THE CHOWLA PROBLEM ANDITS GENERALIZATIONS

Siddhi Pathak

A thesis submitted in partial fulfillment of therequirements for the degree of Master of Science in

Mathematics and Statistics

Department of Mathematics and StatisticsQueen’s University

August, 2015

Copyright © Siddhi Pathak, 2015

Abstract

In this thesis, we study the vanishing of certain L-series attachedto periodic arithmetical functions. Throughout the writeup, we letf be an algebraic-valued (at times even rational-valued) arithmeticalfunction, periodic with period q. Define

L(s, f) :=∞∑n=1

f(n)

ns.

We discuss a conjecture of Sarvadaman Chowla, made in the early1960’s of the non-vanishing of L(1, f) for rational-valued functions f .We further discuss the theorem of Baker-Birch-Wirsing that classi-fies all odd algebraic-valued, periodic arithmetical functions f withL(1, f) = 0. We then apply a beautiful result of Bass to give a nec-essary condition for even algebraic-valued, periodic arithmetical func-tions f , to satisfy L(1, f) = 0. The sufficiency condition obtained viathis method is unfortunately not very clean. This, in view of a theoremof Ram Murty and Tapas Chatterjee, completes the characterization ofall algebraic-valued periodic arithmetical functions f with L(1, f) = 0.

In the next part of the thesis, we discuss the Lerch-zeta functionand its functional equation. We also deduce the transcendence of itscertain special values. Keeping this in mind, we define a new L-seriesattached to periodic functions and deduce a necessary condition for thevanishing of this L-series at s = 1.

Finally, we mention some research topics that were stumbled uponduring the course of this study and a few partial results regarding thesequestions.

i

Acknowledgment

First and foremost, I offer my sincerest gratitude to my supervisor,Prof. Ram Murty for guiding me through this project. He inspiredme to work, patiently corrected my mistakes and helped me in everyway possible. Besides my advisor, I would like to thank the rest of myexamining committee: Prof. Jamie Mingo and Prof. Francesco Cel-larosi for their insightful comments. I take this opportunity to extendmy gratitude to Prof. Purusottam Rath from Chennai MathematicalInstitute, for bringing to my notice an important hypothesis necessaryfor a result.

I thank my batchmates for stimulating discussions and maintaininga friendly and cheerful atmosphere. I would like to specially thankKannappan Sampath for helping me to learn LaTex and to draw adiagram for this report. I am also grateful to Akshaa Vatwani, whohas been a very helpful and kind friend throughout. Last but not theleast, I am deeply thankful to my parents whose constant support madeit possible for me to stay away from home and complete my studies.

ii

Contents

Abstract i

Acknowledgment ii

Introduction iv

Chapter 1. Preliminaries 11.1. L-series attached to a periodic function 11.2. Expressing L(1, f) as a linear form in logarithms of

algebraic numbers 21.3. Baker’s Theorem 31.4. Linear forms in logarithms of positive algebraic numbers 4

Chapter 2. The Chowla-Siegel theorem 62.1. Proof of Chowla’s theorem 82.2. Non-vanishing for even functions 112.3. Resolution of Chowla’s problem 12

Chapter 3. A vanishing criterion for L(1, f) 143.1. Odd functions 143.2. Even functions 163.3. Combining the even and the odd 20

Chapter 4. The Lerch zeta function 234.1. Analytic continuation 254.2. Functional Equation 284.3. Special values 30

Chapter 5. Generalization of the Chowla problem 335.1. Convergence of L (1, f, α) 335.2. Non-vanishing of L (1, f, α) 34

Chapter 6. Topics for further research 386.1. Transcendence of values of Dirichlet L-functions at 1/2. 386.2. Analogue of the Dedekind determinant 41

Bibliography 44

iii

Introduction

Non-vanishing of L-functions has been very thoroughly studied innumber theory since a long time. One of the most remarkable the-orems of this kind in the past 150 years is due to Dirichlet, namely,L(1, χ) 6= 0 for a non-principal Dirichlet character χ. As a general prin-ciple in mathematics, many concepts become clearer on generalization.To unravel the mystery surrounding Dirichlet’s theorem, SarvadamanChowla [9] made the following conjecture in the early 1960s.

Conjecture. Let f be a rational-valued arithmetical function,periodic with prime period p. Further assume that f(p) = 0 and

p∑a=1

f(a) = 0.

Then,∞∑n=1

f(n)

n6= 0,

unless f is identically zero.

Chowla proved this conjecture in the case of odd functions i.e, f(p−n) = −f(n) based on an outline of the proof by Siegel [9].

The complete resolution of Chowla’s conjecture in a wider settingwas given by Baker, Birch and Wirsing in 1972 [5]. They proved thefollowing general theorem:

Theorem. If f is a non-vanishing function defined on the integerswith algebraic values and period q such that (i) f(n) = 0 whenever1 < (n, q) < q and (ii) the qth cyclotomic polynomial Φq is irreducibleover Q(f(1), f(2), · · · , f(q)), then

∞∑n=1

f(n)

n6= 0.

Let us observe that in the case of Chowla’s conjecture, condition(i) is vacuous as q is prime and f(q) = 0. The condition (ii) is alsosatisfied as f is rational-valued and Φq is irreducible over Q. Thus, theBaker-Birch-Wirsing theorem implies Chowla’s conjecture.

In their paper [5], the authors mention that Chowla communicatedto them that he had also solved the problem “to the extent mentionedabove”. It is although unclear as to what exactly Chowla had in mind.

iv

Using Baker’s theory of linear forms in logarithms in a minimal wayand the methods known to Chowla, one can prove Chowla’s conjecturefor even functions. This was done by Ram Murty [17] recently. Onecan therefore guess that this was the proof that Chowla had in mindwhen he wrote to Baker, Birch and Wirsing. We mention the proofs ofChowla [9] and Ram Murty [17] in the first chapter. This gives us anidea of the intricacies of the topic at hand.

Chowla’s question can be asked in more generality as follows: Fix apositive integer q. Let f be an algebraic-valued arithmetical function,periodic with period q. It is useful to define an L-function associatedto the function f , namely,

L(s, f) =∞∑n=1

f(n)

ns.

Using the analytic continuation of the Hurwitz zeta function, we candeduce that L(1, f) exists if and only if

∑qa=1 f(a) = 0. Thus, we can

ask the following question: if f is not identically zero, then is it truethat

∞∑n=1

f(n)

n6= 0 ?

The answer to this in such a generality turns out to be negative [5].As an example, consider the function f defined such that

∞∑n=1

f(n)

ns= (1− p(1−s))

2ζ(s). (0.1)

In particular,

f(n) =

1 if (n, p) = 1,

1− 2p if p|n, p2 - n,(p− 1)2 otherwise.

Note that the function f is periodic with period p2. Taking limit of theright hand side of (0.1) as s→ 1, we get

L(1, f) = 0,

because ζ(s) has a simple pole at s = 1.In their paper, [5], Baker, Birch and Wirsing also gave a charac-

terization of all odd algebraic-valued periodic arithmetical functions fthat satisfy L(1, f) = 0. Since their argument is short and elegant,we describe it in the first part of the second chapter. Their approachsuggests a change in perspective. Instead of trying to prove the non-vanishing of an expression, we will try to characterize the functions ffor which L(1, f) = 0.

Let f be any function. Then it can be written as the sum of aneven function and an odd function as follows. Define fo(a) := [f(a)−f(−a)]/2 and fe(a) := [f(a) + f(−a)]/2. Then clearly f = fo + fe,

v

where fo is the odd part of f and fe is the even part of f . In 2014,Ram Murty and Tapas Chatterjee [8] made an important observation.They proved,

Theorem. For a periodic, algebraic-valued arithmetical functionf ,

L(1, f) = 0 ⇐⇒ L(1, fo) = 0 and L(1, fe) = 0.

In the view of this theorem, it is enough to consider even algebraic-valued periodic arithmetical functions to understand the vanishing ofL(1, f). In the next part of the second chapter, we apply a beautifulresult of Bass [6] to obtain a set of functions that act as buildingblocks for even algebraic-valued periodic arithmetical functions f withL(1, f) = 0. This completes to an extent the characterization we wereafter.

In the history of number theory, various zeta functions have played acrucial role. A couple of the most extensively studied ones are probablythe Riemann zeta and the Hurwitz zeta functions. As a generalizationof these, Lerch (in 1887) and Lipschitz (in 1889) independently intro-duced another zeta function, which is more commonly known as theLerch zeta function. The Hurwitz zeta function has one real parameterwhereas the Lerch zeta function has two real parameters. We introducethis zeta function in the third chapter and include a very sleek proof ofits functional equation given by B. Berndt [7]. We also mention a resultof Apostol [2] wherein he explicitly computes the special values of theLerch zeta function at negative integers and zero. As a consequence ofthis and the functional equation, we derive the transcendence of valuesof the Lerch zeta function at strictly positive integers when both thereal parameters involved are rational.

Motivated by the definition of the Lerch zeta function, we define anew L-series attached to periodic arithmetical functions in the fourthchapter. Given such a function f , we define L (s, f, α) for 0 < α < 1.We obtain a condition for the convergence of this series at s = 1.Using a simple observation, we express L (1, f, α) as a linear form inlogarithms of algebraic numbers. Finally, using Baker’s theory of linearforms in logarithms, we obtain a necessary condition for L (1, f, α) = 0.

In the course of this study, we stumbled upon a few unsolved andinteresting questions. We were able to partially answer a few. Wemention these questions and the partial answers as topics for furtherresearch in the last chapter.

vi

CHAPTER 1

Preliminaries

The aim of this chapter is to introduce notation and some funda-mental results that will be used in the later part of the thesis.

1.1. L-series attached to a periodic function

Let us first recall a few facts about the Hurwitz zeta function. For0 < x ≤ 1, define the Hurwitz zeta function, ζ(s, x) as

∞∑n=0

1

(n+ x)s.

This series converges absolutely for <(s) > 1. In 1882, Hurwitz [12]obtained the analytic continuation and functional equation of ζ(s, x).He proved:

Theorem 1.1. The Hurwitz zeta function, ζ(s, x) extends analyti-cally to the entire complex plane except for a simple pole at s = 1 withresidue 1. In particular,

ζ(s, x) =1

s− 1−Ψ(x) +O(s− 1),

where Ψ is the Digamma function.

Let q be a fixed positive integer. Consider f : Z→ Q, periodic withperiod q. It is useful to define an L-function attached to f , namely

L(s, f) =∞∑n=1

f(n)

ns.

Let us observe that L(s, f) converges absolutely for <(s) > 1. This istrue because f(n) is bounded and

∑∞n=1

1ns

converges for <(s) > 1.

Since f is periodic,

L(s, f) =

q∑a=1

f(a)∞∑k=0

1

(a+ kq)s

=1

qs

q∑a=1

f(a)ζ

(s,a

q

),

(1.1)

1

where ζ(s, x) is the Hurwitz zeta function. Using Theorem 1.1, we canconclude that L(s, f) can be extended analytically to the entire com-plex plane except for a simple pole at s = 1 with residue 1

q

∑qa=1 f(a) .

Thus,∑∞

n=1f(n)n

exists whenever∑q

a=1 f(a) = 0, which we will assumehenceforth. Hence, L(s, f) is an entire function. Moreover, L(1, f)can be expressed as a combination of values of the Digamma function.Theorem 1.1 implies

L(1, f) = −1

q

q∑a=1

f(a)Ψ

(a

q

).

1.2. Expressing L(1, f) as a linear form in logarithms ofalgebraic numbers

Let us first define the Fourier transform of f . Given a function fwhich is periodic with period q,

f(x) :=1

q

q∑a=1

f(a)ζ−axq ,

where ζq = e2πi/q. This can be inverted using the identity

f(n) =

q∑x=1

f(x)ζxnq . (1.2)

Thus, the condition for convergence of L(1, f) derived in the earlier

section, i.e,∑q

a=1 f(a) = 0 can be interpreted as f(q) = 0.

Substituting (1.2) in the expression for L(s, f) we have,

L(s, f) =∞∑n=1

1

ns

q∑x=1

f(x)ζxnq

=

q∑x=1

f(x)∞∑n=1

ζxnqns.

Recall the partial summation or the Abel summation formula that says:

Theorem. Let an be a sequence of complex numbers and f be C1

function on R>0. For x > 0, if A(x) :=∑

n≤x an, then∑1≤n≤x

anf(n) = A(x)f(x)−∫ x

1

A(t)f ′(t)dt.

2

The Abel summation formula, along with the assumption that f(q) =0, helps us to conclude that

L(1, f) = −q−1∑x=1

f(x) log(1− ζxq ), (1.3)

where log is the principal branch.

1.3. Baker’s Theorem

Regarded as one of the most remarkable theorems of the 20th cen-tury number theory, Baker’s theorem on linear forms in logarithms isa very useful tool in proving the transcendence of many complex num-bers. We will use it to prove that a certain linear form in logarithmsof algebraic numbers is transcendental and hence non-zero. The state-ment of the theorem is

Theorem 1.2. If α1, α2, · · · , αn are non-zero algebraic numbers,such that logα1, logα2, · · · , logαn are Q-linearly independent, then 1, logα1, · · · ,logαn are Q-linearly independent.

A re-statement of Baker’s theorem is

Theorem 1.3. Any linear form in logarithms of algebraic numbersis either zero or transcendental. In other words, for any non-zero al-gebraic numbers α1, · · · , αn and any algebraic numbers β0, β1, · · · , βnwith βo 6= 0 we have,

β0 + β1 logα1 + · · ·+ βn logαn 6= 0. (1.4)

A useful corollary of this statement is the following.

Corollary 1.4. If α1, α2, · · · , αn are algebraic numbers differentfrom 0 and 1, and β1, β2, · · · , βn are Q-linearly independent algebraicnumbers then,

β1 logα1 + β2 logα2 + · · ·+ βn logαn

is non-zero and hence, transcendental.

Proof. We will prove the above statement by induction on n.Since α1 is algebraic, not equal to 1 and β1 6= 0, the statement istrue for n = 1.

Suppose that the statement is true for all positive integers m,m < n. If logα1, · · · , logαn are linearly independent over Q, thenthe statement is a consequence of Theorem 1.2. Hence, suppose thatthe numbers logα1, · · · , logαn satisfy a relation over the rationals. Inparticular, there exist rational numbers b1, · · · , bn with say br 6= 0, suchthat

b1 logα1 + · · ·+ bn logαn = 0.

3

Thus,

br(β1 logα1 + β2 logα2 + · · ·+ βn logαn)

= β′1 logα1 + β′2 logα2 + · · ·+ β′n logαn, (1.5)

where

β′j = brβj − βrbj,for 1 ≤ j ≤ n. The right hand side of (1.5) is a linear form in logarithmsof at most n− 1 algebraic numbers as β′r = 0. Since β1, · · · , βn are Q-linearly independent, the set {β′j|1 ≤ j ≤ n, j 6= r} is also Q-linearlyindependent. Thus, by induction, the right hand side of (1.5) is notzero. �

1.4. Linear forms in logarithms of positive algebraic numbers

The following lemma is an immediate consequence of Baker’s theo-rem and will be used many times.

Lemma 1.5. Let α1, · · · , αn be positive algebraic numbers. Ifc0, · · · , cn are algebraic numbers and c0 6= 0, then

c0π +n∑j=1

cj logαj

is non-zero and hence, a transcendental number.

Proof. Suppose that T := {logα1, · · · , logαr} is a maximal Q-linearly independent set of {logα1, logα2, · · · , logαn} after re-labelingif necessary. Let us chose a branch of log such that

iπ = log(−1). (1.6)

On multiplying by i, the given linear form can be re-written as

c0 log(−1) +r∑j=1

dj logαj, (1.7)

for some algebraic numbers dj.

Suppose b0, b1, · · · , br are rational numbers such that b0 6= 0 and thelinear form

b0iπ +r∑j=1

bj logαj = 0.

Thus,

b0iπ = −r∑j=1

bj logαj. (1.8)

4

Observe that the right hand side of the above equation is purely realwhereas the left hand side is purely imaginary. This is possible only ifboth the sides are zero. In this case,

r∑j=1

bj logαj = 0. (1.9)

This contradicts the Q-linear independence of the set T .Therefore, the set {log(−1), logα1, · · · , logαr} is Q-linearly indepen-dent. By Baker’s Theorem 1.2, {1, log(−1), logα1, · · · , logαr} is Q-linearly independent. Thus, (1.7) is non-zero and hence transcenden-tal. �

Remark. This proof also goes through when the branch of loga-rithm chosen in (1.6) is the principal branch. In that case, we replaceiπ by 2 log i and proceed as before.

5

CHAPTER 2

The Chowla-Siegel theorem

In the early 1960s, Sarvadaman Chowla [9] considered the followingproblem: Let p be a prime number and f be a rational-valued func-tion on the natural numbers that is periodic with period p and notidentically zero. Assume further that f(p) = 0 and that

p∑n=1

f(n) = 0. (2.1)

Then is it true that

∞∑n=1

f(n)

n6= 0 ?

In this context, let us define

L(s, f) :=∞∑n=1

f(n)

ns.

Chowla’s question can then be reformulated as determining the non-vanishing of L(1, f) under the conditions f(p) = 0 and (2.1). As seen

earlier, (2.1) implies the convergence of the series∑∞

n=1f(n)n

.Following an argument outlined by Siegel, Chowla proved this con-

jecture in the case that f is an odd function, i.e, when f(p−n) = −f(n).We present his proof in this chapter. We will also observe that his ar-gument can be generalized with minor variations. The main idea hereis to use the well-known cotangent expansion

π cot πz =∑n∈Z

1

n+ z, (2.2)

where z /∈ Z. We interpret the right hand side as

1

z+ lim

N→∞

N∑n=1

1

z + n+

1

z − n.

Chowla proved:

Theorem 2.1. Let f be a rational-valued arithmetical function,periodic with prime period p. Suppose f is not identically zero and

6

f(p) = 0. Further assume that f is odd. Then,

∞∑n=1

f(n)

n6= 0.

It was observed by Ram Murty [17] that the argument for the proofof the above theorem can be applied to a wider context. It can be usedto prove:

Theorem 2.2. Let q be a positive integer. Let K be an an algebraicnumber field which is disjoint from the qth cyclotomic field. Let f be aK-valued arithmetical function, periodic with period q. Further supposethat f is odd and that f(n) = 0 whenever (n, q) > 1. Then,

∞∑n=1

f(n)

n6= 0

unless f is identically zero.

In a paper in 2011, Ram Murty [17] answered Chowla’s questionwhen f is an even function. In this paper, he tried to give a shortargument towards the resolution of Chowla’s problem which may ap-proximate the proof Chowla himself had in mind when he had writtento Baker, Birch and Wirsing that he had solved the problem “to theextent stated above”. The ideas used in the proof were familiar to himand draw upon his earlier work. We include Ram Murty’s proof ofChowla’s conjecture when f is an even function.

These theorems put together settle Chowla’s problem completelydue to the lemma below. Before proceeding, let us fix the followingnotation: Let f be any given function. The even part of f is definedas

fe(x) =f(x) + f(−x)

2.

Similarly, define the odd part of f as

fo(x) =f(x)− f(−x)

2.

Observe that fe is even, fo is odd and that f = fe + fo.

Lemma 2.3. Suppose f is a rational-valued arithmetical functionthat is periodic with period p. Suppose f(p) = 0 and

∑pn=1 f(n) = 0.

Let fo denote the odd part of f and fe denote the even part of f . Then,

L(1, f) = 0 ⇐⇒ L(1, fe) = 0 and L(1, fo) = 0.

Remark. We will prove a similar result in the next chapter forfunctions that are periodic with period q, where q is not necessarilyprime and f(q) is not necessarily zero.

7

2.1. Proof of Chowla’s theorem

Let us observe that for q = 2, the conditions f(2) = 0 and f(1) +f(2) = 0 imply that f is identically zero. Thus, we assume q > 2.

Proof. Since f is odd and f(q) = 0,

2

q∑n=1

f(n) =

q∑n=1

[f(n) + f(q − n)] = 0.

Thus L(1, f) exists.Now, suppose L(1, f) = 0. For a positive integer k such that

(k, q) = 1, define

Sk =∑n∈Zn 6=0

f(kn)

n,

where the sum is to be interpreted as

limN→∞

N∑n=1

f(kn)

n+f(−kn)

(−n).

Thus,

S1 = 2∞∑n=1

f(n)

n. (2.3)

We will show that

S1 = 0⇒ f is identically zero.

Since f(q) = 0,

Sk =

q−1∑a=1

f(ka)∑

n≡a(modq)n6=0

1

n.

We view the inner sum as∑n≡a(modq)

n 6=0

1

n=

1

a+ lim

N→∞

N∑n=1

1

a+ nq+

1

a− nq

=1

a+

1

qlimN→∞

N∑n=1

1

a/q + n+

1

a/q − n

=π

qcot

(πa

q

).

Now,

π

qcot

(πa

q

)=iπ

q

(ζaq + 1

ζaq − 1

)= −2π

iq

(1

2+

1

ζaq − 1

),

8

where ζaq = e2πia/q. Thus,

−iSk =

q∑a=1

f(ka)2π

q

(1

2+

1

ζaq − 1

).

Since f is supported only on the co-prime residue classes mod q,

q∑a=1

(a,q)=1

f(a) =

q∑a=1

(a,q)=1

f(ka) = 0,

for (k, q) = 1. Hence,

−iq2π

Sk =

q∑a=1

f(ka)

ζaq − 1.

In particular, for k = 1,

−iq2π

S1 =

q∑a=1

f(a)

ζaq − 1. (2.4)

Let σk be a Galois automorphism that sends ζq to ζk′q for kk′ ≡ 1

(mod q). Observe that the right hand side of (2.4) is algebraic. Thus,applying σk to the expression in (2.4), we have

σk

(−iq2π

S1

)=

q∑a=1

f(a)

ζk′aq − 1

=

q∑a=1

f(k′ka)

ζk′aq − 1

=

q∑a=1

f(ka)

ζaq − 1

= − iq2πSk.

Therefore, S1 = 0 implies that Sk = 0 for all positive integers k thatare coprime to q. Note that we use the hypothesis of K being disjointfrom the qth cyclotomic field. Also observe that this computation canalso be used to calculate L(1, χ) when χ is an odd Dirichlet charactermod q. Thus,

− iq2πL(1, χ) =

q∑m=1

χ(m)

ζmq − 1. (2.5)

Now, Sk = 0 for all positive integers k such that, (k, q) = 1. Thisimplies that

0 =

q∑k=1

χ(k)Sk,

9

as χ is supported on coprime residue classes mod q. Thus,

0 =

q∑k=1

χ(k)Sk

=

q∑k=1

χ(k)

q∑a=1

f(ka)

ζaq − 1

=

q∑k=1

q∑a=1

f(ka)χ(ka)χ(a)

ζaq − 1.

Now put ka ≡ b (mod q) to get

0 =

q∑b=1

f(b)χ(b)

q∑k=1

χ(k′b)

ζk′bq − 1.

As k runs over all coprime residue classes mod q, so does k′b. Thus, by(2.5),

0 = L(1, χ)

( q∑b=1

f(b)χ(b)

).

The non-vanishing of L(1, χ) for non-principal Dirichlet characters leadsus to conclude that

q∑b=1

f(b)χ(b) = 0, (2.6)

for odd characters χ mod q. Let us observe that (2.6) holds for evenDirichlet characters too. This is because

q∑b=1

f(b)χ(b) =

q∑b=1

f(−b)χ(−b)

= −q∑b=1

f(b)χ(b).

Since a Dirichlet character is either odd or even, for all Dirichlet char-acters χ mod q,

q∑b=1

f(b)χ(b) = 0.

Hence,

0 =∑

χ mod q

χ(a)

( q∑b=1

f(b)χ(b)

)

=

q∑b=1

f(b)

( ∑χ mod q

χ(a)χ(b)

).

10

By orthogonality relations, the inner sum is φ(q) whenever b = a and0 otherwise. Thus,

0 = φ(q)f(a), for all a ∈ Z.Therefore, f is identically 0. �

2.2. Non-vanishing for even functions

In this section, we prove the following theorem about the non-vanishing of L(1, f) for even functions, f .

Theorem 2.4. Let f be an even rational-valued arithmetical func-tion, periodic with prime period p. Further assume that f(p) = 0.Then,

∞∑n=1

f(n)

n6= 0.

Let us note as before that we only need to consider the case whenp > 2.

Proof. By (1.3), we know that

∞∑n=1

f(n)

n= −

p−1∑x=1

f(x) log(1− ζxp ).

Since f is even, so is f , i.e, f(p− x) = f(x). Thus,

L(1, f) = −2

(p−1)2∑

x=1

f(x) log |1− ζxp |.

Now, 0 = f(p) =∑p

x=1 f(x) implies that∑(p−1)/2

x=1 f(x) = 0 as f iseven. Thus,

(p−1)2∑

x=1

f(x) log |1− ζp| = 0.

Hence,

L(1, f) = −2

(p−1)2∑

x=1

f(x) log

∣∣∣∣1− ζxp1− ζp

∣∣∣∣.Now observe that∣∣∣∣1− ζxp1− ζp

∣∣∣∣ =

∣∣∣∣e−iπx/p − eiπx/pe−iπ − eiπ

∣∣∣∣ =

∣∣∣∣sinπx/psinπ/p

∣∣∣∣.Therefore,

L(1, f) = −2

(p−1)2∑

x=1

f(x) log

(sin πx/p

sin π/p

). (2.7)

11

Remark. This expression for L(1, f) also holds in the case whenthe period of the function is not necessarily prime. Suppose f is anarithmetical function, periodic with period q and f(q) = 0. Then,

L(1, f) = −2

b (q−1)2 c∑

x=1

f(x) log

(sin πx/q

sin π/q

).

By Lemma 8.1 from [22], the algebraic numbers{sin πx/p

sin π/p: 1 ≤ x ≤ p− 1

2

}are Q-linearly independent. By Baker’s Theorem 1.2, they are Q-linearly independent. Thus, the right hand side of (2.7) is not zero.

Hence, L(1, f) = 0 ⇐⇒ f(x) = 0 for all 1 ≤ x ≤ p. This proves thetheorem. �

2.3. Resolution of Chowla’s problem

We have proved the non-vanishing of L(1, f) when f is either anodd or an even function. We can thus resolve Chowla’s question withthe help of the following lemma:

Lemma 2.5. Suppose f is a rational-valued function that is periodicmod p. Suppose f(p) = 0 and

∑pn=1 f(n) = 0. Let fo denote the odd

part of f and fe denote the even part of f . Then,

L(1, f) = 0 ⇐⇒ L(1, fe) = 0 and L(1, fo) = 0.

Proof. Given the function f , we can write it as a sum of an evenfunction and an odd function as seen before. Let f = fe + fo. Thus,

∞∑n=1

f(n)

n=∞∑n=1

fe(n)

n+∞∑n=1

fo(n)

n.

If L(1, fe) = 0 and L(1, fo) = 0, then clearly L(1, f) = 0.Now, suppose L(1, f) = 0. By the Chowla-Siegel theorem, the

first sum is an algebraic multiple of π. By (2.7), the second sum is alinear form in logarithms of positive algebraic numbers. If L(1, fo) 6= 0,then by Lemma 1.5, L(1, f) 6= 0. Thus, L(1, fo) = 0 and in turn,L(1, fe) = 0. �

This completes the solution to Chowla’s problem when the func-tion f is periodic with prime period p and f(p) = 0. The case whenf(p) 6= 0 can also be dealt with as follows.

Define a function g that is periodic with period p such that

g(1) = g(2) = · · · = g(p− 1) = 1

12

and g(p) = −(p− 1). Thus, for 1 ≤ a ≤ p− 1,

g(a) =1

p

p−1∑b=1

ζ−abp − (p− 1) = −1.

Thus by (1.3),

L(1, g) =

p−1∑a=1

log(1− ζap ).

Since,p−1∏a=1

(1− ζap ) = p,

we conclude that L(1, g) = log p.

Suppose f is an arithmetical function that is periodic mod p with∑pn=1 f(n) = 0 and f(p) 6= 0. Assume further that L(1, f) = 0. Now

definef = (p− 1)f + f(p)g.

Note that f(p) = (p− 1)f(p)− f(p)(p− 1) = 0. Also observe that

L(1, f) = f(p)L(1, g) = f(p) log p 6= 0.

On the other hand, L(1, f) = L(1, fe) + L(1, fo). By the Chowla-

Siegel theorem, L(1, fo) is an algebraic multiple of π. Also, L(1, fe) isa linear form in logarithms of multiplicativly independent units in thepth cyclotomic field. By Baker’s Theorem 1.4, L(1, f) 6= 0 implies that

L(1, f) is transcendental and hence, the logarithms of multiplicativelyindependent units, log(−1) and log(p) are Q-linearly dependent. Thus,p can be written as a product of units. This is a contradiction. Hence,if L(1, f) = 0 then f(p) = 0. We are now reduced to the cases provedearlier. This completes the discussion of Chowla’s problem.

13

CHAPTER 3

A vanishing criterion for L(1, f)

As seen in the last chapter, Chowla had resolved the non-vanishingof f when f is odd and f(p) = 0, where p is the period of f and pis prime. Chowla’s problem was resolved in more generality by Baker,Birch and Wirsing in 1972. In the same paper, they proved anotherremarkable theorem. They derived a necessary and sufficient conditionfor odd, algebraic-valued periodic functions f that satisfy L(1, f) = 0.This suggests a change in perspective. Instead of proving the non-vanishing, another approach to Chowla’s problem is to characterize allfunctions f for which L(1, f) = 0. Pursuing this approach, we presentthe characterization of odd functions as given by Baker, Birch andWirsing in the first section. In the next, we apply a theorem of Basstowards the characterization of even functions. In the last section, weprove a theorem similar to the one in the first chapter which says thatit is enough to consider only the even and the odd functions in orderto obtain a complete characterization.

Let us first observe that for q = 2,

L(1, f) = f(1) log(2),

which is non-zero if and only if f(1) 6= 0. Since, f(1) + f(2) = 0, thismeans that L(1, f) = 0 if and only if f is identically zero. Hence, wewill assume q > 2.

3.1. Odd functions

In this section, we reproduce a simple argument of Baker, Birchand Wirsing [5] that gives us a necessary and sufficient condition onthe function f such that L(1, f) = 0.

Theorem 3.1. Let f be an odd algebraic-valued arithmetical func-tion, periodic with period q. Then

L(1, f) = 0 ⇐⇒q−1∑x=1

xf(x) = 0.

Here f denotes the Fourier transform of f as defined in chapter one.

Proof. Let us note that

1− ζxq = −(ζx/2q − ζ−x/2q )ζx/2q = −2i

(sin

(xπ

q

))exπi/q, (3.1)

14

and so the principal value of the logarithm is

log(1− ζxq ) = log

(2 sin

xπ

q

)+

(x

q− 1

2

)πi, (3.2)

for 1 ≤ x < q. Substituting (3.2) in the expression for L(1, f) as alinear form in logarithms (1.3), we get

L(1, f) = −q−1∑x=1

f(x)

[log

(2 sin

xπ

q

)+

(x

q− 1

2

)πi

]

= −q−1∑x=1

f(x) log

(2 sin

xπ

q

)− iπ

q

q−1∑x=1

xf(x) +iπ

2

q−1∑x=1

f(x).

(3.3)

Since f is an odd function, f is also an odd function. Hence,

2

q−1∑x=1

f(x) =

q−1∑x=1

[f(x) + f(q − x)] = 0.

Therefore, the last term of (3.3) is zero. Now, note that sin(π − θ) =sin(θ). Thus, sin(xπ/q) is an even function. Hence, log(2 sin xπ

q) is even

which implies that f(x) log(2 sin xπq

) is an odd function.

Therefore, the first term of (3.3),

q−1∑x=1

f(x) log

(2 sin

xπ

q

)= 0.

The result is immediate from here. �

Remark 1. The condition obtained above is on the Fourier trans-form of the function and not the function itself. Using the Fourierinversion formula, we can deduce a condition on the function. Thecondition obtained in Theorem 3.1 can be simplified to

q−1∑x=1

xf(x) =1

q

q−1∑x=1

x

q∑n=1

f(n)ζ−nxq

=1

q

q∑n=1

f(n)

q−1∑x=1

xζ−nxq

=1

q

q−1∑n=1

f(n)

q−1∑x=1

xζ−nxq ,

as f(q) = f(0) = f(−q) = −f(q) = 0 since f is odd. The innermostsum can be evaluated as follows. Let T be an indeterminate. Observethat

q−1∑x=0

T x =T q − 1

T − 1. (3.4)

15

Differentiating (3.4) with respect to T , we have

q−1∑x=1

xT x−1 =qT q−1

T − 1− T q − 1

(T − 1)2 .

Multiplying the above equation by T and substituting T = ζ−nq , we get

q−1∑x=1

xζ−nxq =q

ζ−nq − 1.

This observation along with Theorem 3.1 gives: for f odd,

L(1, f) = 0 ⇐⇒q−1∑n=1

f(n)

1− ζnq= 0. (3.5)

Let us note that

1

1− ζnq=i

2cot

(nπ

q

)+

1

2

and that since f is odd,

2

q−1∑n=1

f(n) =

q−1∑n=1

[f(n) + f(q − n)] = 0.

Hence, the condition (3.5) can be translated as: For odd algebraic-valued periodic functions f ,

L(1, f) = 0 ⇐⇒q−1∑n=1

f(n) cot

(nπ

q

)= 0.

We would like to mention that Baker, Birch and Wirsing also ob-tained a basis for the Q-vector space of odd algebraic-valued arithmeti-cal functions f , periodic with period q and L(1, f) = 0 in their paper[5]. Thus, the characterization of odd functions is complete.

3.2. Even functions

As seen earlier, the non-vanishing of L(1, f) when f is odd andalgebraic valued has been settled. It still remains whether the sameis true for even periodic functions. In this section, we will use a the-orem of Bass that characterizes all the multiplicative relations amongcyclotomic numbers modulo torsion ([6], [11]). We will give a neces-sary condition for even algebraic-valued periodic functions f to satisfyL(1, f) = 0.

16

We define the following functions that serve as building blocks foreven algebraic-valued functions, periodic with period q whose L-seriesvanish at s = 1. Fix a divisor d of q. For 1 ≤ c ≤ d−1, we define Fd,c :=

F(1)d,c − F

(2)d,c where the two functions, F

(1)d,c and F

(2)d,c are arithmetical

functions, periodic with period q. They are defined as follows:

F(1)d,c (x) =

{1/2 if x ≡ c mod d,

0 otherwise.

F(2)d,c (x) =

{1/2 if x ≡ ( q

d)c mod q,

0 otherwise.

We will prove the following result in this section:

Theorem 3.2. Let f be an algebraic valued, even function whichis periodic with period q. If L(1, f) = 0, then f is an algebraic linear

combination of the functions {Fd,c| any divisor d of q, 1 < d < q, 1 ≤c ≤ d− 1}.

Here, Fd,c denotes the Fourier transform of Fd,c which can be com-puted as follows: For 1 ≤ y ≤ q,

F(1)d,c (y) =

1

q

q∑a=1

F(1)d,c (a)ζ−ayq

=1

2q

qd−1∑j=0

ζ−(c+dj)yq

=ζ−ycq

2q

qd−1∑j=0

ζ−djyq

=ζ−ycq

2q

qd−1∑j=0

ζ−jyqd.

Note that the sumqd−1∑j=0

ζ−jyq/d =

{qd

if y ≡ 0 mod qd,

0 otherwise.

Similarly, for 1 ≤ y ≤ q,

F(2)d,c (y) =

1

q

q∑a=1

F(2)d,c (a)ζ−ayq

=1

2qζ− qdcy

q

=1

2qζ−cyd .

17

More precisely,

Fd,c(y) =

{ζ−cyq

2d− ζ−cyd

2qif y ≡ 0 mod q

d,

− 12qζ−cyd otherwise.

Thus, we note that Fd,c ’s are in fact, simple functions.

Proof. Let f be an even algebraic-valued arithmetical, periodicfunction with period q, not identically zero. LetM := Q(f(1), · · · , f(q), ζq).Let {ω1, ω2, · · · , ωr} be a basis for M over Q. Then there exists dj(x)∈ Q such that,

f(x) =r∑j=1

dj(x)ωj.

We can choose an integer N such that ∀ 1 ≤ x ≤ q, ∀ 1 ≤ j ≤ r,cj(x) := Ndj(x) ∈ Z. Hence, Nf(x) =

∑rj=1 cj(x)ωj. Thus,

NL(1, f) = −q−1∑x=1

r∑j=1

ωjcj(x) log (1− ζxq )

= −r∑j=1

ωj

q−1∑x=1

cj(x) log (1− ζxq )

Let

Rj :=

q−1∑x=1

cj(x) log (1− ζxq ).

Therefore,

−NL(1, f) =r∑j=1

ωjRj.

As f is even and ω1, · · · , ωr is a basis,

cj(x) = cj(q − x).

Therefore,

Rj =

b (q−1)2 c∑

x=1

cj(x)[log(1− ζxq ) + log(1− ζ−xq )].

Note that log denotes the principal branch of logarithm. Thus, arg(1−ζxq ) = − arg(1 − ζ−xq ). For 1 ≤ x ≤ q − 1, define ax := log(|1 − ζxq |).Thus,

Rj =

b (q−1)2 c∑

x=1

2cj(x)ax.

18

Since cj(x) = cj(q − x), Rj can be written as

Rj =

q−1∑x=1

cj(x)ax = log

( q−1∏x=1

(|1− ζxq |)cj(x)

).

Let

αj :=

q−1∏x=1

(|1− ζxq |)cj(x).

Therefore,

(−N)L(1, f) =r∑j=1

ωj logαj. (3.6)

Let us note that αj is non-zero, algebraic. Thus, by Corollary 1.4, ifαj 6= 1 for some 1 ≤ j ≤ r, then L(1, f) will be transcendental andhence non-zero. But L(1, f) = 0 by assumption.Hence, αj = 1 and in turn Rj = 0 , ∀ 1 ≤ j ≤ r. Thus, we are led toconsider relations among logarithms of the cyclotomic numbers, 1−ζxq .

In an unpublished paper, Milnor conjectured the complete set ofmultiplicative relations among cyclotomic numbers in the set {1− ζxq :1 ≤ x ≤ q−1} for a fixed positive integer q. Here ζq denotes a primitiveqth root of unity. This conjecture was proved by Hyman Bass [6] in1965. In 1972, Veikko Ennola [11] realized that the conjecture wastrue upto a factor of 2 and not in general. He gave a different proofof the conjecture in his paper. Since his formulation of the theoremis easier to apply in our setting, we will state it here. Let, as before,ax := log(|1 − ζxq |). The following theorem characterizes all additiverelations among the numbers {ax|1 ≤ x ≤ q − 1}.

Theorem 3.3. Consider the following two relations: For 1 ≤ x ≤b(q − 1)/2c,

R1 : ax − aq−x = 0 (3.7)

and for any divisor d of q and 1 < d < q and 1 ≤ c ≤ d− 1,

R2 : a qdc −

qd−1∑j=0

ac+dj = 0. (3.8)

Let R be an additive relation among the ax’s over the integers. Then,2R is a Z-linear combination of relations of the form R1 and R2.

Let R denote the relation 2Rj = 0. By Theorem 3.3, R belongs tothe Z-module generated by relations of the form (3.7) and (3.8). Sincecj(x) = cj(q−x), R belong to the Z-module generated by (3.8). Indeed

all relations R :=∑q−1

x=1Cxax = 0 in the Z-module generated by (3.7)satisfy Cx = −Cq−x, which along with the fact that Cj(x) = Cj(q − x)(which stems from the evenness of f) imply that cj(x) = 0, ∀ 1 ≤ x ≤q − 1. Thus, R is a Z-linear combination of (3.8). Observe that the

19

functions Fd,c are precisely those that represent the relation (3.8). Thisimplies that the functions cj are integer linear combinations of Fd,c, say

cj(x) =∑d|q

1<d<q

d−1∑c=1

mj,d,cFd,c(x),

where mj,d,c ∈ Z. Then,

Nf(x) =r∑j=1

cj(x)ωj.

=r∑j=1

ωj∑d|q

1<d<q

d−1∑c=1

mj,d,cFd,c(x).

=∑d|q

1<d<q

d−1∑c=1

Fd,c(x)r∑j=1

mj,d,cωj.

Taking the Fourier transform of both sides and using the Fourier in-version formula, the result follows. �

Remark. On examining the above proof, we observe that we alsoobtain a sufficiency condition for L(1, f) = 0. More precisely, supposef is a given algebraic linear combination of the functions

{Fd,c | any divisor d of q, 1 < d < q, 1 ≤ c ≤ d− 1},such that f is an even function, then L(1, f) = 0. But it does not seemto be easy to determine whether a given function is an algebraic linearcombination of these particular functions.

As a corollary, we have:

Corollary 3.4. When q is prime, there are no algebraic-valuedeven functions f such that L(1, f) = 0.

Proof. If q if prime, the set

{Fd,c| any divisor d of q, 1 < d < q, 1 ≤ c ≤ d− 1}is an empty set. The conclusion follows from Theorem 3.2. �

3.3. Combining the even and the odd

The previous two sections give a characterization of algebraic-valuedperiodic arithmetic functions, odd and even respectively whose L-seriesvanish at s = 1. In this section, we mention a theorem of Ram Murtyand Tapas Chatterjee [8] which ties the non-vanishing of the odd andthe even functions to give us the characterization required. This issimilar to Lemma 2.5. The proof in their paper is incorrect as carewas not taken about the branch of logarithm. The proof given below

20

follows a different argument.

As before, given any function f , we write it as f = fo + fe, wherefo is the odd part of f and fe is the even part of f .

Theorem 3.5. If f is an algebraic-valued arithmetical function,periodic with period q, then L(1, f) = 0 ⇐⇒ L(1, fo) = 0 andL(1, fe) = 0.

Proof. Let us note that∞∑n=1

f(n)

n=∞∑n=1

fe(n)

n+∞∑n=1

fo(n)

n.

Thus, it is clear that if L(1, fo) and L(1, fe) are both zero, then so isL(1, f).

Now, suppose that L(1, f) = 0. The arguments in the first sectionof this chapter imply

L(1, fo) =−iπq

q−1∑x=1

xfo(x).

We claim that,

L(1, fe) = −2

b (q−1)2 c∑

x=1

fe(x) log

(2 sin

xπ

q

). (3.9)

Indeed, by (1.3), we know that

∞∑n=1

fe(n)

n= −

q−1∑x=1

fe(x) log(1− ζxq ).

Since fe is even, so is fe, i.e, fe(q − x) = fe(x). Thus,

L(1, fe) = −2

b (q−1)2 c∑

x=1

fe(x) log |1− ζxq |.

By (3.1),

|1− ζxq | =∣∣∣∣2 sin

(xπ

q

)∣∣∣∣.Hence,

L(1, fe) = −2

b (q−1)2 c∑

x=1

fe(x) log

∣∣∣∣2 sin

(xπ

q

)∣∣∣∣.Note that, for 1 ≤ x ≤

⌊(q−1)

2

⌋, sin(xπ/q) > 0.

21

Thus (3.3) and (3.9) imply

L(1, f) =−iq

( q−1∑x=1

xfo(x)

)π − 2

b (q−1)2 c∑

x=1

fe(x) log

(2 sin

xπ

q

). (3.10)

Suppose L(1, fo) 6= 0. By Lemma 1.5, L(1, f) 6= 0, which contra-dicts our assumption. Thus, L(1, fo) = 0 and hence, L(1, fe) = 0. �

This gives a characterization of algebraic-valued periodic arithmeti-cal functions f with L(1, f) = 0. But the conditions obtained in theprevious characterizations do not seem to be very easy to verify.

22

CHAPTER 4

The Lerch zeta function

The Riemann zeta function and the Hurwitz zeta function are acouple of very well-studied functions in number theory. The Riemannzeta function was introduced by Euler (as a function of a real variable)in 1737 but was studied extensively by Riemann [20] in his 1859 paper,where he obtained the analytic continuation and functional equationfor it. Riemann zeta function is given by the series

ζ(s) =∞∑n=1

1

ns,

which converges absolutely for <(s) > 1. This function played a re-markable role in the study of the distribution of primes.

The Hurwitz zeta function was introduced by Hurwitz [12] in his1882 paper. It depends on a real parameter, namely, x. For 0 < x ≤ 1,it is defined by the series

ζ(s, x) =∞∑n=0

1

(n+ x)s,

which converges absolutely for <(s) > 1. The Hurwitz zeta function isimportant in the study of Dirichlet L-functions. A major contributionof Riemann and Hurwitz was obtaining the analytic continuation ofthe respective zeta functions to the entire complex plane except for asimple pole at s = 1.

The Lerch zeta function was introduced independently by Lerch[14] in 1887 and Lipschitz [15] in 1889. It is a generalization of boththe Riemann zeta and the Hurwitz zeta functions. It depends on tworeal parameters, λ and α. For λ ∈ R and 0 < α ≤ 1, the function isdefined as

L(λ, α, s) =∞∑m=0

e2πiλm

(m+ α)s, (4.1)

which converges for <(s) > 1 if λ ∈ Z and for <(s) > 0 if λ /∈ Z.In his 1887 paper, Lerch [14] obtained the analytic continuation andfunctional equation for L(λ, α, s). In contrast to the Riemann and theHurwitz zeta functions, the Lerch zeta function was not extensivelystudied, perhaps due to the lack of its evident applications at thattime. But it has caught the attention of mathematicians in the lasttwo decades and new results have started coming up in its theory.

23

The following elementary observations can be made from the defi-nition of L(λ, α, s):1. For λ ∈ Z, L(λ, α, s) = ζ(s, α), the Hurwitz zeta function.2. For λ ∈ Z and α = 1, L(λ, 1, s) = ζ(s), the Riemann zeta function.3. For λ = 1

2and α = 1, L(1

2, 1, s) = ζ(s)(1− 21−s).

4. For λ ∈ Z and α = 12, L(λ, 1

2, s) = ζ(s)(2s − 1).

Thus we see that values of the Lerch zeta function are intricately re-lated to the values of the Riemann and the Hurwitz zeta functions.

A notable fact of the Lerch-zeta function as mentioned before is thefollowing:

Theorem 4.1. For λ /∈ Z, the series (4.1) converges for <(s) > 0.

Proof. Without loss of generality, we can assume that 0 < λ < 1.Fix such a λ. For u > 0, define the partial sums,

S(u) =∑m≤u

e2πiλm.

Since this is a geometric sum and λ /∈ Z,

|S(u)| = |S(buc)| =∣∣∣∣1− e2πiλ(buc+1)

1− e2πiλ

∣∣∣∣ ≤ 2

|1− e2πiλ|. (4.2)

Recall the partial summation or the Abel summation formula that says:

Theorem. Let an be a sequence of complex numbers and f be C1

function on R. For x > 0, if A(x) :=∑

n≤x an, then∑0≤m≤x

amf(m) = A(x)f(x)−∫ x

0

A(t)f ′(t)dt.

This implies∑0≤m≤x

e2πiλm

(m+ α)s= S(x)(x+ α)−s + s

∫ x

0

S(u)

(u+ α)s+1 du. (4.3)

Suppose <(s) > 1. Let x→∞. Then (4.2) and (4.3) imply,

L(λ, α, s) = s

∫ ∞0

S(u)

(u+ α)s+1du. (4.4)

The integral on the right hand side of the above equation convergesuniformly on compact subsets of the half plane <(s) > 0. This, togetherwith (4.3) shows that the series

∞∑m=0

e2πiλm

(m+ α)s

converges uniformly on compact subsets of <(s) > 0. �

24

In this chapter, we will give a proof of the analytic continuationof the Lerch zeta function and its functional equation. We will alsoobserve a few new results about the transcendence of its certain specialvalues. Since the Hurwitz zeta function has already been well-studied,we will only consider the case λ /∈ Z in this chapter. Indeed, we canthen assume 0 < λ < 1.

4.1. Analytic continuation

Theorem 4.2. For 0 < λ < 1 and 0 < α ≤ 1, the Lerch zetafunction, L(λ, α, s) defined by the series (4.1) for <(s) > 0, can beanalytically continued to an entire function of s.

Proof. The proof given here is a reproduction of the proof by B.C. Berndt published in 1972 [7]. Let s > 1 be real. Put λ = b + 1

2.

Hence, |b| < 1/2. Define the function

Fs(z) =πe2πibz

sin(πz)(z + α)s. (4.5)

Observe that Fs is analytic in the the entire complex plane except forsimple poles at −α and the integers. Let k ∈ Z. The residue of Fs atk,

Resz=kFs(z) = limz→k

(z − k)Fs(z)

= limz→k

(z − k)πe2πibz

sin(πz)(z + α)s

= limz→k

πe2πibz

(z + α)s cos(πz)π

=e2πibk

(k + α)s cos(πk).

Since cos(πk) = ±1, 1cos(πk)

= e2πi(k/2),

Resz=kFs(z) =e2πiλk

(k + α)s. (4.6)

Choose a real number c such that −α < c < 0.For a positive integer m, define the contour Cm as the positively ori-ented contour consisting of the right half circle with center (c, 0) andradius (m + 1

2− c) together with the vertical diameter through (c, 0).

Let Γm denote the curved part of the contour.

By Cauchy’s Residue theorem,

1

2πi

∫Cm

Fs(z)dz =m∑k=0

Resz=kFs(z).

25

c−α−1

m+ 12

Figure 1. The contour C

26

By (4.6),

1

2πi

∫Cm

Fs(z)dz =m∑k=0

e2πiλk

(k + α)s. (4.7)

Since we would like to take limit as m tends to ∞, we need to study

the integral along Γm. Since b < 1/2,

∣∣∣∣e2πibz/sin πz

∣∣∣∣ is bounded inde-

pendent of m on Γm. To see this, put y := =(z) and observe that:∣∣∣∣ e2πibz

sin πz

∣∣∣∣ = 2

∣∣∣∣ e2πibz

eiπz − e−iπz

∣∣∣∣≤ 2

∣∣∣∣ e−2πby

e−πy − eπy

∣∣∣∣≤ c′

eπy

|e−πy − eπy|,

for some constant c′. The above function is of the form t2/|1−t2| whichis bounded by 1 for t sufficiently large.

Thus, there exists M such that for z ∈ Γm,∣∣∣∣ e2πibz

sin πz

∣∣∣∣ ≤M.

Thus, ∣∣∣∣ ∫Γm

Fs(z)dz

∣∣∣∣ ≤ πM

∫Γm

1

|z + α|sdz.

Recall that s ∈ R. Note that, on Γm,

|z + α|2 ≥ (c+ α)2 +

(m+

1

2− c)2

≥(m+

1

2

)2

.

Hence, ∣∣∣∣ ∫Γm

Fs(z)dz

∣∣∣∣ ≤ π2M

(m+ 1/2)s

(m+

1

2− c).

The right hand side of the above expression tends to zero as m tendsto ∞ since s > 1. Thus, taking limit as m tends to infinity in (4.7)gives:

−1

2πi

∫ c+i∞

c−i∞Fs(z)dz = L(λ, α, s).

27

Using the identity sinπz = (eiπz − e−iπz)/2i, the above expression canbe re-written as:

L(λ, α, s) =

∫ c+i∞

c

e2πibz−iπz

e−2πiz − 1(z + α)−sdz +∫ c−i∞

c

e−2πibz+iπz

e2πiz − 1(z + α)−sdz. (4.8)

Observe that following an earlier argument, |b| < 1/2 implies that boththe integrals in (4.8) converge uniformly on compact subsets of C andthus define an analytic function. In this way, L(λ, α, s) can be extendedto an entire function. �

4.2. Functional Equation

The Lerch zeta function satisfies a rather curious functional equa-tion. It was first proved by Lerch [14] in 1887. The proof given belowuses elementary contour integration. This was also given by B. Berndtin 1972 [7]. Other interesting proofs of the functional equation aregiven by:(i) T. Apostol (1951) [2] - This proof follows the argument used by Rie-mann to prove the functional equation for the Riemann Zeta function.It uses theta-functions.(ii) Oberhettinger (1959) [19] - This proof uses the Poisson summationformula.(iii) M. Mikolas (1971) [16] - This proof uses the Fourier series expan-sion of

L(λ, α, 1− s)e2πiαλ, 0 < σ < 1,

as a function of α over (0, 1).

Theorem 4.3. For 0 < λ < 1, 0 < α < 1 and for all s ∈ C, wehave

L(λ, α, 1− s) =Γ(s)

2πs

{eiπ

s2−2πiαλL(−α, λ, s) +

e−iπs2

+2πiα(1−λ)L(α, 1− λ, s)}. (4.9)

Before proceeding with the proof, let us note the following interest-ing lemma.

Lemma 4.4. For all <(s) > 1,

Γ(s)L(λ, α, s) =

∫ ∞0

e(1−α)x−2πiλxs−1

ex−2πiλ − 1dx. (4.10)

28

Proof. For <(s) > 1,∫ ∞0

e−(m+α)x+2πiλmxs−1dx =e2πiλm

(m+ α)s

∫ ∞0

e−xxs−1dx =e2πiλm

(m+ α)sΓ(s),

and hence, summing over m, we find

Γ(s)L(λ, α, s) =∞∑m=0

∫ ∞0

e−(m+α)x+2πiλmxs−1dx

=

∫ ∞0

e(1−α)x−2πiλxs−1

ex−2πiλ − 1dx.

The exchange of summation is justified by the absolute convergence ofthe series

∞∑m=0

∫ ∞0

e−(m+α)x+2πiλmxs−1dx = Γ(s)L(λ, α, s),

for <(s) > 1. This proves the lemma. �

We are now set to prove the functional equation of L(λ, α, s).

Proof. Suppose −1 < s < 0. We want to let c tend to −α in (4.8).

Since sin(πα) 6= 0, by continuity, there exists a neighborhood around−α such that for z in that neighborhood,

1

| sin(πα)|≤ 1

|z + α|.

Thus, for z in that neighborhood,∣∣∣∣ e2πibz±iπz

(e±2πiz − 1)|z + α|s

∣∣∣∣ ≤ A|z + α|−s−1,

for some positive constant A.

Since −1 < s < 0,∫ ±1

0|y|−s−1dy < ∞.

Hence, the integrals in (4.8) converge uniformly for −α ≤ c ≤ −α+ ε,for all ε > 0. Letting c→ −α in (4.8), we have

L(λ, α, s) =

∫ −α+i∞

−α

e2πibz−iπz

(z + α)s(e−2πiz − 1)dz +∫ α−i∞

α

e2πibz+iπz

(z + α)s(e2πiz − 1)dz.

29

Put y = z + α in the above equation to get

L(λ, α, s) =

∫ i∞

0

e2πib(y−α)−iπ(y−α)

ys(e−2πi(y−α) − 1)dy +∫ −i∞

0

e2πib(y−α)+iπ(y−α)

ys(e2πi(y−α) − 1)dy.

Now, substitute x = −iy in the first integral above and x = iy in thesecond integral. This gives,

L(λ, α, s) = (−i)e{2πibα+iπα−iπ s2}∫ ∞

0

e−2πbx+πx

xs(e2πx+2πiα − 1)dx +

ie{−2πibα−iπα+iπ s2}∫ ∞

0

e−2πbx+πx

xs(e2πx−2πiα − 1)dx.

Substituting u = 2πx, b = λ− 1/2 and replace s by 1− s to get

L(λ, α, 1− s) = 2π−se{iπs2−2πiα(λ−1)}

∫ ∞0

e−u(λ−1)us − 1

eu+2πiα − 1du +

2π−se{−iπs

2−2πiαλ}

∫ ∞0

euλus − 1

eu−2πiα − 1du. (4.11)

Applying (4.10) to the integrals in (4.11), we have

e2πiα

∫ ∞0

e−u(λ−1)us − 1

eu+2πiα − 1du = Γ(s)L(−α, λ, s) (4.12)

and

e−2πiα

∫ ∞0

euλus − 1

eu−2πiα − 1du = Γ(s)L(α, 1− λ, s). (4.13)

Thus, (4.11), (4.12) and (4.13) imply that the following identity holdsfor s > 1.

L(λ, α, 1− s) =Γ(s)

2πs

{eiπ

s2−2πiαλL(−α, λ, s) +

e−iπs2

+2πiα(1−λ)L(α, 1− λ, s)}.

By analytic continuation, we can conclude that the above identity istrue for all s ∈ C.

�

4.3. Special values

In 1951, Apostol [2] computed the values of the Lerch zeta functionat 0 and negative integers. For λ /∈ Z, 0 < α < 1,

L(λ, α, 0) =1

1− e2πiλ=i

2cot(πλ) +

1

2. (4.14)

30

It is interesting to note that the value at 0 is independent of α.For k ∈ Z>0,

L(λ, α,−k) = −βk+1(α, e2πiλ)

k + 1, (4.15)

where βn(t, u) is defined by the generating function

zetz

uez − 1=∞∑n=0

βn(t, u)

n!zn.

In particular, let us observe that for α ∈ Q ∩ (0, 1), L(λ, α,m) ∈ Qfor m ∈ Z≤0. This information together with the functional equationhelps us to determine the transcendental nature of certain values of theLerch zeta function.

Let k be a nonzero positive integer. Substitute s = k in (4.9) to get

L(α, λ, 1− k) =(k − 1)!

(2π)k

{eiπ

k2−2πiαλL(−α, λ, k) +

e−iπk2

+2πiα(1−λ)L(α, 1− λ, k)

}. (4.16)

Suppose that λ, α ∈ Q∩(0, 1). Let δ1 := ie−2πiλα and δ2 := −ie2πiα(1−λ).Note that δ1 and δ2 are algebraic. Thus,

L(λ, α, 1− k)(2π)k

(k − 1)!= δ1L(−α, λ, k) + δ2L(α, 1− λ, k).

Observe that since α and λ are rational, (4.14) and (4.15) imply thatL(λ, α, 1−k) is algebraic for all nonzero positive integers k. If L(λ, α, 1−k) 6= 0 and both L(−α, λ, k) and L(α, 1 − λ, k) are algebraic, then πk

will be algebraic and we are led to a contradiction. Therefore, we havethe following theorem:

Theorem 4.5. If λ, α ∈ Q ∩ (0, 1) and L(λ, α, 1 − k) 6= 0 , thenatleast one of L(−α, λ, k) and L(α, 1− λ, k) is transcendental.

On specializing further, we conclude that

Theorem 4.6. For α ∈ Q ∩ (0, 1), if L(1/2, α, 1 − k) 6= 0, thenL(α, 1/2, k) is transcendental, for all k ∈ Z>0.

Proof. Substitute λ = 1/2 in (4.16) and observe that

L

(α,

1

2, k

)= L

(− α, 1

2, k

), ∀ k ∈ Z>0.

The result now follows. �

Remark. For λ = 1/2 and k ∈ Z>0, the value of L(1/2, α,−k) asgiven by (4.15) can be determined by the coefficient of zk+1 in

−z eαz

ez + 1.

31

We realize that the above expression is the generating function of Eulerpolynomials, En(x). The real zeros of Euler polynomials in (0, 1) havebeen completely determined. By [10], E2m(x) 6= 0 for 0 < x < 1 andE2m+1(x) 6= 0 for 0 < x < 1 except for x = 1/2. Hence, Theorem 4.6indeed gives us that L(1/2, α, k) is transcendental for all 0 < α < 1when k is an odd positive integer and for 0 < α < 1, α 6= 1/2 when kis an even positive integer.

Thus, we see that many interesting properties of the Lerch zetafunction emerge out of the functional equation. There is scope forfurther study of the nature of values of the Lerch Zeta function atcomplex arguments and other properties of its functional equation. Inparticular, it will be interesting to find out when L(λ, α, 1 − k) = 0for λ 6= 1/2. We believe it can be done using ω-Euler polynomials andrelegate this to further research.

32

CHAPTER 5

Generalization of the Chowla problem

Let q be a fixed positive integer and f be an arithmetical functionwhich is periodic with period q. We will define

L (s, f, α) :=∞∑m=0

f(m)

(m+ α)s, (5.1)

for <(s) > 1 and 0 < α ≤ 1.

As in the case of the Chowla question, we can ask whether L (1, f, α)is non-zero for a periodic function f when f is not identically zero. Wewill study this question in this chapter.

5.1. Convergence of L (1, f, α)

Recall the Fourier inversion formula, namely,

f(m) =

q∑a=1

f(a)ζamq .

Substituting this in the expression for L (s, f, α), we get

L (s, f, α) =∞∑m=0

1

(m+ α)s

( q∑a=1

f(a)ζamq

)

=

q∑a=1

f(a)∞∑m=0

(e2πiam/q

(m+ α)s

)

= f(q)ζ(s, α) +

q−1∑a=1

f(a)∞∑m=0

(e2πiam/q

(m+ α)s

)

= f(q)ζ(s, α) +

q−1∑a=1

f(a)L

(a

q, α, s

),

where ζ(s, x) is the Hurwitz zeta function and L(λ, α, s) is the Lerchzeta function.

Note that the Hurwitz zeta function is analytic in the entire complexplane except for a simple pole at s = 1 with residue 1, and the Lerchzeta function is entire for 0 < λ < 1 and 0 < α ≤ 1. Thus,

33

Theorem 5.1. For a periodic arithmetical function f and 0 < α ≤1, L (s, f, α) is entire ⇐⇒ f(q) = 0.

Thus, we will assume henceforth that f(q) = 0.

5.2. Non-vanishing of L (1, f, α)

We will now investigate the non-vanishing of L (1, f, α). Define

gα(z) :=∞∑m=0

zm

m+ α.

Using summation by parts, we can see that the above series convergesfor |z| ≤ 1, z 6= 1.

Recall that for 0 < α ≤ 1 and λ /∈ Z, the series∑∞

m=0e2πiλm

(m+α)s

converges for <(s) > 0. Thus, L (1, f, α) =∑∞

m=0e2πiλm

(m+α), which can be

written as

L (1, f, α) =

q−1∑a=1

f(a)gα(ζaq ).

Lemma 5.2. For a rational number α = c/d,

gα(z) = −z−c/dd−1∑t=0

ζ−ctd log(1− ζtdz1/d).

Proof. On expanding the definition of gα we have,

gα(z) =∞∑m=0

zm

m+ c/d

= d

∞∑m=0

zm

dm+ c

= dz−c/d∞∑m=0

z(dm+c)/d

dm+ c

= dz−c/d∞∑n=1

n≡c mod d

zn/d

n

Let ζd = e2πi/d. Observe that

d−1∑t=0

ζ(n−c)td =

{d if n ≡ c mod d,

0 otherwise.

34

Hence,

gα(z) = z−c/d∞∑n=1

zn/d

n

( d−1∑t=0

ζ(n−c)td

)

= z−c/dd−1∑t=0

∞∑n=1

zn/dζntd ζ−ctd

n

= z−c/dd−1∑t=0

ζ−ctd

∞∑n=1

(z1/dζtd)n

n

= −z−c/dd−1∑t=0

ζ−ctd log(1− z1/dζtd).

This proves the lemma. �

Thus, for α ∈ Q∩ (0, 1], we can express L (1, f, α) as a linear formin logarithms of algebraic numbers.

Let α = c/d, for d 6= 0 and (c, d) = 1. Therefore,

L (1, f, α) = −q−1∑a=1

f(a)ζ−ac/dq

d−1∑t=0

ζ−ctd log(1− ζtdζa/dq )

= −q−1∑a=1

d−1∑t=0

f(a)ζ−ac/dq ζ−ctd log(1− ζtdζa/dq ).

Hence, Baker’s theorem 1.3 implies that

Theorem 5.3. Let f be an algebraic-valued arithmetical functionwhich is periodic with period q and α ∈ Q ∩ (0, 1]. Then L (1, f, α) iseither zero or transcendental.

We now investigate L (1, f, α) in more detail. By a computationsimilar to section 3.1, we can write the logarithms as a combination oftheir real and imaginary parts, ie,

log(1− ζtdζa/dq ) = log

(2 sin

[(qt+ a)π

qd

])+ iπ

(qt+ a

qd− 1

2

). (5.2)

Substituting (5.2) in the expression for L (1, f, α) as a linear form inlogarithms, we have

L (1, f, α) =

q−1∑a=1

d−1∑t=0

f(a)ζ−ac/dq ζ−ctd log

(2 sin

[(qt+ a)π

qd

])+

iπ

q−1∑a=1

d−1∑t=0

f(a)ζ−ac/dq ζ−ctd

(qt+ a

qd− 1

2

). (5.3)

Let us recall Lemma 1.5 from the first chapter.

35

Lemma. Let α1, · · · , αn be positive algebraic numbers. If c0, c1, · · · , cnare algebraic numbers and c0 6= 0, then

c0π +n∑j=1

cj logαj

is a transcendental number and hence non-zero.

Observe that (5.3) expresses L (1, f, α) in the setup of the abovelemma. Thus, we are lead to conclude that L (1, f, α) 6= 0 if the coeffi-cient of π in (5.3) is non-zero. Therefore, we have that if L (1, f, α) = 0,then

0 =

q−1∑a=1

d−1∑t=0

f(a)ζ−ac/dq ζ−ctd

(qt+ a

qd− 1

2

).

On simplifying the right hand side, we have

0 =

q−1∑a=1

d−1∑t=0

f(a)ζ−ac/dq ζ−ctd

(t

d− 1

2

)+

q−1∑a=1

d−1∑t=0

f(a)ζ−ac/dq ζ−ctd

a

qd. (5.4)

The second sum in the above equation can be further simplified to get

1

qd

q−1∑a=1

af(a)ζ−ac/dq

d−1∑t=0

ζ−ctd .

Let us note that the computations till now hold valid even for α = 1.We now assume 0 < α < 1. Thus, c 6≡ 0 mod d and

∑d−1t=0 ζ

−ctd = 0.

Therefore, the second sum in (5.4) is zero.

Consider the first sum in (5.4). It can be further split as

q−1∑a=1

d−1∑t=0

f(a)ζ−ac/dq ζ−ctd

(t

d− 1

2

)=

1

d

q−1∑a=1

d−1∑t=0

tf(a)ζ−ac/dq ζ−ctd

− 1

2

q−1∑a=1

d−1∑t=0

f(a)ζ−ac/dq ζ−ctd .

The second sum on the right hand side can be re-written as

q−1∑a=1

d−1∑t=0

f(a)ζ−ac/dq ζ−ctd =

q−1∑a=1

f(a)ζ−ac/dq

d−1∑t=0

ζ−ctd ,

which is zero as∑d−1

t=0 ζ−ctd = 0 when c 6≡ 0 mod d. Thus, the equation

(5.4) is simplified to( d−1∑t=0

tζ−ctd

)( q−1∑a=1

f(a)ζ−ac/dq

)= 0.

36

A computation similar to the one in Remark 1 from Chapter 3 helpsus to evaluate the first sum as

d−1∑t=0

tζ−ctd =d

ζ−cd − 16= 0.

Therefore, we have the following:

Theorem 5.4. Let f be an algebraic-valued arithmetical functionwhich is periodic with period q. Let α = c/d ∈ Q ∩ (0, 1), (c, d) = 1. IfL (1, f, α) = 0, then

q−1∑a=1

f(a)ζ−ac/dq = 0. (5.5)

Remark. We can “interpolate” the Fourier inversion formula to geta “continuation” of the given function f to the entire complex planeby setting

f(z) =

q∑a=1

f(a)ζazq .

Then condition (5.5) can be interpreted as f(−α) = 0.

This gives a clean necessary condition for L (1, f, α) = 0. One canfurther study the problem and ask for sufficiency conditions.

37

CHAPTER 6

Topics for further research

In the course of studying the Chowla problem, many questionsarose, to which I was unable to find satisfactory answers. I will statethose questions and the partial progress made. These will help in or-ganizing together, a set of problems for further research.

6.1. Transcendence of values of Dirichlet L-functions at 1/2.

Let us first observe the following:

Lemma 6.1.q−1∑a=1

cos

(π

4− 2πab

q

)=−1√

2, (6.1)

for 1 ≤ b ≤ q − 1.

Proof. Let θa = 2πab/q.Using the identity

cos(θ) =eiθ + e−iθ

2for θ ∈ R, the sum in (6.1) becomes

q−1∑a=1

cos

(π

4− θa

)=

q∑a=1

cos

(π

4− θa

)− 1√

2

=eiπ/4

2

( q∑a=1

eiθa +

q∑a=1

e−iθa)− 1√

2

=eiπ/4

2

(eiθ1

eiθq − 1

eiθ1 − 1+ e−iθ1

e−iθq − 1

e−iθ1 − 1

)− 1√

2

= − 1√2,

since e±iθq = 1. �

This lemma will be useful in our computation.Let ζ(s, x) denote the Hurwitz zeta function with 0 < x ≤ 1. Con-

sider the following functional equation of the Hurwitz zeta function[1].

ζ

(1− s, b

q

)=

2Γ(s)

(2πq)s

q∑a=1

cos

(πs

2− 2πab

q

)ζ

(s,a

q

),

38

where 1 ≤ b ≤ q. Specializing at s = 1/2, we get,

ζ

(1

2,b

q

)=

√2

q

q∑a=1

ζ

(1

2,a

q

)cos

(π

4− 2πab

q

).

On separating the term with a = q, we have

ζ

(1

2,b

q

)=

1√qζ

(1

2

)+

√2

q

q−1∑a=1

ζ

(1

2,a

q

)cos

(π

4− 2πab

q

). (6.2)

Thus, we have the following theorem.

Theorem 6.2. Let Mq be the (q− 1)× (q− 1) matrix whose (a, b)-

th entry is√

2/q cos(π/4 − 2πab/q) for 1 ≤ a, b ≤ q − 1. If v =[ζ

(12, bq

)− ζ(

12

)]q−1

b=1

, then v is an eigenvector of Mq with eigenvalue

1.

Proof. Let λ ∈ R such that

ζ

(1

2,b

q

)− λ√qζ

(1

2

)=

√2

q

q−1∑a=1

[ζ

(1

2,a

q

)− λ√qζ

(1

2

)]cos

(π

4−2πab

q

).

(6.3)By (6.2),

ζ

(1

2,b

q

)− λ√qζ

(1

2

)=

1√qζ

(1

2

)+

√2

q

q−1∑a=1

ζ

(1

2,a

q

)cos

(π

4− 2πab

q

).

On equating (6.3) to the expression obtained above, we get√2

q

q−1∑a=1

[ζ

(1

2,a

q

)− λ√qζ

(1

2

)]cos

(π

4− 2πab

q

)

=1√qζ

(1

2

)+

√2

q

q−1∑a=1

ζ

(1

2,a

q

)cos

(π

4− 2πab

q

)Thus,

−√

2λ

qζ

(1

2

) q−1∑a=1

cos

(π

4− 2πab

q

)=

1√qζ

(1

2

).

The sum on the left hand side is as calculated in Lemma 6.1. Hence,

λ

q=

1√q,

as ζ(1/2) 6= 0. Thus, λ =√q. �

We would like to make the following conjecture on the basis ofSAGE [21] computations:

39

Conjecture 6.3. Mq has eigenvalue 1 with multiplicity (q− 1)/2and eigenvalue −1 with multiplicity (q − 2)/2. Moreover, 1, −1 and1/√q are the only eigenvalues of Mq.

Under this conjecture, we can obtain a bound on the transcen-dence degree of Q({L(1/2, χ)|χ is a Dirichlet character mod q}) overQ as follows.

Let us recall the following basic theorem from linear algebra [3] :

Theorem 6.4. Let K be a field and M be an n × n matrix withentries in K. Suppose M is diagonalizable over K. Then M has neigenvectors with entries in K.

Let v be as in Theorem 6.2. Then, Theorem 6.2 implies that vbelongs to the eigenspace corresponding to eigenvalue 1. Conjecture6.3 along with Theorem 6.4 lead us to conclude that there exist vectorsv1, · · · , vq−1/2 that form a basis for the eigenspace corresponding toeigenvalue 1 and have entries in Q.

Therefore, v =

[ζ

(12, bq

)− ζ(

12

)]q−1

b=1

is a C-linear combination of

v1, · · · , vq−1/2. In particular, there exist complex numbersω1, · · · , ω(q−1)/2 such that,

ζ

(1

2,b

q

)− ζ(

1

2

)=

q−12∑j=1

ηj,b ωj, (6.4)

where ηb,j ∈ Q for all 1 ≤ b ≤ q − 1.

Let χ be a Dirichlet character mod q. Consider χ as a functionon the integers in the canonical way. Let L(s, χ) denote the DirichletL-function attached to χ. Since, χ is a function that is periodic withperiod q, by 1.1,

L(s, χ) =

q−1∑b=1

χ(b)ζ

(s,b

q

).

Thus,

L

(1

2, χ

)=

q−1∑b=1

χ(b)ζ

(1

2,b

q

)

=

q−1∑b=1

χ(b)

[ζ

(1

2,b

q

)− ζ(

1

2

)]+ ζ

(1

2

) q−1∑b=1

χ(b)

=

q−1∑b=1

χ(b)

[ζ

(1

2,b

q

)− ζ(

1

2

)],

40

since,∑q

b=1 χ(b) = 0 for χ 6= χ0 and χ(q) = 0. Here, χ0 denotes thetrivial character mod q, i.e,

χ0(n) =

{1 if (n, q) = 1,

0 otherwise.

Hence, we have the following important lemma.

Lemma 6.5. For a Dirichlet character χmod q and χ 6= χ0, L(1/2, χ)is an algebraic linear combination of {ζ(1/2, b/q)−ζ(1/2) : 1 ≤ b < q}.

The above lemma along with (6.4) lead us to conclude that: L(1/2, χ)is a Q-linear combination of the complex numbers ω1, · · · , ω(q−1)/2.Therefore, we have the following result.

Theorem 6.6. Under Conjecture 6.3,

tq := trdegQ

(Q({L(1/2, χ) : χ- Dirichlet character mod q})

)≤ q + 1

2.

Remark. The number of distinct Dirichlet characters mod q isφ(q). So the trivial bound for tq is φ(q). The bound obtained above isbetter than the trivial bound when φ(q) > (q+1)/2, for example, whenq is prime. But in certain cases, the trivial bound is more effective.

6.2. Analogue of the Dedekind determinant

Let n be a fixed positive integer and G be a group of order n,{x1, · · · , xn}. Let f : G→ C be any function on G. Dedekind studiedn× n matrices of the form D = (di,j) with

di,j = f(x−1i xj).

We will call such matrices as Dedekind matrices. Define

Sχ =∑g∈G

f(g)χ(g),

for each character χ of G. Dedekind computed explicitly the deter-minant of such matrices to be

∏χ Sχ where the product is over all

characters χ of the group G. Ram Murty and Kaneenika Sinha [18]also computed the eigenvalues and eigenvectors of Dedekind matrices.As an analogue of the Dedekind matrix, we can also study matrices ofthe form E = (ei,j) with

ei,j = f(xixj). (6.5)

A motivation to study these kind of matrices is the matrixMq from theprevious section. A computation of the eigenvalues and eigenvectors ofthe matrices of the type mentioned in (6.5) would help to resolve theconjecture made in the previous section.

As a partial solution to the question raised, we would like to like to

41

compute the determinant of such matrices in the case when the groupG is isomorphic to (Z/pZ)∗.

Let f : Z/pZ → C. Let E be the (p − 1) × (p − 1) matrix whose(i, j)-th entry is f(ij), for 1 ≤ i, j ≤ p− 1.

Recall the Fourier inversion formula that says

f(n) =

p∑k=1

f(k)ζnkp .

Thus,

f(ij) =

p∑k=1

f(k)ζ ijkp

= f(p) +

p−1∑k=1

f(k)ζ ijkp

= f(p) +

p−1∑t=1

f(i−1t)ζtjp .

Hence, for all 1 ≤ i, j ≤ p− 1,

f(ij)− f(p) =

p−1∑k=1

f(i−1t)ζtjp . (6.6)

Define the following p − 1 × p − 1 matrices: E0 = (εi,j), with εi,j =

f(ij) − f(p), D0 = (δi,j), with δi,j = f(i−1j) and V0 = (ϑi,j), withϑi,j = ζ ijp for all 1 ≤ i, j ≤ p− 1.

Thus, equation (6.6) can be re-written as a matrix equation as follows:

E0 = D0 × V0. (6.7)

Observe that D0 is a Dedekind matrix and V0 is a Vandermonde ma-trix, both of whose determinants can be computed using known means.Thus, the determinant of E0 can be computed.Define a function

g(n) = f(n)− f(p).

Note that g is also periodic with period p. On taking the Fouriertransform, we have

g(m) =1

p

p∑a=1

f(a)ζ−amp − f(p)

p

p∑a=1

ζ−amp .

Since,∑p

a=1 ζ−amp = 0 unless m ≡ 0 (mod p), the Fourier transform of

g can be re-written as

g(m) = f(m)− f(p)δ0(m),

42

where,

δ0(m) =

{1 if m ≡ 0 (mod p),

0 otherwise.

The arguments of g under consideration are between 1 and p − 1 andhence co-prime to p. Thus, for those arguments m, g(m) = f(m).Therefore, the matrix D0 = G0 = (γi,j), where γi,j = g(i−1j). Thus, ifG = (g(ij)i,j), then

detG = detG0 × detV0, (6.8)

where, G0 is a Dedekind matrix and V0 is a Vandermonde matrix, bothof whose determinants can be computed. Since the dependence on fno longer exists in (6.8), it holds for all functions, periodic with periodp.

43

Bibliography

[1] T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976,Chapter 12.

[2] T. M. Apostol, On the Lerch zeta function, Pacific Journal of Mathematics, Vol.1, No. 2, December 1951.

[3] M. Artin, Algebra, Prentice Hall, 2011.[4] A. Baker, Transcendental Number Theory, Cambridge University Press (1975).[5] A. Baker, B. J. Birch and E. A. Wirsing, On a problem of Chowla, Journal of

Number Theory 5 (1973), 224-236.[6] H. Bass, Generators and relations for cyclotomic units, Nagoya Math Journal,

Volume 27, Part 2 (1966), 401-407.[7] B. C. Berndt, Two new proofs of Lerch’s functional equation, Proc. Amer. Math.

Soc. 32 (1972), 403-408.[8] T. Chatterjee and M. Ram Murty Non-vanishing of Dirichlet series with periodic

coefficients, Journal of Number Theory, Volume 145, December 2014, 1-21.[9] S. Chowla, A special infinite series, Norske Vid. Selsk. Forth. (Trondheim) 37

(1964), 85 - 87.[10] H. Delange, On the real roots of Euler polynomials, Mh. Math., 106 (1988) ,

115 - 138.[11] V. Ennola, On relations between Cyclotomic Units, Journal of Number Theory,

4, (1972) 236-247.[12] A. Hurwitz, Einige Eigenschaften der Dirichlet Funktionen F (s) =∑

(D/n)n−s, die bei der Bestimmung der Klassenzahlen Binarer quadratischerFormen auftreten, Zeitschrift f. Math. u. Physik, 27 (1882) 86-101.

[13] Antanas Laurincikas, Ramunas Garunkstis, The Lerch-Zeta function, KluwerAcademic Publishers, 2010.

[14] M. Lerch, Note sur la fonction R(w, x, s) =∑∞k=0 e

2πikx/(w+k)s, Acta Math.11 (1887) 19-24.

[15] R. Lipschitz, Untersuchung der Eigenschaften einer Gattung von unendlichenReihen, J. Reine Angew. Math. 105 (1889), 127-156.

[16] Mikolas, M. New proof and extension of the functional equality of Lerch’s zetafunction, Ann. Univ. Sci. Budapest. Sect. Math. 14, (1971), 111-116.

[17] M. Ram Murty, Some remarks on a problem of Chowla, Ann. Sci. Math. Que-bec 35, No 2, (2011), 229-237.

[18] M. Ram Murty and Kaneenika Sinha, The generalized Dedekind determinant,to be published in Contemporary Mathematics, American Math. Soc., Provi-dence, Rhode Island, 2015.

[19] Oberhettinger, F. Note on the Lerch-zeta function, Pacific Journal of Mathe-matics, 6, (1956), 117-120.

[20] B. Riemann, Uber die Anzahl der Primzahlen unter einer gegebenen Grosse,Monatsberichte der Koniglich Preußischen Akademie der Wissenschaften zuBerlin, November 1859.

[21] W. A. Stein, et al. Sage Mathematics Software (Version 6.7), The Sage Devel-opment Team, 2015, http://www.sagemath.org.

44

[22] L. Washington, Introduction to Cyclotomic Fields, Graduate Texts in mathe-matics, 83, Springer-Verlag, new York, 1982. xi+389 pp.

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