Generalized Dedekind–Bernoulli Sums

A thesis presented to the faculty ofSan Francisco State University

In partial fulfilment ofThe Requirements for

The Degree

Master of ArtsIn

Mathematics

by

Anastasia Chavez

San Francisco, California

June 2010

Copyright byAnastasia Chavez

2010

CERTIFICATION OF APPROVAL

I certify that I have read Generalized Dedekind–Bernoulli Sums by Anas-

tasia Chavez and that in my opinion this work meets the criteria for

approving a thesis submitted in partial fulfillment of the requirements

for the degree: Master of Arts in Mathematics at San Francisco State

University.

Dr. Matthias BeckProfessor of Mathematics

Dr. Federico ArdilloProfessor of Mathematics

Dr. David EllisProfessor of Mathematics

Generalized Dedekind–Bernoulli Sums

Anastasia ChavezSan Francisco State University

2010

The finite arithmetic sum called a Dedekind sum appears in many areas of math-

ematics, such as topology, geometric combinatorics, algorithmic complexity, and

algebraic geometry. Dedekind sums exhibit many beautiful properties, the most fa-

mous being Dedekind’s reciprocity law. Since Dedekind, Apostol, Rademacher and

others have generalized Dedekind sums to involve Bernoulli polynomials. In 1995,

Hall, Wilson and Zagier introduced the 3-variable Dedekind-like sum and proved

a reciprocity relation. In this paper we introduce an n-variable generalization of

Hall, Wilson and Zagier’s sum called a multivariable Dedekind–Bernoulli sum and

prove a reciprocity law that generalizes the generic case of Hall, Wilson and Zagier’s

reciprocity theorem. Our proof uses a novel, combinatorial approach that simplifies

the proof of Hall, Wilson and Zagier’s reciprocity theorem and aids in proving the

general 4-variable extension of Hall, Wilson and Zagier’s reciprocity theorem.

I certify that the Abstract is a correct representation of the content of this thesis.

Chair, Thesis Committee Date

ACKNOWLEDGMENTS

I would like to thank my thesis advisor, collaborator and mentor, Dr.

Matthias Beck, for his support both academically and in life matters;

professor and mentor Dr. David Ellis, thank you for urging me to con-

tinue this mathematical path; many thanks to the talented and dedicated

professors and staff of the San Francisco State University Mathematics

Department for their support through the years; to my parents, thank

you for everything that was intentionally and unintentionally done, it

was and is exactly what I need to be me; to my husband and best friend,

Davi, your unconditional support and love helps me be the best version

of spirit that I can express, my endless gratitude to you resides in every

breath I take; and last, to the angels who bless my life in the most per-

fect ways, Ayla and Asha, thank you for your medicine and being the

profound teachers that you are.

v

TABLE OF CONTENTS

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 More History, Definitions and Notations . . . . . . . . . . . . . . . . . . . 4

2.1 Bernoulli Polynomials and More . . . . . . . . . . . . . . . . . . . . . 4

2.2 Dedekind and Bernoulli meet . . . . . . . . . . . . . . . . . . . . . . 9

3 The Multivariate Dedekind–Bernoulli sum . . . . . . . . . . . . . . . . . . 12

4 A New Proof of Hall–Wilson–Zagier’s Reciprocity Theorem . . . . . . . . 42

5 The 4-variable reciprocity theorem . . . . . . . . . . . . . . . . . . . . . . 53

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

vi

Chapter 1

Introduction

In the 1880’s, Richard Dedekind developed the finite sum that today is called the

Dedekind sum [12].

Definition 1.1 (Classical Dedekind Sum). For any positive integers a and b, the

classical Dedekind sum s(a, b) is defined to be

s(a, b) =∑

h mod b

B1

(h

b

)B1

(ha

b

), (1.1)

where

B1(x) =

x − 1/2 (x 6∈ Z),

0 (x ∈ Z),

(1.2)

1

2

and we call x− bxc = x the fractional part of x. The sum∑

h mod b is interpreted

as the periodic sum over all values of h modulo b where h ∈ Z.

Dedekind arrived at the Dedekind sums while studying the η−function. Al-

though Dedekind sums are studied in number theory, they have appeared in many

areas of mathematics such as analytical number theory [6], combinatorial geome-

try [4], topology [9], and algorithmic complexity [8]. Since Dedekind sums have

no closed form, calculating their sum can become very time intensive. Thankfully,

Dedekind discovered an invaluable tool that simplifies Dedekind sum calculations

immensely, called reciprocity. Although Dedekind proved the following reciprocity

relation by a transcendental method, Rademacher recognized it’s elementary nature

and developed multiple proofs of Dedekind’s reciprocity theorem [12] addressing it’s

arithmetic signifigance.

Theorem 1.1 (Dedekind Reciprocity [13]). For any positive integers a and b with

(a, b) = 1,

s(a, b) + s(b, a) =1

12

(a

b+

b

a+

1

ab

)− 1

4. (1.3)

Since Dedekind, many mathematicians such as Rademacher [11], Carlitz [5], and

Apostol [1] have introduced Dedekind-like sums that generalize B1(u) to periodized

Bernoulli polynomials, Bk(u) (see Definition 2.4). In suit with Dedekind, reciprocity

relations have been proven for the respective Dedekind-like sums.

3

In 1995, Hall, Wilson and Zagier [7] introduced a generalization of Dedekind sums

(see Definition 2.4) that involve three variables and introduce shifts. They proved

a reciprocity theorem that is a wonderful example of how these relations express

the symmetry inherent in Dedekind-like sums and characterize their potential to be

evaluated so swiftly.

Most recent, Bayad and Raouj [3] have introduced a generalization of Hall, Wil-

son and Zagier’s sum and proved a reciprocity theorem that shows the non-generic

case of Hall, Wilson and Zagier’s reciprocity theorem is a special case.

In this paper we introduce a different generalization of Hall, Wilson and Zagier’s

sum called the multivariate Dedekind–Bernoulli sum (see Definition 3.1) and prove

our main result, a reciprocity theorem (see Theorem 3.1). Along with our main

result, we show that the generic case of Hall, Wilson and Zagier’s reciprocity theorem

follows from the proof of our reciprocity theorem and that one can say more about

the 4-variable version of Hall, Wilson and Zagier’s sum than is covered by the

reciprocity theorem given by Bayad and Raouj, as well as our own.

Chapter 2

More History, Definitions and Notations

We now begin working towards rediscovering reciprocity theorems for Dedekind-like

sums.

2.1 Bernoulli Polynomials and More

Bernoulli polynomials and numbers are a pervasive bunch, which show up in many

areas of mathematics such as number theory and probability [2].

Definition 2.1 (Bernoulli functions). For u ∈ R, the Bernoulli function Bk(u) is

defined through the generating function

zeuz

ez − 1=

∑k≥0

Bk(u)

k!zk. (2.1)

4

5

Definition 2.2 (Bernoulli numbers). The Bernoulli numbers are Bk := Bk(0) and

have the generating function

z

ez − 1=

∑k≥0

Bk

k!zk. (2.2)

It has been proven that Bernoulli functions are in fact polynomials [2, Theorem

12.12].

Theorem 2.1 (k-th Bernoulli polynomial). The functions Bk(u) are polynomials

in u given by

Bk(u) =k∑

n=0

(k

n

)Bnu

k−n. (2.3)

An important property of k-th Bernoulli polynomials is the Fourier expansion

formula [2, Theorem 12.19]

∑k∈Z\0

e2πiu

km= −(2πi)m

m!Bm(u). (2.4)

In the non-absolutely convergent case m = 1, the sum is to be interpreted as a

Cauchy principal value.

Property 2.4 defines the unique periodic function Bm(u) with period 1 that

coincides with Bm(u) on [0, 1), except we set B1(u) = 0 for u ∈ Z [7].

6

Lemma 2.2 (Raabe’s formula [7]). For a ∈ N, x ∈ R,

∑f mod a

Bm

(x +

f

a

)= a1−mBm(ax).

Proof. Begin by embedding the right-hand side into a generating function:

∞∑m=0

a1−mBm(ax)Y m−1

m!=

∞∑m≥0

Bm(ax)(Y/a)m−1

m!

=eax·Y

a

eYa − 1

=exY

eYa − 1

.

Now, embed the left-hand side into a generating function:

∞∑m=0

∑f mod a

Bm

(x +

f

a

)Y m−1

m!=

a−1∑f=0

e(x+ fa)Y

eY − 1

=exY

eY − 1+

e(x+ 1a)Y

eY − 1+

e(x+ 2a)Y

eY − 1+ · · ·+ e(x+a−1

a)Y

eY − 1

=exY (1 + e

1aY + e

2aY + · · ·+ e

a−1a

Y )

eY − 1

7

(applying the geometric series identity∑n−1

k=0 rk = 1−rn

1−rwhere r = e

Ya )

=exY

eY − 1· 1− e

Ya·a

1− eYa

=−exY

1− eYa

=exY

eYa − 1

.

We have deduced the generating functions of the right and left hand sides of

Raabe’s formula are equal, so the coefficients of the generating functions are equal.

That is,

∑f mod a

Bm

(x +

f

a

)= a1−mBm(ax).

Definition 2.3. Let

β(α, Y ) =∞∑

m=0

Bm(α)

m!Y m−1, (2.5)

where α ∈ R.

8

Lemma 2.3.

β(α, Y ) =

12· eY +1

eY −1for α ∈ Z,

eαY

eY −1for α 6∈ Z.

(2.6)

Proof. Recall that Bm(α) coincides with Bm(α) on the interval [0, 1) and is periodic

with period 1. Thus, the cases we are interested in are when α is not in Z and

α = 0. We will first evaluate β(α, Y ) for α 6∈ Z.

β(α, Y ) =∞∑

m=0

Bm(α)

m!Y m−1

=∞∑

m=0

Bm(α)m!

Y m−1

=eαY

eY − 1.

Now, we look at the case α = 0. We first write out the first few terms of β(α, Y )

evaluated at 0, then manipulate the sum to have period 1. We use the well known

Bernoulli polynomial identities B0(α) = 1 and Bm(α) = 0 for odd m ≥ 3, as well as

9

the first few Bernoulli numbers [2].

B0(0)Y −1 + B1(0) + B2(0)Y

2!+ B3(0)

Y 2

3!+ B4(0)

Y 3

4!+ · · · = 1

eY − 1

1− 1

2+

1

6+ 0− 1

30+ 0 + · · · = 1

eY − 1

1 +1

6+ 0− 1

30+ 0 + · · · = 1

eY − 1+

1

2

=2 + eY − 1

2(eY − 1)

=1

2· eY + 1

(eY − 1).

Recall that B1(α) = 0 for all α ∈ Z, and so we have the following identity:

B0(0)Y −1 + B1(0) + B2(0)Y

2!+ B3(0)

Y 2

3!+ B4(0)

Y 3

4!+ · · · = 1

2· eY + 1

(eY − 1).

When α is an integer α = 0, thus the above identity holds for α ∈ Z.

2.2 Dedekind and Bernoulli meet

Dedekind-like sums involving higher-order Bernoulli polynomials have been defined

by Apostol [1], Carlitz [5], Mikolas [10] and Hall, Wilson and Zagier [7]. The lat-

ter introduced the generalized Dedekind–Rademacher sum, the function we wish to

generalize to n variables.

10

Definition 2.4. The generalized Dedekind–Rademacher sum is

Sm,n

a b c

x y z

=∑

h mod c

Bm

(ah + z

c− x

)Bn

(bh + z

c− y

), (2.7)

where a, b, c ∈ Z>0 and x, y, z ∈ R.

The generalized Dedekind–Rademacher sum satisfies certain reciprocity relations

that mix various pairs of indices (m, n) and so are most conveniently stated in terms

of the generating function

Ω

a b c

x y z

X Y Z

=∑

m,n≥0

1

m!n!Sm,n

a b c

x y z

(X/a)m−1(Y/b)n−1, (2.8)

where Z = −X − Y .

Hall,Wilson and Zagier proved the following reciprocity theorem for the gener-

alized Dedekind–Rademacher sum.

Theorem 2.4. [7] Let a, b, c ∈ Z>0 and pairwise relatively prime, x, y, z ∈ R, and

11

X, Y, Z three variables such that X + Y + Z = 0. Then

Ω

a b c

x y z

X Y Z

+ Ω

b c a

y z x

Y Z X

+ Ω

c a b

z x y

Z X Y

=

−1/4 if (x, y, z) ∈ (a, b, c)R + Z3,

0 otherwise.

(2.9)

To clarify, we call the result

Ω

a b c

x y z

X Y Z

+ Ω

b c a

y z x

Y Z X

+ Ω

c a b

z x y

Z X Y

= 0

the generic case of the above reciprocity theorem and this occurs most of the time.

The one exception is when (x, y, z) are chosen in such a way that the involved

Bernoulli periodic polynomials are evaluated at integer values. In the following

chapter we introduce our main theorem which generalizes Hall, Wilson and Zagier’s

reciprocity theorem using a simple combinatorial approach and will make clear how

the exceptional integer values come to light.

Chapter 3

The Multivariate Dedekind–Bernoulli sum

We now introduce our main object of study, the multivariate Dedekind–Bernoulli

sum.

Definition 3.1. For a fixed integer n ≥ 2, we consider positive integers (p1, p2, . . . , pn)

and (a1, a2, . . . , an), and real numbers (x1, x2, . . . , xn). For 1 ≤ k ≤ n, set

Ak = (a1, a2, . . . , ak, . . . , an) , Xk = (x1, x2, . . . , xk, . . . , xn)

and Pk = (p1, p2, . . . , pk, . . . , pn) ,

where ak means we omit the entry ak. Then the multivariate Dedekind–Bernoulli

12

13

sum is

SPk

Ak ak

Xk xk

=∑

h mod ak

n∏i=1i6=k

Bpi

(ai

h + xk

ak

− xi

). (3.1)

We see that if (a1, a2, a3) = (a, b, c), (x1, x2, x3) = (x, y, z), and P3 = (m, n) then

A3 = (a, b), X3 = (x, y) and thus we recover Hall, Wilson and Zagier’s generalized

Dedekind–Rademacher sum,

Sm,n

A3 c

X3 z

=∑

h mod c

Bm

(ah + z

c− x

)Bn

(bh + z

c− y

).

Moreover, by letting (x1, x2, . . . , xn) = ~0 and Pk = (p1, p2, . . . , pk, . . . , pn), we

recover another generalization of Hall, Wilson and Zagier’s generalized Dedekind–

Rademacher sum introduced by Bayad and Raouj in 2009 [3]. Bayad and Raouj’s

multiple Dedekind–Rademacher sum is thus defined

SPk

Ak ak

Pk pk

=∑

h mod ak

n∏i=1i6=k

Bpi

(aih

ak

). (3.2)

Similar to the generalized Dedekind–Rademacher sum, a multivariate Dedekind–

Bernoulli sum reciprocity relation mixes various (n− 1)-tuples of indices and is

most conveniently stated in terms of generating functions. For nonzero variables

14

(y1, y2, . . . , yn) where yn = −y1 − y2 − · · · − yn−1, let Yk = (y1, y2, . . . , yk, . . . , yn).

Then

Ω

Ak ak

Xk xk

Yk yk

=

∑p1

. . .∑pk

· · ·∑pn

1

p1!p2! · · · pk−1!pk+1! · · · pn!SPk

Ak ak

Xk xk

n∏i=1i6=k

(yi

ai

)pi−1

.

The series of summands∑

p1. . .

∑pk· · ·

∑pn

is understood as summations over all

non-negative integers p1, . . . , pk, . . . , pn.

Our main result is the following reciprocity law involving multivariate Dedekind–

Bernoulli sums.

Theorem 3.1. Let (x1, x2, . . . , xn) ∈ R and (p1, p2, . . . , pn) , (a1, a2, . . . , an) ∈ Z>0

where (au, av) = 1 for all 1 ≤ u < v ≤ n. For 1 ≤ k ≤ n, let

Ak = (a1, a2, . . . , ak, . . . , an) , Xk = (x1, x2, . . . , xk, . . . , xn)

and Pk = (p1, p2, . . . , pk, . . . , pn) .

For nonzero variables y1, y2, . . . , yn, let Yk = (y1, y2, . . . , yk, . . . , yn) such that y1 +

15

y2 + · · ·+ yn = 0, then

n∑k=1

Ω

Ak ak

Xk xk

Yk yk

= 0

if xu−hu

au− xv−hv

av6∈ Z whenever 1 ≤ u < v ≤ n and hu, hv ∈ Z.

We now embark on a journey of transforming Ω

Ak ak

Xk xk

Yk yk

into a simplified and

accessible form that will make proving Theorem 3.1 as easy as a matching game.

Before we continue, we first introduce two useful lemmas for the fractional-part

function.

Lemma 3.2. Given a, b, c ∈ R,

a− b − a− c > 0 ⇒ a− b − a− c = c− b.

Proof. Our goal is to show

a− b − a− c − c− b = 0.

By definition x = x − ∗, where ∗ is an integer and x ∈ R. Thus, the above

16

sum can be rewritten as follows,

a− b − a− c − c− b

= a− b− ∗1 − (a− c− ∗2)− (c− b− ∗3)

= − ∗1 + ∗2 +∗3

= ∗,

where all ∗’s are integers.

Given our assumption and the fact that 0 ≤ x < 1, it follows

0 ≤ a− b − a− c < 1

⇒ −1 < a− b − a− c − c− b < 1

⇒ −1 < ∗ < 1

⇒ ∗ = 0 .

Lemma 3.3. Given a, b, c ∈ R,

a− b − a− c < 0 ⇒ a− b − a− c = −b− c.

17

Proof. Our goal is to show

a− b − a− c+ b− c = 0.

By definition x = x − ∗, where ∗ is an integer and x ∈ R. Thus, the above

sum can be rewritten as follows,

a− b − a− c+ b− c

= a− b− ∗1 − (a− c− ∗2) + b− c− ∗3

= − ∗1 + ∗2 −∗3

= ∗,

where all ∗’s are integers.

Given our assumption, 0 ≤ x < 1, and 0 ≤ b− c < 1, it follows

−1 < a− b − a− c < 0

⇒ −1 < a− b − a− c+ b− c < 1

⇒ −1 < ∗ < 1

⇒ ∗ = 0 .

18

We will now begin manipulating Ω

Ak ak

Xk xk

Yk yk

using the following identity for

multivariate Dedekind–Bernoulli sums.

SPk

Ak ak

Xk xk

n∏i=1i6=k

a1−mii =

∑h mod ak

n∏i=1i6=k

Bpi

(ai

h + xk

ak

− xi

)a1−mi

i

(applying 2.2, Raabe’s formula)

=∑H

n∏i=1i6=k

Bpi

(xk + hk

ak

− xi + hi

ai

),

where∑

H =∑

h1 mod a1

∑h2 mod a2

· · ·∑

hn mod anincludes the original summand h

19

we now call hk. Let ri = xi+hi

ai, then

Ω

Ak ak

Xk xk

Yk yk

=

∑p1

. . .∑pk

· · ·∑pn

1

p1!p2! · · · pk−1!pk+1! · · · pn!SPk

Ak ak

Xk xk

n∏i=1i6=k

(yi

ai

)pi−1

=∑H

∑p1

. . .∑pk

· · ·∑pn

1

p1!p2! · · · pk−1!pk+1! · · · pn!

n∏i=1i6=k

Bpi(rk − ri) · ypi−1

i

=∑H

n∏i=1i6=k

β (rk − ri, yi) , (3.3)

where Definition 2.3 is applied in the final equality. It is now clear that (3.3) depends

on the differences rk− ri for 1 ≤ k < i ≤ n and all β (rk − ri, yi) depend on whether

or not these differences are integers (see 2.3). From now on we assume the differences

are not integers, which is analogous to the generic case of Theorem 2.4.

20

Consider the overall sum of Ω

Ak ak

Xk xk

Yk yk

as k ranges from 1 to n.

n∑k=1

Ω

Ak ak

Xk xk

Yk yk

=n∑

k=1

∑H

n∏i=1i6=k

β (rk − ri, yi)

=n−1∑k=1

∑H

n∏i=1i6=k

β (rk − ri, yi) +∑H

n−1∏j=1

β (rn − rj, yj)

(since rk − ri 6= 0 for each 1 ≤ k < i ≤ n, we apply Lemma 2.3)

=n−1∑k=1

∑H

n∏i=1i6=k

erk−riyi

eyi − 1+

∑H

n−1∏j=1

ern−rjyj

eyj − 1

=n−1∑k=1

∑H

n∏i=1i6=k

erk−riyi

eyi − 1· eyk − 1

eyk − 1+

∑H

n−1∏j=1

ern−rjyj

eyj − 1· eyn − 1

eyn − 1

=

∑n−1k=1

∑H

∏ni=1i6=k

erk−riyi · eyk −∑n−1

k=1

∑H

∏ni=1i6=k

erk−riyi

+∑

H

∏n−1j=1 ern−rjyj · eyn −

∑H

∏n−1j=1 ern−rjyj∏n

f=1 (eyf − 1).

21

We will drop the denominator and continue to focus only on the numerator:

n−1∑k=1

∑H

erk−rnyn+yk

n−1∏i=1i6=k

erk−riyi −n−1∑k=1

∑H

erk−rnyn

n−1∏i=1i6=k

erk−riyi

+∑H

n−1∏j=1

ern−rjyj · eyn −∑H

n−1∏j=1

ern−rjyj

=n−1∑k=1

∑H

erk−rn(−y1−···−yn−1)+yk

n−1∏i=1i6=k

erk−riyi

−n−1∑k=1

∑H

erk−rn(−y1−···−yn−1)

n−1∏i=1i6=k

erk−riyi

+∑H

n−1∏j=1

ern−rjyj · e(−y1−···−yn−1) −∑H

n−1∏j=1

ern−rjyj .

Recall that 1−ri − rj = −rj − ri since ri − rj 6∈ Z. Thus the numerator equals

n−1∑k=1

∑H

ern−rkyk

n−1∏i=1i6=k

e(rk−ri−rk−rn)yi

−n−1∑k=1

∑H

e−rk−rnyk

n−1∏i=1i6=k

e(rk−ri−rk−rn)yi (3.4)

+∑H

n−1∏j=1

e−rj−rnyj −∑H

n−1∏j=1

ern−rjyj .

In order to prove Theorem 3.1, we will show that the exponents of opposite signed

terms in the numerator can be paired and so the sum vanishes. Examining (3.4),

22

we can see that only three types of exponents appear, rn − ri,−ri − rn, and

ri − rj − ri − rn. A critical charasteristic of the latter difference is that, by

applying Lemma 3.2 or Lemma 3.3, rj − ri− rj − rn = rn − ri or −ri − rn,

respectively. This means the exponents can be condensed to just the first two forms,

rn − ri and −ri − rn. Moreover, we can represent these numbers by their sign

since the sign of rj−ri−rj−rn determines if it is equal to rn−ri or −ri−rn.

This lends us to the following representations:

rn − ri = +, −ri − rn = − and Cij = sign (rj − ri − rj − rn) .

Thus every term of∑n

k=1 Ω

Ak ak

Xk xk

Yk yk

can be represented by a sign vector, which

we will state explicitly.

The exponent corresponding to k = 1 in

n−1∑k=1

∑H

ern−rkyk

n−1∏i=1i6=k

e(rk−ri−rk−rn)yi

is

rn − r1y1 + (r1 − r2 − r1 − rn) y2 + · · ·+ (r1 − rn−1 − r1 − rn) yn−1

23

and is represented as the sign vector

(+, C12, . . . , C1,n−1) .

Similarly, the (k = 1)-exponent in

−n−1∑k=1

∑H

e−rk−rnyk

n−1∏i=1i6=k

e(rk−ri−rk−rn)yi

is

−r1 − rny1 + (r1 − r2 − r1 − rn) y2 + · · ·+ (r1 − rn−1 − r1 − rn) yn−1,

and the sign vector representation is

(−, C12, . . . , C1,n−1) .

Finally, the terms

∑H

n−1∏j=1

e−rj−rnyj and −∑H

n−1∏j=1

ern−rjyj

24

are represented as the respective sign vectors

(−,−, . . . ,−) and (+, +, . . . , +) .

Note that if the sign of, say, r1−rn and r4−rn are the same, it is not necessarily

true that r1−rn = r4−rn. We will address this futher in the following argument.

To prove that the terms of∑n

k=1 Ω

Ak ak

Xk xk

Yk yk

do in fact cancel, we proceed to

construct two matrices Mn and M′n that consist of the sign vector representations of

the terms of∑n

k=1 Ω

Ak ak

Xk xk

Yk yk

and will show that∑n

k=1 Ω

Ak ak

Xk xk

Yk yk

= 0 equates

to proving Mn = M′n after row swapping.

Let Mn be the matrix of all sign vectors representing the exponents of the positive

terms of∑n

k=1 Ω

Ak ak

Xk xk

Yk yk

and let M′n be the matrix of all sign vectors representing

the exponents of the negative terms. This means the sign vector that represents

every exponent from the positive terms∑n−1

k=1

∑H ern−rkyk

∏n−1i=1i6=k

e(rk−ri−rk−rn)yi

for each k is placed in the kth row of matrix Mn.

Similarly, the sign vector representation for every exponent from the negative

terms −∑n−1

k=1

∑H e−rk−rnyk

∏n−1i=1i6=k

e(rk−ri−rk−rn)yi for each k is placed in the kth

25

row of the matrix M′n.

Finally, place the sign vector representing∑

H

∏n−1j=1 e−rj−rnyj in the last row

of Mn and the sign vector representing −∑

H

∏n−1j=1 ern−rjyj in the last row of M

′n.

Notice that the placement of entry Cki depends on the indices k and i of each

term in∑n

k=1 Ω

Ak ak

Xk xk

Yk yk

and given the symmetry among the terms, Cki lives

in the same row and column in both matrices. This latter fact implies that if

sign (rn − ri) = sign (rn − rj) then rn− ri = rn− rj only when i = j. Thus

we have constructed the following matrices,

Mn =

+ C12 C13 · · · C1,n−1

C21 + C23 · · · C2,n−1

C31 C32 + · · · C3,n−1

......

.... . .

...

Cn−1,1 Cn−1,2 Cn−1,3 · · · +

− − · · · − −

26

and

M′

n =

− C12 C13 · · · C1,n−1

C21 − C23 · · · C2,n−1

C31 C32 − · · · C3,n−1

......

.... . .

...

Cn−1,1 Cn−1,2 Cn−1,3 · · · −

+ + · · · + +

.

Last, we will show that Mn = M′n after row swapping implies

∑nk=1 Ω

Ak ak

Xk xk

Yk yk

vanishes. Assume Mn = M

′n up to row swapping. Then for each sign row vector

(Cs1, . . . , Cs,s−1, +, Cs,s+1, . . . , Cs,n−1) ∈ Mn

there exists

(Ct1, . . . , Ct,t−1,−, Ct,t+1, . . . , Ct,n−1) ∈ M′

n

such that

(Cs1, . . . , Cs,s−1, +, Cs,s+1, . . . , Cs,n−1) = (Ct1, . . . , Ct,t−1,−, Ct,t+1, . . . , Ct,n−1) .

27

Also, we have for some row f in Mn

(Cf1, . . . , Cf,f−1, +, Cf,f+1, . . . , Cf,n−1) = (+, . . . , +)

and for some row g in M′n

(−, . . . ,−) = (Cg1, . . . , Cg,g−1, +, Cg,g+1, . . . , Cg,n−1) .

For each identity, we will show that the sign row vectors correspond to canceling

terms of∑n

k=1 Ω

Ak ak

Xk xk

Yk yk

. Beginning with (Cs1, . . . , Cs,s−1, +, Cs,s+1, . . . , Cs,n−1) ∈

Mn, this corresponds to the exponent

(rs − r1 − rs − rn) y1 + · · ·+ (rs − rs−1 − rs − rn) ys−1 + rn − rsys

+ (rs − rs+1 − rs − rn) ys+1 + · · ·+ (rs − rn−1 − rs − rn) yn−1,

and more specifically, represents the positive term

e

0BBBB@(rs − r1 − rs − rn) y1 + · · ·+ (rs − rs−1 − rs − rn) ys−1 + rn − rsys

+ (rs − rs+1 − rs − rn) ys+1 + · · ·+ (rs − rn−1 − rs − rn) yn−1

1CCCCA.

28

Similarly, (Ct1, Ct2, . . . , Ct,t−1,−, Ct,t+1, . . . , Ct,n−1) ∈ M′n represents the exponent

(rt − r1 − rt − rn) y1 + · · ·+ (rt − rt−1 − rt − rn) yt−1 − rt − rtyt

+ (rt − rt+1 − rt − rn) yt+1 + · · ·+ (rt − rn−1 − rt − rn) yn−1,

and so also the negative term

−e

0BBBB@(rt − r1 − rt − rn) y1 + · · ·+ (rt − rt−1 − rt − rn) yt−1 − rt − rtyt

+ (rt − rt+1 − rt − rn) yt+1 + · · ·+ (rt − rn−1 − rt − rn) yn−1

1CCCCA.

Then

(Cs1, . . . , Cs,s−1, +, Cs,s+1, . . . , Cs,n−1) = (Ct1, . . . , Ct,t−1,−, Ct,t+1, . . . , Ct,n−1) ,

29

implies

sign (rs − r1 − rs − rn) y1 + · · ·+ sign (rs − rs−1 − rs − rn) ys−1

+ rn − rsys + sign (rs − rs+1 − rs − rn) ys+1

+ · · ·+ sign (rs − rn−1 − rs − rn) yn−1

= sign (rt − r1 − rt − rn) y1 + · · ·+ sign (rt − rt−1 − rt − rn) yt−1

− rt − rtyt + sign (rt − rt+1 − rt − rn) yt+1

+ · · ·+ sign (rt − rn−1 − rt − rn) yn−1.

If sign (rs − ri − rs − rn) = sign (rt − ri − rt − rn) then, by Lemmas 3.2

and 3.3, (rs − ri − rs − rn) = (rt − ri − rt − rn). Thus,

(rs − r1 − rs − rn) y1 + · · ·+ (rs − rs−1 − rs − rn) ys−1 + rn − rsys

+ (rs − rs+1 − rs − rn) ys+1 + · · ·+ (rs − rn−1 − rs − rn) yn−1

= (rt − r1 − rt − rn) y1 + · · ·+ (rt − rt−1 − rt − rn) yt−1 − rt − rtyt

+ (rt − rt+1 − rt − rn) yt+1 + · · ·+ (rt − rn−1 − rt − rn) yn−1

30

and so

e

0BBBB@(rs − r1 − rs − rn) y1 + · · ·+ (rs − rs−1 − rs − rn) ys−1 + rn − rsys

+ (rs − rs+1 − rs − rn) ys+1 + · · ·+ (rs − rn−1 − rs − rn) yn−1

1CCCCA

− e

0BBBB@(rt − r1 − rt − rn) y1 + · · ·+ (rt − rt−1 − rt − rn) yt−1 − rt − rtyt

+ (rt − rt+1 − rt − rn) yt+1 + · · ·+ (rt − rn−1 − rt − rn) yn−1

1CCCCA

= 0.

As for the identities

(Cf1, Cf2, . . . , Cf,f−1, +, Cf,f+1, . . . , Cf,n−1) = (+, +, . . . , +)

and

(−,−, . . . ,−) = (Cg1, Cg2, . . . , Cg,g−1, +, Cg,g+1, . . . , Cg,n−1) ,

a similar argument follows.

(Cf1, Cf2, . . . , Cf,f−1, +, Cf,f+1, . . . , Cf,n−1) represents the exponent

(rf − r1 − rf − rn) y1 + · · ·+ (rf − rf−1 − rf − rn) yf−1 + rn − rfyf

+ (rf − rf+1 − rf − rn) yf+1 + · · ·+ (rf − rn−1 − rf − rn) yn−1

31

and thus the term

e

0BBBB@(rf − r1 − rf − rn) y1 + · · ·+ (rf − rf−1 − rf − rn) yf−1 + rn − rfyf

+ (rf − rf+1 − rf − rn) yf+1 + · · ·+ (rf − rn−1 − rf − rn) yn−1

1CCCCA.

The identity

(Cf1, Cf2, . . . , Cf,f−1, +, Cf,f+1, . . . , Cf,n−1) = (+, +, . . . , +)

means rf − ri − rf − rn = rn − ri. Thus

e

0BBBB@(rf − r1 − rf − rn) y1 + · · ·+ (rf − rf−1 − rf − rn) yf−1 + rn − rfyf

+ (rf − rf+1 − rf − rn) yf+1 + · · ·+ (rf − rn−1 − rf − rn) yn−1

1CCCCA

= e

0BBBB@rn − r1y1 + · · ·+ rn − rf−1yf−1 − rf − rnyf

+rn − rf+1yf+1 + · · ·+ rn − rn−1yn−1

1CCCCA

=∑H

n−1∏i=1

ern−riyi .

The exponent of the negative term −∑

H

∏n−1j=1 ern−rjyj is represented by the sign

vector (+, +, . . . , +) which cancels with the above.

32

(Cg1, Cg2, . . . , Cg,g−1, +, Cg,g+1, . . . , Cg,n−1) represents the exponent

(rg − r1 − rg − rn) y1 + · · ·+ (rg − rg−1 − rg − rn) yg−1 − rg − rnyg

+ (rg − rg+1 − rg − rn) yg+1 + · · ·+ (rg − rn−1 − rg − rn) yn−1

and thus the negative term

−e

0BBBB@(rg − r1 − rg − rn) y1 + · · ·+ (rg − rg−1 − rg − rn) yg−1 − rg − rnyg

+ (rg − rg+1 − rg − rn) yg+1 + · · ·+ (rg − rn−1 − rg − rn) yn−1

1CCCCA.

The identity

(−,−, . . . ,−) = (Cg1, Cg2, . . . , Cg,g−1, +, Cg,g+1, . . . , Cg,n−1)

means rg − ri − rg − rn = −ri − rn. Thus

−e

0BBBB@(rg − r1 − rg − rn) y1 + · · ·+ (rg − rg−1 − rg − rn) yg−1 − rg − rnyg

+ (rg − rg+1 − rg − rn) yg+1 + · · ·+ (rg − rn−1 − rg − rn) yn−1

1CCCCA

= −e

0BBBB@−r1 − rny1 − · · · − rg−1 − rnyg−1 − rg − rnyg

−rg+1 − rnyg+1 − · · · − rn−1 − rnyn−1

1CCCCA

= −∑H

n−1∏i=1

e−ri−rnyi .

33

The exponent of the positive term∑

H

∏n−1j=1 e−rj−rnyj is represented by the sign

vector (−,−, . . . ,−), which cancels with the above.

Thus the matching of rows of the matrices Mn and M′n correspond to the match-

ing of like terms of opposite signs and we have∑n

k=1 Ω

Ak ak

Xk xk

Yk yk

= 0. We have

just revealed the potential of our main result,

Mn = M′

n, up to row swapping, impliesn∑

k=1

Ω

Ak ak

Xk xk

Yk yk

= 0.

We must now show that under certain conditions, we can indeed find Mn = M′n up

to row swapping. By proving the following properties of the matrices Mn and M′n,

we are able to quickly show Mn = M′n up to row swapping and thus our reciprocity

theorem follows.

Lemma 3.4. Mn and M′n are of the form such that Mn has all + entries on the

diagonal and the last row has all − entries, and M′n has all − entries on the diagonal

and the last row has all + entries.

Proof. As was explained above, the diagonal entries of Mn (excluding the last row

34

of −’s) represent the terms in the exponent of the sum

n−1∑k=1

∑H

ern−rkyk

n−1∏i=1i6=k

e(rk−ri−rk−rn)yi ;

when k = i, the exponent is rn − rk = +. Likewise, the diagonal entries of M′n

(excluding the last row of +’s) represent the terms in the exponent of the sum

n−1∑k=1

∑H

e−rk−rnyk

n−1∏i=1i6=k

e(rk−ri−rk−rn)yi ;

when k = i, this exponent is −rk − rn = −.

Lemma 3.5. As above, let Cij = sign (ri − rj − ri − rn) where i, j, n ∈ Z>0

and 1 ≤ i < j ≤ n− 1. Then Cij = + if and only if Cji = −.

Proof. Assume Cij = +. Then ri − rj − ri − rn ≥ 0 and by Lemma 3.2

ri − rj − ri − rn = rn − rj. We want to show Cji = − which is to show

rj − ri − rj − rn < 0.

rj − ri − rj − rn = rj − ri − 1 + 1− rj − rn

= −ri − rj+ rn − rj

= −ri − rn

⇔ Cji = −.

35

Assume Cij = −. Then ri− rj−ri− rn < 0 and by Lemma 3.3 ri− rj−ri−

rn = −rj−rn. We want to show Cji = + which is to show rj−ri−rj−rn > 0.

rj − ri − rj − rn = rj − ri+ ri − rj − ri − rn

= 1− ri − rn

= rn − ri

⇔ Cji = +,

recalling that rj − ri+ ri − rj = 1.

By Lemma 3.5, it follows that the sign of one difference of fractional parts ri −

rj − ri − rn is dependent on the sign of another difference of fractional parts

rj − ri− rj − rn. The lemma implies Cij determines Cji. Thus, we will rename

Cji = −Cij, meaning if Cij = + then Cji = − and if Cij = − then Cji = +. Thus,

we can update the sign vector matrices utilizing this new information as

36

Mn =

+ C12 C13 · · · C1,n−1

−C12 + C23 · · · C2,n−1

−C13 −C23 + · · · C3,n−1

......

.... . .

...

−C1,n−1 −C2,n−1 −C3,n−1 · · · +

− − · · · − −

and

M′

n =

− C12 C13 · · · C1,n−1

−C12 − C23 · · · C2,n−1

−C13 −C23 − · · · C3,n−1

......

.... . .

...

−C1,n−1 −C2,n−1 −C3,n−1 · · · −

+ + · · · + +

,

and we can state the following property.

Lemma 3.6. Mn and M′n exhibit antisymmetry about the diagonal.

Lemma 3.7. As above, let Cij = sign (ri − rj − ri − rn) where i, j, n ∈ Z>0

and 1 ≤ i < j ≤ n− 1.

If Cij = + and Cik = − then Cjk = −.

37

Proof. Assume Cij = + and Cik = −. Then ri − rj − ri − rn > 0 and ri −

rk − ri − rn < 0, and by Lemmas 3.2 and 3.3

ri − rj − ri − rn = rn − rj (3.5)

and

ri − rk − ri − rn = −rk − ri. (3.6)

Then the difference (3.5)-(3.6) is positive and we get

ri − rj − ri − rk = rn − rj+ rk − ri.

The final identity is positive, which means the left hand side is positive. Then by

Lemma 3.2

ri − rj − ri − rk = rk − rj,

and we have

rk − rj = rn − rj+ rk − ri.

38

We want to show Cjk = − which reduces to showing rj − rk − rj − rn < 0.

rj − rk − rj − rn = rj − rk − 1 + 1− rj − rn

= −rk − rj+ rn − rj

= −rk − ri

⇒ Cjk = −.

Lemma 3.8. There exists a unique row with k +’s, for each 0 ≤ k ≤ n− 1, in the

matrix Mn.

Proof. We begin by showing that every row of the matrix Mn is unique. Assume on

the contrary that row m and row l of Mn are equal. Then we can view the rows as

follows:

row m: −C1m −C2m · · · + · · · Cml · · · Cm,n−1

row l: −C1l −C2l · · · −Cml · · · + · · · Cl,n−1

.

Then Cml = + and −Cml = +, but by Lemma 3.5, Cml = + implies −Cml = −, a

contradiction. Therefore, the rows of the matrix Mn are unique.

Next, we will show that no two rows contain the same number of +’s. Assume

on the contrary that row m and row l of Mn contain exactly i +’s, are not equal

39

(and thus 1 ≤ i < n−1) and look the same as above. We will examine the elements

Cml of row m and −Cml of row l. Let Cml = +. Since the mth row does not contain

only +’s, there exists a − in column, say, w. Then by Lemma 3.7, the entry Cwl is

−. So, for every − in row m, Lemma 3.7 can be applied to show there is a − in the

same column entry of row l. But row l also contains a − on the diagonal since it is

a row of the matrix M′n. Thus, row l contains i− 1 many +’s, a contradiction. The

same argument holds for the entry −Cml = + of row l. Thus, no two rows can have

an equal number of +’s.

We’ve shown no two rows contain the same number of +’s and that every row

is unique. Thus, for each 0 ≤ k ≤ n− 1, there exists a unique row with k +’s.

Lemma 3.9. There exists a unique row with k +’s, for each 0 ≤ k ≤ n− 1, in the

matrix M′n.

Proof. Mn and M′n share all entries Cij for 1 ≤ i < j ≤ n − 1. Then the k +’s in

the kth row of Mn determine that the kth row of M′n contains k − 1 +’s since the

+ diagonal entry of M′n becomes a −. Therefore, M

′n exhibits the same uniqueness

property of Mn.

We can now prove our main Theorem 3.1.

Proof. Our goal is to show∑n

k=1 Ω

Ak ak

Xk xk

Yk yk

. Assume xu−hu

au− xv−hv

av6∈ Z whenever

40

1 ≤ u < v ≤ n and hu, hv ∈ Z which allows us to implement the following matrix

argument. We’ve seen that∑n

k=1 Ω

Ak ak

Xk xk

Yk yk

gives rise to the following matrices

Mn =

+ C12 C13 · · · C1,n−1

−C12 + C23 · · · C2,n−1

−C13 −C23 + · · · C3,n−1

......

.... . .

...

−C1,n −C2,n−1 −C3,n−1 · · · +

− − · · · − −

and

M′

n =

− C12 C13 · · · C1,n−1

−C12 − C23 · · · C2,n−1

−C13 −C23 − · · · C3,n−1

......

.... . .

...

−C1,n−1 −C2,n−1 −C3,n−1 · · · −

+ + · · · + +

,

41

where Mn represents the exponents of all positive terms of∑n

k=1 Ω

Ak ak

Xk xk

Yk yk

and

M′n represents all negative terms. As was shown above, to prove that the matrices

Mn and M′n contain the same rows is equivalent to showing every positive term of

∑nk=1 Ω

Ak ak

Xk xk

Yk yk

has a canceling negative term. Thus our problem reduces to

showing Mn = M′n after row swapping.

We’ve proved in Lemma 3.4 that Mn has +’s on the diagonal and in row n + 1,

M′n has −’s on the diagonal and in row n + 1, Lemma 3.6 shows both matrices

exhibit antisymmetry, and Lemma 3.7 tells us that two opposite signed entries in

a row off the diagonal imply the sign of another entry in the matrix is a −. These

three lemmas lead to conclude Lemma 3.8 which states that for both matrices Mn

and M′n there exists a unique row of k +’s for every 1 ≤ k ≤ n − 1. We can use

row swapping to place the unique rows of k +’s in the same row of matrix Mn as

they appear in matrix M′n such that Mn = M

′n, which, by our previous argument,

implies∑n

k=1 Ω

Ak ak

Xk xk

Yk yk

= 0.

Chapter 4

A New Proof of Hall–Wilson–Zagier’s

Reciprocity Theorem

The original proof of Hall, Wilson and Zagier’s reciprocity theorem for the general-

ized Dedekind–Rademacher sum ultimately uses cotangent identities to show

Ω

a b c

x y z

X Y Z

+ Ω

b c a

y z x

Y Z X

+ Ω

c a b

z x y

Z X Y

=

−1/4 if (x, y, z) ∈ (a, b, c)R + Z3,

0 otherwise.

42

43

Here we give an example of applying the approach used to prove Theorem 3.1,

simplifying the proof of Theorem 2.4.

New proof of Theorem 2.4. Let (a1, a2, a3) ∈ Z>0 and (x1, x2, x3) ∈ R. For non-zero

variables y1, y2, y3 such that y1 + y2 + y3 = 0 we want to show that

Ω

a1 a2 a3

x1 x2 x3

y1 y2 y3

+ Ω

a2 a3 a1

x2 x3 x1

y2 y3 y1

+ Ω

a3 a1 a2

x3 x1 x2

y3 y1 y2

=

−1/4 if (x1, x2, x3) ∈ (a1, a2, a3)R + Z3,

0 otherwise.

Recall identity (3.3),

n∑k=1

Ω

Ak ak

Xk xk

Yk yk

=n∑

k=1

∑H

n∏i=1i6=k

β (rk − ri, yi) ,

where

β(α, Y ) =

12· eY +1

eY −1for α ∈ Z,

eαY

eY −1for α 6∈ Z,

44

and∑

H =∑

h1 mod a1

∑h2 mod a2

· · ·∑

hn mod an. Then we have

Ω

a1 a2 a3

x1 x2 x3

y1 y2 y3

+ Ω

a2 a3 a1

x2 x3 x1

y2 y3 y1

+ Ω

a3 a1 a2

x3 x1 x2

y3 y1 y2

=

3∑k=1

∑H

3∏i=1i6=k

β (rk − ri, yi) .

We must examine the following cases.

(i) (x1, x2, x3) ∈ (a1, a2, a3)R + Z3;

(ii) (xi, xj) ∈ (ai, aj)R + Z2 for some 1 ≤ i < j ≤ 3 but not (i);

(iii) None of the above.

We will begin with case (i). Let (x1, x2, x3) ∈ (a1, a2, a3)R+Z3. Then xi = λai+zi

for each i where λ ∈ R and zi ∈ Z. Thus

hi + xi

ai

− hj + xj

aj

=hi + λai + zi

ai

− hj + λaj + zj

aj

=hi + zi

ai

− hj + zj

aj

. (4.1)

Since hi + zi only permutes the modular values of hi, we can make a change in

45

indices (letting hi − zi = hi) so that (4.1) becomes

hi

ai

− hj

aj

and

Ω

a1 a2 a3

x1 x2 x3

y1 y2 y3

+ Ω

a2 a3 a1

x2 x3 x1

y2 y3 y1

+ Ω

a3 a1 a2

x3 x1 x2

y3 y1 y2

=

3∑k=1

∑H

3∏i=1i6=k

β (rk − ri, yi)

=3∑

k=1

∑H

3∏i=1i6=k

β

(hk

ak

− hi

ai

, yi

)

=3∑

k=1

∑H

3∏i=1i6=k

β

(hk

ak

− hi

ai

, yi

),

since we sum hi over a complete residue system modai.

Since (ai, aj) = 1, then hi

ai− hj

aj∈ Z occurs only when hi = hj = 0. Thus, we

can split up the sum∑3

k=1

∑H

∏3i=1i6=k

β(

hk

al− hi

ai, yi

)into two parts, a term when all

46

hi = 0 and a sum over the remaining hi ∈ Z where (h1, h2, h3) 6= (0, 0, 0)

3∑k=1

∑H

3∏i=1i6=k

β

(hk

al

− hi

ai

, yi

)

=β (0, y2) β (0, y3) + β (0, y1) β (0, y3) + β (0, y1) β (0, y2)

+∑

H\0,0,0

β

(h1

a1− h2

a2, y2

)β

(h1

a1− h3

a3, y3

)+β

(h2

a2− h1

a1, y1

)β

(h2

a2− h3

a3, y3

)+β

(h3

a3− h1

a1, y1

)β

(h3

a3− h2

a2, y2

) .

Let us first address the term when all hi = 0. We use the following cotangent

identities

cot (y) = ie2πy + 1

e2πy − 1, (4.2)

cot (α) + cot (β) =cot (α) cot (β)− 1

cot (α + β). (4.3)

Combined with the definition of cot y, β(0, y) becomes

β(0, y) =1

icot

y

2i. (4.4)

47

Let y∗k = yk

2i. Then

β (0, y2) β (0, y3) + β (0, y1) β (0, y3) + β (0, y1) β (0, y2)

=1

4

ey2 + 1

ey2 − 1

ey3 + 1

ey3 − 1+

1

4

ey1 + 1

ey1 − 1

ey3 + 1

ey3 − 1+

1

4

ey1 + 1

ey1 − 1

ey2 + 1

ey2 − 1

= −1

4(cot y∗2 cot y∗3 + cot y∗1 cot y∗3 + cot y∗1 cot y∗2)

= −1

4(cot y∗2 (cot y∗3 + cot y∗1) + cot y∗1 cot y∗3)

= −1

4

(cot y∗2

(cot y∗1 cot y∗3 − 1

cot y∗1 + y∗3

)+ cot y∗1 cot y∗3

).

By assumption, y1 + y2 + y3 = 0, so it follows that y∗1 + y∗3 = −y∗2 and

cot (y∗1 + y∗3) = cot (−y∗2) = − cot y∗2.

Thus,

−1

4

(cot y∗2

(cot y∗1 cot y∗3 − 1

cot y∗1 + y∗3

)+ cot y∗1 cot y∗3

)= −1

4

(cot y∗2

(cot y∗1 cot y∗3 − 1

− cot y∗2

)+ cot y∗1 cot y∗3

)= −1

4((− cot y∗1 cot y∗3 + 1) + cot y∗1 cot y∗3)

= −1

4.

It is left to show that the summand over H\0, 0, 0 vanish. But we’ve seen this

48

sum before. It is just

3∑k=1

∑H\0,0,0

3∏i=1i6=k

β (rk − ri, yi)

when (x1, x2, x3) = ~0. Since rk − ri 6∈ Z, we can apply the same matrix argument

from the proof of Theorem 3.1 and the terms cancel.

Assume case (ii). Let (xi, xj) ∈ (ai, aj)R + Z2 for some 1 ≤ i < j ≤ 3 but not

(i). Without loss of generality, we assume (x1, x2) ∈ (a1, a2)R+Z2. Then, as before,

x1 = λa1 + z1 and x2 = λa2 + z2 for λ ∈ R and z1, z2 ∈ Z so that

h1 + x1

a1

− h2 + x2

a2

=h1 + λa1 + z1

a1

− h2 + λa2 + z2

a2

=h1 + z1

a1

− h2 + z2

a2

.

Since z1 and z2 permute the summands over h1 and h2, we introduce a change of

variables and let h1 = h1 + z1 and h2 = h2 + z2. We can rewrite the differences

involving r3 as

hi + xi

ai

− h3 + x3

a3

=

(hi

ai

+ λ− r3

)=

(hi

ai

− h3 + x3 − λa3

a3

)=

(hi

ai

− r3

).

49

Then

Ω

a1 a2 a3

x1 x2 x3

y1 y2 y3

+ Ω

a2 a3 a1

x2 x3 x1

y2 y3 y1

+ Ω

a3 a1 a2

x3 x1 x2

y3 y1 y2

=

3∑k=1

∑H

3∏i=1i6=k

β (rk − ri, yi)

=∑

h1 mod a1

∑h2 mod a2

∑h3 mod a3

β

(h1

a1− h2

a2, y2

)β

(h1

a1− r3, y3

)+β

(h2

a2− h1

a1, y1

)β

(h2

a2− r3, y3

)+β

(r3 − h1

a1− λ, y1

)β

(r3 − h2

a2, y2

)

=∑H

β

(h1

a1− h2

a2, y2

)β

(h1

a1− r3, y3

)+β

(h2

a2− h1

a1, y1

)β

(h2

a2− r3, y3

)+β

(r3 − h1

a1, y1

)β

(r3 − h2

a2, y2

) .

We will again split up the final expression into two parts, one part includes all terms

where h1 = h2 = 0 and the other part are all summands where h1 and h2 are not

50

both zero

∑H

β

(h1

a1− h2

a2, y2

)β

(h1

a1− r3, y3

)+β

(h2

a2− h1

a1, y1

)β

(h2

a2− r3, y3

)+β

(r3 − h1

a1, y1

)β

(r3 − h2

a2, y2

)

=∑

h3 mod a3

β (0, y2) β (−r3, y3)

+β (0, y1) β (−r3, y3)

+β (r3, y1) β (r3, y2)

+∑H

(h1,h2) 6=(0,0)

β

(h1

a1− h2

a2, y2

)β

(h1

a1− r3, y3

)+β

(h2

a2− h1

a1, y1

)β

(h2

a2− r3, y3

)+β

(r3 − h1

a1, y1

)β

(r3 − h2

a2, y2

) .

First, we show

∑H

β

(h1

a1− h2

a2, y2

)β

(h1

a1− r3, y3

)+β

(h2

a2− h1

a1, y1

)β

(h2

a2− r3, y3

)+β

(r3 − h1

a1, y1

)β

(r3 − h2

a2, y2

) = 0.

51

∑H

β

(h1

a1− h2

a2, y2

)β

(h1

a1− r3, y3

)+β

(h2

a2− h1

a1, y1

)β

(h2

a2− r3, y3

)+β

(r3 − h1

a1, y1

)β

(r3 − h2

a2, y2

)

=∑

h3 mod a3

(1

2

ey2 + 1

ey2 − 1· e−r3y3

ey3 − 1+

1

2

ey1 + 1

ey1 − 1· e−r3y3

ey3 − 1+

er3y1

ey1 − 1· er3y2

ey2 − 1

).

After multiplying each term by the common denominator and combining terms, we

get the numerator

∑h3 mod a3

ey1+y2+−r3y3 − ey2+−r3y3 + ey1+−r3y3 − e−r3y3

+ey1+y2+−r3y3 − ey1+−r3y3 + ey2+−r3y3 − e−r3y3

+2er3y1+r3y2+y3 − 2er3y1+r3y2

(writing all exponents in terms of y1 and y2)

=∑

h3 mod a3

er3y1+r3y2 − e−−r3y1+r3y2 + er3y1−−r3y2

−e−−r3y1−−r3y2 + er3y1+r3y2 − er3y1−−r3y2

+e−−r3y1+r3y2 − e−−r3y1−−r3y2

+2e−−r3y1−−r3y2 − 2er3y1+r3y2

= 0.

52

We must now show

∑H

(h1,h2) 6=(0,0)

β

(h1

a1− h2

a2, y2

)β

(h1

a1− r3, y3

)+β

(h2

a2− h1

a1, y1

)β

(h2

a2− r3, y3

)+β

(r3 − h1

a1, y1

)β

(r3 − h2

a2, y2

) = 0.

Since ±(

h1

a1− h2

a2

),±

(hi

ai− r3

)6∈ Z, we see that this is just another special case

of our proof of Theorem 3.1 when (x1, x2, x3) = (0, 0, x3 − λa3). Thus, these terms

vanish.

Last, we apply our Theorem 3.1 to the final case.

Chapter 5

The 4-variable reciprocity theorem

We’ve shown that the generic case of Hall–Wilson–Zagier’s reciprocity theorem can

be generalized to the n-variable case using a simple combinatorial argument. The

remaining conditions left to deal with in the n-variable case cannot be addressed so

easily, as shown by Bayad and Raouj [3] who use involved number theory to prove

a reciprocity theorem for the multiple Dedekind–Rademacher sum. For n = 4, most

of the remaining conditions can be dealt with and the following theorem is revealed.

Theorem 5.1. Let a1, a2, a3, a4 ∈ Z , x1, x2, x3, x4 ∈ R, and y1, y2, y3, y4 be four

nonzero variables such that y1 + y2 + y3 + y4 = 0. Then

4∑k=1

Ω

Ak ak

Xk xk

Yk yk

= − 1

8i

(cot

(y1

2i

)+ cot

(y2

2i

)+ cot

(y3

2i

)+ cot

(y4

2i

))

53

54

if (x1, x2, x3, x4) ∈ (a1, a2, a3, a4) R+Z4. The sum vanishes for all other cases, except

possibly when

(xi, xj, xk) ∈ (ai, aj, ak) R + Z3

and

(hi + xi

ai

− hl + xl

al

,hj + xj

aj

− hl + xl

al

,hk + xk

ak

− hl + xl

al

)6∈ Z3

for all hi, hj, hk, hl ∈ Z, where 1 ≤ i < j < k < l ≤ 4.

In the unknown case we believe the sum has the potential to simplify. After the

proof of Theorem 5.1 we will show the challenge the unknown case poses in proving

the sum has a closed formula and ultimately the difficulty in generalizing all cases

to n-variables.

Proof. First, recall identity (3.3),

n∑k=1

Ω

Ak ak

Xk xk

Yk yk

=n∑

k=1

∑H

n∏i=1i6=k

β (rk − ri, yi) ,

55

where

β(α, Y ) =

12· eY +1

eY −1for α ∈ Z,

eαY

eY −1for α 6∈ Z

and∑

H =∑

h1 mod a1

∑h2 mod a2

· · ·∑

hn mod an.

Our approach to proving Theorem 5.1 will mimic that of our new proof of The-

orem 2.4 outlined previously. Thus, we must check the following cases

(i) (x1, x2, x3, x4) ∈ (a1, a2, a3, a4) R + Z4;

(ii) (xi, xj, xk) ∈ (ai, aj, ak)R + Z3 for some 1 ≤ i < j < k ≤ 4 but not (i);

(iii) (xi, xj) ∈ (ai, aj)R + Z2 for some 1 ≤ i < j ≤ 4 but not (i) and not (ii);

(iv) None of the above.

Cases (i), (iii), and (iv) are covered in the statement of Theorem 5.1 and will be

proven here. Case (ii) will be discussed after the proof.

First, we see that case (iv) follows with Theorem 3.1.

Assume case (i). Let (x1, x2, x3, x4) ∈ (a1, a2, a3, a4) R + Z4. As in our simpli-

fied proof of Theorem 2.4, our concern is if hi

ai− hj

aj∈ Z. Since this occurs only

when hi = hj = 0 after introducing a change in variables, we can split the sum

56

∑4k=1 Ω

Ak ak

Xk xk

Yk yk

into the terms where hi = 0 for all i = 1, 2, 3, 4 and the remain-

ing terms when not all hi = 0

4∑k=1

Ω

Ak ak

Xk xk

Yk yk

=4∑

k=1

∑H

4∏i=1i6=k

β (rk − ri, yi)

=β(0, y2)β(0, y3)β(0, y4) + β(0, y1)β(0, y3)β(0, y4) + β(0, y1)β(0, y2)β(0, y4)

+ β(0, y1)β(0, y2)β(0, y3)

+4∑

k=1

∑H\0,0,0,0

4∏i=1i6=k

β

(hk

ak

− hi

ai

, yi

).

We know that since in the last sum hk

ak− hi

ai6∈ Z, we can apply the idea of the proof

of Theorem 3.1 and

4∑k=1

∑H\0,0,0,0

4∏i=1i6=k

β

(hk

ak

− hi

ai

, yi

)= 0.

The interesting part consists of the remaining terms. We will now show that the

remaining terms in fact simplify to a compact finite sum.

57

Let y∗k = yk

2i. Then

4∑k=1

Ω

Ak ak

Xk xk

Yk yk

=β(0, y2)β(0, y3)β(0, y4) + β(0, y1)β(0, y3)β(0, y4) + β(0, y1)β(0, y2)β(0, y4)

+ β(0, y1)β(0, y2)β(0, y3)

=− 1

8i(cot (y∗2) cot (y∗3) cot (y∗4) + cot (y∗1) cot (y∗3) cot (y∗4) + cot (y∗1) cot (y∗2) cot (y∗4)

+ cot (y∗1) cot (y∗2) cot (y∗3))

=− 1

8i(cot (y∗3) cot (y∗4) (cot (y∗2) + cot (y∗1)) + cot (y∗1) cot (y∗2) (cot (y∗3) + cot (y∗4)))

(applying the cotangent identity to the sums cot(y∗1)+cot(y∗2) and cot(y∗3)+cot(y∗4))

=− 1

8i

cot (y∗3) cot (y∗4)

(cot (y∗2) cot (y∗1)−1

cot (y∗1+Y ∗2 )

)+ cot (y∗1) cot (y∗2)

(cot (y∗3) cot (y∗4)−1

cot (y∗3+y∗4)

)

58

(since y1 + y2 + y3 + y4 = 0, then cot (y∗3 + y∗4) = cot (−y∗1 − y∗2) = − cot (y∗1 + y∗2))

=− 1

8i·

cot (y3∗) cot (y∗4) (cot (y∗2) cot (y∗1)− 1)

− cot (y∗1) cot (y∗2) (cot (y∗3) cot (y∗4)− 1)

cot (y∗1 + y∗2)

=− 1

8i(cot (y∗1) cot (y∗2)− cot (y∗3) cot (y∗4))

(solving the cotangent identity for cot(α)cot(β), we apply this identity to both

cotangent products in the sum above)

=− 1

8i·

cot (y∗1 + y∗2) (cot (y∗2) + cot (y∗1)) + 1

− cot (y∗3 + y∗4) (cot (y∗3) + cot (y∗4))− 1

cot (y∗1 + y∗2)

=− 1

8i·

cot (y∗1 + y∗2) (cot (y∗2) + cot (y∗1)) + 1

+ cot (y∗1 + y∗2) (cot (y∗3) + cot (y∗4))− 1

cot (y∗1 + y∗2)

=− 1

8i

(cot (y∗2) + cot (y∗1) +

1

cot (y∗1 + y∗2)+ cot (y∗3) + cot (y∗4)−

1

cot (y∗1 + y∗2)

)=− 1

8i(cot (y∗1) + cot (y∗2) + cot (y∗3) + cot (y∗4)) .

Next, we assume case (iii). Let (xi, xj) ∈ (ai, aj)R + Z2 for some 1 ≤ i < j ≤ 4

and not case (i) and not case (ii). Without loss of generality, assume (x1, x2) ∈

59

(a1, a2)R + Z2. Then, again, we can divide the sum∑4

k=1 Ω

Ak ak

Xk xk

Yk yk

into terms

where h1 = h2 = 0 and the remaining terms when not both h1 and h2 are zero

4∑k=1

Ω

Ak ak

Xk xk

Yk yk

=∑

h3 mod a3

∑h4 mod a4

β (0, y2) β (−r3, y3) β (−r4, y4)

+β (0, y1) β (−r3, y3) β (−r4, y4)

+β (r3, y1) β (r3, y2) β (r3 − r4, y4)

+β (r4, y1) β (r4, y2) β (r4 − r3, y3)

+∑H

(h1,h2) 6=(0,0)

β(

h1

a1− h2

a2, y2

)β

(h1

a1− r3, y3

)β

(h1

a1− r4, y4

)+β

(h2

a2− h1

a1, y1

)β

(h2

a2− r3, y3

)β

(h2

a2− r4, y4

)+β

(r3 − h1

a1, y1

)β

(r3 − h2

a2, y2

)β (r3 − r4, y4)

+β(r4 − h1

a1, y1

)β

(r4 − h2

a2, y2

)β (r4 − r3, y3)

.

Again, we see that all beta functions are evaluated at non-integer values and so

the latter term vanishes by the proof method of Theorem 3.1. We now employ a

similar matrix argument used in the proof of Theorem 3.1 to prove the first summand

equals zero.

60

∑h3 mod a3

∑h4 mod a4

β (0, y2) β (−r3, y3) β (−r4, y4)

+β (0, y1) β (−r3, y3) β (−r4, y4)

+β (r3, y1) β (r3, y2) β (r3 − r4, y4)

+β (r4, y1) β (r4, y2) β (r4 − r3, y3)

=∑

h3 mod a3

∑h4 mod a4

12· ey2+1

ey2−1· e−r3y3

ey3−1· e−r4y4

ey4−1

+12· ey1+1

ey1−1· e−r3y3

ey3−1· e−r4y4

ey4−1

+ er3y1

ey1−1· er3y2

ey2−1· er3−r4y4

ey4−1

+ er4y1

ey1−1· er4y2

ey2−1· er4−r3y3

ey3−1

=∑

h3 mod a3

∑h4 mod a4

12· ey2+1

ey2−1

· e−r3y3

ey3−1· e−r4y4

ey4−1·(

ey1−1ey1−1

)+1

2· ey

1+1

ey1−1

· e−r3y3

ey3−1· e−r4y4

ey4−1·(

ey2−1ey2−1

)+ er3y1

ey1−1· er3y2

ey2−1· er3−r4y4

ey4−1·(

22( ey3−1

ey3−1))

+ er4y1

ey1−1· er4ry2

ey2−1· er4−r3y3

ey3−1·(

22( ey4−1

ey4−1))

=∑

h3 mod a3

∑h4 mod a4

ey1+y2+−r3y3+−r4y4 − ey2+−r3y3+−r4y4 + ey1+−r3y3+−r4y4

−e−r3y3+−r4y4 + ey1+y2+−r3y3+−r4y4 − ey1+−r3y3+−r4y4

+ey2+−r3y3+−r4y4 − e−r3y3+−r4y4

+2er3y1+r3y2+y3+r3−r4y4 − 2er3y1+r3y2+r3−r4y4

+2er4y1+r4y2+r4−r3y3+y4 − 2er4y1+r4y2+r4−r3y3

2 (ey1 − 1) (ey2 − 1) (ey3 − 1) (ey4 − 1).

61

We write the final equality in terms of y1, y2 and y3 and get the numerator

er4y1+r4y2+(−r3−−r4)y3 − e−−r4y1+r4y2+(−r3−−r4)y3

+ er4y1−−r4y2+(−r3−−r4)y3 − e−−r4y1−−r4y2+(−r3−−r4)y3

+ er4y1+r4y2+(−r3−−r4)y3 − er4y1−−r4y2+(−r3−−r4)y3

+ e−−r4y1+r4y2+(−r3−−r4)y3 − e−−r4y1−−r4y2+(−r3−−r4)y3

+ 2e(r3−r3−r4)y1+(r3−r3−r4)y2+r4−r3y3

− 2e(r3−r3−r4)y1+(r3−r3−r4)y2−r3−r4y4

+ 2e−−r4y1−−r4y2−r3−r4y3 − 2er4y1+r4y2+r4−r3y3 .

Let every term in an exponent be represented by its sign. Then

signr4 = +, sign (−−r4) = −,

sign (−r3 − −r4) = C03, sign (r3 − r3 − r4) = C30,

and let

−r3 − r4 = − r4 − r3 = +.

62

We can construct the following matrices

M4 =

+ + C03

+ − C03

+ + C03

− + C03

C30 C30 +

C30 C30 +

− − −

− − −

and M′

4 =

− + C03

− − C03

+ − C03

− − C03

C30 C30 −

C30 C30 −

+ + +

+ + +

,

where M4 consists of all vectors representing the exponents of the positive terms in

the numerator and M′4 consists of all vectors representing the exponents of the neg-

ative terms in the numerator. We can apply Lemmas 3.2 and 3.3 to the differences

−r3 − −r4 and r3 − r3 − r4,

where a = 0 and b = 0, respectively. Then we can make a statement about the

entries C03 and C30.

Lemma 5.2. C03 = + if and only if C30 = −.

Proof. Assume C30 = +. This means the difference −r3 − −r4 is positive and,

63

by Lemma 3.2,

−r3 − −r4 = r4 − r3.

Then

r3 − r3 − r4 = r3 − 1 + 1− r3 − r4

= −−r3+ r4 − r3

= −−r4.

Thus, r3 − r3 − r4 is negative and so C30 = −.

Thus, C30 = −C03 and the matrices become

M4 =

+ + C03

+ − C03

+ + C03

− + C03

−C03 −C03 +

−C30 −C30 +

− − −

− − −

, M′

4 =

− + C03

− − C03

+ − C03

− − C03

−C03 −C03 −

−C03 −C03 −

+ + +

+ + +

.

64

It is left to show that M4 = M′4 after row swapping.

The matrices M4 and M′4 exhibit some of the same properties as the matrices

constructed in Chapter 3. Just as in Chapter 3, an entry in the i-th colum and

j-th row represents the term in an exponent of the summand evaluated at indeces

i and j. This guarentees that row swapping is allowed and thus comparable entries

will always be found in the same row and column. Furthermore, the argument in

Chapter 3 that shows matrix equality means opposite signed terms cancel holds for

these matrices since the method of representation is the same.

Assume C03 = + and the matrices are equal up to row swapping. The same is

true for C03 = −. This finishes the proof of Theorem 5.1.

Finally, we shall discuss case (ii) and why we cannot find a closed formula in

this case. Assume (xi, xj, xk) ∈ (ai, aj, ak)R + Z3 for some 1 ≤ i < j < k ≤ 4 and

not case (i). Without loss of generality, let (x1, x2, x3) ∈ (a1, a2, a3)R + Z3. Then as

before, we divide the sum into the terms when h1 = h2 = h3 = 0 and the remaining

65

terms when not all h1, h2 and h3 are zero

4∑k=1

Ω

Ak ak

Xk xk

Yk yk

=∑

h4 mod a4

β (0, y2) β (0, y3) β (−r4, y4)

+β (0, y1) β (0, y3) β (−r4, y4)

+β (0, y1) β (0, y2) β (−r4, y4)

+β (r4, y1) β (r4, y2) β (r4 − r3, y3)

+∑H

(h1,h2,h3) 6=(0,0,0)

β(

h1

a1− h2

a2, y2

)β

(h1

a1− h3

a3, y3

)β

(h1

a1− r4, y4

)+β

(h2

a2− h1

a1, y1

)β

(h2

a2− h3

a3, y3

)β

(h2

a2− r4, y4

)+β

(h3

a3− h1

a1, y1

)β

(h3

a3− h2

a2, y2

)β

(h3

a3− r4, y4

)+β

(r4 − h1

a1, y1

)β

(r4 − h2

a2, y2

)β

(r4 − h3

a3, y3

)

.

The ideas of the proof of Theorem 3.1 can be applied to the latter summand

since all β-functions are evaluated at non-integer values. Let us investigate what

66

goes awry in the first summand.

∑h4 mod a4

β (0, y2) β (0, y3) β (−r4, y4)

+β (0, y1) β (0, y3) β (−r4, y4)

+β (0, y1) β (0, y2) β (−r4, y4)

+β (r4, y1) β (r4, y2) β (r4 − r3, y3)

=∑

h4 mod a4

12· ey2+1

ey2−1· 1

2· ey3+1

ey3−1· e−r4y4

ey4−1

+12· ey1+1

ey1−1· 1

2· ey3+1

ey3−1· e−r4y4

ey4−1

+12· ey1+1

ey1−1· 1

2· ey2+1

ey2−1· e−r4y4

ey4−1

+ er4y1

ey1−1· er4y2

ey2−1· er4y3

ey3−1

=∑

h4 mod a4

12· ey2+1

ey2−1· 1

2· ey3+1

ey3−1· e−r4y4

ey4−1·(

ey1−1ey1−1

)+1

2· ey1+1

ey1−1· 1

2· ey3+1

ey3−1· e−r4y4

ey4−1·(

ey2−1ey2−1

)+1

2· ey1+1

ey1−1· 1

2· ey2+1

ey2−1· e−r4y4

ey4−1·(

ey3−1ey3−1

)+ er4y1

ey1−1· er4y2

ey2−1· er4y3

ey3−1·(

44( ey4−1

ey4−1))

.

67

Expanding terms we get the numerator

∑h4 mod a4

ey1+y2+y3+−r4y4 − ey2+y3+−r4y4

+ey1+y2+−r4y4 − ey1+−r4y4

+ey1+y3+−r4y4 − ey3+−r4y4

+ey1+−r4y4 − e−r4y4

ey1+y2+y3+−r4y4 − ey1+y3+−r4y4

+ey1+y2+−r4y4 − ey1+−r4y4

+ey2+y3+−r4y4 − ey3+−r4y4

+ey2+−r4y4 − e−r4y4

ey1+y2+y3+−r4y4 − ey1+y2+−r4y4

+ey1+y3+−r4y4 − ey1+−r4y4

+ey2+y3+−r4y4 − ey2+−r4y4

+ey3+−r4y4 − e−r4y4

+4er4y1+r4y2+r4y3+y4 − 4er4y1+r4y2+r4y3

68

(writing the sum in terms of y1, y2 and y3)

=∑

h4 mod a4

er4y1+r4y2+r4y3 − e−−r4y1+r4y2+r4y3

+er4y1+r4y2−−r4y3 − er4y1−−r4y2−−r4y3

+er4y1−−r4y2+r4y3 − e−r4y1−−r4y2+r4y3

+er4y1−−r4y2−−r4y3 − e−−r4y1−−r4y2−−r4y3

er4y1+r4y2+r4y3 − er4y1−−r4y2+r4y3

+er4y1+r4y2−−r4y3 − er4y1−−r4y2−−r4y3

+e−−r4y1+r4y2+r4y3 − e−−r4y1−−r4y2+r4y3

+e−−r4y1+r4y2−−r4y3 − e−−r4y1−−r4y2−−r4y3

+er4y1+r4y2+r4y3 − er4y1+r4y2−−r4y3

+er4y1−−r4y2+r4y3 − er4y1−−r4y2−−r4y3

+e−−r4y1+r4y2+r4y3 − e−−r4y1+r4y2−−r4y3

+e−−r4y1−−r4y2+r4y3 − e−−r4y1−−r4y2−−r4y3

+4e−−r4y1−−r4y2−−r4y3 − 4er4y1+r4y2+r4y3

.

After canceling like terms, we are left with

=∑

h4 mod a4

er4y1+r4y2−−r4y3 − e−−r4y1−−r4y2+r4y3

+er4y1−−r4y2+r4y3 − e−−r4y1+r4y2−−r4y3

+2e−−r4y1+r4y2+r4y3 − 2er4y1−−r4y2−−r4y3

.

As with all the previous summands, this final sum exhibits an intriguing sym-

69

metry among the exponents. And though it is not yet clear how, it is quite possible

that these few terms combine to equal something simple. The unknown fate of

this extraneous case gives light to the level of complexity that builds when dealing

with higher dimensional Dedekind–Bernoulli sums. Although our work and Bayad

and Raouj’s work shows the extreme cases of Hall, Wilson and Zagier’s reciprocity

law can be generalized, the 4-variable reciprocity Theorem shows us the more vari-

ables involved the more challenging computations become and generalizations of the

intermediate cases are not strait forward. We leave this open question to the reader.

Bibliography

[1] T. M. Apostol, Generalized Dedekind sums and transformation formulae ofcertain Lambert series, Duke Math. J. 17 (1950), 147–157.

[2] Tom M. Apostol, Introduction to analytic number theory, Springer-Verlag, NewYork, 1976.

[3] Abdelmejid Bayad and Abdelaziz Raouj, Reciprocity formulae for general mul-tiple Dedekind–Rademacher sums and enumerations of lattice points, preprint(2009).

[4] Matthias Beck and Sinai Robins, Computing the continuous discretely, Under-graduate Texts in Mathematics, Springer, New York, 2007.

[5] L. Carlitz, Some theorems on generalized Dedekind sums, Pacific J. Math. 3(1953), 513–522.

[6] Ulrich Dieter, Beziehungen zwischen Dedekindschen Summen, Abh. Math. Sem.Univ. Hamburg 21 (1957), 109–125.

[7] R. R. Hall, J. C. Wilson, and D. Zagier, Reciprocity formulae for generalDedekind-Rademacher sums, Acta Arith. 73 (1995), no. 4, 389–396.

[8] Donald E. Knuth, The art of computer programming. Vol. 2, second ed.,Addison-Wesley Publishing Co., Reading, Mass., 1981.

[9] W. Meyer and R. Sczech, Uber eine topologische und zahlentheoretische Anwen-dung von Hirzebruchs Spitzenauflosung, Math. Ann. 240 (1979), no. 1, 69–96.

70

71

[10] M. Mikolas, On certain sums generating the Dedekind sums and their reciprocitylaws, Pacific J. Math. 7 (1957), 1167–1178.

[11] Hans Rademacher, Generalization of the reciprocity formula for Dedekind sums,Duke Math. J. 21 (1954), 391–397.

[12] Hans Rademacher and Emil Grosswald, Dedekind sums, The MathematicalAssociation of America, Washington, D.C., 1972.

[13] Hans Rademacher and Albert Whiteman, Theorems on Dedekind sums, Amer.J. Math. 63 (1941), 377–407.