Montgomery pair-correlation conjecture, see e.g. [CS].
The key to studying the n-level densities of our family F , is a collection of precise
estimates for the number of isomorphism classes of certain cubic fields, with discriminant
bounded by a given magnitude. The main terms in these estimates were originally found in
[DH], by relating cubic fields to certain binary cubic forms. Through recent breakthroughs,
secondary terms necessary for the precise calculations in [CFLS] have been found in [BST]
and [TT] independently. These proofs combine analytic and algebraic methods, and all begin
by counting the corresponding cubic forms. Work has been done to improve the error terms
in the estimates and the current best result is due to [BTT]. In the opposite direction, lower
bounds for the error terms have been found in [CFLS] by studying the one-level density
of the family F , conditional on the GRH. This demonstrates how the zeros of L-functions
provide information about the corresponding mathematical object, not only in the classical
case of the Riemann zeta function.
1.2 Results and outline of the report
In Chapter 2, we begin by giving a more technical introduction to the subject, as well as
describing useful properties of the L-functions we will be studying. The actual computations
begin in Chapter 3, where we study the one-level density. The approach and all results here
2
1. Introduction
are taken from [CFLS, Ch. 3]. Moving on to Chapter 4, we obtain our first new result
by extending the approach of the previous chapter to find an expression for the two-level
density containing lower-order terms. The result is given in Theorem 4.1, and this is the
first time an expression containing the lower-order terms has been found for the two-level
density of this family.
We take a brief pause from the study of low-lying zeros in Chapter 5 to sketch the proof
from [BST] of the main term in the counting function for cubic fields. Compared to the
previous sections, the focus here is not to keep track of error terms, but rather to indicate
the methods used for studying this counting function, as well as explain the shape of the
main term. We remark that this chapter requires more algebraic knowledge compared to
the other sections, where complex analysis is our main tool.
The rest of the report is concerned with applying the Ratios Conjecture. We begin in
Chapter 6 by using it to find the one-level density. As in the unconditional calculations,
the approach and all results are from [CFLS]. In Chapter 7, we study the expression for
the prediction of the one-level density obtained in the previous chapter. First, we follow
[CFLS, Ch. 5] and compare this expression with the one obtained in Chapter 3. Next, by
studying the Ratios Conjecture prediction more closely, we obtain new results. In particular,
we find a so-called phase transition in the secondary term, and a curious symmetry, see the
remarks following Theorem 7.3. Phase transitions have been found in other families, see e.g.
[FPS2], [Rk], [Wa], but the symmetry that we observe seems to have gone unnoticed in the
literature. Together with a brief analysis of the one-level density of several other families,
this symmetry leads us to formulate Conjecture 7.4 for a general symplectic family.
The last two chapters of the report contain our main contributions to the study of the
two-level density. First, in Chapter 8, we apply the Ratios Conjecture to evaluate the two-
level density and find an expression with a power-saving error term, given in Proposition
8.4. Next, in Chapter 9, we compare the expression we found conditional on the Ratios
Conjecture with the unconditional expression from Chapter 4, and find that they agree
quite well, see Proposition 9.1. We also explicitly find the second-order term of the Ratios
Conjecture prediction and uncover several phase transitions in the lower-order terms of the
two-level density, conditional on the Ratios Conjecture, see Theorem 9.3. To the best of
our knowledge, this is the first time a phase transition has been observed in the lower-order
terms of any n-level density with n ≥ 2.
1.3 Notation
We briefly describe some of the notation that will be used throughout the report. First, as
is usual in number theory, the letters p and q always denote primes. Sums, with index p or
q, should be interpreted as sums over the set of primes with the natural ordering. Usually,
these sums will be absolutely convergent, so that the ordering does not matter. Sometimes
we will write the condition pe || m, for some integer m, in our sums or products. This means
that we only consider the e ≥ 0 such that pe is the largest power of p dividing m.
We will sometimes want to study arithmetic functions, i.e. functions f : Z+ → C. The
multiplicative arithmetic functions will be of special interest, which are the functions with
f(1) = 1, and where f(nm) = f(n)f(m) for relatively prime integers n and m. Note that a
multiplicative function is completely determined by its values at prime powers.
Let logN denote the natural logarithm of the number N . Then, Euler’s constant γ is
defined by the limit ∑ 
:= lim  1γ − logN ,
N→∞ n
n≤N
and we will see that this constant naturally arises from certain prime sums, and also when
evaluating expressions related to the gamma function.
If f : R→ C is an absolutely integrable function, say, then we define its Fourier transform
3
1. Introduction
by ∫ ∞
f̂(u) := f(x)e−2πiuxdx.
−∞
The inverse Fourier transform is defined similarly, but with the sign of the exponent reversed.
In particular, for even functions f , the inverse Fourier transform of f is the same as the
Fourier transform of f .
When stating our results, or performing calculations, we will make use of both the big-
O notation and Vinogradov’s notation. We remind the reader that for a complex-valued
function f and a nonnegative function g, we write
f(x) = O(g(x)),
if there exists a constant C such that |f(x)| ≤ Cg(x) for all x in a set S. The constant C is
often referred to as the "implied constant". The set S is usually not specified, but if x is a
real variable, then S can usually be taken to be the set of all sufficiently large real numbers,
often the set of all x ≥ 2. If x is a complex variable, then S will usually be specified more
explicitly. Furthermore, we write
f(x) = h(x) +O(g(x)), when f(x)− h(x) = O(g(x)).
Sometimes it will be more convenient to use Vinogradov’s notation and write f(x) ≪ g(x)
when f(x) = O(g(x)). If the implied constant depends on a variable, say ϵ, then this variable
will be indicated by a subscript, i.e. as Oϵ or ≪ϵ.
Finally, we will make occasional use of little-o notation. If g(x) > 0 is a positive function,
then by f(x) = o(g(x)), we mean that the quotient f(x)/g(x) converges to 0 as x tends to
some limit a. The limit a can usually be inferred from the context and is often equal to
either 0 or ∞.
4
2
Preliminaries
We introduce the Dedekind zeta function, and the Artin L-function of a cubic field. The
latter function will be our main object of study throughout the report. Next, we give a more
technical presentation of the low-lying zeros of L-functions than in the introduction, and of
the Katz-Sarnak prediction. Lastly, we define the one-level density, which will be studied in
the next chapter.
2.1 The Dedekind zeta-function of a non-Galois cubic
field
Let K be a field extension of Q. We say that K is a number field if its dimension [K : Q]
as a vector space over Q is finite. One may consider such a field to be a subfield of C, and
we will usually do so. To every number field, we may consider the associated Galois group
G(K/Q), which consists of all field automorphisms of K fixing Q pointwise. One can show
that the Galois group of a number field is always finite and that its cardinality divides the
dimension [K : Q]. If the cardinality of the Galois group is equal to [K : Q], we say that
the field extension is Galois. Every number field K is contained in a unique smallest field
L, the Galois closure of K in C, such that Q ⊆ K ⊆ L ⊆ C, where L is Galois over Q.
Furthermore, G(L/Q) is isomorphic to a subgroup of the symmetric group Sn, n := [K : Q].
The discriminant DK of a number field K will be a central object in our future studies.
To define it we need some preparation, taken from [N, Ch. I]. In a number field K, one may
consider the set OK of all integral elements. Here, an element x ∈ K is called integral if
it is the root of some polynomial with integer coefficients. It turns out that OK is a free
Z-module of dimension n := [K : Q], and thus has a Z-basis {α1, ..., αn} of integral elements.
One can then define
DK := det((TrK(αiαj))i,j=1,...,n),
where Tr(x), x ∈ K, denotes the trace of the linear map Tr : K → K given by y →7 xy.
Without much effort, one can show that the discriminant is a nonzero integer, independent
of the choice of basis. With more effort, one can show that there are only finitely many
number fields whose discriminant is of bounded magnitude, and that Q is the only field
whose discriminant is strictly less than 2 in absolute value [N, Thms. III.2.16, III.2.17].
We will be interested in studying L-functions associated to cubic number fields. It turns
out that the non-Galois fields are the most common, see e.g. [C], and thus we decide to focus
our efforts on these fields, following [CFLS]. By the discussion above, the Galois extensions
have Galois group isomorphic to the cyclic group C3, while the non-Galois extensions have
a Galois group only containing the identity element. We may sometimes refer to the non-
Galois extension fields as S3-fields as the Galois group of their Galois closures is isomorphic
to S3. For the reader unfamiliar with the results about fields stated so far, we refer to any
introductory text on Galois theory.
We will now introduce the Dedekind zeta-function ζK associated to a number field K.
We begin with the special case ζQ, or simply ζ, the Riemann zeta-function, defined by the
equation ∑∞
ζQ(s) =
1
,
ks
k=1
5
2. Preliminaries
for complex s with Re(s) > 1. From the classical theory of the Riemann zeta-function, see
e.g. [D3], we know that ζ(s) can be meromorphically continued to the entire complex plane,
with a simple pole at s = 1 with residue equal to 1.
In the general case, we define ∑
( ) = 1ζK s ( ) , (2.1.1)N I s
I
for s with Re(s) > 1. Here, the sum is over all ideals I contained in the ring of integers
OK ,∏and N(I) denotes the∏ir norm, which is defined as the cardinality of the ring OK/I.The norm is completely multiplicative in the sense that if one factorises I into prime idealsℓ ℓ
I = j=1 pj , then N(I) = j=1N(pj), where ℓ is some positive integer. We also point out
that the norm of an ideal generated by a rational integer has a particularly simple expression.
If k ∈ Z, then N((k)) = kn, where again n = [K : Q]. For an introduction to the ring of
integers, ideals, and prime factorisation in number fields, see e.g. [N, Ch. 1].
The summation in (2.1.1) has no specific ordering attached to it, which leads us to the
question of convergence. We will show that for Re(s) > 1, the series is absolutely convergent
so that the ordering does not matter. Indeed, we may formally rewrite the series as
∑∞ |{I ⊆ OK : N(I) = k|}
.
ks
k=1
To estimate this expression we need a few facts from algebraic number theory. First, from
the proof of [N, Theorem I.3.1] it follows that every nonzero prime ideal p divides some ideal
(p), where p is a rational prime. Thus, by multiplicativity p | N(p) | pn, n = [K : Q]. In
particular, this implies that (p) splits into at most n prime factors. Combined with the fact
that every integral ideal admits a unique factorisation into prime ideals, this is enough to
show that the number |{I ⊆ OK : N(I) = k|} is finite. To show convergence we follow an
approach similar to [La, Ch. 8.2].
For Re s > 1, consider the infinite product
∏ ∏(∑∞ )
(1−N(p)−s)−1 = N(p)−es ,
p p e=0
where p runs over the prime ideals in OK . For an introduction to infinite products, and
some results concerning their convergence, see [A, Ch. 5.2.2], or the summary in Appendix
A.1. We remark that a product over primes such as the one above is often referred to as an
Euler product. Formally, multiplying out the second expression for the products yields the
series defining ζK(s) using multiplicativity of the norm. This formal manipulation can be
justified using the absolute convergence of the product, and the fact that a finite product of
absolutely convergent series is itself an absolutely convergent series.
We now show that the product is absolutely convergent, which follows if we show
∑∑∞ 1
Re( <∞.N(p)e s)
p e=1
Order the sum by which rational prime p that p divides, and use N(p) ≥ p to bound the
sum from above by ∑∑∞ ∞n ∑ n
pRe(s)
≤ Re <∞,
=1 k
s
p e k=1
for Re(s) > 1 as desired. In fact, the argument above also implies that the convergence is
uniform on any compact subset of Re(s) > 1, and thus that ζK(s) is holomorphic for such
s as a uniform limit of holomorphic functions. Also, as no factor in the infinite product is
0 and the product converges absolutely, we find that ζK(s) ̸= 0 for Re(s) > 1. We will not
provide a proof, but it is possible to show that ζK can be meromorphically continued to C
with a simple pole at s = 1 [N, Cor. VII.5.11].
6
2. Preliminaries
Now we specialize to the case when K is a non-Galois cubic number field. The discussion
below is essentially contained in [Ya, Ch. 2.2.2]. It turns out that we may write ζK(s) =
ζ(s)L(s, fK), for a certain function L(s, fK). Indeed, this can be shown by considering the
Euler product above. First, by lettin∏g K = Q above, we see that ζ(s) has the productexpansion
(1− p−s)−1.
p
Now, ζK(s) has a product expansi∏on ∏
(1−N(p)−s)−1. (2.1.2)
p p|(p)
In particular, this shows that the Dedekind zeta function is completely determined by the
splitting behaviour of the rational primes (p) in K. To find how many ways a rational prime
can split over a cubic field, we consider its norm. If (p) = p1...pk, then p3 = N((p)) =
N(p1)...N(pk). From the definition, we see that the norm of a prime ideal is strictly greater
than 1 and thus k ≤ 3. More precisely N(p ) = prii and thus by considering the exponents
we see that a prime p can split in five different ways in K.
To distinguish between the five different cases we assign every splitting type a symbol
out of (111), (21), (3), (121), and (13), where the symbols correspond to the splittings (p) =
p1p2p
2 3
3, p1p2, p1, p1p2 and p1 respectively. We will also use the symbols T1,T2,..., T5 to refer
to these splitting types. Here all pi are distinct primes. In the (111) case we see that all
prime ideals have norm p, while in the (21) case, one has norm p2 and the other norm p.
The rest of the splitting types can b∏e analysed similarly.This allows us to find
(1−N(p)−s)−1,
p|(p)
as a function of the splitting type of (p), and s. We call this the local factor of ζK at p.
It turns out that every such local factor is a product of the local factor of ζ at p, with
some other factor. This allows us to factor ζK into a product of ζ, and another function
which we will call L(s, fK). It turns out that L(s, fK) is easier than ζK to study, and we
therefore take the time to find some of the properties of L(s, fK). We first remark that we
will consider the symbol L(s, fK) simply as notation for the function defined by ζK/ζ, but it
can also be given a separate meaning as the L-function associated to a certain automorphic
form fK . Alternatively, L(s, fK) is the Artin L-function associated to the unique irreducible
two-dimensional representation of the Galois group, which is isomorphic to S3, of the Galois
closure of K. We will not go into this further.
The table below summarises the different local factors we can obtain. In the third column
we have removed the factor coming from ζ. Here, ω is a primitive third root of unity.
Splitting type Local factor in ζK Local factor in ζK/ζ
T −s −3 −s −21 := (111) (1− p ) (1− p )
T := (21) (1− p−s)−1(1− p−2s)−1 (1− p−2s)−12
T3 := (3) (1− p−3s)−1 (1− ωp−s)−1(1− ω2p−s)−1
T := (121) (1− p−s)−2 (1− p−s)−14
T5 := (13) (1− p−s)−1 1
2.2 General L-functions
The function ζK turns out to be an L-function. There is no precise definition of what
constitutes an L-function, but we can give some important properties that many L-functions
satisfy, see also [IK, Ch. 5.1].
First, as mentioned in the introduction, every L-function should be representable by an
absolutely convergent Dirichlet series for R∑e(s) > 1, that is a series of the form∞ an
,
ns
n=1
7
2. Preliminaries
for some coefficients an. This has the consequence that L-functions are holomorphic for
Re(s) > 1, as a locally uniform limit of the partial series. Furthermore, this Dirichlet series
should in turn be representable by an absolutely convergent Euler product, in the same
domain, i.e. a product o∏f t(he form − )−1 ( )1− α (p)p s1 · ... · 1− αd(p)p−s −1 ,
p
for some |αi(p)| < p. The number d is called the degree of the L-function.
It should be possible to meromorphically continue an L-function to the entire complex
plane C, with the only possible pole at s = 1. Furthermore, an L-function should satisfy a
functional equation, that is an equation relating the values of the L-function at some s ∈ C,
to the values at 1− s of either the same L-function, or of the dual L-function, mentioned in
the introduction. For the self-dual L-functions, the functional equation provides a symmetry
about the point s = 1/2. An example of a functional equation, relevant to our purposes,
is given below in (2.4.1). Lastly, every L-function has a conductor, which is related to
the functional equation. We will not go into the general definition of a conductor, but we
mention that L(s, fK) has the conductor |DK |.
We have seen that the Dedekind zeta function has a Dirichlet series representation, as
well as an Euler product. One can show that it also obeys a functional equation. The same
also holds for L(s, fK), which indicates that both L(s, fK) and ζK are L-functions, and this
is indeed the case. Furthermore, it can be shown that L(s, fK) is an entire function.
The Euler product representation of an L-function ensures that it has no zeros with a
real part strictly larger than 1, as an absolutely convergent product of nonzero numbers is
nonzero. The functional equation then implies that an L-function does not have any zeros
with a real part strictly less than 0, except for possible trivial zeros, which are a consequence
of other factors present in the functional equation. The nontrivial zeros are thus all located
in the critical strip, i.e. the strip in C where the real part is neither smaller than 0, nor
larger than 1. In the introduction, we already mentioned the utility of studying zeros of
L-functions, and in particular why the low-lying zeros are of interest. This brings us to the
topic of the next section.
2.3 The low-lying zeros of L-functions
Any fixed L-function has a fixed set of zeros, and in particular, has a finite number of such
zeros lying within distance, say 1, of the point 1/2. Thus, it is uninteresting to study the
distribution of the low-lying zeros of a single L-function, and one should instead consider
an entire family of L-functions. One such family is the family of Dedekind zeta functions
associated to S3-cubic fields introduced earlier in this chapter.
Katz and Sarnak conjectured that every natural family of L-functions has one of five
symmetry types, which determines the distribution of the low-lying zeros, see [KS]. This
conjecture is motivated by analogues in the study of the eigenvalues lying close to 1 of
random matrices, and by analogues in the study of the central zeros of certain families of
polynomials associated to so-called function fields. What constitutes a natural family is
a deep question, see e.g. [SST], which we will not investigate further here. To make the
conjecture more precise, we need some notation from [KS], which we modify slightly.
Let F be a family of L-functions, and ϕ : R → R a Schwartz function. Here, a
Schwartz function is an infinitely differentiable function with derivatives admitting the bound
ϕ(k)(x)≪A,k (1+|x|)−A. For every L-function L(s, f) ∈ F we let cf denote the conductor of
L(s, f). We assume that all sets FX := {L(s, f) ∈ F : cf ≤ X} are finite. The distribution
of low-lying zeros can then be studied∑thr∑ough (the one-lev)el density, defined as1 γf log cf
ϕ 2 , (2.3.1)|FX | π
cf≤X γf
where the inner sum is over all nontrivial zeros ρf = 1/2 + iγf of L(s, f). If the GRH is
true for the family F , then all γf ∈ R so that the expression above is well-defined for ϕ with
8
2. Preliminaries
domain R. As ϕ decays fast, the presence of cf in the argument of ϕ ensures that only the
zeros ρ close to 1/2 give a significant contribution to the one-level density. We remark that
even if one does not assume the GRH, it can still make sense to study the expression above,
as long as ϕ can be extended to a function defined in the horizontal strip {|Im(s)| ≤ 1/2},
decaying quickly.
The conjecture of Katz and Sa∫rnak is that as X → ∞, the one-level density convergesto one of the integrals
ϕ(x)w(G(F))(x)dx.
R
Here G(F) denotes the symmetry type of F . The five symmetry types (or groups) are
unitary, orthogonal, special orthogonal (even), special orthogonal (odd), and symplectic.
The expression w(G(F))(x) denotes a density function associated to the symmetry type
G(F). In particular, it will be important later that the symplectic density is given by
1− sin 2πx2 .πx
This choice of naming comes from the random matrix theory analogue, mentioned earlier,
where Katz and Sarnak studied the eigenvalues of random matrices from the groups men-
tioned above. They then found that the distribution of these eigenvalues converged to the
integrals given above if one let the dimension of the matrices tend to infinity, see [KS, Ch.
2] for a summary.
Aside from the one-level density, one may also study the n-level density for any integer
n ≥ 2, by replacing the inner sum of (2.3.1) with a sum over certain n-tuples of zeros, and
by replacing ϕ with an appropriate multivariate function. A similar conjecture as above
can be stated for these n-level densities, but we will not do this in the general case. The
important point is that once again, we expect the n-level density to converge to one of five
integrals. See Chapter 4 for the study of one specific two-level density.
As we have already mentioned, the conjecture above, often called the Katz–Sarnak pre-
diction, has been partially confirmed for several families. By the conjecture being partially
confirmed, we mean that the limit of the one-level density can only be computed for a spe-
cial class of Schwartz functions ϕ, with strong conditions on the support of ϕ̂, the Fourier
transform of ϕ.
2.4 The low-lying zeros of L(s, fK)
We want to study the one-level density associated to the family of Dedekind zeta functions
corresponding to S3-cubic fields. It turns out that the zeros of ζK are difficult to study
directly because of the influence of the zeros of ζ. Fortunately, these zeros can be removed
by factoring ζ from ζK , as we saw above. As ζ is a fixed function with fixed zeros, removing
these zeros from consideration will not majorly change the one-level density. In particular,
it is possible to show that the symmetry type will remain unchanged, at least assuming the
Riemann Hypothesis, but for the sake of brevity, we leave out this calculation.
It is not clear that the function L(s, fK) = ζK/ζ is a nice function as it could possibly
have poles wherever ζ has a zero. However, as we have briefly mentioned, this is not the
case and L(s, fK) is actually an L-function. Furthermore, it turns out that any Dedekind
zeta function can be decomposed into a product of ζ, and other functions, known as Artin
L-functions. In general, unlike L(s, fK), these Artin L-functions are not known to satisfy
all of the properties for L-functions that we outlined earlier. We will not need a general
definition of an Artin L-function, nor shall we provide one. We will instead think of L(s, fK)
simply as the factor of ζK obtained by dividing with ζ. The interested reader may consult
[IK, Ch. 5.13] for a general definition.
In the case of cubic non-Galois K, the Artin L-functions L(s, fK) are known to be entire.
Furthermore, they satisfy the functional equation
Λ(s, fK) = Λ(1− s, f ), where Λ(s, f ) := |D |s/2K K K Γ±(s)L(s, fK), (2.4.1)
9
2. Preliminaries
see [IK, Ch 5.10, 5.13] for all these prope(rtie)s. H(ere, )
Γ±( ) = −sΓ
s s+ δ−(DK)
s π 2 Γ 2 , (2.4.2)
where δ−(x) = 0 if x ≥ 0, and else δ−(x) = 1. As usual Γ(s), denotes the gamma function,
and as before, DK denotes the discriminant of K. For a description of some important
properties of the gamma function, see Appendix A.2. By definition, the nontrivial zeros of
L(s, fK) are the zeros of Λ(s, fK), while the nontrivial zeros are the zeros of Γ±(s). We see
that if DK is positive, then the nontrivial zeros are located at the non-positive even integers,
while they are located at all non-positive integers if DK < 0.
We are interested in studying the one-level density for the Artin L-functions correspond-
ing to non-Galois cubic fields. In this and all following chapters, we will consider our fields
to be subsets of C. Note that by our earlier discussion L(s, fK) only depends on the splitting
behaviour of the ideals generated by rational primes. In particular, L(s, fK) only depend
on the isomorphism class of K.
Define two sets F+(X) and F−(X) by
F±(X) = {Non-Galois cubic K : 0 < ±DK < X},
where we only include one field from each isomorphism class in F±(X). As the field ex-
tensions are non-Galois, each isomorphism class contains exactly three elements. Both sets
are finite, by our earlier remarks about number fields of bounded discriminant. Define
N±(X) to be the cardinality of F±(X). For a fixed K, denote the nontrivial zeros of
L(s, fK) by ρK , and write ρK =: 1/2 + iγK . We will let ϕ : R → R be a fixed non-zero,
real-valued, even Schwartz function whose Fourier transform has compact support, and we
define σ = sup supp(ϕ̂), where supp(f) denotes the support of a function f . As in [CFLS],
we note that any such ϕ can be extended to an entire function on C by means of the inverse
fourier transform. Holomorphicity is seen by differentiating under the integral sign, justified
by the compact support of ϕ̂. This extension of ϕ means that we can investigate our version
of (2.3.1) without assuming the GRH for L(s, fK).
The one-level density is then given b∑y1 ∑ ( )L
±( ) ϕN X 2 γK , (2.4.3)π
K∈F±(X) γK
where L = log(X/(2πe)2). Now, L is not really the logarithm of the conductor of L(s, fK),
but it turns out that replacing L by the logarithm log|DK | of the conductor yields expressions
that are hard to calculate explicitly. Furthermore, for most elements in F±(X), log|DK |
is approximately equal to logX. We remark that this change does not alter the symmetry
type, which one can confirm by summing Theorem 3.1 over dyadic intervals for X, and
using that log x is nearly constant in such an interval. This particular choice of L is made
in an attempt to renomalise the spacings of the γK to essentially be constant, cf. [IK, Thm.
5.8]. All zeros above are counted with multiplicity. The next chapter will be concerned with
evaluating this one-level density.
10
3
The one-level density
We will now study the one-level density of the family of Artin L-functions L(s, fK) corre-
sponding to non-Galois cubic number fields. The approach, and most of the notation, are
taken from [CFLS, Ch. 3] with very few modifications. The calculation begins by relating
the zeros of L(s, fK) to an expression involving certain prime sums, through an explicit
formula.
The key to studying this expression is a precise estimate, given in (3.2.3), of the number
of cubic fields where some prime p has a given splitting type, and where the discriminant
is bounded by a given magnitude X. Using this estimate, we find the one-level density in
Theorem 3.1. Next, we analyse the result of Theorem 3.1 and confirm that the main term
of the one-level density is of symplectic symmetry type, at least for σ = sup supp(ϕ̂) strictly
less than 2/7. We end by briefly describing a result from [CFLS], bounding the error term
in the counting function from (3.2.3) from below.
3.1 The explicit formula
We now begin studying the one-level density, given in (2.4.3). The approach is identical to
the one found in [CFLS, Lemma 3.1]. First, as we have seen, L(s, fK) satisfies the functional
equation
Λ(s, fK) = Λ(1− s, fK), where Λ(s, fK) := |D |s/2K Γ±(s)L(s, fK). (3.1.1)
Here, Γ±(s) is defined in connection to (2.4.1). Now, the gamma function is meromorphic
with its poles at the non-positive integers, and L(s, fK) is entire with no zeros with Re(s) >
1. This implies that Λ(s, fK) is entire, and that
Λ′
Λ (s, fK)
has poles precisely at the nontrivial zeros ρK of L(s, fK). For T ≥ 1 let R(T ) denote the
boundary of the rectangle in the complex plane with corners at −1/2+ iT, −1/2− iT, 3/2−
iT, 3/2 + iT orient(ed coun)terclockwise. T∫hen, by(the r(esidue )th)eorem, we find∑ L 1 L 1 Λ′
ϕ 2 γK = lim 2 ϕ s− (s, fK)ds.π T→∞ πi
γ R(T ) 2πi 2 ΛK
Here T is chosen to not coincide with the ordinate of any zero of Λ(s, fK).
Now, using the inverse Fourier transform, the support of ϕ̂, and integration by parts, we
have f(or an)y int∫eger k ≥ 1 and r ̸= 0 (cf. ∫[CFLS, Eq. (4.17)])
Lr σ (−1)k σ σL|Re(r)|
ϕ = eLrxϕ̂(x)dx = eLxRe(r)+iLx Im(r)ϕ̂(k)2 (x)dx≪
e
k k k
.
πi k k−σ L r −σ L |r|
(3.1.2)
The implied constant also depends on ϕ, but for the sake of conciseness, we do not indicate
that above, nor will we do so in the rest of the report. In particular, this calculation shows
that ϕ decays fast in any vertical strip.
11
3. The one-level density
By choosing T sufficiently far from the ordinate of any zero of Λ(s, fK) using [IK, Prop.
5.7, Thm. 5.8 ], we may bound the growth of Λ′/Λ on horizontal lines with imaginary part
equal to ±T polynomially in T, where the gamma factors are bounded using Stirling’s for-
mula. Letting T →∞ above thus makes the contribution from the horizontal line segments
vanish, as ϕ decays faster than any polynomial.
Thus, we ha
∑ (
ve show)n (∫ ∫ ) ( ( ))
L = 1 − L 1 Λ
′
ϕ 2 γK ϕ s− (s, fK)ds,π 2πi
γ (3/2) (−1/2) 2πi 2 ΛK
where (δ) denotes the vertical line with Re(s) = δ, oriented upwards. We wish to apply
the change of variables r = 1 − s to the second integral. We find that dr = −ds, and that
(−1/2) is mapped to (3/2) with the orientation reversed. Now, by the functional equation
Λ′ ′
Λ (
Λ
s, fK) = − Λ (1− s, fK),
so that (
1 ∫ ∫ ) ( ( ))− L − 1 Λ′2 ϕ 2 ∫ s 2( Λ((s, fK))d)sπi (3/2) (−1/2) πi (3.1.3)′
= 1 Lϕ 2 s−
1 Λ (s, fK)ds,
πi (3/2) πi 2 Λ
where we also used that ϕ is even.
To continue, we need to study the integrand further. As Re(s) = 3/2 > 1, the function
Λ(s, fK) is nonzero and thus has a holomorphic logarithm in any simply connected set, and
further, the derivative of any such logarithm is
Λ′
Λ (s, fK),
which is then naturally called the logarithmic derivate of L(s, fK). By the definition of
Λ(s, fK) we see that one such logarithm is simply
s
2 log|DK |+ log Γ±(s) + logL(s, fK),
where log f(s) denotes any logarithm of f . Differentiating this expression yields
log|DK | Γ′±(s) ′
2 +
L
Γ±(s)
+ (s, fK).
L
Before proceeding with our evaluation of the integral, we require a more explicit form of the
last term. To find such a form, we begin by observing that
L′ ′ ′( ) = ζs, f KK (s)−
ζ (s).
L ζK ζ
We will now expand the right-hand side into a Dirichlet series. We begin by taking the
logarithm of the defining pr∏odu∏ct and write ∑∑
log ζ (s) = log (1−N(p)−s)−1 = − log(1−N(p)−sK ).
p p|(p) p p|(p)
The attentive reader may notice that the second equality is not necessarily true for an arbi-
trary logarithm. To solve this problem, we choose the logarithm in the two first expressions
as the one defined by the right-hand side, where the logarithm in the right-hand side is the
principal branch of the logarithm.
12
3. The one-level density
Differentiate the sum by interchanging the order of differentiation and summation. This
can be justified by the uniform absolute convergence of the resulting series for Re(s) ≥ 1+ϵ,
for any ϵ > 0. We find that the derivative equals
ζ ′ ∑∑ N(p)−s logN(p) ∑∑∑∞K ( ) = − = − logN(p)s 1 ( )− ( ) , (3.1.4)ζ −N p s N p esK p p|(p) p p|(p) e=1
where the last equality is the formula for a geometric sum. To continue we will need the
von-Mangoldt function Λ(n), defined on prime powers by Λ(pk) = log p, and else as 0. Also,
recall that N(p) is a power of p so that we can write the sum as
∑∞
− Λ(n)bK(n) ,
=1 n
s
n
for some coefficients bK(n). Note that we need not concern ourselves with the values of
bK(n) when n is not a prime power, as then Λ(n) = 0.
We first find the coefficients bK , when K = Q. Then the prime ideals p are exactly (p)
so that N(p) = p, and thus bK is constantly equal to 1. We turn to the case when K is a
non-Galois cubic field and study b eK(p ) for some prime p and e ≥ 1. From the right-hand
side of (3.1.4), we see that bK(pe) only depends on the splitting type of (p) and on e.
We begin with the totally split case, when (p) has splitting type (111), and (p) = p1p2p3.
Then N(pi) = p, and for a fix p and e the sum over p in (3.1.4) has 3 terms so that
b eK(p ) = 3. For the splitting type (12) the sum only has 2 terms, whence b (peK ) = 2, and
similarly for the splitting type (13) we have b (peK ) = 1. In the case (21), we have one prime
factor of norm p2 and one of norm p. Thus, b (peK ) = 1+2δ2|e. Here, δP is a function which
equals 1 if P is true, and 0 otherwise. Lastly, for splitting type (3) we find b (peK ) = 3δ3|e.
Using these results, we have shown that for Re(s) > 1
∞
L′ ∑( ) = − Λ(n)aK(n)s, fK ,
L s
n=1 n
where aK(pe) is given in the following table, see also [CFLS, Ch 2.]:
Splitting type a eK(p )
T1 2
T2 2δ2|e
T3 3δ3|e − 1
T4 1
T5 0
From these result∫s, we ca(n exp(and (3.)1.)3) into ∫ ( ( ))log|DK | L 1 1 L 1 Γ′
2 ϕ
±
πi (3/2)∫ 2πi( s−(2 ds)+) ϕ 2 s− 2 Γ (s)dsπ∑i (3/2) πi ±
− 1 L 1
∞ Λ(n)aK(n)
ϕ s− ds.
πi s(3/2) 2πi 2 n=1 n
To proceed we will need a technique that is often referred to as ’shifting the contour’.
Consider a vertical strip defined by a ≤ Re(s) ≤ b, for some a < b, and let f(s) be
holomorphic in an open set containing this strip. Further, assume that |f(s)| decreases to 0
uniformly in the strip, as |Im(s)| → ∞. Then, the conclusion is that the integral of f over (b)
is equal to the integral of f over (a), assuming that one of them exists. Indeed, this follows
by applying Cauchy’s theorem to the rectangle with corners in a− iT, a+ iT, b− iT, b+ iT ,
and then letting T →∞ so that the horizontal contributions tend to 0.
13
3. The one-level density
We now study the third integral above. By absolute convergence, we may interchange
the order of summation and integration∫ and th(us find1 ∑∞ ( ))− Λ(n√)aK(n) L 1ϕ s− 1/2−s
πi =1 n (3/2) 2πi 2
n ds.
n
Now shift all three integrals to the line (1/2), and parametrise the integrals by s = 1/2+ iu.
Next, make∫ the change o∫f variables(t = Lu/(2π)). The result islog|D | 2 Γ′K ( ) + ( ) ± 1 2 ∞+ πit 2 ∑− Λ(n√) ∫aK(n)ϕ t dt ϕ t −2πit/L
L R L R Γ± 2
dt ϕ(t)n dt.
L L n
n=1 R
To simplify further, we apply the definition of the Fourier transform to the first and third
integr(al and fi)nd that∑ ( ) ( )L ϕ̂(0) log ∫| ′ ∑∞DK | 2 Γ
ϕ 2 γK = + ϕ( )
± 1 + 2πit − 2 Λ(n√)aK(n) lognt Γ 2 dt ϕ̂ ,π L L R ± L L n LγK n=1
(3.1.5)
which is also essentially the content of [CFLS, Lemma 3.1].
3.2 Summing over K
We wish to find
1 ∑ ∑ ( )L
N±( ϕ γK ,X) 2π
K∈F±(X) γK
which means that we must sum (3.1.5) over K. The calculations are all taken from [CFLS,
Ch. 3]. We begin by observing that the second term in (3.1.5) does not depend on K, and
thus summing over K simply multiplies the term by N±(X). The only K-dependence in
the first term comes from log|DK |. We use Stieltjes integration and integration by parts,
see Appendix A.3, to write
∑ ∫ X ∫ X ±
log| | = log ±( ) = ±( ) log − N (t)DK t dN t N X X dt. (3.2.1)
t
K∈F±(X) 1 1
Here we made use of the fact that no nontrivial field extension of Q has a discriminant of
modulus 1. To evaluate the integral in the right-hand side we require a precise estimate of
the function N±(t), and therefore we postpone this calculation until later in this section.
For the sum of the third term in (3.1.5) over K, we interchange the order of summation
over K and n. This is allowed as all sums only contain finitely many nonzero terms. The
result∑is∞ ( )2 Λ(n) logn ∑ 2 ∑∑∞ ( ) ∑− √ ϕ̂ aK(n) = − log p e log p2 ϕ̂ aK(pe),L =1 n L L e/n K∈F±(X) p e=1 p L K∈F±(X)
where the last step is that the summand is nonzero only when n is a prime power. By our
earlier investigations, for a fixed p and e ≥ 1, a (peK ) only depends on how (p) splits in K,
and on e. Thus, if we let N±p (X,Ti) count the fields in F±(X) where p has splitting type
Ti we find that the sum above is
2 ∑∑∞ ( )(− log p e log p2 ϕ̂ 2N±p (X,T1) + 2δ ±2|e)Np (X,T2)L pe/=1 Lp e (3.2.2)
+ (3δ3|e − 1)N± ±p (X,T3) +Np (X,T4) .
14
3. The one-level density
We see that in order to study the one-level density, we need not only estimates for
N±(X), but also for the more precise counts N±p (X,Ti). Obtaining these estimates is
highly nontrivial, and these results are the crux of our method for evaluating the one-level
density. Using notation from [CFLS, Ch. 1], it has been shown that
N±(X) = C±X + C±X5/61 2 +Oϵ(Xθ+ϵ),
± ± ± 5 6 + (3.2.3)N (X,T ) = A (T )X +B (T )X / +O (pω θ ϵp i p i p i ϵ X ),
holds for some θ < 5/6, ω ≥ 0, and all ϵ > 0. This was first shown independently in [TT]
and [BST], and the current best result, proved in [BTT], allows us to take θ and ω to be
2/3. We will not fix θ or ω, but instead only require 5/6 > θ ≥ 0 and ω ≥ 0, to allow our
results to be improved in case (3.2.3) is proven to hold wi(th)smaller values of θ and ω. Here,1
C+
1 4ζ
1 = 12 (3) , C
+
2 = ( ) 3 ( ) ,ζ 5Γ 2 3 53 ζ 3
√
and C− = 3C+, C− = 3C+1 1 2 2 . We now explicitly provide the values of Ap(Ti) and Bp(Ti),
cf. the summary in [CFLS, Ch. 2]. Let,
= 1 = 1− p
−1/3
xp 1 + −1 + −2 , yp .p p (1− p−5/3)(1 + p−1)
Then,
A±p (Ti) = C±1 xpci(p), B±p (Ti) = C
±
2 ypdi(p),
where ci(p) and di(p) are given in the table below.
Splitting type ci(p) di(p)
T 1 (1+p
−1/3)3
1 6 6
1 (1+p−1/3)(1+p−2/3T )2 2 2
1 (1+p−1T )3 3 3
T 1 (1+p
−1/3)2
4 p p
1 (1+p−1/3T )5 p2 p2
For a proof sketch of the estimates above, see Chapter 5.
Using the estimates in (3.∫2.3), we can see that (3.2.1) isX ±
log ±( )− C1 t+ C
± 5/6
2 t +Oϵ(tθ+ϵ)XN X dt
1 t
± ( )
= log ±( )− ± − 6CXN X C X 2 X5/6 +O Xθ+ϵ (3.2.4)1 5 ϵ
±
= logXN±
( )
(X)−N±(X)− C2 5/6 θ+ϵ5 X +Oϵ X .
We should also divide this expression by N±(X). We use
1 1 1
N±(X) = ± ·C1 X (1 +X−1/6C±/C±(+O)(Xθ−1+ϵ2 1 ϵ ) )± 2 (3.2.5)
= 1 1− C2 −1/6 + C
± ( )
X 2 X−1/3± ± ± +O X
θ−1+ϵ
ϵ +X−1/2 ,
C1 X C1 C1
and find that dividing (3.2.4) by N±(X) yields
± ± 2
log CX − 1− 2 −1/6 + (C2 )X X−1/3 +O (Xθ−1+ϵ +X−1/2).
5C± ± 2 ϵ1 5(C1 )
15
3. The one-level density
Note that if θ ≥ 1/2, then the second term in the error can be absorbed in the first.
The only remaining term in the one-level density is (3.2.2). By (3.2.3), and a calculation,
we conclude that
± ± ± ± ± ( −1)2Np (X,T1) + 2δ2|eNp (X,T2) + (3δ3|e − 1)Np (X,T3) +Np (X,T4) = C1 X θe + p xp
+ C±X5/6(1 + p−1/32 )(κe(p) + p−1 + p−4/3)y +O (pωXθ+ϵp ϵ ), (3.2.6)
where
θe = δ2|e + δ3|e, κe(p) = θe(1 + p−2/3) + (1− δ )p−1/33|e .
Set γe(p) = (1 + p−1/3)(κe(p) + p−1 + p−4/3)yp, and notice that θe, γe(p)≪ 1.
Us(ing the results abo(ve, we)can conclude t(hat dividing (3.2.)2))by N
±(X) yields
± ± 2
− 2 1− C2 X−1/6 + C2 X−1/3 +O Xθ−1+ϵ± ± +X
−1/2
∑L ∑ C1 C1∞ ( )
ϵ
± (
× xp(θe + 1/p) log p e log p 2 C
±
2 −1 (/6
2 ϕ̂ − ±X 1−
C2 X−1/6 +O Xθ−1+ϵ
pe/
ϵ
=1) (L ) L C C
±
p e 1 1∑∑∞ ∑∑ ∣ ( ) ∣ e
+X−1
) ∣ ∣
/3 γe(p) log p e log p + 1O  θ−1+ϵ log p ∣∣ log pϕ̂ X ϕ̂ ∣pω .pe/2 ϵL L pe/2 ∣
p e=1 Lp e≥1
We begin by handling the error terms above. First, L ≤ logX so that we only need
to consider pe ≤ Xσ in the sums, by the support condition on ϕ̂. The support condition
also implies that ϕ̂ is bounded, which means we may bound ϕ̂ by the indicator function of
pe ≤ Xσ multiplied by a constant.
To estimate the last sum above we will use a version of the Prime Number Theorem, cf.
[D3, Ch. 18], which asserts that∑ ( )′ 1/2
θ(x) := log p = x+O xe−c (log x) , (3.2.7)
p≤x
for some constant c′ > 0. In particular, θ(x)≪ x, which will be a sufficient bound for now.
We then want to bound the sum ∑
Xθ−1+ϵ
log p
pω.
pe/2
pe≤Xσ,e≥1
It suffices to consider the case when e = 1, as the sub-sums obtained by fixing other values
of e give a smaller contribution, and the sum over all large enough e, e.g. all e ≥ 2ω + 3,
converges even without the restriction pe ≤ Xσ so that the contribution from such e is≪ 1.
Thus, we find tha∑t the third sum is ∫ Xσ
≪ θ−1+ϵ log pX ≪ Xθ−1+ϵ 1 dθ(u)≪ Xθ+σ(1/2+ω)−1+ϵ, (3.2.8)
p1/2−ω u1/2−ω
p≤Xσ 1
where the last step is integration by parts.
To handle the rest of the error terms, we note that θ = 0, γ (p)≪ p−1/31 1 so that all the
remaining error terms ar∑e dominated by the bound∑we just found, or by the bound for
−1/2 γe(p) log pX ≪ X−1/2 log p ≪ Xσ/6−1/2, (3.2.9)
pe/2 p5/6
pe≤Xσ p≤Xσ
where the first step is a restriction to e = 1, and the last step is a Stieltjes integration as
above.
Combining our results from this and the preceding section, and using
logX 2 2= 1 + log 4π e
L L
we have proven (cf. [CFLS, Theorem 1.2]).
16
3. The one-level density
Theorem 3.1 (CFLS). Let 5/6 > θ ≥ 1/2, and ω ≥ 0 be such that (3.2.3) holds. Then if ϕ
is a real even∑Schw∑artz f(unction,)whose Fourier transform is supported in [−σ, σ], we have1 L
N±(X) ϕ γK( 2πK∈F±(X) γK ) ( )
log(4π2e) C± X−1/6 (C±)2 −1/3 ∫ Γ′= (0) 1 + − 2 + 2 X + 2 1 2πitϕ̂ ( ± ) ± 2 ( ϕ(t)
±
Γ) 2 + dtL 5C1 L ∑5(C1 ) L L R ± L± ± 2 e
− 2 1− C2 X−1/6 + (C2 ) −1/3 xp(θe + 1/p) log p log pX ϕ̂
L C±1( (C± 21 ))∑ (p
e/2
p,e
−1 6 ± ) L
− 2C X
/ e
2 1− C2 X−1/6 γe(p) log p log p
( )
± 2 ϕ̂ +Oϵ X
θ−1+σ(ω+1/2)+ϵ .
C1L C pe/1 Lp,e
We remark that θ − 1 + σ(ω + 1/2) < 0 is required to make the error term smaller than
the main term. In particular with ω = θ = 2/3, which is currently the best value for which
(3.2.3) holds, one needs σ < 2/7. The restriction to θ ≥ 1/2 is explained in Section 3.4.
3.3 Interpreting Theorem 3.1
We want to compare the main term of the expression above to the expected symplectic
main term from the Katz-Sarnak prediction. First, the term involving the integral is≪ 1/L
by the fast decay of ϕ, and Stirling’s formula. Estimating the rest of the terms requires a
careful study of the two sums. We follow the proof of [CFLS, Lemma 3.4]. We will primarily
concern ourselves with the constant terms, or the terms of size 1/L in the one-level density.
We begin by studying the first of the sums in Theorem 3.1. The idea is to split the sum
into one ’convergent’ and one ’divergent’ part, where divergent means that it would diverge
if the sum did not involve the compactly supported ϕ̂. Recall that θ1 = 0, θ2 = 1, and
x −1p − 1 ≪ p , so that the divergent part is contained in the sum restricted to e = 2. We
write the sum∑as ( ) ( )log p 2 log p ∑∑+ xp(θe + 1/p) log p e log pϕ̂
p L ( (pe/2
ϕ̂
p ∑p e ̸=2 ) ) (L )
+ log p xp 1 +
1 − 1 2 log pϕ̂ .
p p L
p
The last two sums are estimated with a Taylor expansion of order zero of ϕ̂, with error
≪ L−1. There is a technical subtlety involving the implied constant of the error term in
this expansion. What we do is to first consider the function ϕ̂(u) restricted to |u| ≤ σ.
Then, by the compactness of the domain, a Taylor expansion of any order holds with a
fixed implied constant. Now, as ϕ̂(u) is zero outside this domain, we may conclude that the
Taylor expansion also holds in the entirety of R, but with a possibly larger constant, as the
size of the error term will dominate the size of the Taylor polynomial for large u.
To handle the remaining sum( , we use t)he Prime Number Theorem (3.2.7), but with
the error term weakened to O x(log x)−AA , which holds for any A ≥ 1. Using Stieltjes
integr∑ation, and(the com)pact∫support(of ϕ̂, w)e calculatelog p 2 log p ∞= 1 2 log uϕ̂ ϕ̂ dθ(u)
p L u L
p ∫ 1∞ ( ) ( )
= 1 2 log
∫
u ∞ 1 2 log ∫u L ∞
ϕ̂ ∫ du+ ϕ̂( ( d(θ(u))− u) = 2( ϕ̂(u))du1 u L 1 u L 0∞ )
− ( (1)− 1) (0)− ( ( )− ) − 1 2 log u 1θ ϕ̂ ∫θ u u ϕ̂ + ϕ̂′
2 log u
u2 L u2
du
1 L L
∞
= L4 ϕ(0) + (0) + (0)
θ(u)− u
ϕ̂ ϕ̂ 2 du+O(L
−1),
1 u
17
3. The one-level density
where the last step is a Taylor expansion of ϕ̂. Note that the integral converges by the Prime
Number Theorem as stated above, with say A = 2.
In conclusion, we have shown
−2∑∑ ( ) (xp(θe + 1/p) log ∫p log pe ∞= −ϕ(0) − 2 (0) 1 + θ(u)− uϕ̂ ϕ̂ du
L pe/2 L
p∑e≥1∑ ∑ (
2( L ) )) 1 u2 (3.3.1)
+ xp(θe + 1/p) log p + log p 12 x 1 + − 1 .pe/ pp p
p e ̸=2 p
It is possible to estimate the second sum in Theorem 3.1 using the same method. Similar
terms as those calculated above∫are found, but we also find one term of the form∞
L ϕ̂ (u) eLu/6du≪ Xσ/6+ϵϵ . (3.3.2)
0
We are content with remarking that if one carries out the calculations, then one finds the
entire sum to be≪ϵ Xσ/6+ϵ. In particular, if σ < 1, one sees that the contribution from the
term involving this sum in Theorem 3.1 is ≪ X−δ for some δ > 0, as the sum is multiplied
by X−1/6.
As a consequence of our calculations, we can see that for σ < 2/7, the main term in
Theorem 3.1 is indeed the expected symplectic main term
ϕ̂(0)− ϕ(0)2 (3.3.3)
from the Katz-Sarnak prediction. Indeed, the Fourier transform of
sin 2πx
2 ,πx
is 1/2 multiplied by the characteristic function of the interval [−1, 1], so that an application
of Planchere∫l’s formula(, and the defi)nition of the Fou∫rier transform, yields∞
( ) 1− sin 2πx
1
ϕ x 2 dx = ϕ̂(0)−
1
2 ϕ̂(u)du = ϕ̂(0)−
ϕ(0)
−∞ πx −1 2
,
where we used σ < 1 in the last step. In Chapter 7.2, we will also analyse the secondary
term of size L−1, albeit in a slightly different context.
3.4 An application to counting cubic fields
We end the chapter by presenting a theorem [CFLS, Thm. 1.1], which uses the calculation
of the one-level density to find a lower bound for the error term in (3.2.3). It turns out that
the calculations leading up to Theorem 3.1 can be used to prove that if the estimates
N±p (X,Ti) = A±p (Ti)X +Bp(Ti)X5/6 +Oϵ(pωXθ+ϵ)
holds with the same value of ω, θ ≥ 0 for all splitting types Ti, and all primes p, then
conditional on the Generalised Riemann Hypothesis for ζK , one must in fact have θ+ω ≥ 1/2.
The idea of the proof is to show that if θ+ ω < 1/2, then we may fix a Schwartz function ϕ
that in a sense makes the left-hand side of (3.3.2) too large, which leads to a contradiction.
We leave out the details.
This theorem is currently the only known result giving a lower bound for ω and θ.
Moreover, as is pointed out in connection to [CFLS, Thm. 1.1], numerical evidence from
[CFLS] indicates that this bound is sharp in the sense that (3.2.3) appears to hold with
θ = 1/2, and any ω > 0. We remark that this result is why we restrict to θ ≥ 1/2, ω ≥ 0 in
Theorem 3.1.
18
4
The two-level density
We now extend the results of the previous chapter to the two-dimensional case by finding the
two-level density. The same two-level density has previously been studied in a more general
setting in [CK, Thm. 4.26], but then only the main term was found and the error term
was of size (logX)−1. The main result of this chapter is Theorem 4.1, where the two-level
density is found with a power-saving error term. We remark that this is the first time any
n-level density, with n ≥ 2, is calculated to this precision for this family. In other families,
such results are also relatively rare.
The methods of the previous chapter essentially suffice to prove Theorem 4.1. The only
new theoretical result we will need is an estimate for a counting function of cubic fields,
specifying the splitting behaviour of two rational primes, instead of just one. Some of the
methods and notation are taken from [Rb], where the n-level density of another symplectic
family is studied.
4.1 The explicit formula
Consider the zeros ρK = 1/2+iγK of the Artin L-function L(s, fK). We want to count the ze-
ros with multiplicity, so letmρ denote the multiplicity of a zero ρK . By the functional equa-K
tion, any zero 1/2+iγk has a corresponding zero 1/2−iγK , and a possible zero at 1/2 has even
multiplicitym1/2. To every zero, we may associate a collection of pairs (ρK , 1), ..., (ρK ,mρ ).k
In the case when γK = 0, it is for technical reasons advantageous to instead consider the two
collections of pairs (+0, 1), ..., (+0,m1/2/2), and (−0, 1), ..., (−0,m1/2/2), where +0, and −0
are symbols.
The two-level density is then defined as the sum
∑ ∑ ( )1 L L
ϕ γ , γ′±( ) 2 KN X ′ π 2π
K .
K (γK ,i)̸=(±γ ,i)K
In the sequel, we will simply write the condition (γK , i) ̸= (±γ′K , i) as γK ̸= ±γ′K . Here,
1/2 + iγK and 1/2 + iγ′K ranges over the zeros of L(s, fK), and 1 ≤ i ≤ mρ . Also, ϕ isK
a function with compactly supported Fourier transform, and with ϕ(u1, u2) = ϕ1(u1)ϕ(u2),
with ϕ1, ϕ2 both real, even Schwartz functions. Define the sets Bx = {u ∈ R2 : |u1|+ |u2| ≤
x}, and let σ = inf{x : supp(ϕ̂) ⊆ Bx}. The definition implies that ϕ̂(u1, u2) is nonzero only
if |u1|+ |u2| < σ, and furthermore that with σi := sup(supp(ϕ̂i)), we have σ1 + σ2 = σ.
To study the two-level density, we begin by rewriting the inner sum above to remove the
condition γK ̸= ±γ′K , as in [Rb], which will allow us to make use of the methods from the
previous chapter. By our remarks on the consequences of the functional equation above, we
may write the inner sum ab(ove as∑ ) ∑ ( )L L ′ L Lϕ 2 γK ,π 2 γπ K − 2 ϕ 2 γK , 2 γK , (4.1.1)
γ ,γ′
π π
K γK K
by adding and removing the terms when γK = ±γ′K , and using that ϕi is even. All zeros
are counted with multiplicity in the sum above.
19
4. The two-level density
We can already estimate ∑∑ ( )
− 2 L L
N±( ) ϕX 2 γK ,π 2 γK , (4.1.2)π
K γK
using the results of the previous section. Indeed, define the function ϕ3(u) = ϕ(u, u). Then,
(4.1.2) is equal to the one-level density multiplied by −2, with the function ϕ replaced by ϕ3.
In particular, it can be estimated by using the right-hand side in Theorem 3.1, if one replaces
ϕ by ϕ3. Henceforth, we refer to this right-hand side as R3. As the Fourier transform of a
product is the convolution of the Fourier transforms, we find that ϕ̂3(u) is nonzero only if
|u| < σ, cf. [Rb, Claim 1]. In particular, the σ in R3 is the same as the σ above.
∑Now(, by the explici)t for∑mula, ( ) ( )L L ′ L ∑ Lϕ 2 γK , 2 γK = ϕ1 2 γK ϕ γ′2( π π π 2π KγK ,γ′ γ ( γ′K K K ) ( ))
= log
∫
|DK | 2 Γ′
∞
ϕ̂ (0) + ϕ (t) ± 1
∑
( 1 1 Γ 2 +
2πit
dt− 2 Λ(n√)aK(n) lognϕ̂1
L L R ± L L =1 n Ln )
log ∫|D | 2 Γ′ ( )1 ∑∞ ( )× K ϕ̂2(0) + ϕ2(t) ±Γ 2 + 2πit 2 Λ(n)aK(n) logndt− √ ϕ̂2 .L L R ± L L n=1 n L
Similarly as in [Rb], we write this as (C1,K +E1 +D1,K)(C2,K +E2 +D2,K), with obvious
notation.
4.2 A refined counting function for the number of cubic
fields
We have previously seen the estimates (3.2.3) for N±p (Ti;X). These estimates were enough
to find the one-level density but to find the two-level density we will need a slightly refined
counting function. For p and q distinct primes, defineN±p,q(X,Ti, Tj) as the counting function
for isomorphism classes of non-Galois cubic fields whose discriminant is bounded byX, where
(p) has splitting type Ti, while (q) has splitting type Tj . Then, using the notation of [CFLS,
Eq. (2.3)], we have
N± (X,T , T ) = A±
( )
p,q i j p,q(Ti, T )X +B 5/6j p,q(Ti, Tj)X +Oϵ pωqωXθ+ϵ , (4.2.1)
with the same values of θ and ω as before, and with
A± ± ± ±p,q(Ti, Tj) = C1 (xpci(p))(xqcj(q)), Bp,q(Ti, Tj) = C2 (ypdi(p))(yqdj(q)).
We end by remarking that one obtains similar results not only for a pair of primes but
also for any n-tuple of primes, with the same θ and ω. The best such result is proven in [BTT]
along with (3.2.3) and (4.2.1). Specifically, if we let p = (p1, ..., pn) be an n-tuple of distinct
primes, and k = (k1, ..., kn) be an n-tuple of indices of splitting types, i.e. ki ∈ {1, 2, 3, 4, 5}.
Then we have
∏n ∏ (n ∏ )n
N±(X,T ) = C±X x c (p ) + C±X5/6 y d (p ) +O Xθ+ϵ pωp k 1 pi i i 2 pi i i ϵ i , (4.2.2)
i=1 i=1 i=1
where the left-hand side counts fields where each pi has splitting type Ti.
4.3 Finding the two-level density
The goal of this section will be to find the two-level density by proving the following theorem.
20
4. The two-level density
Theorem 4.1. Let 5/6 > θ ≥ 1/2, and ω ≥ 0 be such that (4.2.1) holds. Let ϕ(u1, u2) =
ϕ1(u1)ϕ2(u2) be a product of real even Schwartz functions, with Fourier transform ϕ̂ sup-
ported in |u1| + |u2| < σ < (1 − θ)/(ω + 1/2) ≤ 1, and set ϕ3(u) = ϕ1(u)ϕ2(u). Then we
have
1 ∑ ∑ ( ) [L L 2
ϕ γ , γ′±( ) 2 K 2 K = ϕ̂1(0)ϕ̂2(0) 1 +
2 log(4π e)
N X ′ π π ( )LK γK ̸=±γK ]
+ 4(log
2(2π) + log(2π)) + 2 + ( ) − 2 + 2− 20 log(2π)( F2 XL2 )( 5L 25)L2 ( )
+ (0) 1 + 2 log(2πe) ( ) (2) + ( ) (2) − ϕ̂1(0)ϕ̂1 ( )(F1 X S1 F2 X S2 ) (F1( ) (2) +
6
X S1 5F2(X)S2(2)L L )
+ (0) 2 log(2πe) ϕ̂2(0) 6ϕ̂2 (1 + F1()X)S1(1) +(F2(X)S2(1) − ) F1(X)S1(1) + 5F2(X)S2(1)L L
+ F1(X) S1(1)S1(2)− S3 + S5 + F(2(X) S2(1)S2(2)−)S4 + S6
−1+ (1 2+ )+ (4.3.1)+ E1R2 + E2R1 − E1E2 − 2R3 +O Xθ σ / ω ϵ .
Here, F1(X) and F2(X) are defined in (4.3.5), while S1(i) and S2(i) are defined in (4.3.6).
The sums Sj for j ≥ 3 are defined i∫n (4.3.11) a(nd (4.3.14)). The integral term Ei is givenby
= 2
Γ′
( ) ± 1 + 2πitEi ϕi t dt.
L R Γ± 2 L
Lastly, Ri is defined as the right-hand(side in Theorem 3.1, exc)luding the error term, i.e.2
= (0) 1 + log(4π e) − F2(X)Ri ϕ̂i
L 5L
+ Ei + F1(X)S1(i) + F2(X)S2(i).
Remark. The expression for the two-level density may seem daunting at first, and we
therefore provide some advice on how to discern the largest of the terms. First, F2(X) ≪
X−1/6 so that, at least for σ < 1, F2(X) multiplied by Sj or S2(1)S2(2) is bounded by X to
some negative power.
Furthermore, F1(X) = 1 + O(X−1/6) so that one may substitute every occurence of
F1(X) with 1, if only the largest terms are of interest. The precise estimates of the sums
S1(i), S3, S5 are carried out in section 4.4, but we remark that all of them are ≪ 1. Lastly,
the integrals Ei are all ≪ L−1.
Theorem 4.1 is the first time the two-level density of this family has been investigated past
terms of size L−1. Using similar methods as in the previous chapter, the sums S1(1), S1(2), S3
and S5 can be expanded into ascending powers of L−1. In particular, Theorem 4.1 allows
us to explicitly find all terms of size ≫ L−k for any k ≥ 1.
To prove the theorem, we∑must evaluate1
±( ) (C1,K + E1 +D1,K)(C2,K + E2 +D2,K) (4.3.2)N X
K∈F±(X)
Let Ri be the right-hand side of Theorem 3.1, with ϕ = ϕi. Then, by the calculations in the
previous chapter, and the factthat Ei does not depend on K, we see that (4.3.2) is∑ 
E1R2+E2R1−
1
E1E2+ ±( ) (C1,KC2,K +D1 ,KD2,K + C1,KD2,K +D1,KC2,K) .N X
K∈F±(X)
Thus, we must evaluate four different sums to find the two-level density.
21
4. The two-level density
4.3.1 Summing C1,KC2,K over K
We begin by studying the first and easiest sum above, where the calculations can essentially
be reduced to integrating the function log2 t := (log t)2. We have
∑ (0) (0) ∑ (0) (0) ∫ϕ̂1 ϕ̂ ϕ̂ ϕ̂ X2 2 1 2C 2 ±1,KC2,K = 2 log |DK | = 2 (log t)dN (t).L L
K∈F±(X) K∈F±(X) 1
To simplify, use integration by parts, and (3.2.3). The result is that the integral above equals
N±(X) log2X − 2 logXN±(X)− 2 logXC± 5/6 ± 22 ± 5/6 θ+ϵ5 2 X + 2N (X) + 25C2 X +Oϵ(X ).
Thus, by combining this calculation with((3.2.5) we conclude that
1 ∑ ϕ̂1(0)ϕ̂2(0)
±( ) C1,KC2,K = 2 log
2X − 2 logX + 2
N X L
K∈F±(X)
± ( ± )( ))
+ C2 X−1/6 1− C2 −1 6 22 2
( )
X / − logX +O Xθ−1+ϵϵ
C± ±1 ( C1 25( 5 )
2 log(4π2e) 4 log2(2π) + log(2π) + 2= ϕ̂1(0)ϕ̂2(0) 1 + +
L L2
C±
(
C±
)( ))
+ 2 X−1/6 1− 2 −1
( )
/6 2
± ±X −5 +
2− 20 log(2π) +O Xθ−1+ϵ2 ϵ .C1 C1 L 25L
4.3.2 Summing Cℓ,KDi,K over K
We interchange the order of summation to calculate
1 ∑ ∞ ( )2ϕ̂ ∑ℓ(0) Λ(n) logn ∑
±( ) Cℓ,KDi,K = −N X L2 ±
√
( ) ϕ̂i aK(n) log|DK |.N X
K∈F±(X) n=1
n L
K∈F±(X)
(4.3.3)
As mentioned earlier aK(pe) only depends on e, and how (p) splits inK. Thus, if we separate
the summands depending on the splitting type, then we are left with the task of evaluating
∑ ∫ X
log|DK | = (log t) dN±p (t, Tk),
K∈F±( 1X),
(p) has splitting type Tk in K
for a fixed p and Tk. Applying integration by parts yields
N±(X,T ) logX −N±( 1
( )
X,T )− B±(T )X5/6 +O pωXθ+ϵp k p k 5 p k ϵ . (4.3.4)
Before continuing, we introduce convenient notation. Define
C± (C± 2 ±
( ± )
F 2 −1/6 2
) −1/3 C2 −1/6 C2 −1/6
1(X) = 1− ±X + ± X , F (X) = X 1− X , (4.3.5)C1 (C1 )2
2
C± C±1 1
and let ∑ ( ) ∑ ( )e e
S1(i) = −
2 xp log p log p ( + 1 ) ( ) = − 2 log p log pϕ̂i θe /p , S2 i ϕ̂i γe(p).
L pe/2 L L pe/2 L
p,e p,e
(4.3.6)
The point of the former notation is that
C±X ± 5/61
±( ) = F1(X) +O
θ−1+ϵ
ϵ(X ) and
C2 X
±( ) = F2(X) +O (X
θ−1+ϵ
ϵ ), (4.3.7)
N X N X
22
4. The two-level density
at least for θ ≥ 1/2. After summing over K in (4.3.3), by using (3.2.6) and (4.3.4), we see
that
1 ∑ ( )( )= (0) 1 + 2 log(2πe)±( ) Cℓ,KDi,K ϕ̂ℓ F1(X)S1(i) + F2(X)S2(i)N X L
K∈F±(X)( )
ϕ̂ ( )− ℓ(0) F1( 6X)S1(i) + 5F2(X)S (i) +O Xθ−1+σi(1/2+ω)+ϵ2 .L
4.3.3 Summing D1,KD2,K over K
We now turn∑to the remaining sum. We write1 4
N±(X) D1,KD2,K = L2N±(X)
K∈F±∑(X) ∑∑ ( ) ( )
× (log p)(log q) e log p f log
(4.3.8)
q
2 2 ϕ̂1 ϕ̂2 aK(p
e)a (qfK ).
pe/ qf/ L L
K∈F±(X) p,e q,f
The key to evaluating this sum is to move the sum over K inside, and then note that the
coefficients are constant when both p, and q have fixed splitting types. In total, there are
16 different pairs of splitting types that produce nonzero coefficients aK . When p = q, their
splitting types must be equal. In this special case when p = q we define
N±p,p(X,Ti, Tj) := N±p (X,Ti)δi=j .
Write a ek(p ) for the value of aK(pe) when p has splitting type Tk in K. The bound
aK(pe)≪ 1 will be useful when estimating the error terms below. After separating by pairs
of splitting types in (4.3.8),∑we∑are left with ( ) ( )4 (log p)(log q) e log p f log q
ϕ̂1 ϕ̂2
L2N±(X) pe/2qf/2 L L
p,e q,∑f
× N±p,q(X,Tk1 , Tk2)a ek1(p )ak2(qf ).
1≤k1,k2≤4
No matter if p = q or not, N±p,q(X,Tk1 , Tk2) has the form of an expression containing a main
term, a secondary term, as well as an error that in both cases is ≪ (pq)ωXθ+ϵ. Thus, using
the same method as when bounding the error in the previous chapter, we can split the sum
over p and q and find that the contribution of the error term to (4.3.8) is
≪ Xθ−1+(σ1+σ2)(ω+1/2)+ϵϵ = Xθ−1+σ(ω+1/2)+ϵ. (4.3.9)
Here, recall that σi has the property that ϕ̂i is zero outside [−σi, σi].
(Now, the main term and secondary term of N) ±p,q(X,T(i, Tj) are given by
C±
)
1 Xxpci(p)x (c (q) + C±q j 2 X5/6ypdi(p)yqdj(q) δ ±p ̸=q + C1 Xx)pci((p) + C±X5/62 ypdi(p)
× δ ±p=qδi=j = C1 Xxpci(p)xqcj(q) + C
± 5/6
2 X ypdi(p)yqd (q) − C
±
j 1 Xxpci(p)xpcj(p)
+ C±X5 6
) ( )
/
2 ypdi(p)ypdj(p) δp=q + C
±
1 Xxpci(p) + C
±X5/62 ypdi(p) δp=qδi=j . (4.3.10)
Thus, to find the sum of D1,KD2,K we must evaluate three essentially different sums. Note
that the first term in the left-hand side is the difference of the first and second term in the
right-hand side.
To find the sum coming from the first term of (4.3.10) first separate the X and X5/6
term, then split both sums over p, e, q, f , k1 and k2. The result is that we simply find
the sums we have already studied in the previous chapter, but multiplied by each other.
Specifically, also using (4.3.7), we obtain the main terms
F1(X)S1(1)S1(2) + F2(X)S2(1)S2(2).
23
4. The two-level density
We also get a contribution coming from the error terms in (4.3.7), but these can be bounded
from above by the expression in (4.3.9), after splitting the sums as before.
We turn to the second term in (4.3.10). The presence of the factor δp=q means that we
only have one sum over the primes. Split the inner sum over k1, k2. The sum of ci(p)ai(pe)
over i = 1, 2, 3, 4 was calculated in the previous chapter to equal θe + 1/p, while the sum
over di(p)ai(pe) equals γe(p). Thus, the contribution from the second term is, excluding the
error terms coming from (4.3.7)(which a)re(bounde)d as(before,∑ 2 log2 ) ( )4 xp p− 1F1(X) 2 2+ 2 θe + θf + 1 e log p f log pϕ̂ ϕ̂L pe/ f/ 1 2p p L L
p,e,f ∑ 2 ( ) ( ) (4.3.11)
− 4 log p e log p f log pF2(X)
L2 pe/2+f/2
γe(p)γf (p)ϕ̂1 ϕ̂2
L L
p,e,f
=: −F1(X)S3 − F2(X)S4,
where S3 and S4 denote 4/L2 multiplied by the first and second sum respectively.
We turn to the last term of (4.3.10). Here the δ-factors restrict the sum to p = q and
k1 ∑= k2. Hence, we see that to evaluate the sum we must calculate
ck(p)a e fk(p )ak(p ) = 1 + 2
ιe,f (p)
δ2|eδ2|f + 3δ3|eδ3|f − δ3|e − δ3|f + 1/p =: , (4.3.12)
x
1≤ pk≤4
an∑d (
d ek(p)ak(p )ak(pf ) = (1 + p−1/3) 1 + p−1/3 + p−2/3 + p−1 + p−4/3 + 2δ δ −2/32|e 2|f (1 + p )
1≤k≤4 )
+ (3δ3|eδ3|f − δ3|e − δ3|f )(1− p−1/3 + p−2/3) =:
ξe,f (p)
. (4.3.13)
yp
The contribution from the last sum, exclu(ding the)erro(r term, i)s thus∑
( ) 4 log
2 p ( ) e log p f log pF1 X 2 2+ 2 ιe,f p ϕ̂1 ϕ̂e/ f/ 2L p L L
p,e,f ∑ 2 ( ) ( ) (4.3.14)
+ F2(X)
4 log p
2 2+ ξ (p)
e log p f log p
ϕ̂ ϕ̂
L pe/ f/2
e,f 1 2
L L
p,e,f
=: F1(X)S5 + F2(X)S6,
which concludes the proof of Theorem 4.1.
4.4 Interpreting Theorem 4.1
In this section, we find the main term in Theorem 4.1, and confirm that it matches the
expected symplectic Katz–Sarnak main term. We remark again that this has already been
proven in a more general context in [CK, Theorem 4.26].
First, the main term coming from the first two rows of (4.3.1) is simply ϕ̂1(0)ϕ̂2(0).
Furthermore, by the comments in connection to (3.3.2), we know F (X)S (i)≪ X−δ2 2 , for a
δ > 0, if σi < 1. This bound also holds for F2(X)S2(1)S2(2) if σ = σ1 + σ2 < 1. Moreover,
by the previous chapter, we have F1(X)S1(i) = −ϕi(0)/2 +O(L−1), so that the main term
coming from the third and fourth row of (4.3.1) is
−ϕ1(0)ϕ̂2(0) − ϕ2(0)ϕ̂1(0)2 2 .
In addition, the contribution from F1(X)S1(1)S1(2) on the fifth row is ϕ1(0)ϕ2(0)/4.
24
4. The two-level density
To continue, we need to estimate S3, S4, S5 and S6. We will actually investigate these
expressions up to error ≪ L−2, as we will be interested in the term of size L−1 later. We
begin by consider∑ing S3, i.e2 log2 ( )( ) ( ) ( )4 xp p + 1 + 1 e log p f log p
L2 pe/2+f/2
θe θf ϕ̂1 ϕ̂2 .
p p L L
p,e,f
We can see that the contribution from the terms with e+ f > 2 to the sum is ≪ 1, so that
they contribute ≪ L−2 to S3. Indeed, this follows by bounding the absolute value of the
summand using the formula for a geometric sum, and the fact that the sum over 1/pδ is
finite for δ > 1. However, as θ1 = 0, we see that the sum over p, when e = f = 1 also
contributes≪ L−2 to S3. Thus, we have shown S −23 ≪ L . By a very similar method, using
γ1(p)≪ p−1/3, we can show S4 ≪ L−2.
Recall that S5 is give
4 ∑
n by
2 ( ) ( )log p ( ) e log p f log p2 ιe,f p ϕ̂1 ϕ̂2 ,L pe/2+f/2 L L
p,e,f
where ιe,f (p) is defined in (4.3.12). By the same reasoning as above, we only need to consider
the terms where e = f = 1. Now, ι1,1(p) = xp(1+p−1) = 1+O(p−1). Thus, we may replace
ι1,1(p) by 1 in S5 at the cost of an error of size ≪ L−2. The only term in S5 of interest here
is therefo
4 ∑re log2 ( ) ( ) ( ) ( )p log p log 4 ∫p ∞= log u log u log u2 ϕ̂1 ϕ̂2 2 ϕ̂1 ϕ̂2 dθ(u),L p L L L 1 u L Lp
where θ(u) is as in (3.2.7). Writing θ(u) = u + θ(u) − u, and using the error term from
(3.2.∫7), as well as(integr)ation(by pa)rts, allows ∫us to rewrite(the in)tegra(l as )4 ∞ log u log u log u 4 ∞ log u log u log u
2 ∫ ϕ̂1 ϕ̂2 du+ ϕ̂1 ϕ̂2 d(θ(u)− u)L 1 u L ( ) L ( )L2 1 u L L
= 4
∞ log u log u log u
2 ϕ̂1 ϕ̂2 du+O(L
−2).
L 1 u L L
Finally, after making the change o∫f variables t = L−1 log u and using that both ϕ̂i are even,we find that ∞
S = 2 |u|ϕ̂ −25 1(u)ϕ̂2(u)du+O(L ).
−∞
By using the same method, one finds S6 ≪ 1, and thus F2(X)S6 ≪ X−1/6.
Lastly, as Ei ≪ L−1, the only main-term contribution from the last line in (4.3.1) is
from −2R3. The main term from R3 is calculated in (3.3.3), if one replaces ϕ by ϕ3 = ϕ1ϕ2.
Thus, we have shown that for ϕ, with
1− θ
σ < 1/2 + ≤ 1,ω
the two-level density equals
(0) (0∫)− ϕ1(0)ϕ̂2(0) − ϕ2(0)ϕ̂1(0) + ϕ1(0)ϕ2(0)ϕ̂1 ϕ̂2 2 2 4∞ (4.4.1)
+ 2 |u|ϕ̂1(u)ϕ̂2(u)du− 2ϕ̂1ϕ2(0) + ϕ1(0)ϕ2(0) +O(L−1).
−∞
Using the notation of [Rb], the Katz-Sarnak prediction for the main term of this symplectic
∫family is
ϕ(x1, x2) det((K−1(xi, xj))i,j=1,2)dx1dx2, with ( ) =
sin π(x− y) sin π(x+ y)
K−1 x, y
R2 π(x− y)
−
π(x+ y) .
This is exactly (4.4.1), see e.g. [M, Thm. 5.9].
25
4. The two-level density
26
5
Counting cubic fields
In this chapter we indicate how to obtain the estimates, given in (4.2.2), for the number of
cubic fields with discriminant bounded by a given magnitude and with prescribed splitting
behaviour for a finite set of primes. The main terms in these estimates were found already
in [DH], while the secondary terms were found independently in [BST] and [TT], through
two different approaches. In this chapter essentially all arguments are taken from [BST].
To avoid too long of a digression from our main goal of studying low-lying zeros, we will
focus on studying the main terms in the estimates and not the secondary terms. Despite
this, we are using the method in [BST], instead of [DH], because this method lays the
groundwork for the computation of the secondary term. The interested reader is therefore
well-prepared for studying the rest of the argument in [BST] after having read this chapter.
Furthermore, we will not focus on the error terms arising from the various calculations, as
our goal is simply to motivate how one obtains the main term, as well as how to take into
account the splitting types of a finite number of primes.
We begin by relating cubic fields to certain cubic rings, which are themselves related to
integral cubic forms. We then show how to count the number of cubic forms with discrimi-
nant bounded by a given magnitude. Finally, we use a sieve argument to relate the estimate
for the number of cubic forms to the number of cubic fields. We end the chapter by briefly
describing how to generalise the calculation for the main term to also obtain a secondary
term.
5.1 Cubic rings and forms
Following [BST, Ch. 2-3], we want to relate cubic fields to (binary) cubic forms. To ac-
complish this, we first relate cubic fields to certain cubic rings. Here, a cubic ring is a
commutative ring, that is also a free Z-module of rank 3. We say that a cubic ring R is
nonmaximal at p, if it is contained in some other cubic ring R′, with index divisible by p.
We say that the cubic ring R is maximal if it is not nonmaximal at any p. We remark that
by the theory of modules over a PID, any containment R ⊆ R′ of cubic rings, implies that
R has finite index in R′, cf. [N, Thm I.2.12]. Thus, a maximal cubic ring is not strictly
contained in any cubic ring.
We define the discriminant of a cubic ring in a similar manner as the discriminant of
a cubic field, i.e. as the determinant of any matrix representing the bilinear trace form
(x, y) 7→ Tr(xy), with respect to some Z-basis of R. Finally, we say that a cubic ring R is a
cubic order if it is an integral domain.
We are now ready to state an introductory lemma that relates cubic fields to maximal
cubic rings. We remark that this lemma is not contained in [BST].
Lemma 5.1. There is a discriminant preserving bijection between isomorphism classes of
cubic fields and isomorphism classes of maximal cubic orders.
Proof. Given a cubic field K, we consider its ring of integers R := OK . By algebraic number
theory, we know that this is a cubic ring, see e.g. [N, Ch. I.2]. We claim that R is also
maximal. Indeed, if R′ is any cubic ring containing R with a finite index, then R′ can
naturally be considered as a subset of Frac(OK) = K. Further, as R′ is a free Z-module of
rank 3, it is in particular finitely generated over Z, which means that all elements of R′ are
27
5. Counting cubic fields
integral, whence R′ ⊆ OK = R, so that R is maximal. It follows that we may associate a
maximal cubic order to any cubic field.
Conversely, suppose that R is a maximal cubic order, so that we may consider K :=
Frac(R). Let x1, x2, x3 generate R over Z. Then Q[x1, x2, x3] is a field containing R, as all
xi are algebraic, and thus it is equal to K. It follows that K is a cubic field extension of Q,
as any Z-basis of R is a Q-basis of K. Thus, to any maximal cubic order, we can associate
a cubic field.
The maps we have constructed induce maps on the appropriate sets of isomorphism
classes, and it is clear that the maps are each other’s inverses. It is also clear from the
definitions that the discriminant is preserved.
We conclude that instead of counting cubic fields, we may count maximal cubic rings.
Next, we relate cubic rings to cubic forms using the Delone-Faddeev correspondence. Here,
an integral cubic form is a homogeneous polynomial f(x, y) = ax3+ bx2y+ cxy2+dy3, with
a, b, c, d ∈ Z. On the set of integral cubic forms, we define a group action of GL2(Z) by
letting γ ∈ GL2(Z) act on f by
( 1γf)(x, y) = det( )f((x, y)γ),γ
where (x, y) is considered as a row vector in the right-hand side. We now prove [BST, Thm.
9].
Theorem 5.2 (BST). There is a bijection between the set of GL2(Z)-orbits of integral
binary cubic forms and the set of isomorphism classes of cubic rings.
Proof. We give a direct proof from [BST], with some extra details provided. Let R be a
cubic ring. We begin by finding an appropriate basis of R, which in particular will include 1.
Begin by choosing any basis x1, x2, x3 of R, and write 1 = λ1x1 + λ2x2 + λ3x3, λi ∈ Z. Let
d0 be the greatest integer dividing all λi. Then 1/d0 ∈ R, and we must thus have d0 = 1, as
R is a Z-module of finite type. This implies that R/Z is a finitely generated and torsion-free
module over the PID Z; whence it is also free. Hence, we may pick a basis {1, ω, θ} of R.
Furthermore, we may require ωθ ∈ Z by possibly subtracting some element of Z from each
of ω, θ. We call a basis {1, ω, θ} such that ωθ ∈ Z a normal basis.
Using our normal basis, we write
ωθ = n, ω2 = m− bω + aθ, θ2 = ℓ− dω + cθ, (5.1.1)
for some a, b, c, d, n,m, ℓ ∈ Z. We then associate the binary cubic form f(x, y) = ax3 +
bx2y + cxy2 + dy3 to the cubic ring R. Conversely, to a binary cubic form f(x, y) as above,
we associate a cubic ring R with basis {1, ω, θ} with multiplication laws defined by (5.1.1).
If we set
n = −ad, m = −ac, ℓ = −bd, (5.1.2)
then one can confirm that the multiplication becomes associative, so that the multiplication
laws really define a ring. In particular, this is the unique choice of values for n,m, ℓ which
makes the multiplication associative.
To show that this map induces a map between isomorphism classes of rings, and or-
bits under the GL2(Z)-action, we require another description of the map, using the wedge
product. It turns out that if ∧we start with a cubic ring R, then the associated cubic form
represents the map 2R/Z → (R/Z) ∼= Z defined by r 7→ r ∧ r2. Indeed, if we write
r = xω + yθ, then a direct calculations shows that r ∧ r2 = f(x, y)(ω ∧ θ).
Now, any ring R′ isomorphic to R must have elements obeying the same multiplication
laws as R does. However, they need not be associated to the same cubic form, as the form
depends on the choice of basis {ω, θ} of R/Z, where a denotes the image of a ∈ R under
the natural map R 7→ R/Z. Here ω, θ are chosen in the unique way making {1, ω, θ} a
normal basis. Any change of basis of R/Z =∼ Z2 is given by some γ ∈ GL2(Z). Specifically,
28
5. Counting cubic fields
the coordinates (β1, β2)T with respect to t(he old ba)sis are transformed into γ(β1, β
T
2) with
respect to the new basis. Write
= α1 αγ 2 .
α3 α4
Then, under the map r 7→ r ∧ r2 above, changing the basis of R/Z by γ results in changing
f(x, y)(ω ∧ θ) into
f((x, y)γ)
det (e1 ∧ e2),γ
where e1 = α1ω + α3θ, e2 = α2 + α4θ. It follows that the given map between cubic forms
and cubic rings, induces a bijection between isomorphism classes of cubic rings, and GL2(Z)
orbits of cubic forms, as desired.
The correspondence above, called the Delone-Faddeev correspondence, has several nice
properties. First, it is discriminant preserving [BST, Prop 10], where we define the dis-
criminant Disc(f) of the form f(x, y) = ax3 + bx2y + cxy2 + dy3 as b2c2 − 4ac3 − 4b3d −
27a2d2 + 18abcd. Indeed, this follows by simply calculating the discriminant of the ring R,
with multiplication table (5.1.1). We remark that this discriminant is the usual univari-
ate discriminant of both f(x, 1) and f(1, y), whence it is zero precisely when f(x, y) has a
multiple root as a homogeneous polynomial.
Second, if we denote by R(f) the cubic ring corresponding to the cubic form f , then the
stabiliser of f is isomorphic to the group of ring automorphisms of R(f), [BST, Prop. 12].
Indeed, this follows immediately from the connection between base changes of R/Z, and
the GL2(Z)-action on the space of binary cubic forms. Lastly, the ring R(f) is an integral
domain iff the cubic form f is irreducible [BST, Prop 11]. We leave out the proof for the
sake of brevity. As in [BST], we remark that the Delone–Faddeev correspondence and the
properties above, can be generalised to hold for other base rings than Z. In particular, we
will be interested in the base rings Fp and R. We leave out the precise details of these
generalisations.
We now give a criterion for the cubic ring R being maximal at p [BST, Lemma 13].
Lemma 5.3 (BST). Suppose that a cubic ring R is nonmaximal at p. Then there is a
normal Z-basis {1, ω, θ} of R such that at least one of
Z+ Z(ω/p) + Zθ and Z+ Z(ω/p) + Z(θ/p)
is a ring.
Proof. As R is nonmaximal at p, there exists a cubic ring R′ containing R with finite
index, divisible by p. By a natural identification, we may consider R′ as being contained in
(Z∗)−1R, i.e. the ring of fractions created by inverting all nonzero integers in R. Next, let
S = {p, p2, p3, ...} and consider the ring of fractions S−1R ⊆ (Z∗)−1R, i.e. the result of only
inverting powers of p.
We define the ring R1 = R′ ∩S−1R and note that by applying the structure theorem for
finite abelian groups to R′/R, we see that R1 is nonempty, as R′/R must contain elements
of order p. Clearly, R1 contains R, and furthermore, the index [R1 : R] must equal some
power of p, again by the structure theorem for finite abelian groups. Note also that R1 is a
cubic ring, as it contains the cubic ring R, and is contained in the cubic ring R′.
The existence of the Smith normal form, see [J, Thm. 3.8], implies that we may pick a
basis {1, ω, θ} of R so that
R1 = Z+ Z(ω/pi) + Z(θ/pj), (5.1.3)
with i ≥ j, say. A calculation using (5.1.1) shows that the above holds even if we first
normalise the basis. If i = 1 above, then the proof is completed, so we assume i > 1. We
will now "reduce" the numbers i and j, until i = 1, which will prove the lemma.
Using the multiplication laws (5.1.1), we see that (5.1.3) being a ring is equivalent to the
conditions
a ≡ 0 mod p2i−j , b ≡ 0 mod pi, c ≡ 0 mod pj , d ≡ 0 mod p2j−i, (5.1.4)
29
5. Counting cubic fields
with the convention that any congruence relation holds if the exponent of p is less than or
equal to 0. If j = 0, then we can immediately replace the pair (i, j) by (i − 1, j) with the
conditions above still holding. If instead j ≥ 1, then we can replace (i, j) by (i − 1, j − 1).
Applying this procedure a finite number of times, we will have reduced to i = 1, and j = 0
or j = 1, as desired.
Let R(f) be a cubic ring, and f the corresponding binary cubic form under the Delone–
Faddeev correspondence, i.e. under the correspondence given in Theorem 5.2. Then, (5.1.4)
with i = j = 1, or i = 1, j = 0, implies that R(f) is nonmaximal at p if and only if p divides
all coefficients of f , or if there is some g(x, y) = a′x3 + b′x2y + c′xy2 + dy3 ∈ Orb(f) such
that p2 | a′, and p | b′.
If we let Up denote the set of cubic forms which do not fulfil any of the conditions just
mentioned, then we have proved [BST, Thm. 14].
Theorem 5.4 (BST). The cubic ring R(f) is maximal at p if and only if f ∈ Up. Further-
more, R(f) is maximal if and only if f ∈ Up for all primes p.
We now introduce a connection between certain binary cubic forms and the splitting
type of a prime p in the ring of integers of a cubic field. We will let Fp denote the field with
p elements, and Fp some algebraic closure of this field. Let f be any binary cubic form,
which is nonzero when reduced modulo p. Then, by considering f(x, 1) and f(1, y), the fact
that Fp is algebraically closed implies that the homogeneous polynomial f has exactly three
roots, counting with multiplicity, in the projective space P1F . Here, for any commutativep
ring R, the projective space P1R is defined as the set of equivalence classes of pairs (r1, r2),
with r1, r2 ∈ R not both 0, where (r1, r2) is related to (r3, r4) if and only if there is some
unit s ∈ R such that (r1s, r2s) = (r3, r4). We denote the equivalence class of (r1, r2) by
[r1, r2].
Now, to each prime p, and cubic form f with nonzero reduction modulo p, we want to
associate a symbol depending on the roots of f modulo p. If f mod p has three distinct
roots in Fp, we define this symbol as (111), while if it has one root of multiplicity exactly
equal to 2 in Fp, we set the symbol to (121), and if it has a root of multiplicity 3 we define
the symbol as (13). In the case when f mod p has a root in some quadratic extension of Fp,
we define the symbol as (21), and lastly, if it has a root in some cubic extension, we set the
symbol to (3).
The reader may note the similarity to the symbols we have used in earlier chapters to
denote the splitting type of (p) in some cubic extension. The similarity is not a coincidence;
indeed, the symbol we have just defined is equal to (fe1fe21 2 ...), with all fi, ei ∈ {1, 2, 3}, if
and only if
R(f) ∼ F= p
f1 [t1] Fpf2 [t2]
(p) (te11 )
⊕ ( e2) ⊕ · · ·. (5.1.5)t2
In particular, by using the Chinese Remainder Theorem and lifting representatives of the
prime ideals of the right-hand side, we see that (p) splits as pe11 p
e2
2 · ...., where the norm of
pi is given by pfi .
We illustrate the proof of (5.1.5) for the case (111). Some points of the argument are
taken from [Wr, Ch. 2]. We begin with some preparation. Up to scaling, GL2(Z) acts on the
space of binary cubic forms by a change of variables. Thus, we may essentially consider the
action as a linear transformation of the roots. Furthermore, for the sake of studying the ring
in (5.1.5), we may study GL2(Fp)-orbits of binary cubic forms, instead of GL2(Z)-orbits.
Now, define PGL2(Fp) := GL2(Fp)/D, where D is the group of all matrices λI2, λ ∈ Fp.
This is the projective general linear group of dimension 2. It is well-known, see e.g [DM,
Exercises 2.8.4, 2.8.7], that this group is triply transitive, i.e. for any three projective points
there is some element mapping these to any other three projective points. It follows that f
reduced modulo p lies in the GL2(Fp)-orbit of the form uv(u+ v) modulo p.
Using the above, and the Delone–Faddeev correspondence over Fp, we see that R(f)/(p)
is isomorphic to R(xy(y− x))/(p). Now, again by (5.1.1) and (5.1.2), the ring R(xy(y− x))
30
5. Counting cubic fields
is determined by the multiplication laws ωθ = 0, θ2 = θ, and ω2 = ω. Thus,
R(xy(y − x)) ∼
( ) = (Fp)
3,
p
as desired. An explicit isomorphism is given by 1 →7 (1, 1, 1), [ω] 7→ (0, 1, 0) and [θ] 7→
(0, 0, 1).
Before ending this section, we want to investigate how common the various splitting
types are, and in particular how common they are among the maximal forms. Let T be
some splitting type, and define Tp(T ) as the set of all integral binary cubic forms f , with
nonzero reduction f̄ modulo p, splitting according to T . We define Up(T ) as Up ∩ Tp(T ).
Now, if f is some integral binary cubic form, then membership in the various sets Tp(T )
only depends on the coefficients modulo p, while maximality at p depends on the coefficients
modulo p2. Motivated by this, we define a subset U ′p(T ) of the set of binary cubic forms
with coefficients in Z/p2Z. Here we let f ∈ U ′p(T ) if and only if its roots modulo p splits
as T , and if f is the reduction of some integral binary cubic form f modulo p2, where f is
maximal at p. We define U ′p as the union over i of all U ′p(Ti).
The set of binary cubic forms with coefficients in Z/p2Z is naturally identified with
(Z/p2Z)4, and on this set, we consider the measure µp, defined as the counting measure
divided by p8. We then have the following lemma [BST, Lemmas 18-19].
Lemma 5.5 (BST). We have
2
µ (U ′p p(111)) =
1 · (p− 1) p(p+ 1)6 p4 ,
(U ′ (12)) = 1 · (p− 1)
2p(p+ 1)
µp p 2 ,p4
2
µ (U ′ (3)) = 1 · (p− 1) p(p+ 1)p p 3 ,p4
2
µ ′ 2p(Up(1 1)) =
(p− 1) (p+ 1)
p4
,
(U ′ (13)) = (p− 1)
2(p+ 1)
µp p ,p5
so that we in particular have
3 2
µp(U ′ ) =
(p − 1)(p − 1)
p 5 .p
Proof. The last equality follows by adding together all the other densities. We show how to
compute the density for T = (111) and T = (13), and refer to [BST] for a complete proof.
We begin by considering T = (111). Note that the discriminant of any form f with
distinct roots modulo p, cannot be divisible by p. Furthermore, if R = R(f) ⊆ R′, for some
cubic ring R′, we have Disc(R) = (R′ : R)2Disc(R′), cf. [N, Thm I.2.12], which together
with p ∤ Disc(R) implies that R must be maximal at p. Thus to find µp(U ′p(111)), it suffices
to find the corresponding density for forms with three distinct roots. In particular, we can
reduce our forms modulo p, and compute the density in (Fp)4. Counting such forms is
the same as counting the number of ways to choose 3 different roots, and then choosing a
nonzero scalar. As there are p+ 1 elements to(choose)from in PF , we havep
µp(Up(111)) =
1 · p+ 14 3 · (p− 1),p
as desired.
We turn to the case T = (13). The density of forms in (Fp)4 with a triple root, is
(p− 1)(p+1)/p4, by a similar argument as above. Now, not all forms f in (Z/p2Z)4, whose
reductions modulo p has a triple root, correspond to maximal integral forms, but as we
31
5. Counting cubic fields
will see most of them do. First, as the forms are nonzero modulo p, not all coefficients
can be divisible by p, so we only need to check the second condition for being nonmaximal.
By the action of a GL2(Z) matrix on f (or on some lifting f), we may send the root of
f mod p to [1, 0], which brings f to the form ax3 + bx2y + cxy3 + dy3, where a, b, c are
divisible by p. Exactly 1/p of these forms also have p2 | a, and we thus arrive at the density
(p− 1)(p+ 1)p−4(p− 1)p−1, as desired.
5.2 Counting integral binary cubic forms
We now want to find an estimate for the number of integral binary cubic forms, with dis-
criminant bounded by X. These results will be used in the next section, where we sieve
these forms for maximality.
The goal of this section will be to partially prove the following theorem [BST, Thm. 5],
using the methods of [BST, Ch. 5,6].
Theorem 5.6 (BST). Let N(ξ, η) denote the number of GL2(Z)-equivalence classes of ir-
reducible integral binary cubic forms f satisfying ξ < Disc(f) < η. Then,
√
π2 3ζ(2/3)Γ(1/3)(2π)1/3 ( )
N(0;X) = X + 5/672 30Γ(2 3) X +O X
3/4+ϵ
ϵ ,
/
(− 0) = π
2
+ ζ(2/3)Γ(1/3)(2π)
1/3 ( )
N X, X 5/6 3/4+ϵ24 10Γ(2 X +Oϵ X ./3)
Our goal will not be to rigorously prove these estimates, but rather to indicate how to
obtain the main terms above.
To begin counting these forms, we need to not only consider integral binary cubic forms,
but also real binary cubic forms. We denote the space of integral binary cubic forms by VZ,
and the corresponding real space by (0)VR. Further, let VR be the subset of VR consisting of
forms with positive discriminant, and (1)VR be the set of forms with negative discriminant.
Using the generalisation of the Delone–Faddeev correspondence mentioned in the previous
section, we see that both these sets consist of one GL2(R)-orbit each, as the corresponding
ring is either R3 or R⊕ C.
It turns out that we will need to study the group GL2(R) a bit closer before we can start
counting forms. In particular, we will need to study the natural action of GL2(Z) on GL2(R).
Our goal is to find a natural fundamental domain, i.e. some subset of GL2(R) containing
one representative from each GL2(Z)-orbit. This will be carried out in the subsection below
5.2.1 The group GL2(R)
We now study the action of GL2(Z) on GL2(R) and find a fundamental domain. First, we
define
K1 = {γ ∈ GL2(R) : γ is orthogonal}, ( −1 )
A+ = {a(t) : t ∈ R+}, where a(t) =
t 0
( 0 ) ,t
N = {n(u) : n ∈ R}, where ( ) = 1 0n u ,
u( 1 )
Λ = {L(λ) : λ ∈ R+}, where
λ 0
L(λ) = 0 .λ
We will often write λ instead of L(λ). By considering the QR factorisation of a matrix, and
taking inverses, we see that each element in GL2(R) can be uniquely written as a product
nTakλ, with n ∈ N , a ∈ A+, k ∈ K1 and λ ∈ Λ. Here nT denotes the transpose of n.
32
5. Counting cubic fields
We now find a fundamental domain, essentially following the approach in [Ls, Lemma
3.33], with some modifications. First, as the(matrix)
:= 0 1P 1 0 (5.2.1)
is an element of GL2(Z) with determinant −1, we need only consider elements with a positive
determinant for our fundamental domain. As an invertible matrix with integral entries, has
a determinant equal to ±1, this means that we may consider the action of
SL2(Z) on GL2(R)+, and find a fundamental domain for this action instead. Here GL2(R)+
denotes the elements in GL2(R) with positive determinant, and these elements can also be
written uniquely as a product nTakλ as above, but now k ∈ SO2(R).
Next, we consider the group act(ion of)GL2(R)
+ on the upper half-plane H of C, given
by
a b = az + bz
c d cz + .d
That this is really an action on H follows after a straightforward calculation, using the
positivity of the determinant. Now, the matrix (n(u))Ta(t−1) maps z to t2z + u, which
shows that this is a transitive group action. Now, consider the element i ∈ H. A calculation
shows that this element is stabilised by all of Λ and SO2(R). No other element in GL2(R)+
can stabilise i, as (n(u))Ta(t−1)kλ acts on i by mapping it to t2i + u. Thus, by the orbit
stabiliser theorem, we have the equivalence
GL2(R)+
(R)Λ =
∼ H, (5.2.2)
SO2
as G-sets, where G := GL +2(R) . Here, the left-hand side denotes the left cosets. Now, if
we let I denote the identity matrix, then ±I ∈ SO2(R). In particular, this equivalence also
holds if one instead considers the action of the modular group PSL2(Z) = SL2(Z)/(±I) on
either side of the above.
We recall that our immediate goal is to find a fundamental domain for the action of
SL2(Z) on GL2(R)+. We first find a fundamental domain for the action of PSL2(Z) on the
left-hand side of (5.2.2), or equivalently the action of PSL2(Z) on H. A classical result, see
[S, Thm. VII.1], asserts that such a fundamental domain is given by all z with |Re(z)| < 1/2,
|z| > 1, at least up to a Lebesgue null-set. As the purpose of finding a fundamental domain
will be to integrate over it later, we may disregard any null sets. As the isomorphism above
is given by acting on i, this gives the conditions |u| < 1/2, t4 + u2 > 1 on u and t of
(n(u))Ta(t−1).
Lastly, we must extend the fundamental domain for the action on the cosets to a funda-
mental domain for the action on all of(GL2(R)+. Let)K ′ be the subset of SO2(R) containingmatrices of the form
cos θ − sin θ
sin cos , (5.2.3)θ θ
with π < θ ≤ 2π. This is a set of lifted representatives of the cosets SO2(R)/(±I).
We claim that up to a null set (a null set when viewing GL2(R) as a subset of R4 with the
Lebesgue measure), a fundamental domain of the action of SL2(Z) on GL2(R)+, and thus
also for the action of GL2(Z) on GL2(R), is given by F ′ = {(n(u))Ta(t−1)kλ : k ∈ K ′, λ ∈
Λ, |u| < 1/2, t4+u2 > 1}. Indeed, by our investigation of the action on cosets above, almost
every element ofGL2(R) is represented by some element in F ′. Furthermore, no two elements
in the set above lie in the same SL2(Z) orbit. We see this by noting that if g ∈ SL2(Z) is
such that g(n(u1))Ta(t−11 )k1λ1 = (n(u2))Ta(t
−1
2 )k2λ2, then by reducing modulo SO2(R)Λ,
we find that u1 = u2, t1 = t2 and that g stabilises the point (n(u ))Ta(t−11 1 )i ∈ H. It follows
that g = ±I by [S, Thm. VII.1.(3)]. This shows that ±k1λ1 = k2λ2; whence λ1 = λ2 and
k1 = k2, as K ′ only contains rotations with π < θ ≤ 2π, which proves the claim.
The fundamental domain F ′ that we have found is not the one used in [BST]. We,
therefore, show how to modify F ′ to obtain a fundamental domain more similar to the one
33
5. Counting cubic fields
from [BST]. Recall that we defined a matrix P in (5.2.1). As P ∈ GL2(Z), we see that
the transformed domain F := PF ′P−1 is also a fundamental domain. A straightforward
calculation shows that F is the set {n(u)a(t)kλ : k ∈ K,λ ∈ Λ, |u| < 1/2, t4+u2 > 1}, where
K is now the subset of SO2(R) containing matrices of the form (5.2.3), with 0 ≤ θ < π.
We end this section by introducing a (Haar) measure dg on GL2(R). Indeed, such a
measure is given by dg = t−3λ−1dudtdkdλ, where n(u), a(t), λ, k are as in the decomposition
n(u)a(t)kλ above. The measure dk is a measure on the real orthogonal group, induced by
the parametrisation using θ, that we have seen for SO2(R) above. We normalise dk to give
the set K measure 1. The most important property of dg is that it is GL2(R)-invariant, but
we leave out the proof of this fact for the sake of brevity, see e.g. [Ls, Lemma 3.22].
5.2.2 Reducing to an integral
We now show how to apply the results of the previous subsection to count integral binary
cubic forms, following [BST, Ch. 5.1]. Fix any vector (i)v ∈ VR , i = 0, 1, and consider the
set F0v, where F0 is some fundamental domain of the action of GL2(Z) on GL2(R). Recall
that the set F is not quite an actual fundamental domain, but rather a fundamental domain
up to a set of measure zero. Therefore, for these initial algebraic arguments, we work with
F0 instead.
Let ni be the size of the stabiliser of v in GL2(R). By the generalised Dalone–Faddeev
correspondence, n0 = 6 and n1 = 2. We claim that F0v is the union of ni fundamental
domains for the action of GL (Z) on (i)2 VR . Indeed, this follows from the fact that gv = g′v
is equivalent to g−1g′ ∈ Stab(v), and that F0 is a fundamental domain for GL2(Z) acting
on GL2(R). An element in F0v may belong to several of the ni fundamental domains.
We therefore view F0v as a multiset, where the multiplicity of a point x is given by the
cardinality of {g ∈ F0 : gv = x}. If we let GL2(R) act on v instead of F0, then every
such multiplicity would be ni. Using this fact, a calculation shows that for x ∈ VZ the
multiplicity above in F0v must instead be ni/mi(x), where mi(x) is the size of the stabiliser
of x in GL2(Z). We mention that if x corresponds to an order under the Delone–Faddeev
correspondence, then mi(x) either equals 1 or 3, and we have mi(x) = 3 if and only if the
corresponding fraction field is a C3-field. Indeed, this follows from Q[R] = Frac(R).
Write (i)N(VZ ;X) for the number of GL2(Z)-orbits of irreducible integral binary cubic
forms in (i)VZ with absolute discriminant less than X. Then, the discussion above has shown
that ni · ( (i)N VZ ;X) is equal to the number of integral points in F0v, as long as we weigh
the point x with mi(x) = 3 counted in ( (i)N VZ ;X) with a factor 1/3. This is not a problem
as [BST, Lemma 22] shows that such points are very rare, and for our purposes we may
therefore ignore this weighting when it is convenient to do so.
We remark as in [BST], that counting points from a single F0v0 is very hard. The method
of proof will therefore instead be based on calculating averages of sets F0v, where v varies
over points in some compact subset of (i)VZ . As mentioned earlier, in the actual calculations
we will use the set F in place of an actual fundamental domain. Specifically, the previous
discussion implies th∫at irr −1
(i) #{x ∈ Fv ∩ VZ : |Disc(x)| < X}|Disc(v)| dv
( (i)N VZ ;X) =
v∈B∩VR ∫
−1 , (5.2.4)ni v∈B∩ (i) |Disc(v)| dvVR
where the set in the right-hand side is a multiset. This equality requires some explanation.
First, we have identified VR with R4 in the obvious way, and we have let dv denote the
Lebesgue measure. The equality holds because the number#{x ∈ Fv∩V irrZ : |Disc(x)| < X}
is equal to (i)niN(VZ ;X), at least if the points x with mi(x) = 3 are counted with a weight
1/3 in the left-hand side. Also, V irrZ is the subset of VZ consisting of irreducible forms, and
B is some suitable compact subset of VR. The factor |Disc(v)|−1 is not important for the
equality, but it will be important for the calculations that will follow, as we shall see in the
proposition below.
34
5. Counting cubic fields
The starting point for estimating ( (i)N VZ ;X) will be (5.2.4). Our immediate goal is to
replace the integral over VR with an integral over GL2(R), which will simplify the integration
by allowing us to use the precise definition of F given in the previous subsection. To
accomplish this, we need the following lemma [BST, Proposition 23].
Proposition 5.7 (BST). For i = 0, 1 let f be continuous on V iR, and let vi be any element
of (i)∫VR . Then, we have
( ) = 1
∫ ∫
ni
f gv1 dg 2 f(v)|Disc(v)|
−1dv = f(v)|Disc(v)|−1dv.
g∈GL2(R) π v∈GL2(R) 2π ∈ (i)vi v VR
We leave out a rigorous proof, and instead give the same proof sketch as in [BST]. The
first equality can be shown by changing coordinates in GL2(R) to those induced by its
natural embedding in R4, and then calculating the Jacobian of the map g 7→ gvi. Here, the
natural embedding is given by ( )
a b 7→ (a, b, c, d) ∈ R4.
c d
The second equality follows from our remark above, that the multiset GL2(R)vi contains
every point of (i)VR with multiplicity ni. Here, the integral over a multiset is defined in the
obvious way, by splitting the integral over the various multiplicities. We will make use of
a slight generalisation of the proposition above, where we allow f to be multiplied by the
indicator function of some reasonable subset of V iR.
5.2.3 Integrating over GL2
The content of this subsection is essentially contained in [BST, Ch. 5.3], however, we also
provide some details not found in [BST]. First, we specify the compact set we are integrating
over in (5.2.4) by letting B = {w = (a, b, c, d) ∈ VR : 3a2+ b2+ c2+3d2 ≤ C, |Disc(w)| ≥ 1},
for some fixed C ≥ 1. The precise value of C will not matter much. A calculation confirms
that this set is SO2(R)-invariant, in the sense that kB = B for all k ∈ SO2(R), which will
be important later.
We now move the integration in (5.2.4) to GL2(R), using Proposition 5.7. Consider the
map T : GL2(R)× VR given by T (g, v) = (i)gv. Fix a vi ∈ VR , and let H(i) = T (·, v )−1i (B),
so that the multiset H(i)vi contains every point of ∩ (i)B VR with multiplicity ni. Then, using
Tonelli’s theorem, we h∑ave th∫at the numerator of (5.2.4) is equal to
#{g ∈ F : x = gv}|Disc(v)|−1dv.
∈ (i)v B∩V
x∈V irr RZ
|Disc(x)|<X
The count in this integral is constant with respect to v so that we may apply (the generali-
sation of) Proposition 5.7 to r
2π ∑
ewrite ∫it as
#{g ∈ F : x = ghvi}dh. (5.2.5)
ni
x∈V irr h∈H
(i)
Z
|Disc(x)|<X
The set H(i) need not be very nice, which motivates us to move the integral to F instead.
The details of this move are essentially taken from the proof of [Ls, Prop. 3.17]. Begin by
rewriting the integra∫l as ∑ ∑ ∫
1 dh = 1{h∈F−1g′}dh, (5.2.6)
h∈H(i) g∈F g′
(i)
∈GL (R) h∈H2
x=ghvi x=g′vi
35
5. Counting cubic fields
where we used that multiplication by a group element is bijective, and where F−1 denotes the
set of inverses of the elements in F . Denote the inner integral above by Vol(H(i) ∩ F−1g′),
where the volume is taken with respect to the GL2(R) invariant measure dh. Using this
invariance, as well as the fact that the measure is also invariant under inversion, which can
be confirmed by a computation, we see that this volume is equal to Vol(g′(H(i))−1 ∩ F).
After interchanging summation and integration again, we find that the right-hand side of
(5.2.6) c∫an be wri∑tten as ∫ ∑ ∫
1 dg = 1 dg = #{h ∈ H(i) : x = ghvi} dg
g∈F ′ (i) −1 g∈Fg∈g (H ) g′∈gH(i)
g∈F
x=g′vi x=g′vi
Moving the sum from (5.2.5) back inside, we see that the expression in (5.2.5), i.e. the
numerator in (5.2.4), is e∫qual to2π #{x ∈ V irr ∩ gH(i)Z vi : |Disc(x)| < X} dg.ni g∈F
To proceed, we will analyse the integrand carefully. First, recall that by definition of H(i),
we have (i)H(i)vi = B ∩ V∫R , where each point has multiplicity ni. Hence, the above is equalto
2 (i)π #{x ∈ V irrZ ∩ gB ∩ VR : |Disc(x)| < X} dg, (5.2.7)
g∈F
where the set inside the integral is no longer a multiset.
Next, we write g as a product n(u)a(t)kλ ∈ F , and use that B is SO2(R)-invariant,
to see that gH(i) = (i)vi gB ∩ VR = n(u)a(
(i)
t)λB ∩ VR . We write, with a slight change in
notation compared to [BST], (i)B(u, t, λ,X) = n(u)a(t)λB∩{x ∈ VR : |Disc(x)| < X}. Using
this notation, we see that we should investigate the number of irreducible integral forms in
B(u, t, λ,X). This will be done using the following proposition, essentially due to Davenport
[D1], [D2]. The statement is modified from [BST, Prop. 24].
Proposition 5.8 (Davenport). Let R be a bounded semi-algebraic subset of Rn, defined by
at most k polynomial inequalities, each having degree at most ℓ. Then the number of integral
lattice points contained in R is ( )
Vol(R) +O max{Vol(R), 1} ,
where Vol(R) denotes the largest d-dimensional volume of the projection of R onto any d-
dimensional coordinate subspace, with d < n. The implied constant depends only on n, k
and ℓ.
Using this proposition we can count the integral lattice points in B(u, t, λ,X). Before
doing this, however, we must find a way to exclude the reducible forms. Consider a form
ax3 + bx2y + cxy2 + dy3 in VR, and note that if a = 0, then the form is certainly reducible,
as it has a factor y. Now, [BST, Lemma 21] asserts that reducible forms with a ̸= 0 are very
rare, and their contribution can therefore be absorbed in the error terms.
As mentioned earlier, we will not include the precise estimation of the error terms involved
in the calculations. We do however mention that by only having to consider points with
|a| ≥ 1, one can control the error term arising from Proposition 5.8. We are left with the
main term from (5.2.7) equa∫lling ∫
2π 1{ (i) dvdg,g−1v∈B∩V }
g∈F |Disc(v)|<X R
where the inner integral is the volume from Proposition 5.8. Interchanging the order of
integration, applying Proposition 5.7, using inversion invariance of dg and then interchanging
the order of integration a∫gain shows∫that this double integral is equal to
1 du{Disc(v)<X}dv .
∈ (i)u B∩V v∈Fu |disc(u)|R
36
5. Counting cubic fields
Applying Proposition 5.7 to the inner integral, and making a change of variables g 7→ gh for
a suitable h ∈ Λ the∫n shows th∫at the above is equal to
2π 1 2 −1{(det g)2<X} · (det g) dg|Disc(u)| du,
u∈B∩VR(i) g∈F
where we also made use of the equality Disc(gv) = (det g)2Disc(v), which follows from a
straightforward computation. Note that the inner integral does not depend on u, whence
we can separate the two integrals. The integral over u is exactly the denominator of (5.2.4),
and they thus cancel each other.
We have arrived at ∫
( (i); ) = 2πN VZ X 1
2
{(det g)2<X} · (det g) dg + Error. (5.2.8)
ni g∈F
The error term turns out to be ≪ X5/6. It remains to compute the integral above to find
the explicit value of the main term. If we write g = n(u)a(t)kλ, then det(g)2 = λ4, so that
we should integrate over the part of F where λ < X1/4. From our explicit description of F ,
and the definition of dg, and the fact that dk gives K measure 1, we see that the main term
of (5.2.8) is ∫ X1/42 ∫ ∫π λ3 π2
dndtdλ = X.
ni 0 −1/2≤u≤1/2 t≥(1−u2)1/4 t
3 12ni
As n0 = 6 and n1 = 2, this is indeed the desired main term.
5.3 Sieving for maximality
We now want to obtain estimates for the number of isomorphism classes of cubic fields,
with discriminant bounded by X. The results of the previous section do not quite suffice
to accomplish this, as we need to remove the nonmaximal forms from the count, and to
accomplish this we need to place certain congruence conditions on the coefficients of the
forms modulo p2. We make this more precise below.
Recall that ( (i)N VZ ;X) counts irreducible, integral binary cubic forms, with discriminant
less than X, up to GL2((Z) equivalence. S)imilarly, for any GL2(Z)-invariant set ⊆
(i)
S VZ
we can let N(S;X) count GL2(Z⋂)-orbits of irreducible forms in S. We are of course par-
tic(ularly in(terested in
(i)
) N )VZ ∩ p Up, X , which counts orbits of forms corresponding tocubic fields.⋂To study this counting function, it will be necessary to study the partial counts(i)
N VZ ∩ p<Y Up , X , for Y > 2. We will use the following theorem, which is a special
case of [BST, Thm. 26].
Theorem 5.9 (BST). Let S be a GL2(Z)-invariant subset of VZ, defined by congruence
conditions modulo p2 for primes p ∈ P, where P is a finite subset of the set of primes.
Then, we have
( ∩ (i)lim N S VZ ;X) = π
2 ∏
12 µp(Sp).X→∞ X ni
p∈P
Here µp denotes the counting measure on (Z/p2Z)4, divided by p8, and Sp ⊆ (Z/p2Z)4
denotes the reduction of S modulo p2.
Proof. We will only provide a rough sketch of the proof, taken from [BST, Ch.∏5.5]. The main
point is that S is a union of finitely many translates L1, ..., Lk of the lattice ( 2p∈P p )VZ =:
mVZ, by the Chinese Remainder Theorem. Again by t∏he Chinese Remainder Theorem, we
have that the number of such translates k is equal to ( p8p∈P µp(Sp)).
Next,∏apply the procedure of the previous section to the lattice mVZ in place of VZ.The main difference is that the volume from Davenport’s lemma gets scaled by a factor
m−4 = −8p∈P p , and accordingly, the result of the previous section gets scaled by the
same factor. Multiplying this by the number of translates yields the main term above.
37
5. Counting cubic fields
We remark that it is very important for the proof that the number of congruence condi-
tions is finite, as the error term grows rapidly with the number of congruence conditions.
Now we are ready to begin the calculations for finding the main term C(±1 X of((⋂3.2.3),)fol-)
lowing [BST, Ch. 8]. We will be studying the limit of the quotient −1 (i)X N VZ ∩ p Up ;X ,
as X → ∞. Recall that(U is the(intersec)tion )of all Up. We begin by bounding this fromabove. Trivially,
( ) ⋂i (∩ U ; ≥ (i) )N VZ p X N VZ ∩ U ;X ,
p<Y
as the former counting function imposes fewer restrictions on the forms. Next, by Theorem
5.9, the Chinese Remai(nder Th(eorem, an)d Le)mma 5.5, we h(ave⋂ 2 ∏ )( )
lim −1 (i)X N VZ ∩ Up ;X =
π
12 1−
1 1− 1 ,
X→∞ ni p2 p3
p<Y p<Y
for all Y . Letting Y →∞, and using the Euler product of ζ(s), it follows that
( ) 2
lim sup (i)X−1N VZ ∩ U ;X =
π = 1
X→∞ 12niζ(2)ζ(3) 2niζ(3)
.
Here we made use of the identity ζ(2) = π2/6, which is the conclusion of the well-known
Basel problem.
We now want to show that the right-hand side above is also a lower bound for the lim inf.
LetWp be the set of integral form⋂s, which a(re⋃nonmax)imal at p. Then, we have the inclusion
Up ⊆ Wp ∪ U ,
p<Y p≥Y
whence
( ) ( ( ⋂ ) ) ∑ ( )(i) (i) (i)
N VZ ∩ U ;X ≥ N VZ ∩ Up ;X − N VZ ∩Wp;X .
p<Y p≥Y
We are done if we can show that the second term, when divid(ed by X, tend)s to 0 as Y →∞.
This follows from [BST, Prop. 29], which implies that (i)N VZ ∩Wp;X ≪ Xp−2. The
proof of this requires more information about subrings and overrings of cubic rings than
what has been presented so far, so we omit the proof.
By using the fact that the sum over p−2 is summable, we can let Y → ∞ and thus
conclude that
N±(X) = C±1 X + o(X).
The attentive reader may remember that when we defined N±(X), we only included S3-
fields and not all cubic fields. However, we have a priori only shown the equality above for
the counting function counting all cubic fields, (but a result )due to Cohn [C] asserts that the
C3-fields only contribute a term ≪ 1/2 to (i)X N VZ ∩ U ;X , so that we may disregard this
contribution.
So far we have indicated how one obtains the main term of the first equality in (3.2.3).
Finally, we indicate how to obtain the main terms when imposing splitting conditions on
finitely many primes. Suppose that we pick primes p1, ..., pn and splitting types Tj1 , ..., Tj .n
Then, to count cubic fields satisfying these conditions, we can repeat the proof above, but
replace the sets Up1 , ...,Up by the sets Un p1(Tj1), ...,Up (Tj ). The argument goes throughn n
as before, but the constant C±1 should be multiplied by a factor∏n µpi(U ∏npi(Tji))
µp (Up )
= xp c (p ),i ji i
i=1 i i i=1
where the equality follows from Lemma 5.5. This gives the main term of (4.2.2).
38
5. Counting cubic fields
5.4 The secondary term
We provide a very brief sketch of how to modify the previous arguments to also provide a
secondary term in the counting functions for cubic fields, using the approach from [BST]. A
more complete description, even avoiding the study of the error terms, cannot be included
without sacrificing the brevity of this chapter.
The first step is to obtain a secondary term in the counting function for irreducible binary
cubic forms. Specifically, we will need an improvement of Theorem 5.9, which also provides
a secondary term. The main new idea needed is a technique referred to as "slicing" in
[BST]. The starting point is essentially (5.2.7), but instead of directly applying Davenport’s
lemma, we split the integral into two parts, depending on the size of t compared to λ. One
of the integrals is then sliced into a sum of several integrals by first separating the forms
in B(u, t, λ,X) depending on the value of their first coefficient a and then summing over
a. This sum over different integrals can then be transformed into a single integral that is
analysed by using residue theory, which allows us to separate the secondary term from the
main term.
Once we have refined the counting function for cubic forms in this way, the next step is
to find second-order densities for maximal forms with different splitting types, analogously
to Lemma 5.5. These second-order densities do not have the same obvious geometric inter-
pretation as the first-order densities do, but they can be given meaning by their connection
to certain lattices. We leave out the details and refer the interested reader to [BST, Ch. 7].
Finally, the last step is to sieve for maximality as above. For this, a sieve based on the
inclusion-exclusion principle is used. The first step is to count all binary cubic forms. Next,
for each prime p, we subtract the count for forms that are nonmaximal at p. Now, we have
subtracted too large a number, as forms that are nonmaximal at more than one prime have
been subtracted twice. Thus, we add back the counts for forms nonmaximal at at least two
primes, and so on. This is formalised using the Möbius function and Möbius inversion but
we leave out the details. The point is that maximal forms are hard to count, but it is easier
to count forms nonmaximal at some finite number of primes. Applying this sieve together
with the second-order densities in a refined Theorem 5.9 finally yields the secondary terms
in (4.2.2). The details can be found in [BST, Ch. 9].
39
5. Counting cubic fields
40
6
The one-level density through
the Ratios Conjecture
In Chapter 3, we analysed the one-level density for relatively small values of σ = sup(supp(ϕ̂)).
In this chapter, we describe and apply, the L-functions Ratios Conjecture which will allow
us to study the one-level density for any finite σ, conditional on this conjecture. We ob-
tain Conjecture 6.3, which leads to the estimate of the one-level density in Proposition 6.4.
Throughout the chapter, we will need to assume the Generalised Riemann Hypothesis for
the Dedekind zeta functions ζK(s), with K a non-Galois cubic field. We remark that essen-
tially all arguments of this chapter are taken from [CFLS, Ch. 4], but the presentation may
differ.
6.1 The Ratios Conjecture recipe
In previous chapters, we have studied both the one-level density and the two-level density
through a study of certain prime sums. Unfortunately, the prime sum S2(i), became harder
to estimate if σ was large. The purpose of this chapter is essentially to replace S2 with a
nicer sum, which does not grow with σ, which will allow us to obtain a prediction for the
one-level density for any finite σ. To accomplish this, we will need to accurately estimate
certain averages of ratios, of the form
1 ∑ L(1/2 + α1, fK) · ... · L(1/2 + αn, fK)
N±(X) L(1/2 + γ1, fK) · ... · L(1/2 + γn, fK)
, (6.1.1)
K∈F±(X)
for some complex values α1, ..., αn, γ1, ..., γn, whose real part is relatively small, and n ≥ 1.
Specifically, to evaluate the one-level density we will need to consider such ratios with n = 1,
while the two-level density requires us to set n = 2.
Unfortunately, calculating this sum of ratios turns out to be a very hard problem, and has
not been done for any family of L-functions associated to number fields. Therefore, instead
of calculating the sum above rigorously, we will use a heuristic due to Conrey, Farmer and
Zirnbauer [CFZ], sometimes referred to as the Ratios Conjecture recipe, to conjecture a
reasonable estimate for (6.1.1). The heuristic applies to ratios of any family of L-functions,
but we will only describe it for our family of Artin L-functions L(s, fK). We will also restrict
to the case n = 1 for simplicity, but the calculations for larger n are very similar, and we
will see explicit calculations for n = 2 in later chapters when we study the two-level density.
We will not describe the recipe exactly as given in [CFZ, Ch. 5.1], but instead describe a
slightly modified version, essentially due to [CFLS]. The reason is that the secondary term
in N±(X) will produce an error term that is quite large if one follows the original Ratios
Conjecture recipe, a phenomenon first observed in [CFLS].
We now give the recipe for evaluating
1 ∑ L(1/2 + α, fK)
±( ) (1 2 + ) . (6.1.2)N X L / γ, f
K∈F±( ) KX
41
6. The one-level density through the Ratios Conjecture
The first step is to rewrite the summand. Begin by formally expanding the denominator
into its Dirichlet series,
1 ∑∞= µK(h)
L(1/2 + ,γ, f ) h1/2+γK
h=1
where the explicit coefficients µK can be found by studying the Euler product of L(s, fK).
Recall that L(s, fK) obeys the functional equation
Λ(s, fK) = Λ(1− s, fK), where Λ(s, fK) := |D |s/2K Γ±(s)L(s, fK),
which implies that L(s, fK) has an approximate functional equation of the form
∑ λ (m) |D |(1−s)/2( ) = K + K Γ±(1− s) ∑ λK(m)L s, fK + Error ,
ms |DK |s/2Γ±(s) m1−sm≤N m≤M
for certain values of N,M , cf. [IK, Thm. 5.3]. In (6.1.2) we replace the numerator by its
approximate functional equation, excluding the error term, while formally extending both
summations to ∞.
The first step of the Ratios Conjecture recipe has thus transformed the summand of
(6.1.2) into
∑ λK(m)µ ∑K(h) Γ±(1/2− α) λK(m)µK(h)
1 2+ 1 2+ + |D |
−α
m / αh / γ
K Γ±(1/2 + α) m1 2− 1 2+
.
/ αh / γ
h,m h,m
In Section 6.3 below, we will see that µK(n) and λK(n) are ≪ϵ nϵ. In particular, this
implies that the first sum converges absolutely for Re(α),Re(γ) > 1/2, while the second
sum converges absolutely for Re(α) < −1/2,Re(γ) > 1/2. In particular, we cannot directly
guarantee that the sums converge absolutely for the same α, but this will be remedied in the
last step of the recipe. Until then, we will treat both sums over h and m purely symbolically.
The next step of the recipe is to change the order of summation so that the sum over
K is turned into the innermost sum. We then replace this inner sum, including the factor
( −1N±(X)) , with a suitable estimate of
1 ∑ 1 ∑ −α
±( ) λK(m)µK(h) and ±( ) |DK | λK(m)µK(h), (6.1.3)N X N X
K∈N±(X) K∈N±(X)
respectively. This leaves us with the sum
∑ M1(X) Γ±(1/2− α)∑ M2(X)
m1/2+αh1/2+
+
γ Γ±(1 2 + ) 1
,
/ α m /2−αh1/2+γ
h,m h,m
whereM1(X) andM2(X) denotes an estimate of the first and second term in (6.1.3) respec-
tively.
The last step of the recipe is to find a meromorphic continuation of each term above,
to a domain where both terms are well-defined at the same time, containing the point
α = γ = 0. We denote the terms above, and their meromorphic continuations, byR1(α, γ;X)
and R2(α, γ;X) respectively. The Ratios Conjecture is then that (6.1.2) is equal to
R1(α, γ;X) +R2(α, γ;X) + Error, (6.1.4)
in some suit(able (restr)ictio)n of this domain. Following the original recipe, one would expectan error O X−1/2+ϵϵ , see [CFZ, Ch. 5.1], but we will see that a more reasonable error in
this family is O Xθ−1+ϵϵ , with θ ≥ 1/2 as in (3.2.3).
42
6. The one-level density through the Ratios Conjecture
6.2 The one-level density
We now apply the Ratios Conjecture to obtain a conjectural improvement of Theorem 3.1.
This is done in [CFLS, Ch. 4-5], and essentially all arguments of this chapter are taken from
there. As mentioned above, we will assume the Generalised Riemann Hypothesis for ζK
throughout the entire chapter, in particular, this implies the regular Riemann Hypothesis.
We now begin our calculations. As L(s, fK) is entire, we have similarly as in Chapter 3
that ∑ ∑ ( ) ∑ (1 1 1 ∫ ∫ )L = − L′(1/2 + r, fK)ϕ γK
N±(X) ( ) 2π (
±
∫ N ∫(X)) ( )2πi (a) (−a) L(1/2 + r, f∈F±( ) γ ∈F±( K)K X K K X)
× Lr = 1 Lr 1
∑ L′(1/2 + r, fK)
ϕ 2 drπi 2 − ϕ ± dr.πi (a) (−a) 2πi N (X) L(1/2 + r, f )K∈F±(X) K
(6.2.1)
for any 1/2 > a > 0. Note that we made use of the Generalised Riemann Hypothesis to
ensure that a can be chosen as close to 0∑as we like. This motivates us to study the average1 L(1/2 + α, fK)
N±( ) , (6.2.2)X L(1/2 + γ, fK)
K∈F±(X)
using the Ratios Conjecture. Once we have estimated this expression, we can differentiate
the estimate with respect to α, and set α = γ = r to estimate the sum inside the integral
in (6.2.1). The validity of differentiating an estimate will be discussed near the end of the
next section.
We apply the first steps of the Ratios Conjecture recipe as described in the previous
section and formally w∑rite1 L(1/2 + α, fK)
±( ) (1 2 + ) = R
′
1(α, γ;X) +R′2(α, γ;X) + Error,N X L / γ, fK
K∈F±(X)
where
′ ( ; ) = 1
∑ ∑ λK(m)µK(h)
R1 α, γ X N±( ,X) m1/2+αh1/2+γ
K∈F±(X) h,m
and ∑ ∑
′ ( ; ) = 1 | |−αΓ±(1/2− α) λK(m)µK(h)R2 α, γ X N±( ) DKX Γ±(1/2 + ) 1 2− 1 2+
. (6.2.3)
α m / αh / γ
K∈F±(X) h,m
Recall that for now, we treat these sums purely symbolically, as they are perhaps not both
convergent at the same time. To finish the recipe we need to interchange the order of
summation in R′1 and R′2, and then find the estimates R1 and R2 from (6.1.4).
6.3 Averages over K
In this section, we will replace R′1 with an approximation R1 which will be defined in a
larger domain. The first step of finding such an approximation is to determine λK and µK
explicitly. The existence of an Euler product for L(s, fK) implies that the coefficients are
multiplicative, so it suffices to determine them at prime powers. As before, we will see that
the splitting type influences the shape of the coefficients.
Using the table from the end of Section 2.1 we find the coefficients of L(s, fK). First, fix
a prime p, e ≥ 1, and suppose p has splitting type T1 = (111). By expanding out the Euler
product we see that to find( the coe)fficient(λK(p
e) we need onl)y study the local factor at p,i.e. the factor
1− −s −2p = 1 + p−s + 2p−2s + ... .
43
6. The one-level density through the Ratios Conjecture
A combinatorial argument then implies that the coefficient in front of p−es is
∑e
λ eK(p ) = 1 = e+ 1.
ℓ=0
One can use a similar method for the other splitting types. For the T3 splitting type, it is
useful to first write the local factor as (1− p−3s)−1(1− p−s). We summarise the coefficients
in the table below. See also [CFLS, Ch. 2], where the same table is provided.
Splitting type λK(pe)
T1 e+ 1
T2 δ2|e
T3 τe
T4 1
T5 0
Here 1 e ≡ 0 mod 3,
τe = −1 e ≡ 1 mod 3,0 e ≡ 2 mod 3.
Finding µK is even simpler. Indeed, (L(s, fK))−1 also has an Euler product, which is
simply the reciprocal of the Euler product of L(s, fK). In particular, µK is a multiplica-
tive function. Furthermore, we find µK(pj) = 0 for j ≥ 3. To find the behaviour for
j = 1, 2 we study the local factors in the table from Section 2.1 more closely. If we for
instance suppose that p has splitting type (111), then the local factor at p of (L(s, fK))−1
is (1− p−s)2 = 1 − 2p−s + p−2s, whence we find that µ 2K(p) = −2 and µK(p ) = 1 for
this splitting type. One handles the other splitting types analogously. We summarise the
behaviour for j = 1, 2 in the table below.
Splitting type µK(p) µ 2K(p )
T1 −2 1
T2 0 −1
T3 1 1
T4 −1 0
T5 0 0
With these calculations∑completed, we are ready to find∑an approximation of R
′
1. Write
1 1
R′1(α, γ;X) = m1/2+αh1 2+
· ±( ) λK(m)µ/ γ K(h).N X
h,m K∈F±(X)
The next step of the original Ratios Conjecture is to replace the inner sum with the main
term of its average. We remark, as in [CFLS, Ch. 4], that doing so would lead to inaccurate
results because of the secondary term in N±(X). Instead, we continue to follow [CFLS] and
replace the inner sum with a more accurate approximation. Indeed, we have the following
lemma, and proof from [CFLS, Lemma 4.1].
Lemma 6.1 (CFLS). Let m,h ∈ Z+, and 1/2 ≤ θ < 5/6, ω ≥ 0 be such that (4.2.2) holds.
Then, for cubefre∑e h, we have1 ∏
±( ) λK(m)µK(h) = F1(X) f(e, s, p)xpN X
K∈F±(X) e∏ p ||m,p
s||h ∏ 
+ F θ−1+ϵ ω2(X) g(e, s, p)y p +O ϵ X (2e+ 5)p  ,
pe||m,ps||h p|hm,pe||m
44
6. The one-level density through the Ratios Conjecture
where
e+ 1 δ( 0 ) = + 2|e + τe 1f e, , p 6 2 3 + ,p
f(e, 1, p) = −e+ 13 +
τe 1
3 − ,p
δ
f( 2 e+ 1 2|e τee, , p) = 6 − 2 + 3 ,
−1/3 3 δ (1 + p−1/3)(1 + p−2/3) −1 −1/3 2
( 0 ) = (e+ 1)(1 + p ) + 2|e + τe(1 + p ) (1 + p )g e, , p 6 2 3 + ,p
(e+ 1)(1 + p−1/3)3 τ −1 −1/3 2( 1 ) = − + e(1 + p ) (1 + p )g e, , p 3 3 − ,p
( 2 ) = (e+ 1)(1 + p
−1/3)3 δ2|e(1 + p−1/3)(1 + p−2/3)− + τe(1 + p
−1)
g e, , p 6 2 3 ,
(6.3.1)
and where both Fi(X) are defined in (4.3.5). In the case when both e and s equals 0, then
one should consider the products as empty, equalling 1.
Proof. We begin by remarking that for h that are not cubefree, the sum on the left-h∏and side
above is si∏mply zero. Next, write and as products of distinct prime powers = J em h m p jj=1 j
and = J sh jj=1 pj , where ej , sj ≥ 0. Here, ej = sj = 0 is not allowed, unless m = h = 1.
Th∑en, by multiplicativity ∑ ∏J ∑ ∑ ∏J
( ) e s eλ m µ (h) = λ (p j )µ (p j ) = λ (p j sK K K j K j K j )µK(p
j
j ).
K∈F±(X) K∈F±(X) j=1 k K∈F±(X) j=1
p of type Tk
Here, k = (k1, ..., kJ) where ki ∈ {1, 2, 3, 4, 5}, and the sum is over all possibilities for k.
Further p = (p1, ..., pJ), and that p has splitting type Tk means that all pj have splitting
type Tk in K. Note that the product above is constant for every fixed k. If pj has splittingj
type Tk in K, definej
η1,p (k , e ) = λ (
ej ) ep , η (k , e ) = µ (p j ),
j j j K j 2,pj j j K j
and ∏J ∏J
η1,p(k, e) = η1,pj (kj , ej), η2,p(k, e) = η2,pj (kj , ej).
j=1 j=1
It follows that ∑ ∑
λK(m)µK(h) = η1,p(k, e)η2,p(k, s)N±p (X,Tk).
K∈F±(X) k
Now w∑e use (4.2.2) to write t[he abo∏ve asJ ∏J
± ± 5 6 ( )η /1,p(k, e)η2,p(k, s) C1 X (xpjckj (pj)) + C2 X ypjdkj (pj)
k j=1( ∏ j=1J )]
+Oϵ Xθ+ϵ pωj .
j=1
Next, we separate the expression into three parts and then move the product outside. The
result is ∏ ( ∑5 )
C±1 X xp η1,p(k, e)η
± 5/6
2,p(k, s)ck(p) + C2 X
pe||m,ps∏||h ( k=∑1 )  ∏ 5× y η θ+ϵ ωp 1,p(k, e)η2,p(k, s)dk(p) +Oϵ X (2e+ 5)p ,
pe||m,ps||h k=1 p|hm,pe||m
45
6. The one-level density through the Ratios Conjecture
where we used the explicit form of λK and µK to find the error term. The lemma follows
after evaluating the sums and dividing by N±(X). The functions f(e, s, p) and g(e, s, p)
correspond to the first and second sum above respectively. We show how to evaluate the
sum defining f(e, 1, p), and leave the rest to the interested reader.
To find f(e, 1, p) we should calculate
∑5
( ) ( 1) ( ) = −2(e+ 1)
0 · δ
+ 2|e + 1 · τe + (−1) · 1 + 0 · 0η1,p k, e η2,p k, ck p 6 2 3 ,p p2
k=1
where we used the tables for µK and λK given above and the table for ck(p) given in
connection to (3.2.3). After simplifying we find the desired expression. The remaining cases
are handled in the same way.
Note that the bound for the error term in Lemma 6.1 is not summable over m,h as it
is not even bounded. The philosophy of the Ratios Conjecture is that one should expect
enough cancellation within the error term to make it summable, and make the resulting error
term small. We therefore approximate R′1(α, γ;X) with R1(α, γ;X), where R1 is obtained
by weighting the first two terms in Lemma 6.1 with m−1/2−αh−1/2−γ , and then summing
over m and h. After rewriting the sum as a product, we find (cf. [CFLS, Eq. (4.6)])
R (α, γ;X) = F (X)RM1 1 1 (α, γ) + F2(X)RS1 (α, γ), (6.3.2)
where
∏ ∑ ∑ 
M ( ) = 1 + xpf(e, 0, p) + xpf(e, 1, p) ∑R1 α, γ (1 2+ ) (1 2+ )+(1 2+ ) + xpf(e, 2, p) pe / α pe / α / γ pe(1 ,/2+α)+(1+2γ)
p e≥1 e≥0 e≥0
(6.3.3)
and
∏ ∑ 
S( ) = 1 + ypg(e, 0, p) ∑ ypg(e, 1, p) ∑ ypg(e, 2, p)R1 α, γ pe(1/2+ +α) pe(1/2+α)+(1 +/2+γ) pe(1/2+α)+(1+2γ) .
p e≥1 e≥0 e≥0
(6.3.4)
Note that we are summing over all exponents giving a nonzero contribution to the left-hand
side of Theorem 6.1. Similarly to the sums defining R′1, these products converges absolutely
in Re(α),Re(γ) > 1/2.
We will now find a meromorphic continuation of RM1 and RS1 , beginning with RM1 . We
assume Re(α),Re(γ) > −1/2 so that the summands inside the products converge to 0. By
(6.3.1) we have∑f(1, 0, p) = 1/p, f(2, 0, p) = 1(+ 1/p so that )xpf(e, 0, p) 1 1 1
pe(1 2+ )
= 1+2 +O 3 2+Re( ) + 3 ,/ α p α p / α p /2+3Re(α)
e≥1
where we bounded the tail using a geometric series argument. In particular, if Re(α) >
−1/6 + δ, δ > 0, then the error is ≪ p−1−δδ . Continuing, we have f(0, 1, p) = −1/p,
f(1, 1, p) =∑−1− 1/p, and f(0, 2, p) = 0, whence ( )xpf(e, 1, p) = − 1(1 2+ )+(1 2+ ) 1+ + +O 1 1pe / α / γ p α γ p3/2+Re( +γ) p3/2+Re(2α+γ) ,
e≥0
and ∑ ( )xpf(e, 2, p) 1
pe(1/2+ )+(1+2 )
= O 3 2+Re( +2 .α γ p / α γ)
e≥0
In conclusion, if Re(α),Re(γ) >∏−(1/6 + δ we have ( ))1 1 1
RM1 (α, γ) = 1 + 1+2 − 1+ +O .p α p α+γ δ p1+δ
p
46
6. The one-level density through the Ratios Conjecture
∏
Recall the Euler product of the Riemann zeta function, ζ(s) = p(1 − p−s)−1, valid
when Re(s) > 0. Thus, for Re(α),Re(γ)(> −1/6 + δ,)w(e at least formally∏ −(1+2 ) ∏ (
have ))
ζ(1 + α+ γ) 1− p α
(1 + 2 ) = (( −(1+ + ))=( 1−
1 1
1+)2 1 + 1+) + +O
1
ζ α 1− p α γ p α p α γ p2+Re(2α+2γ)∏p p
= 1− 1 1 + 1 +O(p−1−δ) .
p1+2α p1+α+γ
p
This calculation, and the formula (1− x)(1 + x) = 1− x2, implies that
( ) := ζ(1 + α+ γ)
∏(
A M −1−
)
δ
3 α, γ (1 + 2 ) R1 (α, γ) = 1 +O(p ) , (6.3.5)ζ α
p
for Re(α),Re(γ) > −1/6 + δ. In particular, the error term is absolutely summable, whence
the product defining A3 converges absolutely for these α and γ. Furthermore, the con-
vergence is uniform for a fix δ > 0, whence A3 is holomorphic for Re(α),Re(γ) > −1/6.
Moreover, we have the bound A3(α, γ) ≪δ 1, for Re(α),Re(γ) > −1/6 + δ. We end by
remarking that, by an analogous calculation, one finds that
A4(α, γ) :=
ζ(5/6 + γ)ζ(1 + α+ γ) S
ζ(5 6 + ) (1 + 2 ) R1 (α, γ)/ α ζ α
is holomorphic in the same set and with the same bound as A3, see [CFLS, Eq. (4.9)].
Now that we have found meromorphic continuations of RM1 and RS1 , we can write (6.3.2)
as
R′1(α, γ;X) ≈ (
ζ(1 + 2α)
R1 α, γ;X) = F1(X)
ζ(1 + + )A3(α, γ)α γ
(6.3.6)
+ ( ) ζ(5/6 + α)ζ(1 + 2α)F2 X (5 6 + A4(α, γ).ζ / γ)ζ(1 + α+ γ)
The symbol ≈ indicates that one should not expect an equality, but rather an equality up to
some error. From Lemma 6.1 it appears that one can expect an error≪ Xθ−1+ϵ, in contrast
to the bound X−1/2 that one usually believes holds when applying the Ratios Conjecture.
Nevertheless, one cannot expect equality to hold with this error for all α and γ as the right-
hand side is possibly unbounded for α close to 0, or 1/6. We will therefore only assume
that this bound for the error term holds when one has chosen α, γ to be positive and with a
small enough real part, but not too small, in the sense that Re(α),Re(γ)≫ 1/L. This last
condition is common when applying the Ratios Conjecture, see e.g. [CS].
As we have remarked before, we wish to differentiate this estimate with respect to α. The
reader may notice that one can usually not simply differentiate an estimate of a function,
and expect the error term to remain unchanged. For example, it is easy to find a bounded
function with an unbounded derivative. However, if we assume the equality above holds with
error ≪ Xθ−1+ϵ, then we may formally use Cauchy’s formula to estimate the derivative by
integrating over a circle of radius, say ≪ 1/L2. Thus after possibly modifying ϵ, we find
that we have the same error term in the estimate for the derivative of R′1 with respect to
α. Note that this argument is not rigorous, because the function R′1(α, γ;X) is not defined
for the α and γ we are considering here. The point of the argument is to indicate that a
differentation is justified, and one may consider this assumption as being part of the recipe
for evaluating the ratio we are studying.
6.4 Finding a differentiated estimate
We are now left with the task of differentiating the right-hand side of (6.3.6) with respect
to α, and then setting α = γ = r. We turn back to the case when Re(α),Re(γ) > 1/2 so
47
6. The one-level density through the Ratios Conjecture
that we may differentiate (6.3.2) instead. To perform this differentiation we first need a few
preparations.
A straightforward calculation confirms that
f(e, 0, p) + f(e− 1, 1, p) + f(e− 2, 2, p) = g(e, 0, p) + g(e− 1, 1, p) + g(e− 2, 2, p) = 0,
f(1, 0, p) + f(0, 1, p) = g(1, 0, p) + g(0, 1, p) = 0,
where the first equality holds for e ≥ 2. We use these relations to study the products (6.3.3)
and (6.3.4) defining RM1 (r, r) and RS1 (r, r). By grouping together the term corresponding to
e ≥ 2 from the first sum, with the term corresponding to e−1 from the second sum, and the
term for e−2 from the third sums, we see using the relations on the first row above that the
sum of all these terms is 0. The relations on the second row allow us to group together the
rest of the terms and see that these equal 0 as well. It follows that RM1 (r, r) and RS1 (r, r)
equal 1, by virtue of every factor equalling 1, and also that
A3(r, r) = A4(r, r) = 1. (6.4.1)
Next, if f is a function, then the∣∣logarithmic derivative is given byd ∣∣ ′log ( f (a)f x) = ,dx x=a f(a)
which in turn is simply f ′(a) if f(a) = 1. Thus, instead of differentiating RM and RS1 1 , we
can differentiate their logarithms. Now,∑ ∑ ∑ ∑ 
logRM1 (α, γ) = log1 + xpf(e, 0, p)(1 2+ ) + xpf(e, 1, p) xpf(e, 2, p) pe / α pe(1/2+α)+(1 +/2+γ) pe(1/2+α)+(1+2 .γ)
p e≥1 e≥0 e≥0
As we remarked, setting α = γ makes the argument inside every logarithm equal to 1, and
thus
∣∣∣ ∑  ∑ ∑ ∂ ∣ log 1 + xpf(e, 0, p) + xpf(e, 1, p) ∑+ xpf(e, 2, p) ∂α  pe(1/2+α) pe(1/2+α)+(1/2+γ) pe(1/2+α)+(1+2γ)α=γ=r p∑ ∣ e≥1 e≥0 e≥0 
= ∂ ∣∣∣  ∑ x f(e, 0, p) ∑1 + p + xpf(e, 1, p) ∑+ xpf(e, 2, p) ∂α  pe(1/2+α) pe(1/2+α)+(1/2+γ) pe(1/2+α)+(1+2γ)p α=γ=r∑ ∑e≥1 ∑ e≥0 ∑e≥0 
= − xp log p (1 2+ ) f(e, 0, p)e+ f(e− 1, 1, p)(e− 1) + f(e− 2, 2, p)(e− 2)pe / r
p
∑ e≥1 e≥1 e≥2
= − 
 
xp log p ∑· 1 + xp(f(e, 0, p)− f(e, 2, p)) log p1 2+ (1 .p / r p pe /2+r)
p e≥2
In the last step, we used that
ef(e, 0, p) + (e− 1)f((e− 1, 1, p) + (e− 2)f(e− 2, 2, p) ) ( )
= (e− 1) f(e, 0, p) + f(e− 1, 1, p) + f(e− 2, 0, p) + f(e, 0, p)− f(e− 2, 2, p) ,
for e ≥ 2, together with f(1, 0, p) = 1/p. Next, one can check that f(e, 0, p)−f(e−2, 2, p) =
θe + 1/p so that we can write∣the result as
M ∂ ∣R ∣1,α(r, r) := ∂α ∣ ∑∑
( )
R1(α, γ) = −
xp log p 1
(1 θ + . (6.4.2)pe /2+r) eα=γ=r pp e≥1
This sum converges absolutely for Re(r) > 0, and thus by the uniqueness of analytic con-
tinuations, this formula for the derivative is valid for all such r.
48
6. The one-level density through the Ratios Conjecture
To differentiate RS1 , one proceeds in precisely the same manner, using g instead of f .
Applying the equality ( )
y∣p g(e, 0, p)− g(e− 2, 2, p) = γe(p),one finds
∂ ∣∣∣ ∑∑RS1 (α, γ) = − γe(p) log p∂α pe(1/2+ .r)α=γ=r p e≥1
Now, γ1(p) = p−1/3+O(p−2/3) so that the expression above converges absolutely for Re(r) >
1/6. By using the Dirichlet series for the logarithmic derivative of the Riemann zeta function,
we can subtract∣an appropriate term∑to∑extend the domain of definition. Indeed, we have∂ ∣∣∣ −e/3 ′RS1 (α, γ) = − (γe(p)− p ) log p ζ(1 2+ ) + (5/6 + r),∂α e / rα=γ=r p ζp e≥1
which is now valid for Re(r) > 0, r ̸= 1/6.
In total, we have now m∣∣anaged to differentiate R1 wit∑h r∑espect to α, a∣∣ (
nd have)found
∂
R1,α(r, r;X) := R1(α, γ;X) = −F1(X)
xp log p + 1θe
∂α e(1/2+r)α=∑γ=r∑ p pp e≥1(γ (p)− p−e/3− ( ) e ) log p ζ ′F2 X (1 2+ ) + F2(X) (5/6 + r).pe / r ζ
p e≥1
6.5 Estimating R′2
We now turn to the estimate of R′2, defined in (6.2.3). We begin with an extension of Lemma
6.1 from [CFLS, Corollary 4.2].
Corollary 6.2 (CFLS). Let m,h ∈ Z+, and 1/2 ≤ θ < 5/6, ω ≥ 0 be such that (4.2.2)
holds. Then ∑for cubefree h, and α such that 0 < Re(α) < 1∏/2, we have1 −α| |−α F1(X)X F2(X)X−α±( ) DK λK(m)µK(h) = 1 f(e, s, p)x +N X  − pα 1− 6α/5K∈F±(X∏)   p
e||m,ps||h ∏ 
× g(e, s, p)y +O  θ−1−Re(α)+ϵ ωp ϵ (1 + |α|)X (2e+ 5)p .
pe||m,ps||h p|hm,pe||m
Proof. We choose to only provide the idea of the proof, as the calculations are very similar
to calculations that we have performed several times already. Indeed, one proves the formula
above by using Stieltjes integration
∑ ∫  ∑ X
|DK |−αλK(m)µ (h) = u−αdK λK(m)µK(h) .
K∈F±(X) 1 K∈F±(u)
The result then follows by using integration by parts, as well as the estimate from Lemma
6.1.
We use this corollary to find a reasonable estimate of R′2, after having interchanged
the order of summation so that the sum over K is the innermost sum. Indeed, by similar
reasoning as when we estimated R′1, a reasonable estimate is R2(α, γ;X), defined as the
expression ( )
Γ (1/2− α) X−α −α±
Γ (1 2 + ) 1( F1(X)RM (−α, γ) +
X S Γ±(1/2− α)
± / α − α 1 1− 6α/5
F2(X)R1 (−α, γ) = Γ±(1/2)+ α)
× ζ(1− 2α) X
−α −α
( ) X ζ(5/6− α)(1 + ) 1 F1 X A3(−α, γ) + F− 2(X) A4(−α, γ) .ζ α γ − α 1− 6α/5 ζ(5/6 + γ)
49
6. The one-level density through the Ratios Conjecture
We expect this to be a good approximation of R′2 for the same range of α and γ as when han-
dling R′1. Note that we must work with A3 and A4 instead of RM1 and RS1 , as Re(−α),Re(γ)
cannot both be greater than 1/2 if α = γ.
Now we wish to differentiate R2 with respect to α. The trick to avoiding laborious
calculations is to write R2 as a product, where one of the factors is
ζ(1− 2α)
ζ(1− α+ γ) ,
and then apply the product rule for differentiation. As we want to set α = γ = r after
diffe∣rentiating, we can use that ζ(1− r + r)
−1 = ζ(1)−1 = 0, combined with
∂ ∣∣∣ ∣ζ(1− 2α) ′= ζ (1− α+ γ)ζ(1− 2α)− 2ζ ′(1− 2α)ζ(1− α+ γ) ∣∣∂α ζ(1− α+ γ) ζ(1− α+ γ)2 ∣ = −ζ(1−2r).α=γ=r α=γ=r
For both these results, we used that ζ(s) is meromorphic with a simple pole at s = 1 with
residue equ∣al to 1, see e.g. [D3, p.32]. We thus find
∂ ∣∣∣ R2((α, γ;X) = −ζ(1− 2 )
Γ±(1/2− r)
r
∂α α=γ=r Γ±(1/2 + r) )
× X
−r −r
1 F1(X)A (−r, r) +
X ζ(5/6− r)
− 3r 1− 6 F2(X)r/5 ζ(5 6 + )A4(−r, r) ./ r
6.6 Finding the one-level density
Using the calculations of the preceding sections we can formulate a Ratios Conjecture for
ratios of Artin L-functions, [CFLS, Conjecture 4.3].
Conjecture 6.3 (CFLS). Let 1/2 ≤ θ < 5/6, ω ≥ 0 be such that (4.2.2) holds. Then there
is some δ < 1/6 such that for any fixed ϵ > 0, and r ∈ C with 1/L ≪ Re(r) < δ, and
|r| ≤ Xϵ/2, we have
∑ ′ ( )1 L (1/2 + r, fK) ∑∑
±( ) (1 2 + ) = −F1(X)
xp log p
θe +
1
N X ∑∑ L / r, f p
e(1/2+r) p
K∈F±( KX) p e≥1
(γ (p)− p−e/3− ( ) e ) log p ζ
′
F2 X (1 2+ ) + F2(X) (5
Γ±(1/2− r)
/6 + r)− ζ(1− 2r)
( pe / r ζ Γ±(1/2 + r)p e≥1
X−
)
r X−r× ( ) (− ) + ( )ζ(5/6− r)
(
F X A r, r F X A (−r, r) +O Xθ−1+
)
ϵ
1− 1 3 1− 6 5 2 4 ϵ .r r/ ζ(5/6 + r)
Remark. Some of the conditions above require explanation. First, as we noted above, we
want to keep away from any poles of the expressions we are working with. Thus, we assume
that r has real part smaller than 1/6, as well as a real part which is not too close to 0.
Further, as a safety measure, we only require our estimates to be true for Re(r) < δ, where
δ is some number less than 1/6. Lastly, the condition on |r| is to make sure that the factor
(1 + |α|) from Corollary 6.2 is not too large.
The conditions on r may seem overly restrictive, but we will see that Conjecture 6.3
suffices to evaluate the one-level density. Indeed, we have the following proposition and
proof, [CFLS, Proposition 4.4].
Proposition 6.4 (CFLS). Let 1/2 ≤ θ < 5/6, ω ≥ 0 be such that (4.2.2) holds. Assume the
Generalised Riemann Hypothesis for all ζ ±K(s) with K ∈ F (X) and Conjecture 6.3. Then,
if ϕ is a real, even Schwartz function whose Fourier transform is compactly supported, we
50
6. The one-level density through the Ratios Conjecture
have ∑ ∑ ( ) ( )1 L log(4π2e) C± X−1/6 (C±)2 X−1/32 2
N±(X∫ )
ϕ
( 2
γ = ϕ̂(0) 1 + − +
π)K L 5C± L( 5)(C±)2 LK∈F±(X) γK 1 1
2 Γ′+ ( ) ± 1 + 2πit
∑ e
ϕ t Γ 2 ( d)t−
2 ( ) xp(θe + 1/p) log p log pF1 X ϕ̂
L ± L L pe/2R ∑ Lp,e ( )
− 2 ( ) log p log p
e
( 2
∑ log p log pe
F2 X ( ϕ̂ γ p) + F (X) ϕ̂L ∫ pe/2)[ e 2L L ( p5e/6 Lp,e p,e−r −r
− 1 Lrϕ 2 ζ)(1− 2 )
Γ±(1/2− r) X X
r Γ (1 2 + ) ]1 F1(X)A3(−r, r) + 1 6 5F2(X)πi (1/L) πi ± / r − r − r/
× ζ(5/6− r) ζ
′ ( )
(5 6 + )A4(−r, r) − F2(X) (5/6 + r) dr +O X
θ−1+ϵ
ϵ
ζ / r ζ
Remark. Note that the only condition on the support [−σ, σ] of ϕ̂ is that σ <∞.
Proof. We recall that from (6.2.1), we know that under the Generalised Riemann Hypothesis,
we have
1 ∑ ∑ ( )L
ϕ γK
N±(X) (∫ 2K∈F±(X) γK ∫π ) ( ) ∑ (6.6.1)
= 1 − Lr 1 L
′(1/2 + r, fK)
2 ϕπi ±(1/L) (−1/L) 2πi N (X) L(1 2 + )
dr,
/ r, fK
K∈F±(X)
where we have replaced a with 1/L. We would like to replace the inner sum with the
expression from Conjecture 6.3, but this requires some preparation. First, we can only
apply the conjecture when the real part of r is greater than 0. Therefore, we begin the
calculations by modifying the integral over (−1/L).
First, by logarithmically differentiating the functional equation (3.1.1), we find that
L′(1/2 + r, f ) Γ′ Γ′ ′− K(1 2 + ) = log|DK |+
±
Γ (1/2 + ) +
± (1 2− ) + L (1/2− r, fK)r
L / r, fK ± Γ
/ r (1 , (6.6.2)± L /2− r, fK)
so tha∫t ( )
− 1 Lr 1
∑ L′(1/2 + r, fK)
2 ϕπi (−1/L)∫ 2πi( N±)(X) ∑L(1/2 +
dr
r, f )
K∈F±( ) KX
′
= 1 Lr 1 L (1/2 + r, fK)2 ϕπi ∫ (1/L) (2πi )N±(X) ∑ L((1/2 + r, fK)
dr
K∈F±(X) )
+ 1 Lr 1
Γ′ ′
ϕ log|D ±
Γ±
2 2 ±( ) K |+ Γ (1/2 + r) + Γ (1/2− r) dr.πi (−1/L) πi N X K∈F±( ± ±X)
To analyse the last integral we begin by shifting the contour to (0). A straightforward
calculation, very similar to calculations which we have already performed when studying
similar terms in the proof of Theorem 3.1, then shows that up to an error ≪ Xθ−1+ϵ, we
have ( ) ( )
1 ∫ Lr 1 ∑ Γ′ Γ′
2 ϕ 2 ±( ) log|DK |+
±
Γ (1/2 + r) +
±
Γ (1/2− r)πi (−1/L)( πi N X ∈F±( ) ± ±K X
log(4π2e) C± X−1/6 (C±)2 X−1/3
) ∫ ′ ( )Γ
= (0) 1 + − 2 + 2 + 2 1 2πitϕ̂ ± ± 2 ϕ(t)
± + dt.
L 5C1 L 5(C1 ) L L R Γ± 2 L
It remains to evalu∫ate ( )1 Lr 1 ∑ L′(1/2 + r, fK)
ϕ
πi (1/L) 2πi N±(X)
dr.
L(1/2 + r, f )
K∈F±(X) K
51
6. The one-level density through the Ratios Conjecture
We are almost ready to apply Conjecture 6.3, except for the fact that this is only possible for
relatively small |r|. However, as L′/L grows slowly on (1/2 + 1/L) assuming GRH, see [IK,
Cor. 5.18], while ϕ(Lr/(2πi)) decays very fast in vertical strips by (3.1.2), we may restrict
the integral to |Im(r)| ≤ Xϵ/2 at the cost of an insignificant error ≪ −1ϵ X , say. Indeed, in
(3.1.2), pick k ≥ N/ϵ, for some large enough N .
Next, in this restricted integral we may substitute the inner sum by the expression from
the Ratios Conjecture. We now wish to extend this integral back to (1/L). This can once
again be done after estimating the integrand and using the fast decay of ϕ. Indeed, we have
already shown that both A3 and A4 ≪ 1 for the r we are integrating over. Using Stirling’s
formula we can find a polynomial bound for the ratio of gamma functions, while the various
zeta functions are estimated by using the Riemann Hypothesis, see [MV, Thm. 13.18, 13.23].
Lastly, to find a polyno∑mial bound for the various sums, uselog p 1
(1 2+1 ) ≪ 1 + 1 1 = L ≤ logX,pe / /L /L−
p,e≥2
which∫follows(from)integration by∑parts using the Prime Numbe∫r Theor(em. W)e are left with1 Lr 1 L′(1/2 + r, fK) 1 Lr
ϕ
πi [ 2πi N±(1/L) ∑∑(X) ( L(1/)2 +
dr = ϕ
r, fK)
K∈F±(X)
x log p 1 ∑∑
πi (1/L) 2πi
× − p (γe(p)− p
−e/3) log p
F1(X)
pe(1/2+
θ + − F
r) e 2(X)p pe(1/2+r)
p e≥1 ( p e≥1 (6.6.3)
ζ ′ Γ (1/2− r) X−r+ F2(X) (5/6 + r)− ζ(1− 2r) ±Γ (1 )2 + ) F (X)A(1− 1) ] 3
(−r, r)
ζ ± / r r
+ X
−r ζ(5/6− r) ( )
1 6 5F (X) A (−r, r) +O X
θ−1+ϵ
2 (5 6 + ) 4 ϵ dr +O X
−1
ϵ .− r/ ζ / r ( )
The error term inside the integral simply integrates to O Xθ−1+ϵϵ . To handle the first two
integrands, involving sums, we proceed as in the proof of Theorem 3.1 and shift the contour
to (0) to ob∑tain ( ) ( )
− 2 ( ) xp log p log p
e 2 ∑ log p log pe
F1 X 2 ϕ̂ (θe + 1/p)− F (X) ϕ̂ (γ (p)− p
−e/3
2 e ).
L pe/ L L pe/2 L
p,e p,e
The proposition follows after splitting the last sum into two parts.
52
7
Interpreting the Ratios
Conjecture prediction for the
one-level density
We will now study the result in Proposition 6.4 in two different ways. We begin by comparing
the result to what we obtained unconditionally in Theorem 3.1, for small σ. This has been
done in [CFLS, Ch. 5], and we essentially follow their approach. Once this has been done,
we will study the expression in Proposition 6.4 more closely to extract an explicit main and
secondary term, using methods from [FPS2]. This also uncovers a phase transition in the
lower-order terms of the Ratios Conjecture prediction when σ exceeds 1. This has previously
been done in other families, see Section 7.3 for several examples, but our Theorem 7.3 is
the first result in this direction for the family of Artin L-functions associated to non-Galois
cubic fields.
We end the section by pointing out a relation between the coefficients of the terms of
size L−1 in the one-level density, for the family we have been studying. We show that the
same relation can be found in several other symplectic families, and this evidence leads us
to formulate Conjecture 7.4.
7.1 Two expressions for the one-level density
We have calculated the one-level density in two different ways. First, by unconditional calcu-
lations, but with significant restrictions on the support [−σ, σ] of ϕ̂, in Theorem 3.1. In the
previous chapter, we instead calculated the one-level density conditional on the Generalised
Riemann Hypothesis and the Ratios Conjecture, with a relaxed condition on the support.
Naturally, one would expect that the two different results agree with each other. In fact,
one usually expects the Ratios Conjecture to be able to accurately predict terms up to size
≪ X−1/2+ϵϵ , see [CFZ, Ch. 5.1].
As we can only obtain a nontrivial error term in Theorem 3.1 when
1− θ
σ <
ω + 1 ≤ 1,/2
it only makes sense to compare the two expressions for σ < 1. Note that the second
inequality follows from the result θ+ω ≥ 1/2 that we mentioned at the end of Chapter 3.1.
By comparing the two expressions for the one-level density, excluding the error terms, we
see that they are equal, excep(t for a )term∑ ∫ ( )[2 log p log pe( ) := ( ) − 1 Lr (1− 2 )Γ±(1/2− r)J X( F2 X 5 6 ϕ̂ ϕ ζ rL p e/ L πi (1/L) 2πi ) Γ±(1/2 + r)p,e
X−r −r ′
]
× 1 F1(X)
X ζ(5/6− r) ζ
A3(−r, r) + 1 6 5F2(X) (5 6 + )A4(−r, r) − F2(X) (5/6 + r) dr.− r − r/ ζ / r ζ
(7.1.1)
53
7. Interpreting the Ratios Conjecture prediction for the one-level density
Somewhat surprisingly, it was shown in [CFLS, Ch. 5] that J(X)≫ X−1/3, even for small
values of σ. As the error term in The(orem 3.1 has the f)orm
Oϵ Xθ−1+ϵ+σ(1/2+ω) ,
we remark, as in [CFLS], that if one was able to improve the results in (3.2.3) to hold with
θ < 2/3, then there would be a discrepancy between the Ratios Conjecture prediction and
the actual one-level density, of a larger size than expected. As we have mentioned, numerical
results from [CFLS] indicates that (3.2.3) may hold with θ = 1/2 and any ω > 0. However,
the current best values θ, ω = 2/3 do not suffice to show that a discrepancy exists.
The rest of this section will be dedicated to the proof of the following proposition, which
is essentially [CFLS, Lemma 5.4]. The proof is taken from [CFLS, Ch. 5].
Proposition 7.1 (CFLS). For σ < 1, the difference J(X) from (7.1.1) between the Ratios
Conjecture prediction in Proposition 6.4, and the actual one-level density in Proposition 3.1,
satisfies
J(X)≪ Xσ/6−1/3+ϵ +Xσ/2−1/2+ϵϵ . (7.1.2)
In particular, for sufficiently small σ, the two expressions differ by J(X)≪ X−1/3+ϵϵ .
To begin estimating J(X) we focus on the part of the integrand in (7.1.1) involving
A4(−r, r).∫As A4(−(r, r))is holomorphic for |Re(r)| < 1/6, we may shift
− 1 Lr Γ±(1/2− r) X
−r ζ(5/6− r)
ϕ 2 ζ(1− 2r)πi (1/L) πi Γ±(1/2 + r) 1− 6 5
F2(X)
r/ ζ(5/6 + )A4(−r, r)drr
to the line (1/6 − ϵ). Recall that on this line A4 ≪ϵ 1. Using F (X) ≪ X−1/62 , |X−r| =
X−Re(r), the decay of ϕ specified in (3.1.2), together with a polynomial estimate of the rest
of the factors, as in the proof of Proposition 6.4, we can see that the shifted integral is
≪ Xσ/6−1/3+ϵϵ .
To estimate the remaining part of J(X) we will need to shift the integrand involving
A3(−r, r) further to the right. This will require us to first find a meromorphic continuation
of A3. We prove the following result [CFLS, Lemma 5.1].
Lemma 7.2 (CFLS)(. The exp∏res(sion )
A (−r, r) = ζ(3)ζ 3
)
− 3 1− 1 + 1 1 1 13 2 r p3/2+r p5/2−
−
r p5/2−3
− 3−4 + 9 2−5 ,r p r p / r
p
(7.1.3)
furnishes a meromorphic continuation of A3(−r, r) to |Re(r)| < 1/2, with a simple pole at
r = 1/6 with residue
− ζ(3)3ζ(5/3)ζ(2) .
Furthermore, this is the only pole for |Re r| < 1/2.
Proof. Recall that∏A3(is defi)ned (in (6.3.5) by)the product formula1 −1 1
A3(−r, r) = 1− 1−p p1−2r ∑
p 
× 1 + xpf(e, 0, p)
∑ x ∑+ pf(e, 1, p) xpf(e, 2, p) 
e(1 +/2−r) e(1/2−r)+(1/2+r)
∏( p ) p p
e(1/2−r)+(1+2r)
e≥1 e≥0 e≥0
=ζ(3) 1−
1
p1−2r
p
 ∑ ∑ ∑ × 1 + 1 + 1 + f(e, 0, p) f(e, 1, p) f(e, 2, p) 
p p2 pe(1/2−
+ + ,
r) pe(1/2−r)+(1/2+r) pe(1/2−r)+(1+2r)
e≥1 e≥0 e≥0
54
7. Interpreting the Ratios Conjecture prediction for the one-level density
where we used xp = (1 + 1/p+ 1/p2)−1. As 1 + 1/p is equal to the expression for f(e, 0, p)
evaluated at e = 0, we canwrite this as∏( ) ∑ ( )(3) 1− 1 1 + 1 ( 0 ) + f(e, 1, p)ζ 1−2 2 (1 2− ) f e, , p 1 2+ + f(e, 2, p)  .p r p pe / r p / r p1+2r
p e≥0
The attentive reader may remember that in Lemma 6.1, we said that f(0, 0, p) should be
interpreted as 1. In the expression above, we have used the convention f(0, 0, p) = 1 + 1/p,
which better agrees with the formula for f(e, 0, p) given in the mentioned lemma. This is
simply a matter of notational convenience. From the definition of f(e, s, p), we then see that
the sum e(quals ) ( )
1 1 2∑ e+ 1 1 1 ∑ δ1− 2|e6 (p1/2+r pe(1/2−)r)
+ 2 1− p1+2r ( pe(1/2−re≥0 ∑ e≥0 )
)
∑
+ 13 1 +
1 1 τe 1 1 1
p1
+
/2+r p1+2
+ 1− .
r pe(1/2−r) p p1/2+r pe(1/2−r)
e≥0 e≥0
These sums can all be explicitly evaluated by appealing to the formula for a geometric series,
as well as using ∑ ∑
(e+ 1) d 1xe = xe+1 =
dx (1− x)2 ,
e≥0 e≥0
and t(he definition)τe = δ3|e − δ3|(e−1). The result is
1 1− p−1/2−r
2
1 1− p−1−2r 1 1 + p−1/2−r + p−1−2r 1 1− p−1/2−r
6 · (1− p−1/2+
+
r)2 2
· 1− p−1+2 +r 3 · 1 + p−1/2+
+ · . (7.1.4)
r + p−1+2r p 1− p−1/2+r
Using this expression for the sum, one can simplify and confirm that (7.1.3) holds. This can
be done quickly using a computer algebra system, or more laboriously by hand. The point
is that multiplying by ∏( ( ) )1 1 3
ζ(3/2 = 1− ,− 3r) p1/2−r
p
essentially clears the denominators in (7.1.4). As the product in (7.1.3) converges absolutely
for |Re(r)| < 1/2, we have obtained the desired continuation.
To find the residue at r = 1/6, use that ζ(s) has a simple pole at s = 1 with residue 1,
see [D3, p. 32], to write
ζ(3/2− 3r) = −1 13 · 1 6 +O(1),r − /
and set r = 1/6 in the rest of (7.1.3).
We now estima∑te the rest(of J(X)). We n∫eed to s(tudy)[2 ( ) log p log pe − 1 Lr Γ±(1/2− r)F2 X 5 6 ϕ̂ ϕ ]2 ζ(1− 2r)L p e/ L πi (1/L) πi Γ±(1/2 + r)p,e
X−r× ζ
′
1 F1(X)A3(−r, r)− F2(X) (5/6 + r) dr.− r ζ
The shape of the pole at s = 1 of ζ(s) implies that ζ ′(s)/ζ(s) has a simple pole with residue
−1 at s = 1. Thus, if we shift the contour to the line (1/2− ϵ), we pick up the negative of
the residue ( )( )
−2 L −2ζ(2/3)Γ±(1/3)ζ(3)X
−1/6
ϕ 12πi 5Γ±(2/3)ζ(5/3)ζ(2)
F1(X) + F2(X) ,
55
7. Interpreting the Ratios Conjecture prediction for the one-level density
at s = 1/6.
To simplify this expression,(we)will need the funct(ional)equation of ζ(s), i.e. the relation[D3, p. 59]
π−s/2Γ s2 ζ(s) = π
−(1−s)/2Γ 1−s2 ζ(1− s).
Combining this with the definition of Γ± and C±i , and using the reflection and duplication
formula for the Gamma function, see Appendix A.2, we see that
2ζ(2/3)Γ±(1/3)ζ(3) C±2
5Γ±(2
= .
/3)ζ(5/3)ζ(2) C±1
Also, as
−1/6C
± ( )
X 2 F (X) = F (X) +O X−1/21 2 ,
C±1
we see that the residue is ≪ϵ Xσ/6−1/2 by using (3.1.2).
It remains∑to estimat(e ) ( )[2 ( ) log p log ∫pe − 1 Lr (1− 2 )Γ±(1/2− r)F2 X
L p5e/6
ϕ̂ ϕ ζ r
L πi
p,e (1/2−ϵ) 2πi Γ±(1/2 + r) (7.1.5)
X−r ζ ′
]
× 1 F (X)A (−r, r)− F (X) (5/6 + r) dr.− 1 3 2r ζ
The integral of the term above is estimated by estimating each factor separately and then
using the rapid decay of ϕ, cf. the proof of Proposition 6.4. As |X−r| = X−1/2+ϵ, we can
bound the first integral by Xσ/2−1/2+ϵ multiplied by some appropriate constant, possibly
depending on ϵ.
To handle the second integrand, expand ζ ′/ζ into its Dirichlet series and interchange the
order of summation and integration. Then, similarly as in the proof of Theorem 3.1, move
the integral to (0) and use the definition of Fourier(transfor)m. The result is simply
2 ∑− ( ) log p log peF2 X 5 6 ϕ̂ ,L p e/ L
p,e
which cancels against the sum in (7.1.5). This concludes the proof of Proposition 7.1.
7.2 A phase transition in the one-level density
In the previous section, we studied the expression from Proposition 6.4 for σ < 1, essentially
following [CFLS], and found that it agreed quite well with the unconditional result from
Theorem 3.1. In this section, we will drop the condition on σ, and we will see that for σ > 1
one finds new main terms in the one-level density coming from the term J(X). These terms
have not previously been isolated in this family. More precisely we will prove the following
theorem.
Theorem 7.3. Assume the Generalised Riemann Hypothesis for ζK(s), and Conjecture 6.3.
Then, if ϕ is a real, even Schwartz function whose Fourier transform is compactly supported,
we have
1 ∑ ∑ ( )L
±( ϕ γKN (X) 2πK∈F±∫(X) γK )
= (0)− 1
1
ϕ̂ 2 ϕ̂(r) +
ϕ̂(0)
dr (1−(4 log)2− πδ+ − C)−1 L
+ ϕ̂(1) (−1 + 4 log 2 + πδ+ + C) +O
1
,
L L2
56
7. Interpreting the Ratios Conjecture prediction for the one-level density
where ∑ 2p5/2 + 2p2 + p3/2 − p− p1/2
C = 2 (log p) − 1 ,
(p3 − 1)(p+ p1/2 + 1)
p
and where δ+ = 1 if we are considering positive discriminants, and else it equals 0.
Remark. The main term is the symplectic main term that we expect from the Katz–Sarnak
prediction. The secondary term involves ϕ̂(1), which is possibly nonzero only if σ > 1. Such
behaviour in the one-level density is sometimes referred to as a phase transition, see e.g.
[FPS2], and has been observed in several other families, see the next section for several
examples. The term involving ϕ̂(1) is entirely contained within the term J(X) from (7.1.1),
which explains why we could only obtain a power-saving bound for this term when σ < 1.
Remark. The coefficient in front of L−1ϕ̂(0) is the negative of the coefficient in front of
L−1ϕ̂(1). We will return to this phenomenon in the next section.
The rest of the section is dedicated to the proof of this theorem. The strategy used in
the proof is essentially that of [FPS2], but with the details modified to fit this particular
family. (By Proposition)6.4, th∫e one-level(density is given bylog(4π2 )(0) 1 + e) + 2 Γ′ 1ϕ̂ ∑ (ϕ(t)
±
Γ ) 2 +
2πit
dt
L L R ± L ∑ ( )
− 2 x
e
( ) p log p log p ( + 1 )− 2 ( ) log p log p
e
F1 X ϕ̂ θe /p F2 X ϕ̂ γe(p)
L pe/2∑ ( L) L p
e/2 L
p,e p,e
e
+ 2 log p log pF
∫2
(X)
L (p5e/6p,e )ϕ̂[ L (
− 1 Lr
−r
ϕ ζ(1− 2 )Γ±(1/2− r) Xr F1(X)A3(−r, r)
πi (1/L) 2πi Γ±(1/2)+ r) 1− r ]
+ X
−r
( )ζ(5/6− r)
′
1 6 5F2 X (5 6 + )A4(−r, r) −
ζ
F2(X) (5/6 + r) dr +O(L−2).− r/ ζ / r ζ
We will estimate this expression one term at a time.
7.2.1 An integral of Gamma functions
We begin by estimating
2 ∫ ′ ( )Γ( ) ± 1 + 2πitϕ t Γ dt.L R ± 2 L
After using Stirling’s formula to bound the factor involving Γ± polynomially we may use a
cutoff argument similar to the ones we have already used to restrict the integral to |t| ≤ L,
with an error that is, say ≪ L−2. The precise exponent is not important, as ϕ decays very
fast. In fact, we could have bounded the tail of the integral by any ≪ L−NN . Next, we
perform a zeroth order Taylor expansion
′ (of )Γ± 1
Γ 2 +
2πit
,
± L
to write the cut-off∫integral as (2 Γ′ L ∫ )L± Γ′ ( )· Γ (1/2) ϕ(t) 1 2dt+O 2 |tϕ(t)|dt = · ±Γ (1/2)ϕ̂(0) +O L−2 ,L ± −L L −L L ±
where we extended the(in)tegral back to (−∞,∞( ))in the last step.Now, ( )
Γ′ 1 ′ ′±
Γ 2 = − log π +
1 · Γ 1 + 1 · Γ 3− 2δ+
± 2 Γ 4 2 Γ 4
.
57
7. Interpreting the Ratios Conjecture prediction for the one-level density
By logarithmically differentiating the product definition of Γ(s), one can find Γ′(1)/Γ(1) =
−γ, where γ is Euler’s constant. Then using this, combined with the reflection formula and
duplication formula, i(t fo′ )llows that ( )Γ 1 π Γ′ 3 π
Γ 4 = −γ − 2 − log 8, Γ 4 = −γ + 2 − log 8.
Thus, we have that
Γ′
( )
± 1
Γ 2 = − log 8 − −
π
π γ 2 δ+, (7.2.1)±
whence
2 ∫ ( )Γ′( ) ± 1 + 2πitϕ t Γ 2 dt = ϕ̂(0) (−2 log 8π − 2γ − πδ+) .L R ± L L
7.2.2 Estimating prime sums
We turn to the estimation of th(e sums)over primes. From (3.3.1) we already know
−2∑∑ xp(θe + 1/p) log p log pe
2 ϕ̂ = −
ϕ(0)
L pe/
p e≥[1
2 ∫ ∞ θ(u)− u ∑
L∑ 2 ( ( ) )]
− (0) 1 + + xp(θe + 1/p) log p
∑
+ log p 1ϕ̂ 2 du 2 xp 1 + − 1L u pe/1 p pp e ̸=2 p
+O(L−2), ∑
where θ(u) = p≤u log p. To simplify, we take a look a(t the integral. We have∫ ∞ ∫ ∫ )θ(u)− u N θ(u)− u N θ(u)
u2
du = lim 2 du = lim 2 du− logN . (7.2.2)
1 N→∞ 1 u N→∞ 1 u
Further, we know that this limit exists, as the original integral converges by the Prime
Numbe∫r Theorem. Usin∫g integrat(ion)by parts, and the Prime Number Theorem, we obtainN θ(u) N= − ( ) 1 = −θ(N) ∑+ log p ∑ log p2 du θ u d = − 1 + o(1).
1 u 1 u N p pp≤N p≤N
To study this sum, we use an argument from [Lu, p. 199-200]. Instead of working with θ(n),
it will be more convenient to use the von-Mangoldt function Λ(n). We write
∑ θ(n) ∑= − Λ(n)− θ(n) ∑+ Λ(n)− 1 ∑+ 1 .
n n n n
n≤N n≤N n≤N n≤N
By definition of γ, ∑ 1 = logN + γ + o(1).
n
n≤N
Also by definition, we hav∑e
lim Λ(n)− θ(n)
∑
= log p
∑
= log p .
N→∞ n pe p(p− 1)
n≤N p,e≥2 p
In particular, these calculations, together with the existence of the integral in the left-hand
side of (7.2.2), shows that ∑
lim Λ(n)− 1
N→∞ n
n≤N
58
7. Interpreting the Ratios Conjecture prediction for the one-level density
exists. Using summation by parts, one can show that if a Dirichlet series converges at some,
say real, point x0, then it converges uniformly in the set {(x ∈ R : x ≥ x0}). Thus, we have∑ Λ(n)− 1 ∑= lim Λ(n)− 1 ′= lim −ζ (s)
n s→1+ ns s→1+ ζ(s) − ζ(s) . (7.2.3)
n n
It is well-known, see e.g. [D3, p. 81], that
ζ(s) = 1 1 + γ +O(s− 1) (7.2.4)s−
near s = 1, whence (7.2.3∫) equals −2γ. Putting together all our calculations, we have shownthat ∞ θ(u)− u ∑ log p
2 du = −1− γ − .
1 u p(p− 1)p
Lastly, we should e∑stimate th(e two su)ms ( )2 log p log pe− ( ) ( ) + 2 ∑ eF2 X 2 ϕ̂ γe p F2(X) log p log pϕ̂ .L pe/ L L p5e/6 L
p,e p,e
As γ1(p) = p−1/3 +O(p−2/3), we have
∑ ( ) ( )  log p log pe ∑( ) = log p log pe ∑ log p2 ϕ̂ γe p 5 6 ϕ̂ +O  .pe/ L p e/ L p
p,e p,e p≤Xσ
An application of Stieltjes integration shows that the error term is ≪ L. Thus, as F2(X)≪
X−1/6, we find∑that ( ) ( )2 log p log pe 2 ∑ log p log pe− F2(X) 2 ϕ̂ γe(p) + F2( 1X)L pe/ L L p5 6 ϕ̂ ≪ 2 . (7.2.5)e/ L L
p,e p,e
In fact, the exponent 2 can be made as large as we wish.
7.2.3 Estimating an integral
It remains to est∫imate ( )[ (1 Lr −r− ϕ 2 ζ(1− 2 )Γ±(1/2− r) Xrπi (1/L) πi Γ (1 2 + )) 1 F1(X)A3(−r, r])± / r − r
X−r ′+ ( )ζ(5/6− r) ζ1 6 5F2 X (5 6 + )A4(−r, r) − F2(X) (5/6 + r) dr.− r/ ζ / r ζ
Using (3.1.2), F2(X) ≪ X−1/6, a Laurent expansion of ζ(1 − 2r), as well as estimates for
the other factors that we have made use of before, e.g. when proving Proposition 6.4, we
can directly see th∫at the o(nly r)elevant term for our purposes is1 Lr −r− ϕ 2 ζ(1− 2 )Γ±(1/2− r) Xr Γ (1 2 + ) 1 F1(X)A− 3(−r, r)dr.πi (1/L) πi ± / r r
Indeed, the other terms are ≪ X−1/6+ϵϵ . To estimate this integral, we follow the approach
of [FPS2, Lemma 4.6]. As F (X) = 1 +O(X−1/61 ) we may also replace this factor by 1.
First, w∫e make t(he change)of variables s = Lr/(2πi), which tra(nsforms the in)tegral into
− 2 ( ) 1− 4πis Γ± (1/2− 2πis/L) X
−2πis/L 2πis 2πis
ϕ s ζ
L L L Γ± (1/2 + 2
A3 − , ds,
πis/L) 1− 2πis/L L L
59
7. Interpreting the Ratios Conjecture prediction for the one-level density
where L is the horizontal line given by Im(s) = −1/2π, oriented from left to right. Next,
shift the contour from (1/L) to C = C0 ∪ C1 ∪ C2, defined by
C0 = {s ∈ R : |s| ≥ Lϵ}, C1 = {s ∈ R : η ≤ |s| ≤ Lϵ}, C2 = {s = ηeit : −π ≤ t ≤ 0},
with η > 0, and some fixed ϵ > 0. Note that this contour avoids the pole of ζ. Further, by
a cutoff argument that we have seen several times, we may disregard the contribution from
C0.
Write
X−2πis/L = e−2πis logX/L = e−2πise−4πis log(2πe)/L.
On C1 and C2, s is small enough that we may expand every factor of the integrand, except
ϕ(s) and e−2πis, into its corresponding Laurent series. For our purposes, it will be sufficient
to consider the expansi(on ( ))( ( ) ( ))
− 2
∫
( ) −2πis − L + +O |s| 1− 4πis
Γ′ 2
ϕ s e γ · ± 1 |s|
L ( +OC1∪C2 4πi(s ))( L (L )Γ)±( 2 L2 ( ))
× 1− 4πis log(2πe)
2
+O |s| 1 + 2πis +O |s|
2 2πiCs |s|2
2 2 1 + +O 2 ds,L L L L L L
with
:= d ∣∣C ∣∣ A3(−r, r). (7.2.6)dr r=0
In the Taylor expansion of A3(−r, r) we made use of (6.4.1), which implies that A3(0, 0) = 1.
Now, one∫may m[ultiply toget(her all the factors, and use (7.2.1),)to see (that t)he]above is−2πis
( ) e 1 + 2πis
2
ϕ s (−1 + 4 log 2 + |s|πδ+ + C) +O 2 ds.
C1∪C2 2πis L L
We can immediately see that the error term integrates to be ≪ L−2, whence it can be
disregarded. As ϕ is a Schwartz function, the integral over the remaining terms can be
extended to include C0 once again. We now estimate this extended integral.
We begin with the secon∫dary term, which is the easiest to study. Ind∫eed, we have(−1 + 4 log 2 + πδ+ + C) ( ) −2πis = (−1 + 4 log 2 + πδ+ + C) ∞ϕ s e ds ϕ (t) e−2πitdt
L C L −∞
+ (1) = (−1 + 4 log 2 + πδ+ + C) ϕ̂(1)oη + oη(1),
L
where the last step is the definition of the Fourier transform. The term oη(1) should be
interpreted as a term converging to 0 when η → 0+, corresponding to the contribution from
C2.
We turn to th∫e main term, which requires slightly mo∫re effort. We write it as
( )cos(2πt)− i sin(2πt) 1 e
−2πis
ϕ t 2 dt+ 2 ϕ(s) ds.|t|≥η πit πi C2 s
The last integral is over a half-circle around a pole. One may Taylor expand ϕ(s) and e−2πis
close to 0 and then parameterise the contour to see that the contribution from the last
integral is
ϕ(0)
2 + oη(1).
In the other integral, we may disregard the term containing cos(2πt), as this integrand
is odd, as ϕ is even. Then, by the continuity of the integrand involving sin(2πt) at 0, what
remain∫s is
− ( ) sin(2
∫
πt) ∞= − ( ) sin(2πt)
∫ 1
ϕ t 2 dt ϕ t 2 dt+ oη(1) = −
1
ϕ̂(u)du+ oη(1).
|t|≥η πt −∞ πt 2 −1
60
7. Interpreting the Ratios Conjecture prediction for the one-level density
In the last step, we used Plancherel’s theorem, and the fact that the Fourier transform of
the indicator function of [−1, 1] is
2 · sin(2πt)2 .πt
Letting η → 0+ finishes the analysis.
7.2.4 Completing the proof of Theorem 7.3
Collecting the results of the previous sections, we have proven that the one-level density is
given by ∫ 1
ϕ̂(0)− 12 (ϕ̂(u)du−1 ( ( ) )
+ ϕ̂(0)
∑
2 log p
∑∑ ∑
( 1) − 2
xp(θe + 1/p) log p − 2 log p 12 xp 1 + − 1L p p−) pe/ p pp p e≠ 2 p
+ 1− 4 log 2− + ϕ̂(1)πδ+ (−1 + 4 log 2 + πδ+ + C) +O(L−2).
L
Everything is given quite explicitly above, except for the constant C, defined in (7.2.6). By
Lemma 7.2, A3(−r, r) is(given by( ∏ )
(3) 3
) 1 1 1
ζ ζ 2 − 3r 1− 3 2+ + 5 2− − 5 2−3 −
1 + 13−4 9 2−5 . (7.2.7)p / r p / r p / r p r p / r
p
As A3(0, 0) = 1, we may calculate its logarithmic derivative to find C. Taking the logarithm
of (7.2.7), differentiating with respect to r, and then setting r = 0, yields
∑ (log p)(2p5/2 + 2p2 + p3/2 − p− p1/2= 2 − 1)C
(p3
.
− 1)(p+ p1/2 + 1)
p
∑It remains to sh∑ow∑that ( ( ) )
2 log p − 2 xp(θe + 1/p) log p
∑
− 2 log p( xp 1 +
1 − 1 = −C. (7.2.8)
p p− 1) pe/2 p p
p p e≠ 2 p
This is done by evaluating the sums explicitly, using xp = (1+1/p+1/p2)−1, θe = δ2|e+δ3|e.
In particular, as θe only depends on the congruence class of e modulo 6, one can split every
sum involving θe into six different sums, depending on the congruence class. Together with
the formula for a geometric series, this shows that the part of the second sum in (7.2.8)
involving θe, including the fact∑or −2, is eq(ual to
−2 xp log p
)
2 + 1/p+ p3/23 1 + p ,p −
p
while the part of the second sum involving 1/p is equal to
∑
−2 xp(log p)(p
3/2 + 1)
p2(p− 1) .
p ( )
Adding the two other sums from (7.2.8), and simplifying using xp = 1 + 1/p + 1 2
−1
/p
proves the equality in (7.2.8). Specifically, the pth term in the sum defining C is equal to the
negative of the pth term in the sum obtained by adding together all the sums from (7.2.8).
A computer algebra system may be of use in proving this equality.
61
7. Interpreting the Ratios Conjecture prediction for the one-level density
7.3 A symplectic phenomenon
A curious detail of Theorem 7.3 is that the coefficient of ϕ̂(0)/L is the negative of the
coefficient of ϕ̂(1)/L. It is not clear why this should be the case, and in the proof above,
showing this required quite some effort. The point of this section is to motivate the following
conjecture, extending one part of the Katz–Sarnak prediction.
Conjecture 7.4. Let F be a natural family of L-functions whose symmetry type is sym-
plectic. Then, the one-level d∫ensity correspondi(ng to this fa)mily has the form
(0)− 1
1
ϕ̂ 2 ϕ̂( )
CF
u du+ ϕ̂(1)− ϕ̂(0) + o(M−1),
−1 M
where CF is a constant depending on the family F , and M is an analogue of our L.
We will present a few different sy(mplec)tic families where the one-level density has been
found explicitly, with error term o M−1 , and show that in these cases, the conjecture
above holds. In all the examples that we will study, the relation between the coefficients
cannot be found without some additional effort, which may explain why the phenomenon
in Conjecture 7.4 appears to not have been pointed out before in the literature. When
studying the examples below, we omit most of the details and background and instead focus
on proving the conjecture in these special cases.
We remark that the conjecture does not hold for all families. For example in [DFS], the
one-level density for an SO(even) family is studied, and then the coefficients are not each
other’s negatives, but rather equal to each other. However, in the same article, a related
SO(odd) family is studied, and then the same relation as in the conjecture above holds. It
would be interesting to see if similar relations continue to hold in other families of the same
symmetry type, but that is beyond the scope of this thesis.
7.3.1 Quadratic Dirichlet characters
In [FPS2], a symplectic family related to certain multiplicative functions, known as Dirich-
let characters, was investigated. The results are conditional on the Generalised Riemann
Hypothesis for this family, but not on the Ratios Conjecture, see [FPS2, Theorem 1.1]. The
one-level∫density was found t(o equal∞ ∫ ∞ )
ϕ̂(0) + ( ) + ϕ̂(0)∫ ϕ̂ u du log(2
1−γ) + 2e (0) w(x)(log x)dx1 L ŵ 0
1 ∞
( ) ( )
e−x/2 −3x/2 ( ( )) ∑ −1+ + e (0)− x − 2 log p 1 + 1 2j log p
L ( 1− e−2
ϕ̂ ϕ̂ dx ϕ̂
x
0 L ∫ L pj p Lp>2,j≥1
ϕ̂(1) ′ ∞
)
+ −73 log 2− 1− + 2
ζ
γ (2)− 2 w(x)(log x)dx +O(L−2).
L ζ ŵ(0) 0
Here, the function ϕ is a real, even Schwartz function, and the corresponding σ is assumed
to be less than 2. The function w(x) is a weight function, but it will not be important for
our analysis. Also, the L in the expression above is not quite the same L as we have used,
but they have the same role. This expression was not simplified further in [FPS2], so we
sketch a simplification here.
A Taylor expansion shows that the integral containing the exponential functions is ≪
L−2. Thus, the only term that needs estimation is the sum over primes. We omit the details,
but the calculations are very similar to those of Section 3.3, and those of the previous section.
Carrying o∫ut the calculations(shows that the expression above1 1 (0) 7 ′ 2 ∫
is
∞ )
ϕ̂(0)− ϕ̂ ζ2 ϕ̂((u)du+ 3 log 2 + 1 + γ − 2 (2) +−1 L ζ ∫ ŵ(0) w()x)(log x)dx0 (7.3.1)
+ ϕ̂(1) 7
′ ∞
−3 log 2− 1− γ + 2
ζ (2)− 2(0) w(x)(log x)dx +O(L
−2),
L ζ ŵ 0
62
7. Interpreting the Ratios Conjecture prediction for the one-level density
which means that the expected relation between the two coefficients holds.
7.3.2 Hecke characters
The next symplectic family that we shall concern ourselves with is related to so-called Hecke
characters. As before we will not elaborate further on the setup. In [Wa, Conjecture 1.1], the
one-level density of a certain symplectic family was found, conditional on the Generalised
Riemann Hypothesis and the Ratios Conjecture for the relevant L-functions. The one-level
density in question equals ∫ 1 1 ( )
ϕ̂(0)− ϕ̂(u)du+ cϕ̂(0)− dϕ̂(1) .
−1 M
Here, M plays the same role as L does in our analysis, and c, d are some constants. The
relation
c = d− c1 − γ, (7.3.2)
is also given, where ∫ ∞
= 1 + ψ(t)− t∑ c1 t2 dt1
and ψ(t) := n≤t Λ(n) is the summatory function of the von-Mangoldt function.
The relation (7.3.2) is not simplified further in [Wa], but it is possible to accomplish such
a simplification using the methods of the previous section. Indeed, by the same arguments
as when we found the value of ∫ ∞ θ(t)− t
2 dt,
1 t
one finds that c1 = −γ so that in fact c = d, as expected.
7.3.3 A polynomial analogue
The last example we shall study is slightly different from the examples above, in that the
L-functions are not related to number fields. Instead, the L-functions are related to certain
polynomial rings, but we will not go into this further.
In [Rk, Cor. 3], an analogue of the one-level density is studied, and a symplectic main
term is found. Further(, a sec∑ondary term is found), with co(efficient )
(0) degPϕ̂ 2 1 +
1
2 − ϕ̂(1)
1 + 1 ,
|P | − q − 1 2
P monic, irred
for σ < 2. Here, q > 1 is a prime power, |P | = qdegP , and the sum is over all monic,
irreducible polynomials in the finite field Fq. We remark that the expression above is not
the expression given in [Rk], but instead a slight reformulation taken from [FPS1, Eq. 1.5].
We will show that the two coefficients are related by simplifying the sum. First, by the
formula for a geom∑etric series,deg ∑ ∑∞P = degP ∑= Λ(f) .
|P |2 − 1 |P |2j |f |2
P monic, irred P irred, monic j=1 monic f
Here, Λ(f) is the polynomial von-Mangoldt function defined as being 0 unless f = P k is
the power of an irreducible polynomial, and in this case Λ(f) = degP . To simplify the last
sum, apply the equality ∑
Λ(f) = qn,
monic f of degree n
see e.g. [Ro, Prop 2.1]. Then, by split∑ting the sum over each degree, we find that it equals∞ 1 = 1
n
=1 q q − 1
,
n
as desired.
63
7. Interpreting the Ratios Conjecture prediction for the one-level density
64
8
The Ratios Conjecture
prediction for the two-level
density
In this chapter, we obtain a Ratios Conjecture prediction for the two-level density in the form
of Proposition 8.4, which gives an expression for the two-level density for any σ <∞. The
main difficulty in proving this theorem is that the products that arise are harder to evaluate
explicitly than in the one-level case. As is usual when applying the Ratios Conjecture,
we assume the Generalised Riemann Hypothesis for ζK(s) throughout the chapter. We
emphasise that the results of this chapter are new.
8.1 Sums over K
Recall that by∑(4.1.1),∑the tw(o-level density)is given by ( )1 L L 2 ∑ ∑ L L
±( ) ϕ 2 γK , γ
′ − ϕ γK , γK .
N X π 2π K N±(X) 2π 2π
K∈F±(X) γK ,γ′K K∈F±(X) γK
(8.1.1)
The second sum is just the one-level density, where the function ϕ3(u) = ϕ(u, u) = ϕ1(u)ϕ2(u)
is used in place of the ”ϕ” from (2.4.3). The one-level density has already been predicted
using the Ratios Conjecture in Proposition 6.4, so we need not consider this sum for now.
We write the first sum as
∑ (∑ ( ))∑ ( )1 L L
±( ) ϕ1 2 γK ϕ
′
2 
N X π ′ 2
γ
π K
,
K∈F±(X) γK γK
which by the Re(si∫due theo∫rem, an)d(th∫e Gener∫alised R)iema(nn H)ypoth(esis f)or ζK(s), equals1
(2 )2 − −
Lr Ls
ϕ1 ϕ2
πi (1/L) (∑−1/L) (1/L) (−1/L) 2πi 2πi′ ′ (8.1.2)
× 1 L (1/2 + r, fK) L (1/2 + s, fK)
N(X) (1 2 + drds.L / r, f ) L(1/2 + s, f )
K∈F±( ) K KX
We are thus motivated to study
1 ∑ L(1/2 + α, fK) L(1/2 + β, fK)
( ) ,N X L(1/2 + γ, f
∈F±( ) K
) L(1/2 + δ, fK)
K X
using the Ratios Conjecture. We will then differentiate the result with respect to α and β,
and set α = γ = r, β = δ = s so that we can estimate the sum above with the result. We
proceed very similarly to the one-level density case. The numerators are rewritten using the
approximate functional equation, while the denominators are expanded into their Dirichlet
65
8. The Ratios Conjecture prediction for the two-level density
series. The resulting expression will contain four different terms, instead of only two, as in
Chapter 6. Write
1 ∑ L(1/2 + α, fK) L(1/2 + β, fK) ′
N(X) L(1/2 + γ, f ) L(1/2 + δ, f ) = R1,1(α, β, γ, δ;X)
K∈F±(X) K K
+R′1,2(α, β, γ, δ;X) +R′ ′2,1(α, β, γ, δ;X) +R2,2(α, β, γ, δ;X) + Error,
where
∑ ∑
′ ( 1 λK(m1)µK(h1)λK(m1)µK(h2)R1,1 α, β, γ, δ;X) = ±( ) 1 2+ 1 2+ 1 2+ 1 2+ ,N X / α / γ / β / δ
K∈∑F±(X) h1,h2,m1,m m2 1 h1 m2 h2
R′2,1(
1 Γ±(1/2− α)
α, β, γ, δ;X) = −α
N±
|DK |
∑(X) Γ±(1/2 + α)K∈F±(X)
× λK(m1)µK(h1)λK(m1)µK(h2)1/2−α 1/2+γ 1/2+β 1/2+ ,δ
h ,h ,m ,m m h m h1 2 1 2 ∑1 1 2 2
′ ( 1 Γ±(1/2− β)R1,2 α, β, γ, δ;X) = ±( ) |D |
−β
K
N ∑X Γ±(1/2 + β)K∈F±(X)
× λK(m1)µK(h1)λK(m1)µK(h2)1/2+α 1/2+γ 1 ,/2−β 1/2+δ
h ,h m h m h1 2,m1,m2 1 1 2 2
′ ( ; ) = 1
∑
| |−α−β Γ±(1/2− α) Γ±(1/2− β)R2,2 α, β, γ, δ X N±∑(
DK
X) Γ (1/2 + α) Γ (1/2 + β)
K∈F±( ± ±X)
× λK(m1)µK(h1)λK(m1)µK(h2)1 2− 1 2+ 1 2− 1 2+ ./ α / γ / β / δ
h1,h2,m1,m
m h m h
2 1 1 2 2
We will now find suitable approximations for each of the terms above. We begin by stating
a generalisation of Lemma 6.1.
Lemma 8.1. Let m1,m2, h1, h2 ∈ Z+, and 1/2 ≤ θ < 5/6, ω ≥ 0 be such that (4.2.2) holds.
Then for cubefree h1, h2 we have
1 ∑
±( ) λK(m1)µK(h 1
)λK(m2)µK(h2) = F1(X)P1 + F2(X)P2
N X
K∈F±(X) ∏ ( ) 
+O Xθ−1+ϵ 4(e1 + 1)(e2 + 1) + 3 pωϵ ,
p|m1h1m2h2, pe1 ||m1, pe2 ||m2
where
∏
P1 = f(e1, e2, s1, s2, p)xp,
pe1 ||m1, ps1 ||h∏1, pe2 ||m2, ps2 ||h2
P2 = g(e1, e2, s1, s2, p)yp,
pe1 ||m1, ps1 ||h1, pe2 ||m2, ps2 ||h2
66
8. The Ratios Conjecture prediction for the two-level density
and
( 0 0 ) = (e1 + 1)(e2 + 1)
δ2|e1δ+ 2|e2 + τe1τe 1f e1, e2, , , p 26 2 3 + ,p
f( (e1 + 1)(e2 + 1) τe τe 1e1, e2, 0, 1, p) = − 1 23 + 3 − ,p
( 0 2 ) = (e1 + 1)(e2 + 1)
δ2|e
f e , e , , , p − 1
δ2|e2 + τe1τe21 2 6 2 3 ,
( 1 1 ) = 2(e1 + 1)(e2 + 1) + τe1τe2 + 1f e1, e2, , , p 3 3 ,p
( (e1 + 1)(e2 + 1) τe τef e1, e2, 1, 2, p) = − 3 +
1 2
3 ,
( 2 2 ) = (e1 + 1)(e2 + 1)
δ2|e1δ2|e2 τe1τef e 21, e2, , , p 6 + 2 + 3 ,
(e + 1)(e + 1)(1 + p−1/3)31 2 δ2|e1δ2|e2(1 + p−1/3)(1 + p−2/3)g(e1, e2, 0, 0, p) = 6 + 2
+ τe1τe2(1 + p
−1) −1/3 2
3 +
(1 + p )
,
p
−1/3 3 −1 −1/3 2
( 0 1 (e1 + 1)(e2 + 1)(1 + p ) τe τe (1 + p ) (1 + p )g e1, e2, , , p) = − 1 23 + 3 − ,p
−1/3 3 δ −1/3 −2/3
( 0 2 ) = (e1 + 1)(e2 + 1)(1 + p ) − 2|e
δ2|e (1 + p )(1 + p )
g e1, e2, , , p
1 2
6 2
τ τ (1 + p−1+ e1 e2 )3 ,
( 1 1 ) = 2(e1 + 1)(e2 + 1)(1 + p
−1/3)3 + τe1τe2(1 + p
−1) + (1 + p
−1/3)2
g e1, e2, , , p 3 3 ,p
(e + 1)(e −1/3 3 −1( 1 2 ) = − 1 2 + 1)(1 + p ) + τe1τe2(1 + p )g e1, e2, , , p 3 3 ,
(e + 1)(e + 1)(1 + p−1/31 2 )3 δ δ −1/3 −2/3( 2 2 ) = + 2|e1 2|e2
(1 + p )(1 + p )
g e1, e2, , , p 6 2
+ τe1τe2(1 + p
−1)
3 .
(8.1.3)
Further, both f(e1, e2, s1, s2, p) and g(e1, e2, s1, s2, p) are symmetric in the sense that they
are both invariant under a permutation of e1 and e2, or s1 and s2. In the case when all of
e1, e2, s1, and s2 equal 0 one should consider the products as empty, equalling 1.
Proof. The proposition is proven in the same manner as Lemma 6.1, with the main difference
being that one has to evaluate slightly different sums to find the functions f and g. We
leave out the calculations.
Note that we used the same symbols f and g for the functions above as we did in Lemma
6.1. This will hopefully not lead to any confusion, as the functions have a different number
of arguments.
8.2 Estimating R′1,1
Lemma 8.1 indicates that a reasonable approximation of R′1,1 is given by
R1,1(α, β, γ, δ;X) := F1(X)RM (α, β, γ, δ) + F (X)RS1,1 2 1,1(α, β, γ, δ), (8.2.1)
67
8. The Ratios Conjecture prediction for the two-level density
with ∏( )
RM1,1(α, β, γ, δ) = 1 + ΣM(0 +Σ
M
1 +ΣM2 ,
p
ΣM = 1
∑ xpf(e1, e2, i, 0, p) ∑ xpf(e1, e2, i, 1, p)
i pi(1/2+γ) pe1(1
+
/2+α)+e2(1/2+)β) pe1(1/2+α)+e2(1/2+β)+(1/2+δ)∑ e1,e2 e1,e2
+ xpf(e1, e2, i, 2, p)
pe1(1/2+α)+e2(1/2+β)+2(1/2+δ)
,
e1,e2
(8.2.2)
and ∏( )
RS (α, β, γ, δ) = 1 + ΣS S S1,1 (0 +Σ1 +Σ2 ,p ∑
ΣM = 1 xpg(e1, e2, i, 0, p)
∑
+ xpg(e1, e2, i, 1, p)i pi(1/2+γ) pe1(1/2+α)+e2(1/2+)β) pe1(1/2+α)+e2(1/2+β)+(1/2+δ)∑ e1,e2 e1,e2
+ xpg(e1, e2, i, 2, p)
pe1(1/2+α)+e2(1/2+β)+2(1/2+
.
δ)
e1,e2
The summations over e1, e2 should be interpreted as summing over all e1, e2 ≥ 0, except that
the term containing f(0, 0, 0, 0, p), or g(0, 0, 0, 0, p), should not be included. We remark that
both RM1,1 and RS1,1 are invariant under permuting α and β, or γ and δ, by the invariance of
f and g.
It is apparent that the products above converge as long as e.g. the real parts of all
variables are greater than 1/2. Just as in the one-level calculations, we begin by extending
the domain of these products further. Indeed, by arguments analogous to those establishing
the meromorphic continuation of RM1 and RS1 , one finds that the functions
ζ(1 + α+ δ)ζ(1 + β + δ)ζ(1 + α+ γ)ζ(1 + β + γ)
A1(α, β, γ, δ) := M
ζ(1 + R (α, β, γ, δ),α+ β)ζ(1 + 2α)ζ(1 + 2β)ζ(1 + γ + δ) 1,1
ζ(5/6 + δ)ζ(5/6 + γ)
A2(α, β, γ, δ) :=
ζ(5/6 + α)ζ(5/6 + β)
× ζ(1 + α+ δ)ζ(1 + β + δ)ζ(1 + α+ γ)ζ(1 + β + γ) S
ζ(1 + α+ β)ζ(1 + 2α)ζ(1 + 2β)ζ(1 + + ) R1,1(α, β, γ, δ),γ δ
(8.2.3)
are holomorphic for α, β, γ, δ with real part greater than−1/6. Further by studying the defin-
ing product, we have that both A1(α, β, γ, δ) and A2(α, β, γ, δ) are≪ϵ 1, when Re(α),Re(β),
Re(γ),Re(δ) > −1/6 + ϵ.
Now we want to differentiate these expressions with respect to α and β, but we first need
some preparation. One can confirm that
f(e1, 1, i, 0, p) + f(e1, 0, i, 1, p) = g(e1, 1, i, 0, p) + g(e1, 0, i, 1, p) = 0,
f(e1, e2, i, 0, p) + f(e1, e2 − 1, i, 1, p) + f(e1, e2 − 2, i, 2, p) = 0, (8.2.4)
g(e1, e2, i, 0, p) + g(e1, e2 − 1, i, 1, p) + g(e1, e2 − 2, i, 2, p) = 0,
f(e1, 0, i, 0, p) = f(e1, i, p), g(e1, 0, i, 0, p) = g(e1, i, p),
where f(e, s, p) and g(e, s, p) are the functions from Lemma 6.1. We use the first two rows
above to evaluate RM1,1(r, s, r, s), by matching terms in each ΣMi , as in the argument showing
A3(r, r) = 1. These equalities suffices to eliminate all terms from each ΣMi , except for the
terms in(volving f(e1, 0, i, 0), s)o that ∑ ∑
1 + ΣM +ΣM +ΣM = 1 + xpf(e1, 0, 0, 0, p) + xpf(e1, 0, 1, 0, p)0 1 2 pe1(1/2+r) p(e1+1)(1/2+r)
e∑1≥1 e1≥0
+ xpf(e1, 0, 2, 0, p)
p(
.
e1+2)(1/2+r)
e1≥0
68
8. The Ratios Conjecture prediction for the two-level density
As f(e1, 0, i, 0) = f(e1, i, p), this is exactly the pth factor of the product defining RM1 (r, r),
which we have already shown is equal to 1. Proceeding similarly to evaluate RS1,1, we find
that
RM1,1(r, s, r, s) = RS1,1(r, s, r, s) = A1(r, s, r, s) = A2(r, s, r, s) = 1. (8.2.5)
By a similar calculation, we can also show that
RM M M1,1(r, α, r, γ) = R1,1(α, s, γ, s) = R1 (α, γ), (8.2.6)
with R1 as in the Ratios Conjecture calculation for the one-level density. A similar result
holds for RS1,1. This equality holds factorwise, in the sense that the pth factors of the defining
products are all equal. We also have
A1(r, α, r, γ) = A1(α, s, γ, s) = A3(α, γ), (8.2.7)
by a similar calculation, and an analogous relation holds between A2 and A4.
Instead of calculating the derivative of RM S1,1 and R1,1 directly we will calculate the loga-
rithmic∣ derivative. We begin by studying RM1,1. Now, using (8.2.5) we have
∂2 ∣∣ logRM (α, β, γ, δ) = RM (r, s, r, s)−RM (r, s, r, s)RM (r, s, r, s),
∂α∂β ∣ 1,1 1,1,α,β 1,1,α 1,1,βα=γ=r, β=δ=s
where the subscripts denote that we have differentiated with respect to the indicated variable.
Thus, to find the derivative RM1,1,α,β , we need to add a correction term to the logarithmic
derivative. By (8.2.6), this correction term is equal to
RM M M1,1,α(r, s, r, s)R1,1,β(r, s, r, s) = R1,α(r, r)RM1,α(s, s),
and the right-hand side is given explicitly in (6.4.2).
We are left with the ta∑sk of calcu∣∣ lating∂2 ∣∣ logw(α, β, γ, δ),∂α∂β
p α=γ=r, β=δ=s
where we have written
w(α, β, γ, δ) = 1 + ΣM0 +ΣM +ΣM1 2 ,
and suppress∣ed the p-dependence. Now, as w(r, s, r, s) = 1
∂2 ∣∣∣ logw(α, β, γ, δ) = wα,β(r, s, r, s)− wα(r, s, r, s)wβ(r, s, r, s),∂α∂β α=γ=r, β=δ=s
so that in total ∑ ∑
RM1,1,α,β(r, s, r, s) = w M Mα,β(r, s, r, s)− wα(r, s, r, s)wβ(r, s, r, s) +R1,α(r, r)R1,α(s, s).
p p
To evaluate the expression above we must differentiate the function w, and we begin by
finding the derivative with respect to α evaluated at (r, s, r, s). As w(α, s, γ, s) is equal to
the pth factor defining RM1 (α, γ), we have∑already found(this de)rivative. Indeed, by (6.4.2),we have
wα(r, s, r, s) = −
xp log p 1
pe(1/2+ )
θe + .r p
e≥1
We remark that there is no dependence on s. By symmetry, we find a very similar equality
for the derivative with respect to β, by simply substituting r for s. As before these equalities
are valid for Re(r) > 0 and Re(s) > 0 respectively.
Now we want to differentiate with respect to both α and β. We begin by differentiating
w with respect to β, set β = δ = s, and fin(d that )
w M M Mβ(α, s, γ, s) = Σ0,β +Σ1,β +Σ2,β ,
69
8. The Ratios Conjecture prediction for the two-level density
where ∣ (
ΣM := d
∣∣∣ x log p ∑ΣM = − p e2f(e1, e2, i, 0, p)i,β i +
∑dβ p
i(1/2+γ) pe1(1/2+α)+e2(1/2+s)β=δ=s e1,e2 )
e2f(e1, e2, i, 1, p) ∑ e2f(e1, e2, i, 2, p) xp log p
pe1(1/2+α)+(e2+1)(1/2+s) + pe1(1/2+α)+( = −e2+2)(1/2+s) pi(1/2+γ)e1,e2 ∑ e1∑,e2 ( )f(e1, 1, i, 0, p) f(e , e , i, 0, p)− f(e , e − 2, i, 2, p)× (1 2+ )+(1 2+ ) + 1 2 1 2 pe1 / α / s pe1(1/2+α)+e2(1/2+s) ,
e1≥0 e1≥0,e2≥2
where we used (8.2.4) in the last step, similarly as in the one-level density computations.
To simplify the notation, we define a function h by
h(e1, e2, i) = f(e1, e2, i, 0, p)− f(e1, e2 − 2, i, 2, p),
where we have suppressed the p-dependence in the left-hand side. A straightforward calcu-
lation yields
( 0) = e1 + 1
τe (3δ3|e − 1) 1
h e1, e2, 3 + δ
1 2
2|e1δ2|e2 + 3 + ,p
( 1) = −2(e1 + 1)
τe1(3δ3|e2 − 1)+ − 1h e1, e2, 3 3 ,p
τ (3δ − 1)
h( e1 + 1 e 3|ee1, e2, 2) = 1 23 − δ2|e1δ2|e2 + 3 .
Now we differentiate with respect to α, and set α = γ = r to find 
ΣM = xp log
2 p ∑ e1f(e1, 1, i, 0, p) ∑ e1h(e1, e2, i) 
i,α,β pi(1/2+r) pe1(1/2+r)+(1 2+
+
/ s) pe1(1/2+r)+e2(1/2+s)
e1≥0 e1≥0,e2≥2
=: ΣM,1i,α,β +Σ
M,2
i,α,β .
To find wα,β(r, s, r, s) we must sum all the ΣMi,α,β over i and we begin by summing the Σ
M,1
i,α,β .
This is done using a special case of the second equality of (8.2.4) with the first and second,
and third and fourth arguments interchanged, using symmetry. Specifically, we use
f(e1, 1, 0, 0, p) + f(e1 − 1, 1, 1, 0, p) + f(e1 − 2, 1, 2, 0, p) = 0.
We find  
2 ∑
ΣM,10,α,β+Σ
M,1 +ΣM,1 = xp log p f(1, 1, 0, 0, p)1,α,β 2,α,β 1 2+ 1 2+ + f(e1, 1, 0, 0, p)− f(e1 − 2, 1, 2, 0, p) .p / s p / r pe1(1/2+r)
e1≥2
Now, by the symmetry of f and the definition of h, we have f(e1, 1, 0, 0, p) − f(e1 −
2, 1, 2, 0, p) = h(1, e1, 0) = 1 − δ3|e1 + 1/p. Also, we have f(1, 1, 0, 0, p) = 1 + 1/p =
1− δ3|1 + 1/p.
To simplify the sum of ΣM,2i,α,β , over i, we need a few identities for the function h. Similarly
as with f , we have
h(e1, e2, 0) + h(e1 − 1, e2, 1) + h(e1 − 2, e2, 2) = 0.
This implies that ( ∑
ΣM,20,α,β +Σ
M,2 +ΣM,2 2 h(1, e2, 0)1,α,β 2,α,β = xp log p p(1/2+r)+e2(1/2+s)
e2≥2 ∑ )
+ h(e1, e2, 0)− h(e1 − 2, e2, 2)
pe1(1/2+r)+
.
e2(1/2+s)
e1≥2,e2≥2
70
8. The Ratios Conjecture prediction for the two-level density
A calculation confirms that h(e1, e2, 0)−h(e1− 2, e2, 2) = ιe1,e2(p)/xp, with ι from (4.3.12).
Furthermore, ιe1,1(p)/xp = 1− δ3|e1 +1/p, and similarly for ι1,e2(p)/xp by symmetry. Thus,
we have shown that
∑ ∑ ∑
( xpιe,f (p) log
2 p
wα,β r, s, r, s) =
pe(1/2+
,
r)+f(1/2+s)
p p e,f≥1
valid for Re(r),Re(s) > 0. Finally, by putting together all of our calculations, we can
conclude that
∑ ∑ ι 2 ∑ ∑ 2e,f (p) log p xp(θe + 1/p)(θf + 1/p) log2 p
RM1,1,α,β(r, s, r, s) = − pe(1/2+r)+f(1/2+s) pe(1/2+r)+f(1/2+s)p∑
e,f∑≥1 p∑e,fx (θ ∑
≥1
+ p e + 1/p) log p xq(θf + 1/q) log q
pe(1/2+r) qf(1/2+s)
.
p e≥1 q f≥1
(8.2.8)
Without any significant additional difficulties, one may employ the same method of
calculation using the function g, instead of f , to find RS1,1,α,β(r, s, r, s). We leave out the
details, but the resulting expression is given by
∑ ∑
S ( ) = ξe,f (p) log
2 p ∑ ∑ x2pγe(p)γf (p) log2 p
R1,1,α,β r, s, r, s pe(1/2+r)+f(1
−
/2+s) pe(1/2+r)+f(1/2+s)
∑∑ p e,f≥1  p e,f≥1 (γ (p)− p−e/3) log p ζ ′ ∑∑ −f/3 ′+ e 5(1 2+ ) − ( 6 + r) (γf (q)− q ) log q − ζ ( 5 + s) ,pe / r ζ qf(1/2+s) ζ 6
p e≥1 q f≥1
(8.2.9)
with ξe,f (p) from (4.3.13). This calculation is also valid for Re(r),Re(s) > 0, and this
finishes the estimation of R1,1,α,β(r, s, r, s;X).
8.3 Finishing the estimates
We now turn to the task of estimating the derivatives of R′ , R′1,2 2,1 and R′2,2. First, we must
find appropriate estimates of all these terms, using a generalisation of Corollary 6.2. Indeed,
we have the following corollary of Lemma 8.1.
Corollary 8.2. Let m1,m2, h1, h2 ∈ Z+, and 1/2 ≤ θ < 5/6, ω ≥ 0 be such that (4.2.2)
holds. Then for cubefree h1, h2, and κ with 0 < Re(κ) < 1/2, we have
∑ − ( − )1 κ κ|D |−κ( ) K λK(m1)µK(h1)λK(m2)
X X
µK(h2) = F1(X) P + F (X) P
N X 1− 1 2κ 1− 6 2κ/5
K
 ∏ ( ) +O (1 + |κ|) pω 4(e1 + 1)(e2 + 1) + 3 Xθ−Reκ−1+ϵ ,
p|h1h2m1m2, pe1 ||m1, pe2 ||m2
with P1 and P2 as in Lemma 8.1.
Proof. This is proven from Lemma 8.1 in the same way as Corollary 6.2 is proven from
Lemma 6.1.
The point of the corollary will be to apply it with κ = α, β or α+ β to the estimation of
R′ , R′2,1 1,2 and R′2,2 respectively. We consider one product at a time.
71
8. The Ratios Conjecture prediction for the two-level density
First, the corollary indicates that a r(easonable estimate of R
′
1,2(α, β, γ, δ;X) is
Γ±(1/2− β) X−β
R M1,2(α, β, γ, δ;X) := Γ (1 F1(X) R/2 + β) 1− β 1,1(α,−β, γ, δ)± )
+ ( ) X
−β
F2 X
S
1 R (α,−β, γ, δ) .− 6β/5 1,1
In order to handle this expression, we will need to work with the products A1 and A2 instead
of RM and RS1,1 1,1. We use (8.2.3) to write
( ; ) = Γ±(1/2− β) ζ(1 + α− β)ζ(1 + 2α)ζ(1− 2β)ζ(1 + γ + δ)R1,(2 α, β, γ, δ X Γ±(1/2 + β) ζ(1 + α+ δ)ζ(1− β + δ)ζ(1 + α+ γ)ζ(1− β + γ) )
X−β ζ(5/6 + α)ζ(5/6− β) X−β× F1(X)1 A1(α,−β, γ, δ) +− β ζ(5/6 + δ)ζ(5 F2(X) A2(α,−β, γ, δ) ./6 + γ) 1− 6β/5
Next, we differentiate this expression with respect to β, set β = δ = s and obtain
( ; ) = −Γ±(1/2− s) ζ(1 + 2α)ζ(1 + γ + s)ζ(1 + α− s)ζ(1− 2s)R1,(2,β α, s, γ, s X Γ±(1/2 + s) ζ(1 + α+ s)ζ(1 + α+ γ)ζ(1− s+ γ) )
X−s ζ(5/6 + α)ζ(5/6− s) X−s× F1(X)1 A1(α,−s, γ, s) +− s ζ(5/6 + s)ζ(5/6 + γ)F2(X)1 6 5A2(α,−s, γ, s) ,− s/
(8.3.1)
at least assuming α ≠ s. As usual, a variable as a subscript denotes differentiation with
respect to the indicated variable. We will see later that the apparent singularity at α = s
will not be important. Next, we differentiate w(ith respect to α and set α = β = r and find
Γ (1/2− s) ζ ′ ζ ′ ′
)
±
R1,2,α,β(r,(s, r, s;X) = −Γ (1 2 + )ζ(1− 2s) (1 + r − s) + (1 + 2
ζ
r)− (1 + r + s)
± / s ζ ζ ζ
X−
)
s −s
× F1(X)1 A1(r,−
ζ(5/6− s) X
s[, r, s) + (5 6 + )F2(X)1 6 5A2(r,−s, r, s)− s ζ / s − s/
− Γ±(1/2− s)
−s ′ ( )
Γ (1 2 + )ζ(1− 2 ) ( )
X ζ(5/6− s) ζ
s F1 X 1 A1,α(r,−s, r, s) +
5
/ s − s ζ(5/6 + s) ζ 6 + r±
X−
]
s
× ( ) ( − ) + ζ(5/6− s) X
−s
F2 X 1 6 5A2 r, s, r, s− (5 6 + )F2(X)1 6 A− 2,α(r,−s, r, s) .s/ ζ / s s/5
(8.3.2)
Note that it is possible to simplify the result slightly by using (8.2.7), and the corresponding
equality for A2.
To obtain an estimate for R′2,1,α,β , we can simply exchange the role of r and s above. This
is possible by the symmetry of RM1,1 and RS1,1, induced by the symmetry of the functions f
and g, see (8.2.1), (8.2.2) and the remark after (8.1.3). We find that an appropriate estimate
is (
Γ (1/2− r) ζ ′ ′ ′
)
± ζ ζ
R2,1,α,β(r,(s, r, s;X) = −Γ (1 2 + )ζ(1− 2r) (1 + s− r) + (1 + 2s)− (1 + r + s)± / r ζ ζ ζ )
× ( )X
−r
(− ) + ζ(5/6− r) X
−r
F1 X 1 A1 r, s[, r, s (5 6 + )F2(X)1 6 5A2(−r, s, r, s)− r ζ / r − r/
− Γ±(1/2− r)
−r ′ ( )
Γ (1 2 + )ζ(1− 2 ) ( )
X ζ(5/6− r) ζ
r F1 X 1 A1,β(−r, s, r, s) +
5
6 + s
± / r − r ζ(5/6 + r) ζ
− ]r
× ( ) X (− ) + ζ(5/6− r)
−r
F2 X 1 6 5A2 r, s, r, s (5 6 + )F2(X)
X
1 6 5A2,β(−r, s, r, s) .− r/ ζ / r − r/
(8.3.3)
72
8. The Ratios Conjecture prediction for the two-level density
Finally, we turn to R′2,2. We(define the estimate R2,2(α, β, γ, δ;X) by
Γ±(1/2− α)Γ±(1/2− β) X−α−β
Γ F1(X) R
M
1,1(−α,−β, γ, δ)
±(1/2 + α)Γ±(1/2 + β) 1− α− β
X−α−
)
β
+ F2(X) S1 6( + ) 5R1,1(−α,−β, γ, δ) ,− α β /
which is motivated by Corollary 8.2, with κ = α+ β. Rewriting this in terms of A1 and A2,
we find
R2,2(α, β, γ, δ; ) =
Γ±(1/2− α)Γ±(1/2− β)
X Γ±(1/2 + α)Γ±(1/2 + β) (
ζ(1− α− β)ζ(1− 2α)ζ(1− 2β)ζ(1 + γ + δ) X−α−β×
ζ(1− α+ δ)ζ(1− β + δ)ζ(1− α+ γ)ζ(1− β + γ) F1(X)1− α− β )
× (− ζ(5/6− α)ζ(5/6− β) X
−α−β
A1 α,−β, γ, δ) + (5 6 + ) (5 6 + ) F2(X)1 6( + ) 5A2(−α,−β, γ, δ) .ζ / δ ζ / γ − α β /
We now differentiate with respect to β and set β = δ = s. It will suffice to differentiate the
fraction of zeta functions, which can be done by appealing to (8.3.1), and substituting α by
−α. The result is that
Γ±(1/2− α)Γ±(1/2− s) ζ(1− 2α)ζ(1 + γ + s)ζ(1− α− s)ζ(1− 2s)
R2,2,β(α,(s, γ, s;X) = −Γ±(1/2 + α)Γ±(1/2 + s) ζ(1− α+ s)ζ(1− α+ γ)ζ(1− s+ γ)
X−α−s× F1(X)1 A1(−α,−s, γ, s)− α− s
ζ(5/6− α)ζ(5/6− s) X−α−
)
s
+ F2(X) (5 6 + ) (5 6 + ) 1 6( + ) 5A2(−α,−s, γ, s) .ζ / γ ζ / s − α s /
Next, we differentiate this with respect to α and set α = γ = r. We find that
Γ±(1/2− r)Γ±(1/2− s) ζ(1− r − s)ζ(1− 2r)ζ(1− 2s)ζ(1 + r + s)
R2,2,α,β(r,(s, r, s;X) = Γ±(1/2 + r)Γ±(1/2 + s) ζ(1− r + s)ζ(1− s+ r)
X−r−s× F1(X)1 A1(−r,−s, r, s)− r − s )
+ ( )ζ(5/6− r)ζ(5/6− s) X
−r−s
F2 X
ζ(5/6 + r)ζ(5 6 + ) 1 6( + ) 5A2(−r,−s, r, s) ./ s − r s /
(8.3.4)
8.4 Finding the two-level density
With the calculations complete, we have found the following Ratios Conjecture:
Conjecture 8.3. Let 1/2 ≤ θ < 5/6, ω ≥ 0 be such that (4.2.2) holds. Then there is
some δ < 1/6 such that for any fixed ϵ > 0, and r, s ∈ C with Re(r),Re(s) > 0, 1/L ≪
Re(r),Re(s) < δ, and |r|, |s| ≤ Xϵ/3, we have that
1 ∑ L′(1/2 + r, fK) L′(1/2 + s, fK)
( ) (1 2 + ) (1 2 + ) = F1(X)R
M
1,1,α,β(r, s, r, s)N X L / r, f L / s, f
K∈F±(X) K K
+ F2(X)RS1,1,α,β(r, s, r, s) +(R1,2,α,β()r, s, r, s;X) +R2,1,α,β(r, s, r, s;X)
+R θ−1+ϵ2,2,α,β(r, s, r, s;X) +O X ,
where RM S1,1,α,β, R1,1,α,β, R1,2,α,β, R2,1,α,β, and R2,2,α,β are given in (8.2.8), (8.2.9), (8.3.2),
(8.3.3), and (8.3.4) respectively.
73
8. The Ratios Conjecture prediction for the two-level density
Remark. From the expressions (8.3.2) and (8.3.3) it may appear that the right-hand side
is ill-defined at every point where r = s. The existence of such singularities would be odd,
as the left-hand side is finite, at least assuming the Generalised Riemann Hypothesis, as a
finite sum of finite numbers. However, by a careful study of the sum of (8.3.2) and (8.3.3)
we can see that there is in fact no singularity. Indeed, this follows by first making the change
of variables z = s− r, w = s+ r, a Laurent expansion of ζ ′(1± z)/ζ(1± z), and a complex
multivariate Taylor expansion of the rest of the expression around any point (z, w) = (0, w0).
For a proof of the multivariate Taylor expansion, see e.g. [Le, Ch. 1.2].
Using this conjecture, we can find a prediction for the two-level density. Indeed, we will
prove the following proposition:
Proposition 8.4. Let 1/2 ≤ θ < 5/6, ω ≥ 0 be such that (4.2.2) holds. Assume the
Generalised Riemann Hypothesis for the functions ζK(s), with K ∈ F±(X), Conjecture
6.3 and Conjecture 8.3. Let ϕ(u1, u2) = ϕ1(u1)ϕ2(u2) be a product of real, even Schwartz
functions whose Fourier transform has compact support. Then, we have
∑ ∑ ( )1 L L ′
N±(X) ϕ 2 γK , 2 γ =[ π π
K
K γK ̸=±γ′K ( )
2 4 log2(2 ) + log(2 ) + 2 ( )]π π
ϕ̂1(0)
2 log(4π e) 2 2− 20 log(2π)
ϕ̂2((0) 1 + )(+ 2 ) + F2(X) −L L ( 5 +L 25L2 )
+ (0) 1 + 2 log(2πe) ( ) (2) + ( ) (2) − ϕ̂1(0)ϕ̂1 ( )(F1 X S1 F2 X S2 ) (F1(X)S1(2) +
6
L L 5F2(X)S2(2))
+ (0) 1 + 2 log(2πe)ϕ̂2 F1((X)S1(1) + ( )
ϕ̂2(0)
F2 X S2()1) − F1(X)S1(1) +
6
5F2(X)S2(1)( L ) L
+ (0) ( ) + (0) ( ) 1 + 2 log(2πe)− 1 ϕ̂2(0) ϕ̂1(0)ϕ̂2 J1 X ϕ̂1 J2 X + J ′ (X) + J ′
L L 1 L 2
(X)
+ E1J2(X( ) + E J (X) + J ′′2 1 (X)) ( )
+ F1(X) S1(1)S1(2)− S3 + S5 + F2(X) S2(1)(S2(2)− S−1+ ) 4 + S6+ E θ ϵ1R2 + E2R1 − E1E2 − 2R3 − 2J3(X) +Oϵ X ,
(8.4.1)
where
∑ ( ) ( )e ∫ ′
J ′
( )
i(X) = −
2 · 1 ( ) log p log p − 2 · F2(X) Lr ζ5F2 X
5
( 5 6 ϕ̂i 2 5 ϕi 2 6 + r drL∫ p e/ L πi πi ζp,e) ( (1/L)−r
+ 2 Lr2 ϕi 2 ζ(1− 2 )
Γ±(1/2− r)
r Γ F1( )
rX
X A3(−r, r)
πi (1/L) πi ±(1/2 + r) )(1− r)2−r
+ ( ) · 6 · (1/6 + r)X A4(−r, r)F2 X 5 (1 6 5)2 ·
ζ(5/6− r)
dr,
− r/ ζ(5/6 + r)
(8.4.2)
74
8. The Ratios Conjecture prediction for the two-level density
and  ∑ ( ) ( ) 
′′( ) = ( ) (1) + 2 log p log e 2 ∫p Lr ζ ′ ( )J X F2 X S2 5 6 ϕ̂1 + ϕ 5  e/ 2 1L p L πi 2πi ζ 6
+ r dr
∑ p,e≥1 ( ) ∫
(1(/L) ) 
× 2 log q log q
f 2 Lr ζ ′ (
S 5
)
2(2) + 5 6 ϕ̂2 + 2 ϕ2 + s ds

L q f/ L πi (1/L) 2πi ζ 6q,f≥1
− F2(X)S∫2(1)S2∫(2) ( ) ( )(
+ 4 Lr Ls(2 )2 ϕ1 2 ϕ2 2 ) R1,2,α,β(r, s, r, s;X)+πi (1/L) (1/L) πi πi
R2,1,α,β(r, s, r, s;X) +R2,2,α,β(r, s, r, s;X) drds.
(8.4.3)
Here, Si, Ri and Fi(X) are defined in Chapter 4. Also, Ji(X) is defined as the expression
(7.1.1), with ’ϕ’ replaced by ϕi, with the convention that ϕ3(u) = ϕ(u, u) = ϕ1(u)ϕ2(u).
Proof. We beg∑in by us∑ing (4(.1.1) to see tha)t the two-level d∑ensity∑is eq(ual to )1 L L 2 L L
±( ) ϕ 2 γK , 2 γ
′ −
N X π π K N±′ (X)
ϕ 2 γK , 2 γK .π π
K∈F±(X) γK ,γK K∈F±(X) γK
The second sum can be found directly using Proposition 6.4, with the test function ϕ replaced
by ϕ3(u). By (8.1.2), the first (su∫m is give∫n by∑ ) ′ ( )1 · 1 − L (1/2 + r, fK) Lr(2 )2 (( ) ) (1 2 + )
ϕ1 dr
πi N ∫X ∫ (1/L) (−1/L) L (/ )r, fK 2πiK∈F±(X) (8.4.4)′
× − L (1/2 + s, fK) Lsϕ2 ds,
(1/L) (−1/L) L(1/2 + s, fK) 2πi
which will now be our object of study.
As in the proof of Proposition 6.4, we must first prepare the integral so that the conditions
of Conject∫ure 8.3 app(ly. U)sing (6.6.2), we immediate1 ′(1 2 + ) 1 ∫
ly find th(at )
− Lt L / t, f Lt L
′
K = (1/2 + t, fK)2 ϕiπi (−∫1/L) 2( )((1 2 + ) dt ϕi dtπi L / t, fK 2πi (1/L) 2πi L)(1/2 + t, fK)
+ 1 Lt
Γ′ Γ′
2 ϕi log|D |+
± (1/2 + t) + ±K (1/2− t) dt,
πi (−1/L) 2πi Γ± Γ±
with i = 1, 2. The second integral can be shifted to (0), and we find that it equals Ci,K+Ei,
with no∫tation∫from Ch(apter)4. T(his im)plies that (8.∑4.4) equals4 Lr Ls 1 L′(1/2 + r, fK) L′(1/2 + s, fK)
(2πi)2 ϕ1 ϕ2 ±(1/L) ∫(1/L) (2πi ) 2πi N∑(X) L(1/2 + r, fK) L(1/2 + s, fK)
drds
K∈F±(X)
2 Lr 1 L′+ (1/2 + r, fK)2 ϕ (C + E )drπi ∫ 1(1/L) (2πi) ±( ) (1 2 + 2,K 2N X ∑ L / r, fK)K∈F±(X)
+ 2 Ls 1 L
′(1/2 + s, fK)
2 ϕ2 ± (C1,K + E1)dsπi (1/L) ∑ 2πi N (X) L(1/2 + s, fK)K∈F±(X)
+ 1±( ) (C1,K + E1)(C2,K + E2).N X
K∈F±(X)
(8.4.5)
75
8. The Ratios Conjecture prediction for the two-level density
The last term above has already been simplified in Chapter 3 and 4. The second and
third terms are identical up to a permutation of the variables, so we concentrate on the
second term. The sum involving E2 is easy to calculate, as E2 is independent of both K
and r, so that we can
2 ∫
directly(apply)
Lr 1 ∑ L′(1/2 + r, fK)
2 ϕ1 drπi (1/L) 2πi N±(X) L(1/2 + r(, f∈F±( ) K)K X )
= F1(X)S1(1) + F2(X)S2(1) + J1(X) +Oϵ Xθ−1+ϵ
from (6.6.3), with the sums Si(j) as in Chapter 4.
The sum involving C2,K is harder to estimate. Recall that
C2,K = ϕ̂2(0)
log|DK |
.
L
Let u ≥∑1 be a real variable. We will make use of Stieltjes integration and the resultL′(1/2 + r, fK)
L∑(1/∑2 + r, f∈F±( ) K)K u ( )
= − ± x log p 1
∑∑ (γ (p)− p−e/3p ± 5/6 e ) log pC1 u pe(1 2+ ) θe + − C u/ r p 2 ( pe(1/2+r)p e≥1 p e≥1 (8.4.6)′ −r
+ C± 5/6 ζ Γ±(1/2− r) u2 u (5/6 + r)− ζ(1− 2r)
±
Γ (1 )2 + ) 1 C1 uA3(−r, r)ζ ± / r
u−r ± 5 6 ζ(5/6− r) ( − r+ )+ / θ ϵ1− 6 5C2 u (5 6 + )A4(−r, r) +Oϵ u ,r/ ζ / r
again from (6.6.3). Note that (6.6.3) can only be used when u is large compared to |r|. We
ignore this minor technical point for now an∫d discuss itlater. We compute∑ L′(1/2 + r, f X ∑ ′K)
(1 2 + ) log|DK | = (log u)
 L (1/2 + r, fK)d 
L / r, fK 1 L(1/2 + r, fK)K∈F±(X) ∑ K∈F±(u)′
= log L (1/2 +
∫
r, f X ∑ ′K) 1 L (1/2 + r, fK)
X
L(1/2 + − du.r, fK) 1 u L(1/2 + r, f )K∈F±(X) KK∈F±(u)
Using the same cutoff argument as in the proof of Proposition 6.4, and the result above, we
find that ∫ ( )2 Lr 1 ∑ L′(1/2 + r, fK)
2 ϕ1πi (1/L[) 2
C2,K
πi N±(X) L(1/2 + drr, f )
K∈F±(X) K
= ϕ̂2(0)
( )
( (logX) F1(X)S1(1) + F2(X)S2)(1) + J1(X)L ∑ ( )
− ( ) (1) + 6 ( ) (1) + ( ) − 2 · F2(X) log p log p
e
F1 X S1 5F2 X( S2 ) (J1 X ) ϕ̂∫ 1L 5 p5e/6 Lp,e
− 2 · F (X) Lr ζ
′
2 5
2πi
2 ∫ 5 ( )
ϕ1 2 6 + r d(r(1/L) πi ζ
+ Lr (1− 2 )Γ±(1/2− r) rX
−r
2 ϕ1 2 ζ r Γ (1 2 + ) F1()X) (]1 )2A3(−r, r)πi (1/L) πi ± / r − r
+ ( ) · 6 (1/6 + r)X
−r
· A4(−r, r) · ζ(5/6− r)F2 X 5 (1 6 5)2 dr .− r/ ζ(5/6 + r)
We now discuss the technical issue raised above. As we mentioned, to find (8.4.6), one
needs r to be small in comparison to u so that Conjecture 6.3 can be applied. Hence, we
76
8. The Ratios Conjecture prediction for the two-level density
cannot be sure that this really holds unless u ≥ X1/2, |r| ≤ Xϵ/3, say. However, using e.g
[IK, Thm. 5.17] to bound the size of the summand, we can see that the terms involving
u ≤ X1/2 will not be important to the final result, as they can be absorbed in the error
term so that there is no problem. The cutoff argument also ensures that we only need to
consider small |r|.
The only term left to analyse is the first term of (8.4.5). We use a cutoff argument
so that Conjecture 8.3 can be applied, similar to the argument from Proposition 6.4. For
technical reasons, one may want to move one of the integrals to the line (2/L) before trying
to make this argument, so that it is possible to estimate
ζ ′ (1± (r − s))≪ L,
ζ
and then move the integral back once Conjecture 8.3 has been applied. We leave out the
details for bounding the rest of the integrand for large |r|, |s|, as they have been given before
in similar situations.
The∫result∫is that th(e firs)t ter(m of )(8(.4.5) equals4 Lr Ls M S
(2 )2 ϕ1πi 2 ϕ2 F1(X)Rπi 2πi 1,1,α,β(r, s, r, s) +)F2(X)R1,1,α,β(r, s, r, s)(1/L) (1/L)
+R1,2,α,β(r, s, r, s;X) +R θ−1+ϵ2,1,α,β(r, s, r, s;X) +R2,2,α,β(r, s, r, s;X) drds+Oϵ(X ).
We will only simplify the integral over the first two terms. Recall that these terms are equal
to certain sums, given in (8.2.8) and (8.2.9). To simplify, interchange the order of summation
and) integration, move both contours to ((0), and apply the defin)ition of the(Fourier transform.We find a total contribution of F1(X) S1(1)S1(2) − S3 + S5 + F2(X) S2(1)S2(2) − S4 +
S6 + J ′′(X). The proposition follows.
77
8. The Ratios Conjecture prediction for the two-level density
78
9
Interpreting the Ratios
Conjecture prediction for the
two-level density
In this chapter, we will study the Ratios Conjecture prediction of the two-level density from
Proposition 8.4. We first focus on a comparison to the unconditional result from Theorem
4.1 for σ < 1, and in this direction, Proposition 9.1 implies that the two results agree quite
well. Next, we will study the Ratios Conjecture prediction in the extended range σ < ∞.
For such σ, we explicitly find the main term and the secondary term of the predicted two-
level density in Theorem 9.3. In particular, we detect phase transitions when σ1, σ2, and σ
exceed 1 respectively.
9.1 Two expressions for the two-level density
We now let
σ = σ1 + σ2 < 1,
and compare the results of Theorem 4.1 and Proposition 8.4. In the notation of Proposition
8.4, the two expressions for the two-level(density, excludi)ng the error terms, differ by( ) 2
I(X) := ϕ̂2(0)
log(4π e)
J1(X) + ϕ̂1(0)J2(X) 1 + +
ϕ̂2(0)
J ′1( ) +
ϕ̂1(0)
X J ′2(X)L L L
+ E1J2(X) + E2J1(X) + J ′′(X)− 2J3(X).
We then have the following proposition:
Proposition 9.1. For σ < 1, the difference I(X) between the non-error terms in the Ra-
tios Conjecture prediction from Proposition 8.4, and the actual two-level density given in
Theorem 4.1, admits the bound
I(X)≪ Xσ/6−1/3+ϵ +Xσ/2−1/2+ϵϵ . (9.1.1)
In particular, for sufficiently small σ, depending on ϵ, we have I(X)≪ X−1/3+ϵϵ .
The rest of this section will be dedicated to proving Proposition 9.1. First, we already
have a bound for Ji(X), given in (7.1.2). Recalling that ϕ̂3 is supported in [−σ, σ], we thus
find that ( )
I(X) = O Xσ/6−1/3+ϵ +Xσ/2−1/2+ϵ + ϕ̂2(0) ′ ( ϕ̂1(0)J X) + J ′ϵ 1 2(X) + J ′′(X). (9.1.2)L L
We turn our attention to J ′i(X). Note that the definition of J ′i(X) is very similar to the
definition of J(X), and we can bound J ′i(X) in almost the same way as we bounded J(X),
so we do not provide all the details. First, by using the calculations from when we bounded
J(X), it is not hard to see that the residue at r = 1/6 coming from the integrand of the first
79
9. Interpreting the Ratios Conjecture prediction for the two-level density
integral in (8.4.2) cancels against the residue from the first integrand of the second integral,
up to an error≪ Xσ/6−1/2. Thus, if we shift the contour to (1/2−ϵ) when integrating these
terms, the contribution from the residue can be absorbed in the error term from (9.1.2).
Once the first integral has been moved, we may cancel it against the sum in (8.4.2), while
the second integral that was moved can be bounded as ≪ Xσ/2−1/2+ϵ. Finally, we shift the
contour of the remaining integral in (8.4.2) to (1/6− ϵ) and estimate the resulting integral.
The result is that
J ′(X)≪ Xσ/6−1/3+ϵ +Xσ/2−1/2+ϵi ϵ .
It remains to estimate J ′′(X). We begin by handling the first three rows of the definition
(8.4.3). First, we may move the i∫ntegral ( )2 Lr ζ ′
2 ϕi (5/6 + r)drπi (1/L) 2πi ζ
to the line (1/2 − ϵ). We pick up the negative of the residue at r = 1/6, as we are shifting
the contour to the right, and(the ab)ove thus equals ( )
L 2 ∑ log p log pe2ϕi 12 − ϕ̂i .πi L p5e/6 L
p,e≥1
It follows th(at th(e sum)of th(e first )three rows of ((8.4.3))equal ( ))
F2(X) 4
L L
ϕ1 12 ϕ2 12 + 2 (2)
L
S2 ϕ1 12 + 2S2(1)
L
ϕ2 12 . (9.1.3)πi πi πi πi
This expression cannot be absorbed in the error term of (9.1.2) directly, but we will see that
other terms from J ′′(X) cancels these terms.
It remains to estimate the rest of J ′′(X), namely the integral of R1,2,α,β , R2,1,α,β and
R2,2,α,β . We begin by estimating R2,2,α,β , as this will be quite simple. First, recall that both
A1(α, β, γ, δ) and A2(α, β, γ, δ) are ≪ϵ 1, when |Re(α)|, |Re(β)|, |Re(γ)|, |Re(δ)| < 1/6 − ϵ.
Next, note that in (8.3.4), the expression for R −r−s2,2,α,β contains a factor X . Thus, moving
both contou∫rs to (1∫/6− ϵ/2), we can bound ( ) ( )4 Lr Ls σ/6−1/3+ϵ
(2πi)2 R2,2,α,β(r, s, r, s;X)ϕ1 ϕ2 drds≪ϵ X ,(1/L) (1/L) 2πi 2πi
where we also used (3.1.2) and σ = σ1 + σ2.
It remains to consider the term involving R1,2,α,β and R2,1,α,β . The first part of R2,1,α,β
from (8.3.3) is given by ( )
−Γ±(1/2− r) ζ
′ ζ ′ ζ ′
Γ±((1/2 + )ζ(1− 2r) (1 + s− r) + (1 + 2s)− (1 + r + s)r ζ ζ ζ− − )X r× ( ) (− ) + ζ(5/6− r) X rF1 X 1 A1 r, s, r, s (5 6 + )F2(X)1 6 5A2(−r, s, r, s) .− r ζ / r − r/
We consider the integral of this term together with the integral of the corresponding term
from R1,2,α,β , and merge them into a single integral. The reason for this is to remove the
singularities at s = r.
We begin by separating the terms involving A2 from the terms involving A1, so that
the integral of this part of R1,2,α,β +R2,1,α,β is split into two different double-integrals. We
then begin by considering the integral of the two terms containing a factor A2. The first of
these terms, coming from R2,1,α,β , contains a factor F2(X)X−r, while the term coming from
R1,2,α,β contains a factor F (X)X−s2 . We shift the contour for the r-integral to (1/6 − ϵ)
and the contour for the s-integral to (1/6 − 2ϵ). The reason for shifting to different lines
is that we can bound the integral over each term separately without worrying about the
pole at s = r. The result is that we may bound the integral of these two terms as being
≪ Xσ/6−1/3+ϵϵ , after possibly modifying ϵ.
80
9. Interpreting the Ratios Conjecture prediction for the two-level density
Next, we consider the integral of the terms containing A1. Recall that A1(−r, s, r, s) =
A3(−r, r), and ∫A1(r,−4 ∫
s, r, s) =(A3(−)s, s)(so th)at[we may write this inte(gral as
− Lr Ls Γ
′
±(1/2− r) ζ
(2 )2 ϕ1 2 ) ϕ2 2 Γ (1 2 + )ζ(1− 2r) (1 + s− r)πi (1/L) (1/L) πi πi ± / r ζ
ζ ′ ′ −r+ ( (1 + 2 )−
ζ (1 + + ) ( )X (− ) + Γ±(1/2− s)s r s F1 X 1 A3 )r, rζ ζ − r Γ±(1 ζ(1− 2s)/2 + s) ]
× ζ
′
(1 + − ) + ζ
′ ′ −s
r s (1 + 2r)− ζ (1 + r + s) F1( )
X
X 1 A3(−s, s) drds.ζ ζ ζ − s
We shift the integral over r to (1/2 − ϵ) and afterwards, we shift the integral over s to
(1/2 − 2ϵ). The shifted integral itself can be absorbed in the error term above, as the
integral contains the factors X−r and X−s. What is more interesting is the residues we pick
up at r = 1/6 and s = 1/6 respectively, coming from the pole of A3.
Using the calculation of the residue from the one-level computation, and the relation
±
F1( ) −1/6 ·
2ζ(2/3)Γ±(1/3)ζ(3)
X X −1/6
C2 −1/2
5Γ±(2 3) (5 3) (2)
= F1(X)X ± = F2(X) +O(X ), (9.1.4)/ ζ / ζ C1
we see that th(e cont)ributio∫n from th(e res)id(ues is, up to an acceptable error term, eq)ual to′ ′ ′
− F2(X)
L 4 Ls ζ
ϕ1(12 ) 2 ∫ ϕ2 (2 (5/6 +
ζ
s) + (1 + 2 ζs)− (7/6 + s) ds
πi πi (1/L) πi )(ζ ζ ζ )
− ( ) L 4 Lr ζ
′ ζ ′ ζ ′
F2 X ϕ2 12 ϕ1 (5/6 + r) + (1 + 2r)− (7/6 + r) dr.πi 2πi (1/2−ϵ) 2πi ζ ζ ζ
The second integral can be simplified by expanding ζ ′/ζ into its Dirichlet series, exchanging
the order of summation and integration, shifting the integral to (0), making a change of
variables, and then applying the definition of the Fourier transform. The first integral is
simplified by first shifting to 1/6 + ϵ, picking up a residue at s = 1/6. The shifted integral
is then simplified by the procedure described above for the second integral.
We have studied the integral of the first two rows in (8.3.2) and (8.3.3), coming from
R1,2,α,β and R2,1,α,β respectively. The result is that this integral gives, up to terms ≪ϵ
Xσ/6−1/3+ϵ +(Xσ/2−1)/2+ϵ(, a con)tribution ( ) ( )( ( )
− 4 ( ) L L
∑ e
F2 X ϕ1 ϕ2 − 2F (X)
L · − 2 log p log pϕ ϕ̂
∑ 1(2πi 2 ) 12
2 1
π∑i ( 12)πi) L ( p
5
p,e≥1 )e/6(2 )L
+ log p log p
e
− log p log p
e
− 2 ( ) L 2ϕ̂2 7 6 ϕ̂2 F2 X ϕ( e e/ 2
· −
p L p L 12πi L
p,e≥∑1 ( ) p,e∑≥1 ( 2 ) ( ))
× log p log p
e
+ log p log p
e ∑
− log p log p
e
5 6 ϕ̂1 ϕ̂1 7 6 ϕ̂1 .p e/ L pe L p e/ L
p,e≥1 p,e≥1 p,e≥1
(9.1.5)
Note that the first term above cancels the first term of (9.1.3).
To finish our calculations we must consider the integral of the rest of terms coming from
R1,2,α,β and R2,1,α,β . By symmetry, it will be sufficient to study the terms coming from
R2,1,α,β . ∫The in∫tegral in(quest)ion i(s then) [
− 4 Lr Ls Γ±(1/2− r) X
−r
(2πi)2 ϕ1 ϕ2(1/L) (1/L) 2πi ( 2)πi Γ±(1/2 + )
ζ(1− 2r) F1(X)
r 1 A (−r, s, r, s)− 1,βr
′ −r
+ F2(X)
ζ(5/6− r) ζ 5 X
(5 + s A2(−r, s, r, s)ζ /6 + r) ζ 6 1− 6r/5
− ]r
+ F2( )
ζ(5/6− r) X
X
ζ(5 6 + A2,β(−r, s, r, s) drds./ r) 1− 6r/5
81
9. Interpreting the Ratios Conjecture prediction for the two-level density
As usual, the integrands involving F (X)X−r2 will be unimportant and can be absorbed into
the error term. Here we also make use of the fact that on the contour we are integrating
over, we have that A2,β ≪ 1, which follows from Cauchy’s integral formula for the derivative
of a holomorphic function.(We a)re lef(t with)
− 4F1( )
∫ ∫
X Lr Ls Γ (1/2− r) X−r±
(2 )2 ϕ1 2 ϕ2 2 Γ (1 2 + )ζ(1− 2r)1 A1,β(−r, s, r, s)drds.πi (1/L) (1/L) πi πi ± / r − r
(9.1.6)
We want to shift the integral above to the right, but to accomplish this, we must first
find a meromorphic continuation for A1,β(−r, s, r, s). Recall that A1(−r, β, r, s) is a priori
only defined when the real part of β and s are both greater than −1/6, and the modulus of
the real part of r is less than 1/6. Therefore, we begin by presenting an analogue of Lemma
7.2, extending the domain of definition.
Lemma 9.2. Let |Re(r)|∏< 1(/6, and Re(β)),(Re(s) > −1)/6(. Then, we)h(ave )
A1(−
1 1 1 1
r, β, r, s) = ζ(3) 1− 1+ − 1−( ) p β r p1−2
1− 1−
r p1+2β p1+r+s
p
1 −1
( ) ( )
× 1− 1− 1
−1 1 −1
[ p1+(s−r )p1+ 1−β+s p1+β+r
1 1 (1− p−(1/2+r) 2× 2 +( 6 )2 (
(
1− −(1/2+
)
s))2 (p 1 (1− p−(1+2r)) ( 2 + ) )
)(( −(1+2 ))1− p s )
p 1− p−(1/2−r) 1− p−(1/2+β) 2 1− p−(1−2r) 1− p−(1+2β)
1 (1 + p−(1/2+r) + p−(1+2r)) (1 + p−(1/2+s) + p−(1+2s)+ )3 1(+ p−(1/2−r) + p−(1−2r) 1 + p−(1/2+β) + p−(1+2β)
1 (1− p−(1/2+r)+ ))(( ]1− p−(1/2+ )s)) .
p 1− p−(1/2−r) 1− p−(1/2+β)
For the sake of brevity, we leave out the proof, but we remark that it is very similar to
the proof of Lemma 7.2.
Suppose that both Re(β),Re(s) > 1/2 + ϵ. Then, for |Re(r)| < 1/2 we could write the
expression abov(e a∏ (
s )[ ( )2 ( )
(3) 1− 1 1 1 (1− p−(1/2+r)) 1 (1− p−(1+2r)ζ ( p1−2r p2
+ + )
)6 1−(p−(1/2−r) 2 ) 2] 1− p−(1−2r)p( ) (9.1.7)1 1 + p−(1/2+r) + p−(1+2r)) 1 (1− p−(1/2+r)+ ) ( −1− )ϵ3 1 + p−(1/2−r) ++ p−(1−2r) p 1− p−(1 +O p ./2−r)
Comparing this to the proof of Lemma 7.2, we see that if we multiply this product by the
Euler product of ζ(3/2− 3r)−1, then the result converges absolutely. Thus, we have proven
that A1(−r, β, r, s) is meromorphic in Re(β),Re(s) > 1/2 + ϵ, |Re(r)| < 1/2 − ϵ, with a
singularity whenever r = 1/6.
With this result in mind, we move the integral over s in (9.1.6) to (1/2 + 2ϵ), which
yields
4 ( ) ∫ ∫ ( ) ( )− F1 X Lr Ls Γ±(1/2− r)(2 )2 ϕ1 ϕ2πi (1/2+2ϵ) (1/L) 2πi 2πi Γ±(1/2 + )ζ(1− 2r)r (9.1.8)
X−r× 1 A1,β(−r, s, r, s)drds.− r
The next step will be to move the inner integral over r to (1/2− ϵ). The shifted integral can
then be estimated as usual, and we can see that it will give no significant contribution to
the result. We remark that on the line (1/2− ϵ) the factor A1,β is estimated by combining
82
9. Interpreting the Ratios Conjecture prediction for the two-level density
Cauchy’s integral formula for the derivative with an estimate of A1. The estimate for A1 is
obtained by using (9.1.7) together with an estimate for ζ(3/2− 3r) when the real part of r
is close to 1/2. As the shifted integral is small, it is more interesting to study the residue
we pick up at r = 1/6.
To find the residue at r = 1/6 we will first need to study A1,β more closely. Thus, we
will first need to differentiate A1(−r, β, r, s) with respect to β, and then set β = s. We will
accomplish this by logarithmically differentiating the expression in Lemma 9.2. First, we
multiply by ζ(3/2− 3r)−1 to ensure that all technical conditions for moving differentiation
inside a sum are fulfilled. Next, recall that the pole at r = 1/6 of A1(−r, β, r, s) is simple,
and evidently, it remains so after differentiation with respect to β. Define
g(r, β, s) := A1(−r, β, r, s)ζ(3/2− 3r)−1,
which is then an absolutely convergent product in |Re(r)| < 1/2, Re(β),Re(s) > 1/2 + ϵ.
Recall that A1(−r, s, r, s) = A∣3(−r, r) so that
d ∣∣∣ log A1,β(−r, s, r, s)g(r, β, s) = .dβ β=s A3(−r, r)
The residue is then equal to ∣∣
lim (r − 1 d/6)A1,β(−r, s, r, s) = lim (r − 1/6)A3(−r, r) ∣ log g(r, β, s)
r→1/6 r→1/6 ∣∣ dβ ∣∣ β=s= − ζ(3) d3ζ(5/3)ζ(2) dβ ∣ log g(1/6, β, s),β=s
by Lemma 7.2.
We turn to the task of calcula(ting t(he loga)rithmic(derivative. F)irst,∑ ( )
log g(1 1(/6, β, s) = l)og ζ(3) +( log )1− (+ log 1−
1 + log 1− 1
p p)5/6+β p2/3p
+ log 1− 1( 1+2 )+ log (1−
1 − log 1− 1
p β p7/6+s ) p5/6+s
− log 1− 1 1[ p1+( + − log 1−β s ) ( p7/6+β2
1 1 1 + p−(1/3 1− p−()1/2+s))2 ( ) ( )1 1 + p(−2/3 1− p−)(1+2s)+ log( p2
+ 6 +1)−(p−(1/2+β) 2 2 1− p−(1+2β)−1 3( −2 3 −(1 2+ ) ) −(1+2 )) ( (−1 3) ( ) ])1 1− p / + p / 1 + p / s + p s 1 1 + p / 1− p−(1)/2+s)+ 3 1 + p−(1/2+β) + p−(1+2 +β) p 1− p−(1/2+β) .
Next, we differentiate this with respect to β and set β = s. Note that if we set β = s in
the argument of the last logarithm, then this argument is precisely equal to the reciprocal
of yp(1 +(p−1/3). Using this, the result after differentiation is∑ log p log p log p
p5/6+
+
s − 1( p1+2s − 1
−
p7/6+s − 1
p
+ (1 + −1/3) − (1 + p
−1/3)2 log p (1 + p−2/3− ) log pyp p 3(p1/2(+s − 1) p1+2s − 1 )
(1− p−1/3 + p−2
) )
/3
+ ) log p 1
−1/3
− 3 − (1 + p ) log p3 p1/2+s − 1 p3/2+3s − 1 p(p1/2+
.
s − 1)
Simplifying, and then applying the formula for a geometric series shows that the above is
83
9. Interpreting the Ratios Conjecture prediction for the two-level density
equal to
∑ ( ∑ ∑ ∑
− log − 1 1 1p − + + yp(1 + p−1/3)
( p5e/6+es pe+2es p7e/6+esp e∑≥1 e≥1 e≥1 ))δ
× (1 + −2/3) 2|e
+ δ3| ∑ ∑e
p + p−1/3
1− δ3|e + (p−1 + p−4/32+ 2+ )
1
( pe/ es pe/ es pe)/2+es∑ e≥∑1 ∑ e≥1 ∑ ∑ e≥1
= − log p − 1 1 1 γe(p)
p5e/6+
−
es pe+2
+ + .
es p7e/6+es pe/2+es
p e≥1 e≥1 e≥1 e≥1
Let us denote the expression above by R(s).
In conclusion, our calculations show that the contribution from the residue we pick up
from (9.1.8) is, up to an acceptable error term, equal to
( ) ∫ ( )
−2 L 2 LsF2(X)ϕ1 12 · 2 ϕ2 2 R(s)ds, (9.1.9)πi πi (1/2+2ϵ) πi
where we made use of (9.1.4). Note that we picked up the negative of the residue as we
shift to the right. Next, interchange the order of integration and summation to move the
integral inside every sum of R(s). Then shift all contours to (0) and apply the definition of
the Fourier transform, after making a change of variables. We find that (9.1.9) is equal to
( ) ( )( ( ) ( )
−2 ( ) L · − 2
∑ log p log pe ∑ log p log p2e
F2 X ϕ1
∑12
− ϕ̂ − ϕ̂
πi (L ) p5e/6 2 2L pe Lp),e≥1 ( ) p,e≥1
+ log p log p
e
7 6 ϕ̂2 − 2F2(X)
L
ϕ
e/ 1 12 S2(2).p L πi
p,e≥1
This cancels against the corresponding terms from (9.1.3) and (9.1.5). By symmetry, the
terms coming from R1,2,α,β will cancel the rest of the terms of (9.1.3) and (9.1.5). Thus, all
our calculations have shown that
J ′′(X)≪ Xσ/6−1/3+ϵϵ +Xσ/2−1/2+ϵ,
as desired. This proves Proposition 9.1.
9.2 Phase transitions in the two-level density
We now turn to the problem of investigating the Ratios Conjecture prediction for the two-
level density for any σ. When we calculated the one-level density, we found phase transitions
in both the main term and the secondary term, when the support σ reached 1. In this section,
we will see that for the two-level density, there are several phase transitions, one when each
σi reaches 1, and another when σ = σ1+ σ2 does. Specifically, the purpose of this section is
to prove the following theorem:
Theorem 9.3. Assume the Generalised Riemann Hypothesis for ζK(s), and Conjecture 8.3.
84
9. Interpreting the Ratios Conjecture prediction for the two-level density
Then, we have the estimate
1 ∑ ∑ ( ) ∫L L ′ ϕ̂ (0) 11
±( ) ϕ 2 γK , 2 γK = ϕ̂1(0)ϕ̂2(0)−N X ∫ ′ π∫ π ∫ ∫ 2
ϕ̂2(u)du
K γ ̸=±γ −1K K
ϕ̂ (0) 1 1 1 1− 2∫2 ϕ̂∫1(
1
u)du+ 4 ϕ̂1(u∫)du ϕ̂2(t)dt+ 2 |u(|ϕ̂1(u)ϕ̂2(u)du−1 −1 −1 −11 1−u 1 D±
+ 4 ϕ̂1(u) ∫ ϕ̂2(t)dtdu− 2 ϕ̂
S3
1(u)ϕ̂2(∫u)du+ − 2ϕ̂1(0)ϕ̂2(0)
(9.2.1)
0 0 −1 L
+ ϕ̂1(0)− ϕ̂ (1)
1
1 ( ) + ϕ̂2(0)− ϕ̂2(1)
1
2 ∫ ϕ̂2 u du 2∫ ϕ̂1(u)du+)ϕ̂1(0)ϕ̂2(1)−1 −11 1 ( )
+ ϕ̂2(0)ϕ̂1(1) + 2 ϕ̂1(u)ϕ̂2(u)du− 4
1
ϕ̂1(u)ϕ̂2(1− u)du +O 2 ,
−1 0 L
where
D±S := −1 + 4 log 2 + πδ+ + C.3
Here, δ+ = 1 if we are considering positive discriminants, and else it equals 0.
Remark. The main term is the result of inte(grating ϕ̂1(u1)ϕ̂2(u2) against( )( ) )
δ(u1)− 1
1
2χ
1
[−1,1](u1) δ(u2)− 2χ[−1,1](u2) −2 (1− |u1|)δ(u1 + u2)χ[−1,1](u1)− 2χB(u1, u2) ,
where B denotes the unit L1-ball, and δ the Dirac delta distribution. This is exactly the ex-
pected Katz-Sarnak main term, see [M, Ch. 5], where the calculations leading up to Theorem
5.9 can be used, also for support σ > 1.
Remark. It is interesting to see that the secondary term essentially consists of integral linear
combinations of D±S /L multiplied with appropriate transforms of ϕ1 and ϕ2, which can be3
compared to the result of Proposition 7.2. Also, note that the very last term of (9.2.1) is zero
if ϕ(u1, u2) is supported inside the L1-unit-ball, whence this term gives a phase transition at
σ = 1.
The rest of the section will be focused on proving this theorem. The starting point of all
our calculations is Proposition 8.4. Unlike the case when σ < 1, the sum S2(i) may be very
large, in fact as large as some power of X. Hence, we will need to eliminate a significant
part of these sums. We remark that in this section, an "acceptable error" will always mean
an error that is ≪ L−2.
We now study the Ratios Conjecture prediction in (8.4.1), one row at a time. The second
row of (8.4.1) is already simplified. The r(elevant terms here are
2 log(4π2
)
e)
ϕ̂1(0)ϕ̂2(0) 1 + .
L
We turn to the third row. To el(iminate the influ)ence of the S2 terms, we add the term2
ϕ̂1(0)
log(4π e)
J2(X) 1 + +
ϕ̂1(0)
J ′2(X), (9.2.2)L L
from the fifth row to the third row. We now study the expression that we just added. By
our work in Section 7.2, we kno(w that
2 ∑ )e ∫ 1 ( )( ) = ( ) log p log p + ϕi(0) − 1 ( ) + ± ϕ̂i(1)Ji X F2 X 5 6 ϕ̂i 2 2 ϕ̂i u du DS +O 1 .L p e/ L 3 L L2
p,e −1
We turn to the definition (8.4.2) of J ′i . More specifically we take a look at the integrals
involved in the definition. First, by a similar argument as in the one-level calculations, we
85
9. Interpreting the Ratios Conjecture prediction for the two-level density
may disregard t∫he contri(bution)of any integral involving F2(X). The only remaining integralis 2 Lr −r
2 ϕi 2 ζ(1− 2r)
Γ±(1/2− r) rX
Γ (1 2 + )F1(X) (1 )2A3(−r, r)dr.πi (1/L) πi ± / r − r
We can handle this just as we originally handled J(X), by Taylor expanding all relevant
factors of the integrand. The only difference between this integral and the corresponding
integral from Ji(X) is the presence of the factor r(1−r)−2. As this factor is equal to 0 when
r = 0, a calculation reveals that this integral term is ≪ L−1. As there is already another
factor L−1 in front of J ′2(X) in (9.2.2), we can completely disregard all integral terms from
J ′2(X).
To finish the estimation of the third row, we use (7.2.5), F1(X) = 1 + O(X−1/6), the
estimates in (3.3.1) and Section 7.2 for S1 to see that the result after adding (9.2.2) to the
third row(of (8.4.1) is )( ) ( )
(0) 1 + log(4π
2e) (2 − ) ϕ̂2(0) 1
∫ 1
ϕ̂1 γ C − 2 ϕ̂2(u)du+
± ϕ̂2(1)DS +O
1
.
L L 3−1 L L
2
By symmetry, one has a similar result for the fourth row, and this also eliminates all re-
maining terms of the fifth row.
We skip the sixth row for now and turn to the seventh row of (8.4.1). We begin by
studying the term containing the factor F2(X). By our calculations in Chapter 4, S4, S6 ≪ 1,
so that the only relevant part of this term is F2(X)S2(1)S2(2). To this term, we add the
first three rows of the definition (8.4.3) of J ′′(X), recalling that a term J ′′(X) is present on
row six of (8.4.1). Appealing to (7.2.5), we see that the result of this addition is certainly
≪ L−2.
We move on to the term F1(X)(S1(1)S1(2) − S3 + S5), which up to acceptable error
equa(ls )( )
−ϕ1(0)
∫ ∞
+ (2 − ) ϕ̂1(0) −ϕ2(0)2 γ C 2 + (2 − )
ϕ̂2(0)
γ C + 2 |u|ϕ̂1(u)ϕ̂2(u)du,
L L −∞
using (3.3.1) and the computations from Sections 7.2 and 4.4.
Now, we look at the last row of (8.4.1). It will be convenient to add the two first terms
of the sixth row to this row. Using E ≪ L−1, we are, up to an error ≪ L−2i , left with
E1(R2 + J2(X)) + E2(R1 + J1(X))− 2(R3 + J3(X)).
We may use Theorem 7.3, which tells us the value of Ri + Ji(X), and the calculation of Ei
in Section 7.2.1 to see th(at this(is ) ( ))
(−2 log 8π − 2 ∫ ∫γ − πδ 1 1+) 1 1
ϕ̂1(0) ϕ̂2(0)− 2 ϕ̂∫ 2
(u)du + ϕ̂2(0) ϕ̂1(0)−
L −1 ( 2
ϕ̂1(u)du
) −11 2D±S ϕ̂1ϕ2(0) 2D±− 2ϕ̂ ϕ (0) + ϕ̂ ϕ (u)du+ 3 − S ϕ̂1ϕ2(1)31 2 1 2 +O L−2 .
−1 L L
The only thing that remains is to simplify all rows except the first three of J ′′(X), i.e.
we need∫ to stu∫dy ( ) ( )(4 Lr Ls
(2 )2 ϕ1 2 ϕ2 2 R1,2,α,β(r, s, r, s;X) +R2,1,α,β(r, s, r, s;X)πi (1/L) (1/L) πi πi )
+R2,2,α,β(r, s, r, s;X) drds.
The methods used for studying this integral are very similar to the methods we used for
studying the integral in J(X) in Section 7.2.3, but this case will require quite some additional
effort.
86
9. Interpreting the Ratios Conjecture prediction for the two-level density
We begin by focusing on the integral of R2,2,α,β , defined in (8.3.4). As usual, we may
discard the term containing F2(X), and replace F1(X) with 1. We make a change of variables
s′ = Ls/(2π), r′ = Lr/(2π), and move both of the resulting integrals to (δ), for some small,
fixed δ > 0. We also rename r′ as r, and s′ as s. Using that ϕ1 and ϕ2 are even, we are left
with the integra∫l4 ∫ Γ±(1/2− 2πs/L) Γ±(1/2− 2πr/L)
( )2 ϕ1 (ir)ϕ2 (is)Li (δ) (δ) Γ±(1/2 + 2πs/L) Γ±(1/2 + 2πr/L)
× ζ(1− 2π(r + s)/L)ζ(1 + 2π(r + s)/L)
ζ(1− 2π(r − s)/L) (1 + 2 ( ζ(1− 4πr/L)ζ(1− 4πs/L)ζ π r − s)/L)
X−2π(r+s)/L× 1 2 ( + ) A1(−2πr/L,−2πs/L, 2πr/L, 2πs/L)drds.− π r s /L
We also rewrite
X−2π(r+s)/L = e−2π(r+s) −2π(r+s) log(4π
2e2e )/L.
Next, we proceed similarly to earlier integral calculations. A cutoff argument shows that
we may integrate over a truncated (δ), where the imaginary part of the variable is bounded
in absolute value by Lϵ, say. On this truncated contour, we Taylor expand every factor in
the integrand about (r, s) = (0, 0), except for ϕ , ϕ and e−2π(r+s)1 2 . Then, by another cutoff
argument, we may extend both contours back to the entire (δ).
We turn to the actual calculations. First, we find the relevant Taylor expansion. This has
been done before, so we leave out most of the details, but some words need to be said about
the function A1. To find the Taylor expansion, we must find the partial derivatives, with
respect to both r and s at the origin. By symmetry the partial derivatives are equal, so it
suffices to find one. We know that A1(−r, 0, r, 0) = A3(−r, r), so that this partial derivative
is the constant C, by definition. As A1(r, s, r, s) = 1, we also know that A1(0, 0, 0, 0) = 1.
Thus, by the complex multivariate version of Taylor’s theorem, ( 2 2)
A1(−2
2π(r + s)C |r| + |s|
πr/L,−2πs/L, 2πr/L, 2πs/L) = 1 + +O .
L L2
See [Le, Thm. 1.2.1] for statement and proof of a complex multivariate power series expan-
sion. The proof can be modified to produce an error term for the truncated series.
Taylor expanding the rest of the integrand, except for the factors mentioned above, shows
that the integral in question is the(integral of ϕ1(ir)ϕ)2(is) ag(ainst )
e−2π(r+s)( ±r − s)2 2πDS (r + s)3 P (|r|, |s|)
(2πi)2 ( + )2 1 + +Oδ 2 , (9.2.3)rs r s L L
where P is a polynomial and where the factor −4/L2 in front of the integral has been
included. The specific definition of P does not matter, as (3.1.2) shows that integrating any
polynomial against ϕ1, and ϕ2 gives a result of size ≪ 1.
We now write
(r − s)2 1 4
rs(r + s)2 = −rs (r + .s)2
Then, we see that to integrate (9.2.3) above, it suffices to integrate ϕ1(ir)ϕ2(is) against the
four expressions
e−2π(r+s) e−2π(r+s) e−2π(r+s) e−2π(r+s) re−2π(r+s)+ + se
−2π(r+s)
, , , , (9.2.4)
rs (r + s)2 r s (r + s)2 (r + s)2
over r, s ∈ (δ) and then multiplying by appropriate constants. The first of these expressions
is easily evaluated ∫using
e−2
( )
πr
( ϕi(0) 1
∫ 1
ϕi ir)dr = (2πi) − ϕ̂i(u)du ,
(δ) r 2 2 −1
87
9. Interpreting the Ratios Conjecture prediction for the two-level density
a result we found in Section 7.2.3, albeit formulated slightly differently. The third expression
can be inte∫grated by combining∫the result above with∫
e−2πrϕ (ir)dr = e−2πri ϕi(ir)dr = i e−2πitϕi(t) =
2πi
dt 2 ϕ̂i(1).(δ) (0) R π
The second and fourth expressions are harder to evaluate. We will use a method very
similar to the one used in [MS, Ch. 3] to calculate these integrals. We begin with the second
exp∫ress∫ion and write
e−2π(r+s)
∫ ∫
e−2π(r+s)
∫ ∞
( + )2 ϕ1(ir)ϕ2(ir)drds = ϕ (ir) ϕ̂ (u)e
2πus
( + )2 1 2 dudrds,(δ) (δ) r s (δ) (δ) r s −∞
where we used the inverse Fourier transform. We also replaced u by −u, which is allowed as
ϕ2 is even. Next, we note that all integrals converge absolutely, including the integral over
s, so that we may use∫Fubini’s t∫heorem to interc∫hange the order of integration. The resultis ∞ 2π(u−1)s
ϕ̂ (u) ϕ (ir)e−2πr e2 1 (r + s)2 dsdrdu.−∞ (δ) (δ)
These integrals are all integrals in the Lebesgue sense, so a set of measure zero is not
important. In particular, we will ignore the set where u = 1. For u < 1, we may move the
innermost integral over s to the line (N), and let N → ∞, so that by using the fast decay
of the exponential function here, we see that the inner integral is equal to 0 for such u. If
instead u > 1 we move the integral to the line (−N) and let N → ∞. The shifted integral
also vani∫shes, but we pick∫up the residue at the point s = −r so t∫hat the above is equal to∞ ∞
(2πi)2π (u− 1)ϕ̂2(u) ϕ (ir)e−2πre−2π(u−1)rdrdu = (2πi)21 (u− 1)ϕ̂1(u)ϕ̂2(u)du.
1 (δ) 1
Finally, we turn to the fourth and last expression in (9.2.4). The reason we have split
this expression into two parts is so that the integral over each of these terms converges
absolutely. We concentrate on the term re−2π(r+s)(r + s)−2, as we can find the integral of
the other term using symmetry. The procedure is very similar to the calculation we just
performed. Indeed, the only difference between the integrands is a factor r, so that we may
perform the same steps as abo∫ve until we arriv∫e at the integral∞
(2πi)2π (u− 1)ϕ̂ (u) ϕ (ir)re−2πru2 1 drdu.
1 (δ)
We move the ∫inner integral to (0), and le∫t r = it to see( that th)is integral is∞ 1 ∞ d 1 ′− tϕ (t)e−2πitu1 dt = 2 ϕ1(t) e−2πitu dt = ϕ̂1 (u).−∞ πi −∞ du 2πi
T∫hu∫s, by symmetry( −2 ( + ) −2 ( + ))re π r s π r s ∫ ∞( ) ( ) + se = 2 ( − 1) d ( )ϕ1 ir ϕ2 is 2 2 drds π u ϕ̂1(u)ϕ̂2(u) du,
δ δ (r + s) (r + s) 1 du
which we see is equal to ∫ ∞
−2π ϕ̂1(u)ϕ̂2(u)du,
1
after integrating by parts.
Putting our results(toget)her, (we ha)ve shown that up to an error(≪ L−2, we have )
4 ∫ ∫ Lr Ls ϕ1(0) 1 ∫ 1
(2()2 ϕ1 ϕ2 R2,2,α,β(r, s, r, s;X)drds = − ϕ̂1(u)duπi (1/L) (∫1/L) 2πi) ∫2πi [ 2 2 −1(0) 1 1 ∞ D±× ϕ22 − 2 ϕ̂2(u)du − 4 (u− 1)ϕ̂1(u)ϕ̂2(u)du+ S3 ϕ̂( ) ( ) 2
(1)
−1 1 L ]
× ϕ1(0) − 1
∫ 1 ϕ (0) 1 ∫ 1 ∫ ∞
2 2 ϕ̂1(
2
u)du + ϕ̂1(1) 2 − 2 ϕ̂2(u)du − 4 ϕ̂1(u)ϕ̂2(u)du ,−1 −1 1
88
9. Interpreting the Ratios Conjecture prediction for the two-level density
It remains to calculate the integral of R1,2,α,β and R2,1,α,β . Once again, we need not
take the integrands involving F2(X) into account. We would like to treat R1,2 and R2,1
separately, but recall that there is a possible pole at r = s, unless we treat terms involving
(ζ ′/ζ)(1± (r − s)) together.
We begin with the terms not involving any logarithmic derivative of ζ, which allows us
to consider the terms from R1,2,α,β and R2,1,α,β separately. We begin by focusing on the
terms from R ′2,1,α,β . We make the change of variables s = Ls/(2π), r′ = Lr/(2π), move
both contours to (δ), and then rename s′ to s and r′ to r. We also replace F1(X) by 1, and
absorb the differ∫ence∫ in the error term. We should then compute the integral4 ( ) ( )Γ±(1/2− 2πr/L)2 ϕ1 ir ϕ2 is Γ (1 2 + 2 )ζ(1− 4πr/L)L (δ) (δ) ± / πr/L
X−2πr/L× 1 2 A1,β(−2πr/L, 2πs/L, 2πr/L, 2πs/L)drds.− πr/L
As before we may use a Taylor expansion to see ( )
4 Γ±(1/2− 2πr/L) X−2πr/L −2πr
2 Γ ζ(1− 4πr/L) = −
e +O 1 .
L ±(1/2 + 2πr/L) 1− 2πr/L πrL L2
We will also need a zeroth order Taylor expansion of A1,β , which means we must calcu-
late A1,β(0, 0, 0, 0). Now, we have A1,β(0, 0, 0, 0) = A3,α(0, 0), as A1(r, β, r, δ) = A3(β, δ).
Furthermore, by the definition (6.3.5) and (6.4.2)
∑∑ xp(θe + 1/p) log p ζ ′ ∑∑ (xp(θe + 1/p)− δ( ) = − − (1+2 ) = − 2|e) log pA3,α s, s
pe(1/2+
s
s) ζ pe(1/2+
,
s)
p e≥1 p e≥1
which is now valid when Re(s) > −1/2, as θ1 = 0, θ2 = 1, xp = 1 +O(p−1). Setting s = 0
and comparing with the left-hand side of (7.2.8) shows that A3,α(0, 0) = −C/2.
From the∫cal∫culations above, we see that we need to(integrate−2πr (0) 1 ∫ )C e C ϕ 11
ϕ1(ir)ϕ2(is) 2 drds = − ϕ̂2(0) 2 − 2 ϕ̂1(t)dt ,L (δ) (δ) πr L −1
where the equality follows from our earlier integral calculations in this section. The corre-
sponding term from R1,2,α,β can be found using symmetry.
We now continue with the other terms(of R2,1,α,β , i.e. the terms coming from )
−Γ±(1/2− r) X
−r ζ ′ ′ ′
Γ (1 2 + )ζ(1− 2r)1 A3(−r, r) (1 + s− ) +
ζ (1 + 2 )− ζr s (1 + r + s) ,
± / r − r ζ ζ ζ
(9.2.5)
where we used A1(−r, s, r, s) = A3(−r, r). We will first focus on the term in the middle.
Making the usual change of variables in the integral, and Taylor expanding, yields the
integral ∫ ∫ ( )2 e−2πr 2πD± r ′− ϕ1(ir)ϕ2(is) 1 + S3 ζ (1 + 4πs/L)drds.
L (δ) (δ) 2πr L ζ
Next, by (7.2.4), we have
ζ ′ (1 + z) = −1 + γ +O(|z|).
ζ z
Thus, the integra∫l in∫question is ( )
e−2πr 2πD±S rϕ (ir)ϕ (is) 1 + 3 − 4πγs1 2 2 drds. (9.2.6)
(δ) (δ) (2π) rs L L
89
9. Interpreting the Ratios Conjecture prediction for the two-level density
This can be computed by using o∫ur previous calculations, and the result
ϕ2(is) = 2πids 2 ϕ2(0),(δ) s
which can be shown by a calculation similar to the calculation of one of the integrals in
Section 7(.2.3, where ∫we had an a)dditio(nal factor e
−2πs
( . We conclude tha)t (9.2.6) equals )
−ϕ2(0) ϕ (0) 1
1
1 − ( ) 1 ϕ1(0) 1
∫ 1 ϕ2(0)
2 2 2 ϕ̂1 u du + 2γϕ̂2(0) 2 − 2 ϕ̂1(u)du −D
± ϕ̂ (1) .
L S3
1
−1 −1 2
It remains to integrate the rest of the terms in (9.2.5) against ϕ1 and ϕ2. Here, we
must add the corresponding terms from R1,2,α,β and integrate these at the same time, as
otherwise there is a pole at r = s. We then make the usual change of variables and shift
the integral over r to (δ/2) and the integral over s to (δ). We point out that if the same
contour had been chosen, then it would not have been possible to control the error in the
Taylor expansion, as it would not have been possible to bound |r − s| from below.
T∫aylor∫expanding as be[fore sho(ws that we ne)ed(to evaluate2 e−2πr 2πD± r ζ ′ ′ )− ϕ1(ir)ϕ2((is) 2 1 +
S3
) (1 + 2π(s− r) )−
ζ
/L (1 + 2π(r + s)/L)
L (δ/2) (δ) πr ( L ζ ζ± )]
+ e
−2πs 2πD ′ ′
1 + S
s
3 ζ
2 (1 + 2π(r − s)
ζ
/L)− (1 + 2π(r + s)/L) dsdr.
πs L ζ ζ
Also, by the Laurent expansion of ζ ′/ζ that we found above, we h(ave ( ))
ζ ′ ′(1 + 2π(s− ζr)/L)− (1 + 2 ( + ) ) = − 2rL 1 +O Q1(|r|, |s|)π r s /L 2 ( )( + ) δ 2 ,ζ ζ π s− r r s L
for some polynomial Q1. Hence, the integral under consideration is, up to the usual error
te∫rm, eq∫ual to [ ( ) ( )2 ]πD± r 2πD± s
2 ϕ (ir) 2ϕ (is) e−2πr 1 + S3 − e−2πs1 2 (2 )2( )( + ) 1 +
S3 dsdr.
(δ/2) (δ) π s− r s r L L
(9.2.7)
We first calculate the m∫ain te∫rm coming from(the expression)above, i.e. the integral4 e−2πr − e−2πs
(2 )2 ϕ1(ir)ϕ2(is)π (δ/2) (δ) (s− r)( + )
dsdr.
s r
We begin by expanding ϕ1(ir) into an integral involving its Fourier transform, and inter-
change the order of
4 ∫
integration∫to fi∫nd∞ (e2πr(u−1) − e2πru−2 )πs
2 ϕ̂1(u) ϕ2(is) (9.2.8)(2π) −∞ (δ) (δ/2) (
drdsdu.
s− r)(s+ r)
For u < 0, we may move the integral over r very far to the right (i.e. to the line (N), and
then let N →∞), which makes the shifted integral vanish. The contour for r lies to the left
of the contour for s, so there is a possible pole where r = s, but the residue here is 0, so this
is in fact not a pole, whence we can discard all contributions from u < 0.
For u > 1, we instead move the integral over r very far to the left. We pick up the
residue at r = −s, whi∫ch then giv∫es the con(tribution2i ∞ e−2 )πs(u−1) − e−2πs(u+1)
(2 ) ϕ̂1(u) ϕ2(is) dsdu.π 1 (δ) s
90
9. Interpreting the Ratios Conjecture prediction for the two-level density
As there is no pole at s = 0, we can move the integral over s to the imaginary axis and write
s = it, and use∫that ϕ2 is∫even, whe(nce the above equals∞ ∞ )
2 ( ) ( ) sin(2πt(u− 1))ϕ̂1 u ∫ ϕ2 t ∫ 2 −
sin(2πt(u+ 1))
1 −∞ πt 2
dtdu
πt
∞ u+1
= −2 ϕ̂1(u) ϕ̂2(t)dtdu,
1 u−1
where we used Plancherel’s theorem and the scaling property of the Fourier transform.
We have yet to analyse (9.2.8) when 0 < u < 1. We handle the two terms separately.
First, s∫hift the co∫ntou∫r over r far to the right to find4 1 ( ∫ ∫( ) ϕ2 is)e2πr(u−1) 2i 1 e2πs(u−1)(2 )2 ϕ̂1 u ( )( + ) drdsdu = 2 ϕ̂1(u) ϕ2(is) dsdu,π 0 (δ) (δ/2) s− r s r π 0 (δ) s
where we picked up the negative of the residue at r = s, as we shifted the contour to the
right. Next, by instead shifting to the far left we see that
−4 ∫ 1 ∫ ∫ ϕ (is)e2π(ru−s) ∫ 1 ∫ −2πs(u+1)
(2 )2 ϕ̂1(u)
2 −2i ϕ2(is)e
π 0 (δ) (δ/2) (s− r)( + )
drdsdu = 2 ϕ̂1(u) dsdu,s r π 0 (δ) s
after picking up the residue at r = −s. We add together both of these integrals, shift the
contour to (0), and∫apply Pl∫ancherel’s formula to∫ see that∫their sum is equal to1 1−u 1 u+1
2 ϕ̂1(u) ϕ̂2(t)dtdu− 2 ϕ̂1(u) ϕ̂2(t)dtdu.
0 0 0 0
Thus, after simplif∫ying, we s∫ee that the terms of∫constant ∫size coming from (9.2.7) is1 1−u ∞ u+1
4 ϕ̂1(u) ϕ̂2(t)dtdu− 2 ϕ̂1(u) ϕ̂2(t)dtdu.
0 0 0 u−1
We turn to the rest of (∫9.2.7), i.e.4D± ∫ re−2πr − se−2πsS3 ϕ1(ir)ϕ2(is) (9.2.9)2πL (δ/2) (δ) (s− r)(s+ r) dsdr.
We will need to treat each term separately in order to make use of absolute convergence.
To calculate the integral over the first term, we expand ϕ2 into its Fourier transform and
find that this integral eq∫uals4 ∫ ∫D± ∞ re2π(su−r)S3
2 ϕ̂2(u) ϕ1(ir)πL −∞ (δ/2) (δ) (s− r)(s+ r)
dsdrdu.
If u < 0, we can shift the integral over s to the right, to see that it is zero. We encounter
no poles, as the s-contour lies to the right of the r-contour. For u > 0, we instead shift the
contour to the left picking up the poles at s = r and s = −r, which leaves us with
2iD± ∫ ∞ ∫ ( )S3 ϕ̂ (u) ϕ (ir) e2πr(u−1) − e−2πr(u+1)2 1 drdu.
L 0 (δ/2)
By shifting the contour to (0),∫we see tha(t this is simply2D± ∞ )
− S3 ϕ̂2(u) ϕ̂1(u− 1)− ϕ̂1(u+ 1) du.
L 0
Now we calculate the integral over the other term in (9.2.9) by expanding ϕ1 into its
Fourier transform. We get
4 ∫ ∫ ∫D± ∞ se2π(ru−s)− S32 ϕ̂1(u) ϕ2(is) drdsdu. (9.2.10)πL −∞ (δ) (δ/2) (s− r)(s+ r)
91
9. Interpreting the Ratios Conjecture prediction for the two-level density
For u < 0, we shift the integral over r to the right, and pick up the negative of the residue
at r = s, while for u > 0, we shift the integral to the left and pick up the residue at r = −s.
In total, w∫e obtain2 ∫ ∫ ∫iD± 0 ± ∞
− S
2iD
3 ϕ̂ (u) ϕ (is)e2πs(u−1)dsdu− S3 ϕ̂ (u) ϕ (is)e−2πs(u+1)1 2 1 2 dsdu,
L −∞ (δ) L 0 (δ)
which we can simplify∫further by moving the contour to (0). We obtain2D± 0 2 ± ∫S D ∞3 ϕ̂1(u)ϕ̂2(u− 1)du+ S3 ϕ̂1(u)ϕ̂2(u+ 1)du.
L −∞ L 0
Adding together the results above, and using that both Fourier transforms are even, we
see that (9.2.9) is equal to
2D± 4 ∫S D± 13 (ϕ̂1 ∗ ϕ̂2)(1)− S3 ϕ̂1(u)ϕ̂2(1− u)du,
L L 0
where ∗ denotes the convolution operator.
Wit∫h all in∫tegral ca(lculat)ions(finish)ed, we may conclude that4 Lr Ls ( )
(2 )2 ( ϕ1 2∫ ϕ2 2) R2,1,α,β((r, s, r, s;X)∫+R1,2,α,β(r), s, r, s;X) drdsπi (1/L) (1/L) πi πi
= −C (0) ϕ1(0) 1
1 1
ϕ̂2( 2 −∫ 2 ϕ̂1(t)
C
dt − ϕ̂1(0)
ϕ2(0) − 1 ϕ̂2(t)dt
L −1
1 ) [L ( 2 2 ∫−1ϕ 1 ) ]− 2(0) ϕ1(0) − 1 1 ϕ1(0) 1 ± ϕ2(0)2 ( 2 2
(0) (0) 1 ∫
ϕ̂1(u)du)+ [2γϕ̂2(0)( 2 −−1 L 2 ∫ ϕ̂1(u)du −DS ϕ̂ (1)3 1−1 2
ϕ ϕ 1 1 ϕ (0) 1 1
) ]
− 1 2 2 ϕ1(0)∫2 2∫ − 2 ϕ̂2(u)du∫ + 2γϕ̂1(0) 2 − 2 ϕ̂2(u)du −D
±
S 2 ϕ̂2(1)−1 L 3−1
1 1−u 1
+ 4 ϕ̂1(u) ϕ̂2(t)dtdu∫− ϕ̂1ϕ2(u)du0 0 −12 ± 4 ± 1 ( )D
+ S
D
3 (ϕ̂1 ∗ ϕ̂2)(1)− S3
1
ϕ̂1(u)ϕ̂2(1− u)du+O .
L L 0 L
2
Here, we used the identi∫ty∞ ∫ u+1 ∫ 1
2 ϕ̂1(u) ϕ̂2(t)dtdu = ϕ̂1ϕ2(u)du,
0 u−1 −1
which follows by applying Plancherel’s theorem twice, using the fact that a convolution of
even functions is even, and ∫ u+1
ϕ̂2(t)dt = (ϕ̂2 ∗ χ[−1,1])(u),
u−1
where χA is the indicator function of the set A. Theorem 9.3 now follows from combining
all results of this section.
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96
A
Infinite products, the gamma
function and integration
We present a few results that are used throughout the report, usually without proofs.
A.1 Infinite products
We give a brief description of infinite products, following [A, Ch. 5.2.2]. These products are
frequently used throughout the report.
Let an be complex numbers, whose magnitude is less than 1/2, say for large enough n.
We are then interested in the infinite product
∏∞
(1 + an). (A.1.1)
n=1
To give this expression meaning, we consider the Nth partial product PN and say that the
infinite product converges to P ̸= 0 if the partial products converge to P . We will often
want to take logarithms of infinite products, and this is the reason for excluding the value
P = 0. In this direction, we have the following result, which is essentially [A, Thm. 5.5].
Theorem A.1. Let an be complex numbers such that |an| < 1/2 for all n. The infinite
product ∏∞
(1 + an)
n=1
converges if and only if the series
∑∞
log(1 + an)
n=1
converges, where log denotes the principal branch of the logarithm. Further, if the series
converges to S, then the product converges to eS.
We say that an infinite product is absolutely convergent if the product obtained by
replacing an by |an| is convergent. We have a very simple criterion for determining absolute
convergence, as a consequence of the theorem above [A, Thm. 5.6].
Theorem A.2. Let an be complex numbers with |an| < 1/2 for all n. Then, the infinite
product (A.1.1) converges absolutely if and only if the sum
∑n
|an|
n=1
converges.
I
A. Infinite products, the gamma function and integration
In particular, by combining these theorems with a Taylor expansion, we see that an
absolutely convergent product is convergent. We remark that if an is large in magnitude for
small n, then we can simply exclude these terms when treating convergence.
We are mostly interested in the case when the coefficients an depend on some complex
variable s, and we then write an(s). In particular, if the an(s) are holomorphic functions, we
are interested in knowing whether the function defined by the infinite product (A.1.1) is also
holomorphic. A well-known result from complex analysis asserts that this is the case if the
convergence of the partial products is locally uniform, or equivalently uniform on compact
sets. Both theorems above can be extended to relate the uniform convergence of a product
to the uniform convergence of a series, but we leave out the details. See the remarks after
[A, Thm. 5.6].
A.2 The Gamma function
Now that we have introduced infinite products, we are ready to define the Gamma function
Γ(s), following [A, Ch. 5.2.4]. We define
e−γs ∏∞ ( )−1Γ(s) = 1 + s es/n,
s
n=1 n
where γ is Euler’s constant. A Taylor expansion, combined with the mentioned extensions of
the theorems above, shows that the product is absolutely uniformly convergent on compact
subsets of C \ {0,−1,−2, ...}. In particular, Γ(s) is a meromorphic function, with poles
precisely at the non-positive integers, as (1 + s/n) is only zero on the negative integers.
Furthermore, Γ(s) has no zeros. The gamma function is related to the factorial function by
the relation Γ(n+ 1) = n! for integers n ≥ 0.
We now list several important properties of the Gamma function. First, we have the
relation sΓ(s) = Γ(s+ 1), which is an extension of the relation n! = n · (n− 1)!. Moreover,
we have the reflection formula
Γ( )Γ(1− ) = πs s sin ,πs
and the duplication formula ( ) √
Γ(s)Γ s+ 1 = 21−2s2 πΓ(2s).
Lastly, we will need a growth estimate for Γ(s) given by Stirling’s formula. We use the
form from [D3, Ch. 10], namel(y ) ( )
log Γ(s) = s− 12 log − +
1
s s 2 log 2π +Oδ |s|
−1 ,
for s with −π + δ < arg s < π − δ, with δ > 0. Here log Γ(s) denotes a fixed logarithm of
Γ(s), and log s is the principal branch of the logarithm applied to s. Fixing δ < π/2 gives an
estimate that is valid in the right half-plane for |Γ(s)|. In pa(rticula(r, we )h)ave the equality√
|Γ(s)| = 2π|s|Re(s)−1/2e−Re(s)e− Im(s)Arg(s) 1 +O |s|−1 ,
which is useful when estimating an L-function through its functional equation. Note in
particular that Γ(s) decays fast in any vertical strip, as Im(s) and Arg(s) have the same
sign.
A.3 Stieltjes integration
We now introduce Stieltjes integration, essentially following [Rd, Ch. 6], which is an invalu-
able tool for studying sums. One may consider the quite general Lebesgue-Stieltjes integral,
but for our purposes, the Riemann-Stieltjes integral will be enough.
II
A. Infinite products, the gamma function and integration
Let f be a monotonically increasing function, and let g be a function bounded on [a, b],
with only finitely many discontinuities, none of which coincides with the discontinuities of
f . Then, it is possible to define the Rie∫mann-Stieltjes integralb
g(x)df(x).
a
We will not write out the construction here, but it is defined very similarly to the usual
Riemann integral, by considering partitions a = a0 < ... < an = b. Instead of using
Riemann sums where the values of g are weighed against the lengths of the subintervals, one
weighs the values of g against the differences f(ai+1)−f(ai). We remark that the definition
can readily be extended to functions f of bounded variation, by using linearity, as such
functions are differences of non-decreasing functions.
The first important equalit∫y we will need i∫s thatb b
g(x)df(x) = g(x)f ′(x)dx (A.3.1)
a a
if f is continuously differentiable.
The reason we are interested in these integrals is because of their connections to sums.
Let an be a sequence of real numbers, and defi∑ne the function
f(x) = an.
1≤n≤x
Then, assuming that g is contin∑uous, say, it sh∫ould be intuitively clear from the definitionthat we have x
ang(n) = g(x)df(x),
a<n≤x a
as the function f changes value precisely at the integers. If g is in addition continuously
differentiable, we have the ∫summation by parts formula∑ t ∫ t
anf(n) = g(x)df(x) = f(t)g(t)− f(a)g(a)− g′(x)f(x)dx.
a<n≤t a a
That the left-hand side is equal to the right-hand side can actually be proven quite easily
without any Stieltjes integration, by using that f is piecewise constant.
Lastly, we menti∫on that the more general formula for i∫ntegration by partst t
f(x)dg(x) = f(t)g(t)− f(a)g(a)− g(x)df(x),
a a
holds for a wider class of functions than the ones above. However, for our purposes, we
will only need to use it in the case described above, or when both f and g are continuously
differentiable. The latter case reduces to the familiar formula for integration by parts by
using (A.3.1). The point of using Stieltjes integration at all is that it provides a convenient
formalism for manipulating the various sums we encounter throughout the report.
III