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Math. Proc. Camb. Phil. Soc.: page 1 of 43 1
doi:10.1017/S0305004124000306
On a Mertens-type conjecture for number fields
BYDANIEL HU
Department of Mathematics, Princeton University, Fine Hall, Washington Road, Princeton
NJ 08544-1000, U.S.A.
e-mail:danielhu@princeton.edu
IKUYA KANEKO
The Division of Physics, Mathematics and Astronomy, California Institute of Technology,
1200 East California Boulevard, Pasadena, CA 91125, U.S.A.
e-mail:ikuyak@icloud.com
SPENCER MARTIN
Department of Mathematics, University of Virginia, 141 Cabell Drive, Kerchof Hall,
Charlottesville, VA 22904, U.S.A.
e-mail:sm5ve@virginia.edu
AND CARL SCHILDKRAUT
Department of Mathematics, Massachusetts Institute of Technology, 77 Massachusetts
Avenue, Cambridge, MA 02139-4307, U.S.A.
e-mail:carlsc@mit.edu
(Received 20 September 2021; revised 16 January 2023; accepted 02 March 2023)
Abstract
We introduce a number field analogue of the Mertens conjecture and demonstrate its falsity
for all but finitely many number fields of any given degree. We establish the existence of a
logarithmic limiting distribution for the analogous Mertens function, expanding upon work
of Ng. Finally, we explore properties of the generalised Mertens function of certain dicyclic
number fields as consequences of Artin factorisation.
2020 Mathematics Subject Classification: 11N56 (Primary); 11N64 (Secondary)
1. Introduction
For a number field K, we define the Möbius function μ(a)=μK(a) assigning an integer
to each integral ideal a, according to the rule
μ(pk):=⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
1ifk=0
−1ifk=1
0ifk≥2
C
The Author(s), 2024. Published by Cambridge University Press on behalf of Cambridge Philosophical Society. This is an Open Access
article, distributed under the terms of the Creative Commons Attribution licence (https://creativecommons.org/licenses/by/4.0/), which
permits unrestricted re-use, distribution, and reproduction in any medium, provided the original work is properly cited.
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2DANIEL HUet al.
for prime ideals p. This is extended multiplicatively by the unique factorisation of ideals.
The summatory function of the Möbius function is the Mertens function
MK(x):=
N(a)≤x
μ(a),
where μK(x):=N(a)=xμ(a) is replaced with (1/2)μK(x)ifxis an integer. With this con-
vention, the function MK(x) is expressed as the inverse Mellin transform of 1/sζK(s) via the
formula
1
ζK(s)=s∞
1
MK(x)
xs+1dx.(1·1)
This relation can be seen as the definition of MK(x) as a matter of practical convenience.
1·1. The naïve Mertens-type conjecture over a number field.
For K=Q, the classical conjecture of Mertens [Mer97] in 1897 asserts that
|M(x)|≤√xfor all x≥1.
Knowing that the Riemann hypothesis is equivalent to the weaker statement that M(x)=
O(x1/2+ε) for all ε>0[Tit86, theorem 14·25(c)], this seemed to represent a viable avenue
towards the Riemann hypothesis for the Riemann zeta function ζ(s). Before the landmark
work of Ingham [Ing42] in 1942, preliminary calculations of Mertens and von Sterneck even
compelled the hypothesis x−1/2|M(x)|≤1/2 for sufficiently large x. Nevertheless, using
lattice basis reduction algorithms, the conjecture was disproven by Odlyzko and te Riele
[OtR85] in 1985, who obtained explicit bounds larger in absolute value than 1 for x−1/2M(x)
on either side in the limit. The current record in this direction has been achieved by Hurst
[Hur18] in 2018, namely
lim inf
x→∞
M(x)
x1/2<−1.837625 and lim sup
x→∞
M(x)
x1/2>1.826054.
It is now common belief that x−1/2M(x) grows arbitrarily large in both directions, but this has
not yet been proven unconditionally [OtR85]. Indeed, Ingham [Ing42, theorem A] showed
that, assuming the Riemann hypothesis and that the imaginary parts of the nontrivial zeros
of ζ(s)areQ-linearly independent, the claim is true. This work marked the first serious
doubt regarding the Mertens conjecture until its subsequent disproof. In fact, preliminary
computational support for the linear independence hypothesis is supplied in the proof of Best
and Trudgian [BT15], which builds upon a previous result of Bateman et al. [BBH+71].
Several mathematicians have investigated analogues of this conjecture in other set-
tings, some achieving corresponding disproofs. Anderson [And79] studied the case of cusp
forms of sufficiently large weight k≡2 (mod 4) on the full modular group SL2(Z). Grupp
[Gru82] generalised the work of Anderson to cusp forms of any even weight k. Humphries
[Hum14] considered Mertens-type conjectures for function fields of smooth projective
curves over Fq. Our work focuses on the case of number fields and seeks a disproof of
the relevant conjecture in emulation of previous work.
Following the aforementioned works, it is most relevant to formulate a conjecture based
on the limiting behaviour of the arithmetic function MK(x). Let Kbe a number field, and
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On a mertens-type conjecture for number fields 3
write
M−
K=lim inf
x→∞
MK(x)
x1/2,M+
K=lim sup
x→∞
MK(x)
x1/2.
We now state the guiding question of our work.
CONJECTURE (The naïve Mertens-type conjectureover K). −1≤M−
K≤M+
K≤1.
Although this is in some ways a naïve generalisation of the original Mertens conjecture,
we have chosen 1 as the critical constant following tradition. Yet, it can be seen that this
choice will be of strategic importance for our method.
A corresponding conclusion to that of Ingham’s theorem can be shown to hold for the
function MK(x) (see conjecture 1·6, theorem 2·1, and lemma 2·4), which leads to the predic-
tion that the naïve Mertens-type conjecture over Kis indeed false for all K. Towards partial
disproofs of the conjecture, the quadratic fields Kare the most straightforward to consider
and we give certain unconditional results.
THEOREM 1·1. We have the following statements:
(a) let K be an imaginary quadratic extension of Q, with the exception of K =Q(√−3).
Then the naïve Mertens-type conjecture over K is false;
(b) let K be a real quadratic extension of Q, with the exception of K =Q(√5). Then the
naïve Mertens-type conjecture over K is false.
Remark. While the basic statement |MK(x)|≤x1/2for all x≥1 is false for K=Q(√−3) and
K=Q(√5) (an easy computation demonstrates that MQ(√−3)(7) =−3 and MQ(√5)(11) =
−4), the methods of this paper are not immediately capable of demonstrating that the
naïve Mertens-type conjecture fails for these fields (as this requires knowledge of limit-
ing behaviour), unlike the other cases of the previous theorem. It is, however, likely that
computations analogous to those used in [OtR85] to falsify the original Mertens conjecture
would suffice for both of these quadratic fields as well.
The techniques of the proof of Theorem 1·1allow us to arrive at a similar result for general
number fields of degree nK>2. In what follows, the signature (r1,r2) of a number field K
is the ordered pair of non-negative integers that encodes the number r1of real embeddings
and r2of complex conjugate pairs of embeddings of K.
THEOREM 1·2. Fix a signature (r1,r2). There exists some D=Dr1,2r2>0depending only
on (r1,r2)for which the naïve Mertens-type conjecture is false for every extension K of Qof
signature (r1,r2)with absolute discriminant DK>D.
We establish these results in Section 3.
1·2. Logarithmic limiting distributions.
In order to study the Mertens function over a number field, we now define another object
which serves as a chief interest of our work.
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4DANIEL HUet al.
Definition 1·3. We say that a function
φ:[0, ∞)→Rpossesses a limiting distribution ν
on Rif νis a probability measure on Rand
lim
Y→∞
1
YY
0
f(
φ(y)) dy =R
f(x)dν(x)
for all bounded continuous real-valued functions fon R.
Recent years have seen great refinements in the probabilistic methods used to study
number-theoretic functions. The influential work of Rubinstein and Sarnak [RS94]in
establishing the existence of limiting distributions pertaining to various questions on Rényi–
Shanks prime number races, with its extensive generalisations in the work of Ng [Ng04],
Humphries [Hum13], and Akbary–Ng–Shahabi [ANS14], has seen wide applications to the
summatory functions in number theory, including those of the Liouville function and the
Möbius function. Pertaining to the classical Mertens function M(x), Ng [Ng04] established
that the function
φ(y):=e−y/2M(ey)
possesses a limiting distribution νon R, assuming the Riemann hypothesis for and the
following conjecture of Gonek and Hejhal on the discrete moments of ζ(s) at the zeros:
J−1(T)T,(1·2)
where, with the convention ρ=1/2+iγ,
Jk(T):=
0<γ ≤T
ζ(1
2+iγ)=0
|ζ(ρ)|2k.
We prove the following generalisation of Ng’s result.
THEOREM 1·4. Let K be an abelian number field. Assume the Riemann hypothesis for ζK(s),
and the following extension of (1·2):
JK
−1(T):=
0≤γ≤T
ζK(1
2+iγ)=0
1
|ζK(ρ)|2αT1+α,(1·3)
for some 0≤α<2−√3. Then the function
φK(y):=e−y/2MK(ey)
possesses a limiting distribution νKon R:
lim
Y→∞
1
YY
0
f(φK(y)) dy =∞
−∞
f(x)dνK(x)
for all bounded continuous real-valued functions f on R.
This is proven in Section 4.
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On a mertens-type conjecture for number fields 5
To be well-defined, the bound (1·3) implies that all the nontrivial zeros of ζK(s) are sim-
ple. This dictates, in consequence, the non-vanishing at s=1/2ofζK(s), the sign of whose
functional equation always prescribes at s=1/2 a zero of even multiplicity.
The validity of these suppositions is most transparent when K/Qis a Galois extension, due
to Artin factorisation for the Dedekind zeta functions of such number fields, and is markedly
different between abelian and non-abelian Galois extensions. Conjecturally, it emerges in
the abelian case from the conventional hypotheses that no two Dirichlet L-functions share a
nontrivial zero and that no Dirichlet L-function vanishes at the central point s=1/2. Both
of these assertions are weaker consequences of the Grand Simplicity Hypothesis for the
Dirichlet L-functions (cf. [RS94]). The latter conjecture is remarkably well-studied, along
with the work of Balasubramanian–Murty [BM92] and Iwaniec–Sarnak [IS99] showing that
L(1/2, χq)= 0 for a positive proportion of the primitive characters χqof any sufficiently
large modulus q.
On the other hand, for non-abelian number fields, the simplicity hypothesis is known to
be false unconditionally, with the zeta function of any such number field having infinitely
many nontrivial zeros of multiplicity at least 2 (cf. [HKMS]). By example, the non-
vanishing at s=1/2 has also been disproven, with the earliest examples of number fields
with ζK(1/2) =0 known to Armitage and Serre (see Section 6for further discussion). This
renders an obstruction to any extension of Theorem 1·4to the non-abelian case.
The conjecture (1·2) is the special case k=−1 of a conjecture Jk(T)T( log T)(k+1)2that
was first proposed independently by Gonek [Gon89] and Hejhal [Hej89] for all k∈Rin
view of their examinations of the discrete moments Jk(T). Gonek also predicted the asymp-
totic formula J−1(T)∼(3/π3)Tand proved the bound J−1(T)T. It was subsequently
proposed by Keating–Snaith [KS00] that random matrix theoretic heuristics could be used
in connection with conjectures about moments of the zeta function. In particular, they mod-
eled the value distribution of ζ(s) near the critical line by the characteristic polynomial
of a large unitary random matrix. Expanding upon this work, Hughes–Keating–O’Connell
[HKO00] conjectured asymptotic formulae for all Jk(T), k>−3/2, giving explicit numer-
ical constants depending on k. The random matrix approach has proven to produce reliable
conjectures; for instance, the conjecture of Hughes–Keating–O’Connell in the case k=−1
agrees with that of Gonek. It is expected that similar heuristics would yield accurate con-
jectures for positive moments of Dedekind zeta functions as well (see [GHK07,BGM15,
Hea21]). They would also be useful in support of (1·3).
We argue in support of the conjecture in the equation (1·3) in Section 4. For now, we have
not appealed to random matrix theory, instead relying on prior work and a series of generous
but not unreasonable assumptions, by analogy with the case of the Riemann zeta function.
As a direct consequence of the arguments to be presented in Section 4, it behooves us to
mention that Corollary 1·15 of Akbary–Ng–Shahabi [ANS14] yields the following analogue
of the weak Mertens conjecture as stated in [Ng04, theorem 1·3].
THEOREM 1·5. With the same assumptions as in Theorem 1·4, we have
Y
0MK(ey)
ey/22
dy ∼βY,
where β=2γ>0|ρζK(ρ)|−2. The assumption JK
−1(T)αT1+αimplies that the series
defining βis convergent.
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6DANIEL HUet al.
In the spirit of the work of Rubinstein–Sarnak and Ng, the existence of a limiting distri-
bution for φK(y) yields a number of consequences which are conditional on the following
supplemental analogue of the linear independence for ζ(s).
CONJECTURE 1·6. (Linear independence conjecture for ζK(s)). The multiset of the non-
negative imaginary parts of the nontrivial zeros of ζK(s)is linearly independent over the
rationals.
We remark that this conjecture encompasses the simplicity of the zeros of ζK(s), as well as
the non-vanishing of ζK(s)ats=1/2, and is hence a viable conjecture for a normal extension
K/Qonly when the extension is abelian. Its principal use in the original work of Rubinstein–
Sarnak is to obtain an expression for the Fourier transform of the limiting distribution. By
[ANS14, corollary 1·3], the following is a direct consequence of Theorem 1·4.
THEOREM 1·7. Let K be an abelian extension of Q. Assume the Riemann hypothesis for
ζK(s), that ζK(1/2) = 0, and that JK
−1(T)T1+αfor some 0≤α<2−√3, and let νKbe
the limiting distribution associated to φK(y)as in Theorem 1·4. Assume moreover Conjecture
1·6. Then the Fourier transform
νK(ξ)=R
e−ixξdνK(x)
of νKat ξ∈Rexists and is equal to
νK(ξ)=
|γ|>0
ζK(ρ)=0
˜
J02ξ
|ρζK(ρ)|,
where ˜
J0(z)is the Bessel function
˜
J0(z)=1
0
e−iz cos (2πt)dt.
This result can be used in pursuit of logarithmic density results: for instance, that the set
of numbers x≥1 in the set Pβ={x−1/2|MK(x)|≤β}has a logarithmic density, for certain
β>0 of interest.
Definition 1·8. For P⊂[0, ∞), set
δ(P)=lim
X→∞
1
log Xt∈P∩[1.X]
dt
t
If the limit exists, we say that the logarithmic density of Pis δ(P).
Following the arguments of [Hum13, corollary 6·3, lemma 6·4], we deduce that the
Fourier transform νKso constructed is symmetric and observes rapid decay as a function
of ξ. Hence, νKis absolutely continuous with respect to the Lebesgue measure on R. After
a logarithmic change of coordinates, this yields the following extension of the conclusion of
Theorem 1·4to characteristic functions of well-behaved sets.
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On a mertens-type conjecture for number fields 7
COROLLARY 1·9. With the same assumptions as in Theorem 1·7,
lim
X→∞
1
log Xx∈B∩[1,X]
dx
x=νK(B)
for all Borel sets B ⊂Rwith boundary of Lebesgue measure zero.
Thus, the set Pβ={x≥1||MK(x)|≤β√x}has a logarithmic density, under the assump-
tions of RH, linear independence, and JK
−1(T)T1+α. See Section 7for further discussion.
2. Preliminaries
2·1. Notation and conventions
Throughout this paper, we use the following conventions:
(i) Kis a number field;
(ii) nK=[K:Q];
(iii) r1and 2r2are the number of real and complex embeddings of K, respectively;
(iv) Kis the discriminant of K;
(v) DK=|K|is the absolute discriminant of K;
(vi) ζK(s) is the Dedekind zeta function of K.
Some theorems in this paper apply only to real or imaginary quadratic number fields. In
these cases, we will specify any additional hypotheses on K. Otherwise, it is assumed that K
is a general number field.
The Riemann hypothesis for ζK(s) will denote the conjecture that all nontrivial zeros of
ζK(s) lie on the critical line Re(s)=1/2. In light of this, ρwill always denote a nontrivial
zero of ζK(s) with imaginary part γ.
2·2. Analytic properties of Dedekind zeta functions
In this section, we provide some preliminary statements about Dedekind zeta functions
that will be used in the proofs of each of our results. We start with the functional equation
for ζK(s).
THEOREM 2·1. For any number field K, the Dedekind zeta function ζK(s)satisfies
ζK(1 −s)=ζK(s)DK
πnK2nKs−1
2π
2r1
2(s)r2
( sin πs
2)r1(1 −s)r1+r2.
This functional equation extends ζKto a meromorphic function on C, which is analytic
everywhere except for a simple pole at s =1.
The Dedekind zeta function has trivial zeros of order r2at each negative odd integer and
of order r1+r2at each negative even integer, as well as a trivial zero of order r1+r2−1at
s=0. It also possesses nontrivial zeros, each of which lies in the critical strip 0 <Re(s)<1.
By the above functional equation and the reflection principle, these zeros are symmetric
about Re(s)=1/2 and the real line.
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8DANIEL HUet al.
Next, we give a suitable upper bound, due to Chandrasekharan–Narasimhan, on the
Dirichlet series coefficients for ζK(s).
LEMMA 2·2 (Chandrasekharan{Narasimhan [CN63, lemma 9]). Let K be any number field,
and write in Re(s)>1,
ζK(s)=∞
n=1
an
ns.
Then anis bounded by the coefficient bnof n−sin ζ(s)nKand there exists a constant C
depending only on nKsuch that
an≤bnnKn
C
log log nas n →∞.
The same holds for the Dirichlet series coefficients anof 1/ζK(s).
Useful for our purposes will be an estimate for the number of nontrivial zeros of ζK(s)in
unit intervals in the critical strip. Define
NK(T):=#{ρ=β+iγ|ζK(ρ)=0, 0 <β <1, 0 ≤|γ|≤T}.
It is well known that N(T+1) −N(T)log Tin the case of K=Q. An analogous result
holds for any number field K.
LEMMA 2·3. For any T ≥0, we have that NK(T+1) −NK(T)log DK+nKlog T.
Proof. See for instance [KN12] for suitable estimates of the quantity NK(T).
In the explicit formulae to follow in Section 4, we will also require the following result
which gives good upper bounds for 1/ζK(s) on certain horizontal lines near the critical strip.
We note that it is conditional on the Riemann hypothesis for ζK(s),
LEMMA 2·4. Assume that ζK(s)satisfies the Riemann hypothesis. Then there exists a con-
stant C >0such that, for each positive integer n ≥4, there exists some n ≤Tn<n+1such
that
|ζK(σ+iTn)|≥exp −Clog n
log log n
for −1≤σ≤2. We denote by T={Tn}∞
n=4the sequence so constructed.
This result is well known in the case K=Q(see [MV07, theorem 13·22] or [Tit86, theo-
rem 14·16]) and the proof for general Kis mostly analogous. A complete proof can be found
in Section 5.
2·3. -type lemmas for MK(x)and disproofs of the naïve Mertens-type conjecture in degenerate
cases
Fix any number field K. In this section, we provide various conditional -type theo-
rems for MK(x) which will reduce the unconditional disproofs in Section 3to a number of
conditional assumptions. The main result of this section is Theorem 2·10, which provides
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On a mertens-type conjecture for number fields 9
three cases in which the naïve Mertens-type conjecture is guaranteed to fail. Essential in the
following proofs will be the following well-known method of Landau.
LEMMA 2·5 (Landau [MV07, theorem 15·1]). Suppose that A(x) is a bounded Riemann-
integrable function in any finite interval 1≤x≤X, and that A(x)≥0for all x >X0.Letσc
denote the infimum of those σfor which ∞
X0A(x)x−σdx <∞. Then the function
F(s)=∞
1
A(x)x−sdx
is analytic in the half-plane Re(s)>σ
c, but not at the point s =σc.
We shall demonstrate that most cases in which zeros of ζK(s) behave contrary to con-
ventional expectations (i.e. the Riemann hypothesis and simplicity of zeros) will degenerate
into falsity of the naïve Mertens-type conjecture over K. One additional possible obstruction
present in the case of Dedekind zeta functions, which is not so in the case of the Riemann
zeta function, is the presence of nontrivial zeros on the real line. Let
=sup
ζK(ρ)=0{Re(ρ)}=max{,},
where =sup
ζK(ρ)=0
ρ∈(0,1)
ρ, =sup
ζK(ρ)=0
Im(ρ)=0
{Re(ρ)}.
If no zero ρ∈(0, 1) exists, we set =0. In particular, ≥1/2 and the Riemann hypothesis
for ζK(s) is the statement =1/2. The following is our first degenerate case.
PROPOSITION 2·6. If ζK(s)does not satisfy the Riemann hypothesis, and <, then the
naïve Mertens-type conjecture over K is false. More precisely, MK(x)=±(x−)for all
0<<−.
Proof. Suppose that M(x)<x−for all x>X0(). Consider, in view of (1·1), the function
1
s−+−1
sζK(s)=∞
1
(x−−MK(x))x−s−1dx.
Here the left-hand side has a pole at −, but is analytic for real s>−, since ζK(s)
has no zeros to the right of −on the real line. The integrand of the right-hand side is
non-negative for all x>X0(). By an application of Lemma 2·5, the above identity holds
for Re(s)>−, and both sides are analytic in this half-plane. But by definition of , the
function 1/ζKhas poles with real part >−, which is a contradiction. Hence, we deduce
that M(x)=+(x−).
To obtain the −estimate, we argue similarly using the identity
1
s−++1
sζK(s)=∞
1
(x−+MK(x))x−s−1dx.
We then conclude that MK(x)=±(x−). Specifically, this gives the falsity of the naïve
Mertens-type conjecture over K, since >1/2.
If =, then since real zeros must be isolated, there is a zero at s=which prohibits
the application of Lemma 2·5in the proof above. In this case, we give a result in lieu of the
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10 DANIEL HUet al.
previous proposition, which involves the elimination of the polar behaviour of 1/sζK(s)at
the real nontrivial zeros of ζK(s). More precisely, there exist constants ck,α∈C,1≤k≤mα,
for each of the finitely many real zeros α∈(0, 1), such that the function
1
sζK(s)−
ζK(α)=0
α∈(0,1)
mα
k=1
ck,α
(s−α)k(2·1)
is analytic on the segment (0, 1) on the real line, where mαis the multiplicity of the zero at
α. Upon this simplification, we may provide the following result.
PROPOSITION 2·7. Suppose that ζK(s)does not satisfy the Riemann hypothesis, and =
,1
2<
≤.Letc
k,αbe as in the equation (2·1). Then for all 0<<
−1/2,
MK(x)−˜
MK(x)=±(x−),
where
˜
MK(x):=
ζK(α)=0
α∈(0,1)
mα
k=1
ck,αxα( log x)k−1
(k−1)!.(2·2)
In particular, the naïve Mertens-type conjecture over K is false.
Proof. Suppose that MK(x)−˜
MK(x)<x−for all x>X0(). Now
1
s−+−1
sζK(s)+
ζK(α)=0
α∈(0,1)
mα
k=1
ck,α
(s−α)k=∞
1
(x−−MK(x)+˜
MK(x))x−s−1dx
holds for all real s>−, and by Lemma 2·5extends to Re(s)>−. From here, the
proof is similar to that of Proposition 2·6. For the last statement, note that s=is one
of the αin the sum ˜
MK(x). Since >1/2, this sum surpasses the order of x1/2, and since
MK(x)−˜
MK(x) is oscillatory we deduce the falsity of the naïve Mertens-type conjecture
over K.
It remains to address the case where 1/2= <
=. We give the following auxiliary
result, which is independent of RH.
PROPOSITION 2·8. Suppose ,, are as above, and that there is a non-real zero ρof
ζKof multiplicity m ≥1with Re(ρ)=, say ρ= +iγ. Then
MK(x)−˜
MK(x)=±(x( log x)m−1),
where ˜
MK(x)is defined as in (2·2). More precisely,
lim inf
x→∞
MK(x)−˜
MK(x)
x( log x)m−1≤− m
|ρζ(m)
K(ρ)|<m
|ρζ(m)
K(ρ)|≤lim sup
x→∞
MK(x)−˜
MK(x)
x( log x)m−1.
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On a mertens-type conjecture for number fields 11
Proof. Suppose that MK(x)−˜
MK(x)≤cx( log x)m−1for all x>X0. It suffices to prove
that c≥m/|ρζ(m)
K(ρ)|. Consider the function
c(m−1)!
(s−)m−1
sζK(s)+
ζK(α)=0
α∈(0,1)
ck,αxα( log x)k−1
(k−1)!
=∞
1
(cx( log x)m−1−MK(x)+˜
MK(x))x−s−1dx
for real s>
. By Lemma 2·5, this can be extended to Re(s)>
. Denote the function
F(s). Then
F(s)+1
2eiφF(s+iγ)+1
2e−iφF(s−iγ)
=∞
1
(cx( log x)m−1−MK(x)+˜
MK(x))(1 +cos (φ−γlog x)x−s−1)dx
for Re(s)>
. On the right-hand side, the integral from 1 to X0is uniformly bounded,
while the integral from X0to ∞is non-negative. Thus the lim inf of the right-hand side is
bounded below as s→+. As a result, the coefficient of (s−)−min its Laurent series
must be non-negative. On the other hand, the left-hand side has a pole of multiplicity mat
s=, at which the Laurent series expansion contains a term (s−)−mwith coefficient
equal to
c(m−1)!− m!eiφ
2ρζ(m)
K(ρ)−m!e−iφ
2¯ρζ(m)
K(¯ρ).
Choosing φso that
eiφ=ρζ(m)
K(ρ)
|ρζ(m)
K(ρ)|.
Then the above is (m−1)!(c−m/|ρζK(ρ)|). This quantity must be non-negative, otherwise
the left-hand side would tend to −∞ as s→+. Hence c≥m/|ρζ(m)
K(ρ)|. The −case is
similar.
In the scenario where 1/2= <
=, we deduce from the previous proposition that
the naïve Mertens-type conjecture is false over K. This completes the discussion under
falsity of the Riemann hypothesis for ζK(s).
Turning to results that are instead conditional on the Riemann hypothesis for K,
Proposition 2·8in fact yields the following immediate corollary.
COROLLARY 2·9. Assume the Riemann hypothesis for ζK(s), and that ζK(s)has a zero
ρ=1/2+iγ,γ>0, of multiplicity m ≥1. Then
MK(x)−˜
MK(x)=±(x1/2( log x)m−1),
where ˜
MK(x)defined as in (2·2) is the part corresponding to a possible zero at s =1/2.
Proof. In this case, 1/2== (and =, if a real zero exists).
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12 DANIEL HUet al.
Corollary 2·9implies in particular that if ζK(s) satisfies the Riemann hypothesis, but pos-
sesses nontrivial zeros with multiplicity m≥2(ats=1/2 or otherwise), then the naïve
Mertens-type conjecture is false. Therefore, the only non-degenerate case of the naïve
Mertens-type conjecture occurs when ζK(s) satisfies the Riemann hypothesis and has only
simple nontrivial zeros (a priori none at s=1/2).
The following theorem summarizes the results of this section.
THEOREM 2·10. In the following special cases, MK(x)grows more quickly than √x.
(a) If the Riemann hypothesis for ζK(s)fails with = >
, then for all >0,
MK(x)=±(x−).
If there is a zero +iγof ζK(s), then MK(x)=±(x).
(b) If the Riemann hypothesis for ζK(s)fails and =, then
MK(x)=(x)
(in possibly only one direction).
(c) If the Riemann hypothesis for ζK(s)holds but there is a zero ρ=1/2+iγ,γ>0,of
multiplicity m ≥1, then
MK(x)=√x( log x)m−1.
If ζK(1/2) = 0, or if the multiplicity of the zero of ζKat s =1/2is strictly less than m,
then this can be replaced with ±.
In particular, if MK(x)/√xis bounded (both above and below), then the Riemann hypoth-
esis for ζK(s) holds, and ζK(s) has no multiple zero in the critical strip, thereby implying that
ζK(1
2)= 0. We expand on the possibility of growth in only one direction in Section 6.
COROLLARY 2·11. If K/Qis Galois with non-abelian Galois group, then the naïve
Mertens-type conjecture over K is false.
Proof. It is known that the Dedekind zeta function of a non-abelian Galois extension Kof
Qhas infinitely many zeros of multiplicity ≥2 in the critical strip (cf.[HKMS]).
2·4. The explicit formula for MK(x)
We now state the explicit formula for the Mertens function, which, in its compatibil-
ity with analytic methods, is essential to our applications. We must assume the Riemann
hypothesis for ζK(s) and that it has no nontrivial multiple zeros. The error terms in our
explicit formula are presented in different cases depending on their applicability to x>0;
the generality of x>0 is necessary for the unconditional results of Section 3, while more
specific bounds for x>1 are necessary to prove the existence of a limiting distribution in
Section 4.
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On a mertens-type conjecture for number fields 13
PROPOSITION 2·12. Assume the Riemann hypothesis for ζK(s), and that all nontrivial zeros
of ζK(s)are simple. For any real x >0and any T ∈T, where Tis as in Lemma 2·4,
MK(x)=
|γ|≤T
xρ
ρζK(ρ)+∞
k=0
Ress=−k
xs
sζK(s)+E(x,T),
where for any x >0,E(x,T)=Ox(T1−)for any , and for x >1we have
E(x,T)x1+C/log log xlog x
T+x
T1−+xC/log log x,
where C is as in Lemma 2·2. In addition, for x >1,
∞
k=0
Ress=−k
xs
sζK(s)( log x)O(1),
which in this case can be subsumed into the error term E(x,T).
We defer the proof of this formula to Section 5.
For the previous result and others to follow, we also require the following bound on the
residues of xs/sζK(s) at the trivial zeros s=−k. The purpose of this bound is two-fold:
it reveals dependence on kand x, which suffices already to prove the last statement in
Proposition 2.12, and it clarifies dependence on r1,r2, and DK, which is crucial in our proof
of Theorem 1·2.
LEMMA 2·13. Fix non-negative integers r1and r2, and let K be a number field with r1real
embeddings and r2complex conjugate pairs of embeddings. Then there exists some constant
c=cr1,2r2such that, if ζK(s)has a zero at s =−k for a positive integer k, then
Ress=−k
xs
sζK(s)≤c(2π)knK( log (kxDK))r1+r2
kxkDk+1/2
K(k!)nK
.
In addition,
Ress=0
xs
sζK(s)K( log x)r1+r2−1.
See Section 5for the proof. Heavier consideration of the residue at s=0 is necessary to
prove Theorems 1·1and 1·2, wherein dependence on nKand DKwill be significant.
3. Unconditional failure of the naïve Mertens-type conjecture
In light of Theorem 2·10,ifx−1/2|MK(x)|is bounded, then ζK(s) has no zeros off the line
Re(s)=1/2 and has only simple zeros. In this section, we describe some consequences
of these assumptions, which lead to unconditional disproofs of the naïve Mertens-type
conjecture for “most” number fields.
3·1. The general result.
In this section, we give a generic result about the limiting behaviour of x−1/2MK(x) which
holds unconditionally. We follow a method of Jurkat from [Jur73], using his main theorem
directly in our application. In particular, Jurkat has identified a wide class of trigonometric
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14 DANIEL HUet al.
series which are seen to share a key limiting property enjoyed by Bohr-almost periodic
functions. By exploiting this property, we show x−1/2MK(x) approaches certain explicitly
computable values infinitely often and arbitrarily closely.
Definition 3·1. (cf. [Jur73]). A locally integrable function f:R→Ris called almost
periodic in a distributional sense (APD) if there is a Bohr-almost periodic function gso that
f=(dk/dkx)g.
Formally, this means fadmits an expansion of the form
f(x)∼∞
n=1
Re(aneiλnx), (3·1)
where 0 <λ
n∞and anare complex constants such that
∞
n=1
|an|
λk
n
<∞
for some k∈N. When referring to an APD-function, we shall specify k.
The notion of almost periodicity in a distributional sense is useful because such functions
have easily-calculable limit points, as shown by the following main theorem of Jurkat.
PROPOSITION 3·2 (Jurkat [Jur73, pp. 151[152]). Suppose f :R→Ris an APD-function,
and let L(f) denote the set of Lebesgue points of f. Then for any t ∈L(f),
lim inf
x→∞
x∈L(f)
f(x)≤f(t)≤lim sup
x→∞
x∈L(f)
f(x).
Moreover, all values of f at Lebesgue points are limit points of f.
The technical condition on Lebesgue points can be safely ignored for the remainder of
this paper. The functions we work with are sufficiently well-behaved that the inclusion of
their Lebesgue points does not change either limit.
For any real x>0, define
M∗
K(x)=− ∞
k=0
Ress=−k
xs
sζK(s).(3·2)
(We remark that this carries a slightly different normalisation than Jurkat’s M∗(x), as he
makes a few simplifications particular to the case K=Q.) Motivated by Proposition 2·12,
this is a “trivial completion” of MK(x), in the sense that, for T∈T(where Tis given by
Lemma 2·4),
MK(x)+M∗
K(x)=
|γ|≤T
xρ
ρζK(ρ)+Ox1
T1−(3·3)
for all x>0, assuming the preconditions of Proposition 2·12 hold. We first show an analogue
of Jurkat’s main theorem for number fields under these assumptions, as well as some more
stringent assumptions.
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On a mertens-type conjecture for number fields 15
The function f:R→Rdefined by
f(y):=e−y/2(MK(ey)+M∗
K(ey))
is locally integrable. If we assume the Riemann hypothesis for ζK(s) and that all nontrivial
zeros are simple, and let T→∞along Tin the equation (3·3) above (with the necessary
modifications for MK(ey)ifyis the logarithm of an integer), it furthermore has an expan-
sion of the form in (3·1) with an=2(ρnζK(ρn))−1,λn=γn, where γnis the ordinate of
the nth highest nontrivial zero ρn=1/2+iγnin the upper half plane. Here, the sum con-
verges boundedly given this grouping of the terms. Assuming for now that there exists some
constant C>0 such that
1
|ρζK(ρ)|≤C,
the result of Lemma 2·3implies that fis an APD-function with k=2. Hence, Proposition 3·2
functions produces the following.
PROPOSITION 3·3. Let K be a number field. Assume the Riemann hypothesis for ζK(s), that
all nontrivial zeros of ζK(s)are simple, and that there exists some constant C >0for which
1
|ρζK(ρ)|≤C
for every nontrivial zero ρof ζK(s). Then, for any x >0,
MK(x)+M∗
K(x)
x1/2
is a limit point of the function x−1/2MK(x).
We now describe our major application of this result. Recall the definition
M−
K=lim inf
x→∞
MK(x)
x1/2,M+
K=lim sup
x→∞
MK(x)
x1/2.
LEMMA 3·4. Let K be any number field, and let M∗
K(x)be as in (3·2).
(a) Unconditionally, M+
K−M−
K≥1.
(b) If M∗
K(1) >0(resp. M∗
K(1) <−1), then the naïve Mertens-type conjecture for K fails
infinitely often; in particular, M+
K>1(resp. M−
K<−1).
Proof. If the Riemann hypothesis for ζK(s) is false, or ζK(s) has a nontrivial zero of mul-
tiplicity greater than 1, then Theorem 2·10 implies that either M+
K=∞or M−
K=−∞,in
which case both (a) and (b) follow. By the sign of the functional equation for ζK(s)in
Theorem 2·1, any zero at s=1/2 must be of even order, and thus any Kwith ζK(1/2) =0is
covered in the previous cases.
In the case where the Riemann hypothesis holds and all nontrivial zeros are simple, we
may apply Proposition 2·8with =1/2 and m=1 to obtain
M+
K,−M−
K≥1
|ρζK(ρ)|
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16 DANIEL HUet al.
for any nontrivial zero ρof ζK(s). If one of M±
Kis infinite, then (a) and (b) hold as before.
Otherwise, if M±
Kare both finite, then there is some constant Cfor which
1
|ρζK(ρ)|≤C
for every zero ρof ζK(s). So, the preconditions of Proposition 3·3hold, and we have for any
xthat
MK(x)+M∗
K(x)
x1/2
is between M−
Kand M+
K. In particular,
M−
K≤lim
x→1−
MK(x)+M∗
K(x)
x1/2=M∗
K(1) <M∗
K(1) +1=lim
x→1+
MK(x)+M∗
K(x)
x1/2≤M+
K,
where limits to 1−and 1+denote approaching 1 from below and above, respectively. This
inequality chain shows both (a) and (b).
3·2. Imaginary quadratic fields.
In this subsection, we calculate and bound M∗
K(x) explicitly for imaginary quadratic fields.
We pick D>0 such that −Dis a negative fundamental discriminant and let K=Q(√−D)
be the imaginary quadratic field with discriminant −D. Let χ−Ddenote the odd quadratic
character χ−D(n)=(−D/n)of conductor D, so that ζK(s)=ζ(s)L(s,χ−D). Using Theorem
2·1, we may compute
M∗
K(x):=2π
L(1, χ−D)D1/2+∞
k=1D
4π2−k−1
2(−1)kx−k
k(k!)2ζK(k+1).(3·4)
Via the class number formula of Dirichlet, the leading term of (3·4) is always positive. We
show that, for x=1 and sufficiently large D, it exceeds the other terms in absolute value.
LEMMA 3·5. If D >0and −D is a fundamental discriminant, and K =Q(√−D), then
M∗
K(1) ≥2π
D1/21
1
2log D+log log D+2+log 2 −exp 4π2
D+1.
In particular, if D >307, then M∗
K(1) >0.
Proof. We begin by bounding from below the leading term of (3·4). The Euler product for
L(s,χ−D) implies that it is positive for all real s>1, and hence L(1, χ−D)>0, knowing that
it is nonzero. On the other hand,
L(1, χ−D)=
n≤N
χ−D(n)
n+
n>N
χ−D(n)
n≤1+log N+2
N√Dlog D,
where we have used partial summation and the Pólya–Vinogradov bound √Dlog Dfor the
partial sums of the primitive character χ−Dmodulo D, as well as the bound 1 +log xfor
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On a mertens-type conjecture for number fields 17
n≤x
1
n. Setting N=2√Dlog Dgives that
2π
D1/2L(1, χ−D)≥2π
D1/21
2log D+log log D+2+log 2.(3·5)
Now, for any real σ>1, we have ζK(σ)>1 by definition, so ζK(k+1)−1<1 for every
k≥1. So, we have the bound
∞
k=1D
4π2−k−1
2(−1)k
k(k!)2ζK(k+1)≤2π
D1/2
∞
k=14π2
Dk1
k(k!)2ζK(k+1)
≤2π
D1/2
∞
k=14π2
Dk1
k!
≤2π
D1/2exp 4π2
D−1.
Combining this bound with (3·5) gives the desired result.
Lemma 3·5has the following immediate consequence via Lemma 3·4.
THEOREM 3·6. Let K be an imaginary quadratic extension of Qwith discriminant −D
such that D >307. Then the naïve Mertens-type conjecture over K, i.e. that |MK(x)|≤x1/2,
is false. In particular, M∗
K(1) >0, and if the Riemann hypothesis for ζK(s)is true and all of
its nontrivial zeros are simple, then M+
K≥1+M∗
K(1).
By direct computation, we may address the smaller discriminants.
THEOREM 3·7. Let K be an imaginary quadratic extension of Qwith discriminant −D such
that 3<D≤307. Then the naïve Mertens-type conjecture over K is false, with M+
K>1.
Proof. The values of M∗
K(1) are tabulated in Table 1, where the sum has been computed up
to threshold k=50. This implies the falsity of the naïve Mertens-type conjecture over Kfor
all D≤307 where
−D/∈{−3, −4, −7, −8, −11, −15, −20, −23},
since, via Lemma 3·4, we deduce from M∗
K(1) >0 that M+
K>1. The procedure can be reme-
died for most of those exceptional values of −Dby letting x→n+for different choices of
positive integers nwhere
MK(n+)+M∗
K(n)
n1/2>1,
as represented in Table 2(we have taken the right-hand limits, as n−1/2M∗
K(n) is generally
decreasing). This suffices to resolve all cases except for D=3 (the field K=Q(√−3)) and
D=4 (the field K=Q(i)).
In order to resolve the case D=4, we appeal to a method originally due to Ingham [Ing42,
theorem 1] in examining the smoothed sums of |γ|<Teiγt/ρζK(ρ) by a suitable kernel
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18 DANIEL HUet al.
Table 1. Values of M∗
K(1) for imaginary quadratic fields K =Q(√−D)of discriminant
−D, D ≤307.
DM∗
K(1) D M∗
K(1) D M∗
K(1) D M∗
K(1)
3−0.4851 ... 79 0.2427 ... 152 0.2641 ... 235 0.9416 ...
4−0.5751 ... 83 0.4594 ... 155 0.4156 ... 239 0.1076 ...
7−0.5303 ... 84 0.3287 ... 159 0.1492 ... 244 0.2935 ...
8−0.5754 ... 87 0.2050 ... 163 1.8941 ... 247 0.2999 ...
11 −0.4722 ... 88 0.8007 ... 164 0.1926 ... 248 0.2164 ...
15 −0.1839 ... 91 0.7820 ... 167 0.1367 ... 251 0.2471 ...
19 0.2704 ... 95 0.1493 ... 168 0.4310 ... 255 0.1404 ...
20 −0.0301 ... 103 0.2864 ... 179 0.3342 ... 259 0.4537 ...
23 −0.0137 ... 104 0.2218 ... 183 0.2045 ... 260 0.2186 ...
24 0.0995 ... 107 0.5178 ... 184 0.4368 ... 263 0.1300 ...
31 0.1227 ... 111 0.1644 ... 187 0.9182 ... 264 0.2180 ...
35 0.3389 ... 115 0.8403 ... 191 0.1181 ... 267 0.9542 ...
39 0.1300 ... 116 0.2360 ... 195 0.4350 ... 271 0.1561 ...
40 0.4561 ... 119 0.1296 ... 199 0.1815 ... 276 0.2200 ...
43 1.3179 ... 120 0.3915 ... 203 0.4414 ... 280 0.4647 ...
47 0.1296 ... 123 0.8609 ... 211 0.6023 ... 283 0.6232 ...
51 0.5597 ... 127 0.3136 ... 212 0.2891 ... 287 0.1220 ...
52 0.6023 ... 131 0.3007 ... 215 0.1127 ... 291 0.4633 ...
55 0.2377 ... 132 0.4041 ... 219 0.4451 ... 292 0.4664 ...
56 0.2303 ... 136 0.4047 ... 223 0.2483 ... 295 0.2253 ...
59 0.3538 ... 139 0.5526 ... 227 0.3517 ... 296 0.1751 ...
67 1.6238 ... 143 0.1434 ... 228 0.4550 ... 299 0.2204 ...
68 0.2836 ... 148 0.9040 ... 231 0.1369 ... 303 0.1781 ...
71 0.1374 ... 151 0.2231 ... 232 0.9499 ... 307 0.6279 ...
Table 2. Given K =Q(√−D)
for a fundamental discriminant
−D, the smallest positive integer
nforwhichM
K(n+)+M∗
K(n)>
n1/2.
Dn
MK(n+)+M∗
K(n)
√n
7 22 1.2138 ...
8 57 1.1777 ...
11 20 1.2923 ...
15 98 1.0080 ...
20 15 1.0076 ...
24 6 1.0261 ...
function f(y). In particular, we select the kernel function of Jurkat–Peyerimhoff (see [JP76,
OtR85]),
f(y)=⎧
⎨
⎩
(1 −y) cos πy+1
πsin πyif 0 ≤y≤1,
0ify≥1,
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On a mertens-type conjecture for number fields 19
which produces the smoothed sum
h∗
K,T(t)=
|γ|<T
f|γ|
Teiγy
ρζK(ρ)
=
|γ|<T1−|γ|
Tcos π|γ|
T+1
πsin π|γ|
T eiγy
ρζK(ρ).(3·6)
It holds for this sum that for T>0 fixed,
M−
K≤lim inf
t→∞ h∗
K,T(t)≤lim sup
t→∞
h∗
K,T(t)≤M+
K,
and it follows from the fact that h∗
K,T(t) is almost periodic that
lim inf
t→∞ h∗
K,T(t)≤h∗
K,T(t)≤lim sup
t→∞
h∗
K,T(t),
so it suffices to exhibit T>0 and t∈Rfor which h∗
K,T(t)>1orh∗
K,T(t)<−1.
A computation demonstrates that h∗
Q(i),600(72.85) ≤−1.008 and h∗
Q(i),600(−85.15) ≥
1.029. That is, M−
K≤−1.008 <1.029 ≤M+
K. This completes the proof.
Combining Theorem 3·6and Theorem 3·7concludes the proof of Theorem 1·1(a).
3·3. Real quadratic fields.
Next, we calculate and bound M∗
K(x) explicitly for real quadratic fields. Pick a positive
fundamental discriminant D>0 and let K=Q(√D) be the real quadratic field with discrim-
inant D. Then ζK(s)=ζ(s)L(s,χD), where χDis the even quadratic character χD(n)=D
n
of conductor D. We compute
M∗
K(x)=4
L(1, χD)D1/2log xD
4π2−γ+L
L(1, χD)
+2
D1/2
∞
k=14π2
xD 2k
k((2k)!)2ζK(2k+1) log xD
4π2+1
2k+ζK
ζK
(2k+1) +2
(2k+1)
.
(3·7)
Here, γ=0.5772 ...is Euler’s constant. The L/L,ζK/ζK, and /terms arise (in contrast
to (3·4)) since ζK(s) has zeros of multiplicity two when Kis real quadratic. We now only
need to identify which Dsatisfy the estimate M∗
K(1) >0. We first provide a bound on the
L/Lterm that is particular to the case of real characters.
LEMMA 3·8. Let χbe a non-principal real primitive character of conductor D. Then
L
L(1, χ)>−1
2log D
2π−γ+χ(−1)
2log 2.
Proof. Recall [MV07, corollary 10·18] for any Dirichlet character χof conductor Dthe
identity
L
L(s,χ)=−L
L(1, χ)−1
2
s+κ
2−log D
π+
ρ1
s−ρ+1
ρ+γ
2+(1 −κ) log 2,
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20 DANIEL HUet al.
where κ=(1 −χ(−1))/2 and ρruns over all nontrivial zeros of L(s,χ). Specifying to s=1
and χ=χgives
2L
L(1, χ)=−1
2
1+κ
2−log D
π+
ρ
1
ρ(1 −ρ)+γ
2+(1 −κ) log 2.
Since
1
2=−γ−2 log 2
(1) =−γ,
this gives
2L
L(1, χ)=γ+2(1 −κ) log 2 −log D
π+
ρ
1
ρ(1 −ρ).
By grouping ρand ρin the sum we have
ρ
1
ρ(1 −ρ)=1
2
ρ1
ρ(1 −ρ)+1
ρ(1 −ρ)
=
ρ
Re(ρ−ρ2)
|ρ|2|1−ρ|2
=
ρ
β(1 −β)+γ2
|ρ|2|1−ρ|2>0,
where ρ=β+iγand we have used 0 <β<1. This gives
2L
L(1, χ)=γ+2(1 −κ) log 2 −log D
π+
ρ
1
ρ(1 −ρ)>γ +log 2
+χ(−1) log 2 −log D
π,
as desired.
This allows us to establish the following result on M∗
K(1).
THEOREM 3·9. Let K be a real quadratic extension of Qwith discriminant D >269. Then
the naïve Mertens-type conjecture over K, i.e. that −1≤M−
K≤M+
K≤1, is false. In particu-
lar, M∗
K(1) >0, and if the Riemann hypothesis for ζK(s)is true and all of its nontrivial zeros
are simple and not real, then M+
K≥1+M∗
K(1).
Proof. We begin by bounding the contribution of the infinite series in M∗
K(1). After dividing
out by a factor of 2/D1/2, the series under consideration takes the form
=∞
k=14π2
D2k1
k((2k)!)2ζK(2k+1) log D
4π2+1
2k+ζK
ζK
(2k+1) +2
(2k+1).
(3·8)
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On a mertens-type conjecture for number fields 21
We note that ζK(2k+1) ≥1 for all k≥1, and by the triangle inequality,
ζK
ζK
(2k+1)=
ζ
ζ(2k+1) +L
L(2k+1, χD)≤2∞
n=1
(n)
n2k+1
≤2∞
1
log x
x2k+1dx =1
2k2.
We also have the following known identity for the digamma function:
2
(2k+1) =22k
n=1
1
n−γ≤2 log (2k)+(2 −2γ),
where γis Euler’s constant. Finally, we use these bounds to see that, for D>4π2,
≤∞
k=14π2
D2k1
k((2k)!)2log D
4π2+1
2k+1
2k2+2 log (2k)+(2 −2γ)
≤cosh 4π2
D−11
2log D
4π2+1
4+1
4+log 2 +(1 −γ)
=cosh 4π2
D−11
2log D+3
2−log π−γ.
We now turn our attention to the first term of M∗
K(1) in (3·7). Upon dividing out by a factor
of 2/D1/2,wehave
2
L(1, χD)log D
4π2−γ+L
L(1, χD).
To this end, we recall the previously obtained bound from (3·5)
1
L(1, χD)≥1
1
2log D+log log D+2+log 2.
Finally, using Lemma 3·8in the case of an even character, the problem reduces to finding
the values Dat which the inequality
21
2log D−3
2log π−log 2 −γ
2
1
2log D+log log D+2+log 2 >cosh 4π2
D−11
2log D+3
2−γ−log π
holds. This indeed holds for all D>269.
By direct computation, we may address the smaller discriminants.
THEOREM 3·10. Let K be an real quadratic extension of Qwith discriminant 5<D≤269.
Then the naïve Mertens-type conjecture over K is false, with M+
K>1.
Proof. As in the proof of Theorem 3·7, the values of M∗
K(1) are tabulated in Table 3, where
the sum has been computed up to threshold k=50. This implies the falsity of the naïve
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22 DANIEL HUet al.
Table 3. Values of M∗
K(1) for imaginary quadratic fields K =Q(√D)of discriminant D,
D≤269.
DM∗
K(1) D M∗
K(1) D M∗
K(1) D M∗
K(1)
5−0.4857 ... 73 −0.0079 ... 141 0.2217 ... 205 0.2148 ...
8−0.5362 ... 76 0.0126 ... 145 0.0569 ... 209 0.1307 ...
12 −0.5230 ... 77 0.4411 ... 149 0.3829 ... 213 0.6172 ...
13 −0.6401 ... 85 0.0816 ... 152 0.4500 ... 217 0.0825 ...
17 −0.3642 ... 88 0.0506 ... 156 0.1444 ... 220 0.1559 ...
21 −0.5180 ... 89 0.0410 ... 157 0.2747 ... 221 0.3885 ...
24 −0.3364 ... 92 0.2036 ... 161 0.1068 ... 229 0.2186 ...
28 −0.2605 ... 93 0.2430 ... 165 0.3070 ... 232 0.1609 ...
29 −0.3697 ... 97 0.0274 ... 168 0.2387 ... 233 0.1703 ...
33 −0.1707 ... 101 0.3760 ... 172 0.1414 ... 236 0.3216 ...
37 −0.2060 ... 104 0.1804 ... 173 1.2271 ... 237 0.6833 ...
40 −0.1436 ... 105 0.0461 ... 177 0.0981 ... 241 0.0724 ...
41 −0.1227 ... 109 0.1101 ... 181 0.1952 ... 248 0.6696 ...
44 −0.1153 ... 113 0.0921 ... 184 0.1087 ... 249 0.0898 ...
53 0.1152 ... 120 0.1362 ... 185 0.1436 ... 253 0.2825 ...
56 −0.0201 ... 124 0.0851 ... 188 0.5386 ... 257 0.2055 ...
57 −0.0452 ... 129 0.0617 ... 193 0.0742 ... 264 0.2125 ...
60 −0.0198 ... 133 0.2189 ... 197 0.8499 ... 265 0.0824 ...
61 −0.0162 ... 136 0.0951 ... 201 0.0901 ... 268 0.1677 ...
65 −0.0135 ... 137 0.1183 ... 204 0.1653 ... 269 0.5460 ...
69 0.0750 ... 140 0.2922 ...
Mertens-type conjecture over Kfor all D≤269 where
D/∈{5, 8, 12, 13, 17, 21, 24, 28, 29, 33, 37, 40, 41, 44, 56, 57, 60, 61, 65, 73}.
Again, this can be remedied for most of the above discriminants by letting x→n+for other
positive integers nfor which n−1/2(MK(n+)+M∗
K(n)) >1. These findings are presented in
Table 4, which suffices to resolve all cases except for D=5 (the field Q(√5)), D=8 (the
field Q(√2)), and D=12 (the field Q(√3)).
For these fields, we return to the means h∗
K,T(t) as defined in (3·6). A computation shows
that t=17.32 at T=200 suffices for K=Q(√3) with a bound of M+
K>1.027. Moreover,
t=−24.64 at T=150 suffices for K=Q(√2) with a bound of M+
K>1.049.
This concludes the proof of Theorem 1·1(b).
3·4. General number fields.
In this subsection, we show that M∗
K(1) >0 for all number fields Kwith sufficiently large
discriminant and fixed signature. In light of Lemma 3·4, this implies that the naïve Mertens-
type conjecture is false for such K. We first show that the contribution of the residues at
strictly negative integers is relatively small.
LEMMA 3·11. Fix a signature (r1,r2), and consider a number field K with r1real embed-
dings, 2r2complex embeddings, and varying discriminant Kof magnitude DK. Then for
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On a mertens-type conjecture for number fields 23
Table 4. Given K =Q(√−D)
for a fundamental discriminant
−D, the smallest positive integer
nforwhichM
K(n+)+M∗
K(n)>
n1/2.
Dn
MK(n+)+M∗
K(n)
√n
13 100 1.0670 ...
17 38 1.0439 ...
21 35 1.2558 ...
24 146 1.1272 ...
28 94 1.0049 ...
29 49 1.0038 ...
Dn
MK(n+)+M∗
K(n)
√n
33 82 1.0991 ...
37 33 1.4651 ...
40 159 1.0713 ...
41 215 1.2509 ...
44 917 1.0343 ...
56 65 1.1612 ...
Dn
MK(n+)+M∗
K(n)
√n
57 146 1.1290 ...
60 35 1.2919 ...
61 39 1.1080 ...
65 26 1.0031 ...
73 9 1.1778 ...
all x ≥1,
∞
k=1
Ress=−k
xs
sζK(s)( log (xDK))nK
xD3/2
K
.
Proof. Apply Lemma 2·13 and sum the results.
The remainder of this section will involve bounding the residue at s=0. In our
application, we are taking x=1, and it is easy to see that
Ress=0
xs
sζK(s)x=1=Ress=01
sζK(s).
We first give a result about the Laurent series coefficients of ζK(s)/ζK(s)ats=1.
LEMMA 3·12. Let K be a number field of fixed degree, and assume the Riemann hypothesis
for ζK(s). Then, for a fixed integer j ≥0,
lim
s→1
dj
dsj1
s−1+ζK(s)
ζK(s)( log log DK)j+1
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24 DANIEL HUet al.
Proof. We mimic the proof of [MV07, theorem 13·13]. Fix some real x,y≥2, to be cho-
sen later. Define K(n) to be the coefficient of n−sin the Dirichlet series expansion of
ζK(s)/ζK(s), and define
w(u)=w(x,y;u):=⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
11≤u≤x
1−log u/x
log yx≤u≤xy
0u≥xy.
By Riesz summation ([MV07,(5·20)]; cf. [MV07, theorem 13·13]), we have
n≤xy
w(n)K(n)
ns=−ζK(s)
ζK(s)+(xy)1−s−x1−s
(1 −s)2log y−
ρ
(xy)ρ−s−xρ−s
(ρ−s)2log y,(3·9)
where the sum runs over all zeros of ζK(s), both trivial and nontrivial, and a zero of multi-
plicity mappears mtimes. Now, as the nontrivial zeros occur only at non-positive integers
and with multiplicity bounded by nK
ρtrivial
(xy)ρ−s−xρ−s
(ρ−s)2log y≤nK
∞
k=0
(xy)−k−s−x−k−s
(k+s)2log y2x−3/4
log y,(3·10)
where the last bound follows from |s−1|≤1/4. Also,
ρnontrivial
(xy)ρ−s−xρ−s
(ρ−s)2log y≤2∞
T=0
T≤γ<T+1
x1/2−σyρ−s−1
1
2−σ+i(γ−t)2log y
≤4x−1/4
log y
∞
T=0
N(T+1) −N(T)
1
16 +T2x−1/4log DK
log y,
where we have used Lemma 2·3in the last simplification. As a result,
f(s):=ζK(s)
ζK(s)−⎡
⎣−
n≤xy
w(n)K(n)
ns+(xy)1−s−x1−s
(1 −s)2log y⎤
⎦Ox−1/4log DK
log y
for |s−1|≤1/4. Now, the Taylor series expansion
ζK(s)
ζK(s)=− 1
s−1+∞
j=0
aj(s−1)j
j!,aj=lim
s→1
dj
dsj1
s−1+ζK(s)
ζK(s)
converges for |s−1|<1/2, since ζK(s) has no zeros strictly within distance 1/2ofs=1.
Furthermore, we have the expansions
n≤xy
w(n)K(n)
ns=∞
j=0
(s−1)j
j!
n≤xy
w(n)K(n)( −log n)j
n
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On a mertens-type conjecture for number fields 25
and
(xy)1−s−x1−s
(1 −s)2log y=− 1
s−1+∞
j=0
(s−1)j
j!(−1)jlog (xy)j+2−( log x)j+2
(j+1)(j+2) log y.
Now, select x=( log DK)4and y=e, so that f(s)=O(1). Writing
f(s)=∞
j=0
bj(s−1)j
j!,
we have
bj=j!r
f(s)(s−1)−j−1ds =1
2π2π
0
f1+reiθr−j−1e−i(j+1)θdθ
for any r≤1/4, where ris a counterclockwise circular contour of radius raround 1. For
fixed jand r, bounding the integrand in the second integral gives bj=O(1). So, recalling our
choices of xand yand using partial summation gives
aj=
n≤xy
w(n)K(n)( −log n)j
n+(−1)jlog (xy)j+2−( log x)j+2
(j+1)(j+2) log y
+O(1) ( log log DK)j+1.
This completes the proof.
We are now ready to bound the residue at 0.
LEMMA 3·13. In the same setting as Lemma 3·11,
−Ress=01
sζK(s)=2r1+r2πr2( log DK)r1+r2−1
D1/2
Klims→1(s−1)ζK(s)(1 +o(1)) r1,r2
( log DK)−r2
D1/2
K
.
Proof. Write
Ress=01
sζK(s)=lim
s→0
dr1+r2−1
dsr1+r2−1sr1+r2−1
ζK(s).
Using the functional equation, we may write
sr1+r2−1
ζK(s)=2r2πnK
D1/2
KDK
(2π)nKs1
sζK(1 −s) s
sin πs
2r1s
sin πsr21
(1 −s)nK
.
Define
f(s)=sr1+r2−1
ζK(s)
g1(s)=slog DK
(2π)nK
g2(s)=log (sζK(1 −s))
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26 DANIEL HUet al.
g3(s)=log s−log sin πs
2
g4(s)=log s−log sin πs
g5(s)=log (1 −s)
so that
g(s):=log f(s)=log 2r2πnK
D1/2
K+g1(s)−g2(s)+r1g3(s)+r2g4(s)−nKg5(s).
Since f(s)=g(s)f(s), we have via induction on jthat
f(j)(s)=f(s)Pjg(1)(s), g(2)(s), ...,g(j)(s)
for some polynomial Pj(t1,...,tj)injvariables with positive integral coefficients, which is
homogeneous of degree jwhen tiis assigned degree i. In addition, the coefficient of tj
1in
Pj(t1,...,tj) is 1. We now bound the derivatives of gat 1.
For each 1 ≤j≤r1+r2−1, the derivatives g(j)
i(s)fori∈{3, 4, 5}near s=0 depend only
on r1and r2, so we can treat the terms corresponding to these gias constants. This gives
that, for j≥1,
g(j)(s)=−g(j)
2(s)+O(1) +⎧
⎨
⎩
log DKj=1
0j>1.
Now, note that
g(j)
2(s)=dj−1
dsj−11
s−ζK(1 −s)
ζK(1 −s),
so Lemma 3·12 shows that g(j)
2(0) ( log log DK)jfor all j≥1. This implies that, using the
properties of Pj,
lim
s→0Pr1+r2−1g(1)(s), g(2)(s), ...,g(r1+r2−1)(s)=( log DK)r1+r2−1(1 +o(1)).
Also,
lim
s→0f(s)=2r2πnK
D1/2
Klim
s→0
1
sζK(1 −s)2
πr11
πr2
=−2r1+r2πr2
D1/2
K
lim
s→1
1
(s−1)ζK(s).
By a theorem of Louboutin [Lou00a, theorem 1],
lim
s→1(s−1)ζK(s)( log DK)nK−1,
and is clearly positive. This gives the desired result.
Using this bound, we finish the proof that M∗
K(1) >0 for all sufficiently large DK.
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On a mertens-type conjecture for number fields 27
Proof of Theorem 1·2. By Lemma 3·11,
M∗
K(1) =−Ress=01
sζK(s)+O( log DK)nK
D3/2
K.
Now, Lemma 3·13 implies that the first term is positive, and that it grows at least as quickly
as
( log DK)−r2
D1/2
K
.
For large DKdepending on r1and r2, this outpaces the other terms, and so M∗
K(1) >0. An
application of Lemma 3·4finishes the proof.
4. Proof of Theorem 1·4
In this section, we fix a number field Kwhose Dedekind zeta function satisfies the
Riemann hypothesis and JK
−1(T)αT1+αfor some 0 ≤α<2−√3. We expect these
hypotheses to hold for all Kwith K/Qan abelian Galois extension, and provide support
for the second condition later in this section.
Ng [Ng04] demonstrates the existence of a limiting distribution for e−y/2M(ey)by
achieving bounds for the error term E(x,T) in the truncated explicit formula
M(x)=
|γ|≤T
xρ
ρζ(ρ)+E(x,T)
for M(x). He then follows the method of Rubinstein–Sarnak [RS94] to prove the exis-
tence of a limiting distribution by first constructing measures for the main term at height
T, then bounding the error term E(x,T) to show that they converge to the desired probability
measure.
Later work of Akbary–Ng–Shahabi in [ANS14] allows for substantial generalisation of
this method by proving that a more general class of functions of interest possess limiting
distributions; we will use their framework in our proof. First, it requires the definition of a
rather broad class of almost periodic functions.
Definition 4·1. Denote by Tthe class of all real-valued trigonometric polynomials
PN(y)=
N
n=1
rneiλny,y∈R
where rn∈Cand λn∈R. We say that a locally square-integrable function φ∈L2
loc([0, ∞))
is B2-almost periodic if for any >0, there is a function f(y)∈Tsuch that
φ(y)−f(y)B2<,
where
φB2:=lim sup
Y→∞
1
YY
0|φ(y)|2dy1/2
.
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28 DANIEL HUet al.
All B2-almost periodic functions possess a limiting distribution [ANS14, theorem 2·9].
This allows for concrete descriptions of almost periodic functions in the line of Rubinstein–
Sarnak, as well as the later Mertens-specific work of Ng.
THEOREM 4·2 (Akbary-Ng-Shahabi [ANS14, corollary 1·3]). Let φ:[0, ∞)→Rbe a
scalar-valued function and let y0be such that φis square-integrable on [0, y0]. Assume
there exists a non-decreasing sequence of positive numbers {λn}n∈Nwhich tends to infinity,
a sequence of complex numbers {rn}n∈Nand a real constant c such that for y ≥y0,
φ(y)=c+Re ⎛
⎝
λn≤X
rneiλny⎞
⎠+E(y,X)(4·1)
for any X ≥X0>0, where E(y,X)satisfies
lim
Y→∞
1
YY
y0|E(y,eY)|2dy =0. (4·2)
Assume that
T<λn≤T+1
1log T,(4·3)
and that either
(a) rnλ−β
nfor some β>1/2, or that
(b) for some 0≤θ<3−√3,
λn≤T
λ2
n|rn|2Tθ.(4·4)
Then φ(y)is a B2-almost periodic function and therefore possesses a limiting distribution.
Therefore, to establish the existence of a logarithmic limiting distribution for x−1/2MK(x),
it will suffice to provide a suitable expression for
φK(y):=e−y/2MKey
as above with error small in the sense of (4·2), as well as to verify that (4·4) holds for some
θ<3−√3. Our first step will be to extend the explicit formula (Proposition 2·12)forM(x)
to all heights T≥2, assuming the analogue of the Gonek–Hejhal conjecture for K,asin
(1·3).
LEMMA 4·3. Assume the Riemann hypothesis for ζK(s)and that JK
−1(T)αT1+α.Forx≥2
and arbitrary T ≥2,
MK(x)=
|γ|≤T
xρ
ρζK(ρ)+E(x,T), (4·5)
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On a mertens-type conjecture for number fields 29
where
E(x,T)x1+C/log log xlog x
T+x
T1−+xC/log log x+xlog T
T1−α1/2
for any >0where the constant C is as in Lemma 2·2.
Proof. Suppose T≥2 satisfies n≤T≤n+1 for some positive integer n, and let Tnbe as
in Lemma 2·4. Suppose without loss of generality that n≤Tn≤T≤n+1. Proposition 2·12
gives that
MK(x)=
|γ|≤T
xρ
ρζK(ρ)−
Tn≤|γ|≤T
xρ
ρζK(ρ)+E(x,Tn),
where
E(x,Tn)x1+C/log log xlog x
T+x
T1−+xC/log log x.
We use the assumption JK
−1(T)T1+αto bound the second sum. The Cauchy–Schwarz
inequality gives that, under the Riemann hypothesis,
Tn≤|γ|≤T
xρ
ρζK(ρ)≤x1/2⎛
⎝
Tn≤|γ|≤T
1
|ρζK(ρ)|2⎞
⎠
1/2⎛
⎝
Tn≤|γ|≤T
1⎞
⎠
1/2
.
Since
Tn≤|γ|≤T
1
|ρζK(ρ)|21
T2
Tn≤|γ|≤T
1
|ζK(ρ)|2JK
−1(T)
T2Tα−1,
and the number of zeros of ζKwith ordinates between Tnand Tis log Tby Lemma 2·3,
Tn≤|γ|≤T
xρ
ρζK(ρ)xlog T
T1−α1/2
,
which is enough. The case T≤Tnis treated similarly; all that differs is the sign on the zeros
with ordinates between Tand Tn.
We are ready to establish the main theorem.
Proof of Theorem 1·4. Setting x=eyin Lemma 4·3, where y≥e, we may define
φK(y)=e−y
2MK(ey)=
|γ|≤T
eiγy
ρζK(ρ)+E(ey,T),
where
E(y,T):=e−y
2E(ey,T)e
Cy
log y+y
2( log y)
T+ey
2
T1−+( log T)1/2
T(1−α)/2+e
Cy
log y−y
2.
This is of the form required in Theorem 4·2; if we order the nontrivial zeros of ζK
nondecreasingly by magnitude of ordinate (both with positive and negative ordinate) as
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30 DANIEL HUet al.
ρn=1/2+iγn, then our definition of φK(y) agrees with (4·1) with
λn=γn,rn=1
ρnζK(ρn).
As a result, we need only to verify (4·2), (4·3) and (4·4) for some θ<3−√3. The second
of these is implied by Lemma 2·3, and the third holds for θ=1+αsince
λn≤T
λ2
n|rn|2=
|γ|≤T
4γ2
|ρζK(ρ)|2JK
−1(T)T1+α,
using our assumption on JK
−1(T), having chosen 0 ≤α<2−√3=0.2679 .... Finally, to
verify (4·2), we first have, using the Cauchy–Schwarz inequality, that
Y
e|E(y,eY)|2dy Y
e⎛
⎝e
2Cy
log y+y( log y)2
e2Y+ey
e2Y(1−)+Y
eY(1−α)+e
2Cy
log y−y⎞
⎠dy.
The integral of the third term is bounded as Y→∞when α<2−√3, and so is that of the
second term when <1/2. Moreover, an application of L’Hôpital’s rule shows that the first
and fourth terms are o(Y). This gives
lim
Y→∞
1
YY
e|E(y,eY)|2dy =0,
as desired. We may conclude that φK(y)=e−y/2MK(ey) possesses a limiting distribution by
Theorem 4·2.
4·1. Support for the conjecture (1·3)
By the discussion following the statement of Theorem 1·4, the picture is most transpar-
ent when K/Qis an abelian Galois extension. In this case, the irreducible characters on
Gal(K/Q) are all 1-dimensional, and therefore by abelian reciprocity the functions appear-
ing in the Artin factorisation of ζK(s) are moreover distinct primitive Dirichlet L-functions,
among them ζ(s). Label the functions L1(s), ...,Ln(s), where L1(s)=ζ(s), say, and n=
[K:Q]. We assume the generalised Riemann hypothesis for the Dirichlet L-functions, sim-
plicity of all nontrivial zeros (a priori none at s=1/2), and that no two Dirichlet L-functions
share a nontrivial zero. Let k∈R. Then,
0<γ ≤T
ζK(1
2+iγ)=0
|ζK(ρ)|2k=
n
i=1
0<γi≤T
Li(1
2+iγi)=0
|Li(ρ)|2k
j=i|Lj(ρ)|2k,
where we maintain the convention ρ=1/2+iγi. Upon normalising the ith summand by the
factor Ni(T)=#{0≤γi≤T|Li(1/2+iγi)=0}, we could anticipate that
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On a mertens-type conjecture for number fields 31
1
Ni(T)
0<γi≤T
Li(1
2+iγi)=0
|Li(ρ)|2k
j=i|Lj(ρ)|2k
⎛
⎜
⎜
⎜
⎝
1
Ni(T)
0<γi≤T
Li(1
2+iγi)=0
|Li(ρ)|2k⎞
⎟
⎟
⎟
⎠
j=i
⎛
⎜
⎜
⎜
⎝
1
Ni(T)
0<γi≤T
Li(1
2+iγi)=0
|Lj(ρ)|2k⎞
⎟
⎟
⎟
⎠.
Noting its similarity to the sum Jk(T), and recalling the known asymptotic Ni(T)T( log T)
(see [MV07, theorem 14·5]), we may posit that
1
Ni(T)
0<γi≤T|Li(ρ)|2k( log T)k(k+2),
as in the case of the Riemann zeta function (cf. [Gon89]). Meanwhile, to address the inner
sums
1
Ni(T)
0<γi≤T
Li(1
2+iγi)=0
|Lj(ρ)|2k,
where j= i, it would be most favorable to impose some restriction on the proximity of
the zeros γiof Lito lower values of the Lj; this is because we are taking k=−1. In this
case, we expect that these normalised sums are Oε(Tε) for all ε>0. This would suggest
that JK
−1(T)αT1+αfor all α>0, which would be sufficient for Theorem 1·4. In fact, we
believe that JK
−1(T)KTholds for all abelian number fields K.
5. Proofs of analytic results
In this section, we give proofs of some of the technical statements that we have used in
prior sections.
5·1. Bounds in horizontal strips.
In this section, we outline a proof of Lemma 2·4. This proof is heavily based on the proof
of [MV07, theorem 13·22], the corresponding statement for K=Q. We will run through the
intermediate statements needed for this result as it is proven in [MV07], and in each step
explain the necessary modifications in order to generalise to the setting of Dedekind zeta
functions.
LEMMA 5·1(cf.[MV07, corollary 10·5]). Suppose −1≤σ≤1
2. Then
ζK(σ+iT)
ζK(1 −σ−iT)|T|(1/2−σ)n2
K.
This follows from the reflection formula for Dedekind zeta functions and Stirling’s
approximation. By the Hadamard product for the completed Dedekind zeta function, one
gets the following formula for the logarithmic derivative of ζK.
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32 DANIEL HUet al.
LEMMA 5·2(cf.[MV07, lemma 12·1]). We have
ζK(s)
ζK(s)=−1
s+1
s−1+
|γ−t|≤11
ρ+1
s−ρ+O( log τ)
uniformly for −1≤σ≤2, where τ=|t|+4.
Define K(n) to be, as in the proof of Lemma 3·12, such that
ζK(s)
ζK(s)=∞
n=1
K(n)n−s.
We will use the bound
0≤K(n)≤nK(n)(5·1)
for all positive integers nin many of the following results. It may be proven by a simple
consideration of the powers of prime ideals of OKwith given norm; for the upper bound,
equality is reached for na power of a prime pwhich splits completely in K. The first result
needed for Lemma 2·4is the following.
LEMMA 5·3(cf.[MV07, theorem 13·13]). Assume the Riemann hypothesis for K. Then
ζK(s)
ζK(s)≤
n≤(logτ)2
K(n)
nσ+O(( log τ)2−2σ)
uniformly for 1/2+1/log log τ≤σ≤3/2,|t|≥1.
This lemma is proven essentially identically to the corresponding result from [MV07],
and a proof along very similar lines is given in this paper for Lemma 3·12. In particular,
the only modification to the proof in [MV07] that is needed results from the location and
multiplicity of trivial zeros (see (3·10)). We remark that, in light of (5·1), this lemma implies
the weaker bound
ζK(s)
ζK(s)≤nK
n≤(logτ)2
(n)
nσ+O(( log τ)2−2σ), (5·2)
in which a multiplicative constant is sacrificed for conceptual simplicity.
LEMMA 5·4(cf.[MV07, corollary 13·16]). Assume the Riemann hypothesis for ζK(s). Then,
for |t|≥1,
|log ζK(s)|≤nK·⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
log 1
σ−1+O(σ−1) if 1+1
log log τ≤σ≤3
2
log log log τ+O(1) if 1−1
log log τ≤σ≤1+1
log log τ
log 1
1−σ+O(logτ)2−2σ
(1−σ)loglogτif 1
2+1
log log τ≤σ≤1−1
log log τ.
The proof of this corollary is identical to the proof of [MV07, corollary 13·16], with (5·2)
used in the place of [MV07, theorem 13·13]. We also need two additional results about the
zeros of ζK(s), direct analogues of [MV07, lemma 13·19] and [MV07, lemma 13·20], with
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On a mertens-type conjecture for number fields 33
analogous proofs. Firstly, the bound in Lemma 5·3can be used to show that
NKT+1
log log T−NK(T)log T
log log T,(5·3)
a refinement of Lemma 2·3. Secondly, an application of Lemma 5·2combined with the
counting result from the preceding equation, shows that, for |t|≥1 and |σ−1/2|≤
1/log log τ, then
ζK(s)
ζK(s)=
|γ−t|≤1/log log τ
1
s−ρ+O( log τ). (5·4)
These results are enough to show Lemma 2·4following the proof of [MV07, theorem 13·22].
Proof of Lemma 2·4. Lemma 5·1implies that we need only to consider 1/2<σ <2. From
here, Lemma 5·4shows the result for σ>1/2+1/log log T, and integrating (5·4) over a
short horizontal segment gives a bound of the desired form, but in terms of the distances
between Tand various ordinates of the zeros of ζK(s). Integrating this bound over n≤T<
n+1 for any ngives that the result holds on average, and thus for some fixed T=Tn,as
desired.
5·2. Bounds on residues at the trivial zeros
We now prove Lemma 2·13, following the method of the proof of Lemma 2·13.
Proof of Lemma 2·13. First, the residue at 0 can be written as
lim
s→0
dr1+r2−1
dsr1+r2−1
sr1+r2−1xs
ζK(s)=lim
s→0
dr1+r2−1
dsr1+r2−1xsf(s)
for some findependent of xwith neither a pole nor a zero at s=0. Evaluating the derivative
explicitly yields that this residue is polylogarithmic in x, with exponent r1+r2−1.
For k>0, by Theorem 2·1, we write
xs
sζK(s)=2r2πnK
D1/2
KxDK
(2π)nKs1
sζK(1 −s) 1
sin πs
2r11
sin πsr21
(1 −s)nK
,
(5·5)
where we have used the reflection formula (s)(1 −s)=π/ sin πs.Ats=−k, none of
these terms are zero. The fourth term has a pole of order r1if kis even, and the fifth term has
a pole of order r2regardless. Define to be 1 if kis even and 0 otherwise, so that xs/sζK(s)
has a pole of order r1+r2at s=−k. Define
f(s)=(s+k)r1+r2xs
sζK(s)
g1(s)=slog xDK
(2π)nK
g2(s)=log (sζK(1 −s))
g3(s)=log (s+k)−log sin πs
2
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34 DANIEL HUet al.
g4(s)=log (s+k)−log sin πs
g5(s)=log (1 −s)
so that
g(s):=log f(s)=log 2r2πnK
D1/2
K+g1(s)−g2(s)+r1g3(s)+r2g4(s)−nKg5(s),
and so that
Ress=−k
xs
sζK(s)=lim
s→−k
dr1+r2−1
dr1+r2−1sf(s).
We may show that, for ≤nK,
g()
1(k)≤log xDK
(2π)nK
g()
2(k),g()
3(k),g()
4(k)r1,r21
g()
5(k)log k.
This shows that g()(s)r1,r2log (kxDK). In addition,
lim
s→−kf(s)=2r2πnK
D1/2
KxDK
(2π)nK−k
lims→−ksζK(1 −s)2
πr11
πr21
k!nK
r1,r2
(2π)knK
xkDk+1/2
Kk!nK·1
|lims→−ksζK(1 −s)|(2π)knK
kxkDk+1/2
Kk!nK
.
As in the proof of Lemma 3·13, since f(s)=g(s)f(s), we have that f(j)(s)/f(s)isa
polynomial of degree at most jin the first jderivatives of g. This finishes the proof.
5·3. Proof of the explicit formula.
To prove Proposition 2·12, the explicit formula for MK(x), we need a rather precise trun-
cated form of Perron’s formula. This can be found by combining the statements of [Tit86,
lemma 3·12] and [Ng04, lemma 2], and a similar result can be found in [MV07, corollary
5·3].
LEMMA 5·5. Suppose f (s)=∞
n=1ann−sis absolutely convergent for Re(s)>1, and let
(n)be a positive and non-decreasing function such that an(n). Assume that
∞
n=1
|an|
nσ=O1
(σ−1)α
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On a mertens-type conjecture for number fields 35
as σtends to 1 from above. Pick w =u+iv and c >0such that u +c>1, and pick T >0.
Then, for x >0,
n≤x
an
nw=1
2πic+iT
c−iT
f(w+s)xs
sds +Oxc
T(u+c−1)α+(2x)x1−ulog x
T+E1(x,T),
where
E1(x,T)(2x)x−u
if x >1, and, if N is the nearest integer to x (besides possibly x itself),
E1(x,T)(N)x1−u
T|x−N|
for any x >0.
We now give a bound on the integral found in our application of Perron’s formula. This is
done separately from the main proof of the explicit formula in order to elucidate the locations
where the appropriate error terms arise.
LEMMA 5·6. Fix a number field K, and let Tbe as in Lemma 2·4. For any x >0,0<T∈T,
and 1<c≤2, write
1
2πic+iT
c−iT
xs
sζK(s)ds =
|γ|<T
xρ
ρζK(ρ)+∞
k=0
Ress=−k
xs
sζK(s)+E2(x,c,T).
Then, for any >0,E
2(x,c,T)=Ox(1/T1−). When x >1,E
2(x,c,T)=O(xc/T1−)uni-
formly.
Proof. Pick a sufficiently large positive half-integer U(we will let U→∞), and integrate
around the rectangular contour with vertices c±iT and −U±iT. This gives
1
2πic+iT
c−iT
xs
sζK(s)ds
=
|γ|<T
xρ
ρζK(ρ)+U
k=0
Ress=−k
xs
sζK(s)+1
2πi−U−iT
c−iT +−U+iT
−U−iT +c+iT
−U+iT xs
sζK(s)ds.
(5·6)
We first bound these integrals using the functional equation for ζK(s) (Theorem 2·1). First,
we recall the identities and bounds
(s)(1 −s)=π
sin πs
(s)−1eσ−(σ−1/2) log σ+π|t|/2
|sin πs|−1e−π|t|
sin πs
2±1e±π|t|/2
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36 DANIEL HUet al.
ζK(s)−11forσ>2,
which hold for sbounded away from integers. As the paths of the three integrals in (5·6)
always stay a distance of 1/4 away from each integer, we have for son such a path, and with
real part not between −1 and 2, that
xs
sζK(s)K
(xDK/(2π)nK)σ
TenK[1−σ−(1/2−σ)log(1−σ)].(5·7)
Thus, the middle integral is asymptotically at most xUe−Ulog U+O(U)uniformly in xand Tas
U→∞. For the two horizontal integrals, we first bound the portion where Re(s) runs from
−1to−U. Here, we may use (5·7). Since the integral in the right-hand side of
−U+iT
−1+iT
xs
sζK(s)ds 1
T−U
−1xDK
(2π)nKσ
enK[1−σ−(1/2−σ)log(1−σ)]dσ
converges as U→∞, this is O(1/T) for fixed x, and for x>1 the dependence is O(x−1). As
a result, we may write
1
2πic+iT
c−iT
xs
sζK(s)ds
=
|γ|<T
xρ
ρζK(ρ)+∞
k=0
Ress=−k
xs
sζK(s)+1
2πi−1−iT
c−iT +c+iT
−1+iT xs
sζK(s)ds +E3(x,T),
where E3(x,T)=Ox(1/T)forx>0, and E3(x,T)=O(1/xT )forx>1. Finally, for the
remaining integrals, we use the definition of T. Since |ζK(σ±iT)|−1Tuniformly in
−1≤σ≤2forT∈Tand any >0, we have
xs
sζK(s)max (xc,1/x)
T1−
for any son one of our remaining line segments. This bounds our integrals by
max (xc,1/x)/T1−, which is enough.
We are now ready to show the explicit formula for MK(x).
Proof of Proposition 2·12. First, note that |μK(n)|is bounded by the coefficient of n−sin the
Dirichlet series expansion of ζK(s). As a result,
∞
n=1
|μK(n)|
nσ=O1
σ−1as σ→1+,
since the pole of ζK(s)ats=1 is simple. Define
(x)=x
C
log log (x+10) ,
where Cis as in Lemma 2·2, so that (x) is positive, non-decreasing, and satisfies μK(n)
(n). Now, apply 5.5 to f(s)=1/ζK(s), w=0, α=1, T>0, and with 1 <c≤2 to be chosen
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On a mertens-type conjecture for number fields 37
later. This gives
MK(x)=1
2πic+iT
c−iT
xs
sζK(s)ds +Oxc
T(c−1) +(2x)xlog x
T+E1(x,T),
where E1(x,T) is as in the statement of Lemma 5·5. By Lemma 5·6,wehave
MK(x)=
|γ|≤T
xρ
ρζK(ρ)+∞
k=0
Ress=−k
xs
sζK(s)
+Oxc
T(c−1) +(2x)xlog x
T+E1(x,T)+E2(x,c,T).
This is enough to show our inequality on the error term when x>0, as each term in the
above is asymptotically bounded by T−(1−)(any 1 <c≤2 will suffice). When x>1, we
have
E(x,T)xc
T(c−1) +(2x)xlog x
T+xc
T1−+(2x).
Setting c=1+( log x)−1gives
E(x,T)x1+C/log log xlog x
T+x
T1−+xC/log log x.
Finally, Lemma 2·13 implies that the sum of the residues at negative integers is convergent
as a series in k, and its dependence on xis O(( log x)r1+r2) depending on K.
6. Speculations on one-sided growth of Mertens functions
We have mostly avoided the discussion of the fields Kfor which ζK(s) has nontrivial
non-simple zeros. Although these fields are not problematic if the goal is to disprove the
naïve Mertens-type conjecture over K, they pose a distinct challenge when trying to study
properties like limiting distributions of normalisations of MK(x). In this setting, stronger
hypotheses are needed. In this section, we posit and motivate a particularly strange property
that Mertens functions of non-abelian number fields may satisfy.
CONJECTURE 6·1. There are infinitely many number fields K for which the Mertens
function MK(x)changes sign only finitely many times.
The heuristics behind this conjecture are covered in Section 6·3.
6·1. Complexities of non-abelian number fields
A main tool in studying the limiting distribution of MK(x)/√xis the explicit formula
(Proposition 2·12). This formula was derived via Perron’s formula, which allows one to
express MK(x) as a sum of residues of xs/sζK(s). If ρis a simple zero of ζK(s), then this
residue has a particularly simple formula and is of order x1/2as a function of x.However,
if ords=ρζK(s)=n>1, then this residue is a linear combination of terms x1/2( log x)kwhere
k<n. In particular, the coefficients of these terms involve large positive and negative powers
of higher derivatives of ζK(x), quantities whose moments are not well understood.
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38 DANIEL HUet al.
Assuming some standard conjectures, the zeros of ζK(s) for a non-abelian number field K
(with the possible exception of a zero at s=1/2) have multiplicity at most the largest dimen-
sion of an irreducible complex representation of Gal(K/Q). This motivates the following
question.
QUESTION 6·2. If the largest irreducible representation of Gal(K/Q)has dimension n, then
does MK(x)/(√xlogn−1(x)) have a limiting distribution?
Moreover, assume that the ζK(s) has a zero at s=1/2 of multiplicity larger than that
of any other nontrivial zero. As described in Proposition 2·7, this term does not oscillate
between large positive and negative values (like terms associated with non-real zeros), and
will dominate all others in the explicit formula for MK(x). Hence, one may ask the following
question.
QUESTION 6·3. If ords=1/2ζK(s)>ords=ρζK(s)for all ρ= 1
2, does MK(x)change sign
finitely many times?
One may ask if such number fields satisfying the hypothesis of this question even exist.
This is answered in the affirmative, assuming appropriate standard conjectures, by the work
of Louboutin [Lou00b].
It now remains to identify number fields Kof interest, as well as give heuristic bounds on
certain sums over zeros, which we turn to now. In what follows, it will be relevant to state the
following assumptions on Artin L-functions, for a Galois extension of number fields L/K:
CONJECTURE 6·4. (Artin holomorphy conjecture (AHC)). For any character χof a
nontrivial irreducible representation of Gal(L/K), the function L(s,χ,L/K)is holomorphic.
CONJECTURE 6·5. (Simplicity hypothesis (SHK/Q)). If χis an irreducible character of
Gal(K/Q), then the nontrivial zeros of L(s,χ,K/Q)are simple.
CONJECTURE 6·6 (Independence conjecture (ICK/Q)). If χ1,χ2are distinct irreducible
characters of Gal(K/Q), then L(s,χ1,K/Q)and L(s,χ2,K/Q)share no nontrivial zeros,
except possibly for a zero at s =1/2.
6·2. Candidate number fields.
The first appearance of a number field Ksatisfying ζK(1
2)=0 occurred in a paper of
Armitage [Arm72], who studied a degree 48 extension of Qwhich had first appeared in
aworkofSerre[Ser71]. Later, Serre (unpublished, see [Ng00, chapter 2]) discovered the
simpler example
K=Q (5 +√5)(41 +√205),
which is Galois with quaternionic Galois group Q8.
In emulation of this latter construction, we identify as candidates for Gal(K/Q) the
following well-known class of groups to which Q8belongs. These groups possess conve-
nient representation-theoretic properties, which allow control over the behaviour of ζK(s)at
s=1/2.
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On a mertens-type conjecture for number fields 39
Definition 6·7. Let n≥2 an integer. We define the dicyclic group Q4nby the
presentation
Q4n=a,b:a2p=1, ap=b2,b−1ab =a−1.
This is a solvable non-abelian group of order 4n.
In what follows, let p≥3 be an odd prime, and let N/Qbe a normal extension with Galois
group G=Q4p, called a dicyclic number field of degree 4p. We recall the Galois theory of
such fields, which in be found in the introduction to [Lou00b].
If pis not totally ramified in N/Q, then by [Lou00b, theorem 1], the root number
WN/Q(ψ)=WN/L(χ) does not depend on χbut just on the field N, so either there are
(p−1)/2 two-dimensional characters with root number −1 or there are none. Hence, assum-
ing SHN/Q, the order of the zero at s=1/2 of the Dedekind zeta function is either p−1or
zero. Moreover, there exist infinitely many examples of dicyclic fields Nwhich produce
either case. If pis totally ramified in N/Q, which can only occur when p≡1 (mod 4), and
if L=Q(√p), then half of the (p−1)/2 quaternionic characters have root number −1 and
half have root number +1. Hence, assuming SHN/Q, the order of the zero at s=1/2is
(p−1)/2.
Now that we have built up the multiplicity of the zero at s=1/2, we invoke the condi-
tions SHN/Qand ICN/Qin order to tame the multiplicities of all the other nontrivial zeros.
Equipped with these assumptions, the largest dimension of an irreducible representation of a
dicyclic number field would be 2, so that all other zeroes besides s=1/2 should have order
at most 2.
THEOREM 6·8. Let p ≥3be a prime and let N be a dicyclic number field of order 4p.
Assume SHN/Q.
(a) If p is not totally ramified in N/Q, and W (χ)=−1for all characters χon H of order
2p, then the order of the zero of ζN(s)at s =1/2is p −1.
(b) If p is totally ramified in N/Q, and L =Q(√p), then the order of the zero of ζN(s)at
s=1/2is (p−1)/2.
Assuming moreover ICN/Q, and for psufficiently large, this order is higher than that of
any other nontrivial zero.
Moreover, there are infinitely many examples of Nas in case (a) assuming SHN/Q.
6·3. Heuristics in support of one-sided growth of MK(x)
From now on, we shall specialise to the scenario of Theorem 6·8, where K:=N. Hence,
Kis a dicyclic number field satisfying SHK/Qand ICK/Q, with the function ζK(s) having a
zero of multiplicity μ≥3ats=1/2 and all other nontrivial zeros of multiplicity equal to
either 1 (simple) or 2 (double). In the notation of Proposition 2·12, for all Tn∈Tand x>1
we have, under the Riemann hypothesis for ζK(s),
MK(x)=Ress=1/2
xs
sζK(s)+
|γ|≤Tn
ρsimple
Ress=ρ
xs
sζK(s)+
|γ|≤Tn
ρdouble
Ress=ρ
xs
sζK(s)+E(x,Tn).
(6·1)
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40 DANIEL HUet al.
As noted already, the leading term above grows one-sidedly in xas x1/2( log x)μ−1,sowe
seek to restrain the growth of the remaining terms. We have
Ress=ρ
xs
sζK(s)=xρ
ρζK(ρ)and Ress=ρ
xs
sζK(s)=2xρ
ρζK(ρ)log x−1
ρ−ζK(ρ)
ζK(ρ),
when s=ρis a simple and double zero, respectively. In the former case, one suitable bound
would be a generalised form of (1·3) which is applicable to Dedekind zeta functions having
nontrivial zeros of multiplicity, but which targets only the simple ones, namely
0<γ ≤T
ρsimple
1
|ζK(ρ)|2T,
give or take the addition of a low power of log Tto the right-hand side. In this case, we may
replace the Tnin the second term of (6·1) with any T>0, upon inserting an additional error
term of
xlog T
T1/2
.
A similar estimate pertaining to the double zeros,
0<γ ≤T
ρdouble
1
|ζK(ρ)|2T,
and an estimate of the form
0<γ ≤T
ρdouble
|ζK(ρ)|
|ρ||ζK(ρ)|2( log T)a,
with aa reasonably small integer, would also come in handy, once again allowing for the
replacement of Tnwith any T>0in(6·1) in exchange for residual error terms. Finally,
following the arguments in the proof of [Ng04, theorem 1] would produce the conclusion of
Conjecture 6·1, assuming that the resulting contribution of extra powers of log xhas been
smaller than ( log x)μ−1.
7. Conclusions and conjectures
In this paper, we have disproved the naïve Mertens-type conjecture for most number fields,
and have conditionally shown the existence of limiting distributions for abelian number
fields. There are, however, a number of questions prompted by our results.
First, as discussed in Section 6, it is possible that the Mertens function of certain non-
abelian number fields may change sign only finitely many times – a surprising departure
from the behaviour over Q. Furthermore, the question of existence of limiting distributions
for non-abelian number fields was also posed.
We have also shown in Section 3that for a fixed signature, M+
K>1 for all but finitely many
number fields given conventional assumptions on the zero structure of ζK(s). This has been
achieved by bounding from below the maxima of the function x−1/2(MK(x)+M∗
K(x)), which
suffices for a disproof under the Riemann hypothesis and the simplicity of nontrivial zeros.
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On a mertens-type conjecture for number fields 41
As a companion to these findings, one may ask instead for upper bounds on the minima of
this function, a question which pertains to M−
K.
There are also several questions prompted by the existence of the limiting distributions
νK(Theorem 1·4). Although the Mertens conjecture has been known to be false, numerical
evidence suggests that the bound |MQ(x)|≤√xholds most of the time (in the sense of
logarithmic density). Namely, if νKis the logarithmic limiting distribution of x−1/2MK(x)as
in Theorem 1·4, then preliminary computations suggest that νQ([ −1, 1]) ≥0.99999993366
(cf. [Hum14]). Nevertheless, it has not yet been proven that νQ([ −1, 1]) >0. This motivates
the following two questions.
QUESTION 7·1. How does νK([ −1, 1]) vary with K?
QUESTION 7·2. Let βK=inf{α>0|νK([ −α,α]) ≥0.5}. How does βKvary with K?
In other words, is the statement |MK(x)|≤√xcorrect more or less frequently as the
degree and discriminant of Kincrease? Alternatively, one could study the asymptotic
behaviour of the distributions νK— in particular, the behaviour of its tails. In [Ng04, section
4·3], Ng shows (assuming some conjectures on J−1(T) and J−1/2(T)asin(1·2)) that there
are constants c1,c2so that
exp −exp (c1V4
5)νQ([V,∞)) exp −exp (c2V4
5).
In particular, νQ([V,∞)) >0 for all V, meaning MQ(x)>V√xa positive proportion of the
time. That is, counterexamples to the Mertens conjecture over Qhave positive logarithmic
density. One may ask how these asymptotics generalise to the settings of number fields.
Thus, we pose the following two questions.
QUESTION 7·3. For an abelian number field K, are there c1,c2so that
exp −exp (c1V4
5)νK([V,∞)) exp −exp (c2V4
5)?
QUESTION 7·4. If Question 7·3is answered affirmatively, how should the constants vary
with K?
Acknowledgements. We are deeply grateful to Peter Humphries for supervising this
project and to Ken Ono for his valuable suggestions. We would also like to thank Winston
Heap, David Lowry Duda and Micah Milinovich for helpful discussions. We are grateful
for the generous support of the National Science Foundation (Grants DMS 2002265 and
DMS 205118), National Security Agency (Grant H98230-21-1-0059), the Thomas Jefferson
Fund at the University of Virginia, and the Templeton World Charity Foundation. This
research was conducted as part of the 2021 Research Experiences for Undergraduates at
the University of Virginia.
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