Generalized Dedekind Bernoulli Sums

Transcription

1 Generalized Dedekind Bernoulli Sums A thesis presented to the faculty of San Francisco State University In partial fulfilment of The Requirements for The Degree Master of Arts In Mathematics by Anastasia Chavez San Francisco, California June 2010

2 Copyright by Anastasia Chavez 2010

3 CERTIFICATION OF APPROVAL I certify that I have read Generalized Dedekind Bernoulli Sums by Anastasia 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 Beck Professor of Mathematics Dr. Federico Ardillo Professor of Mathematics Dr. David Ellis Professor of Mathematics

4 Generalized Dedekind Bernoulli Sums Anastasia Chavez San Francisco State University 2010 The finite arithmetic sum called a Dedekind sum appears in many areas of mathematics, such as topology, geometric combinatorics, algorithmic complexity, and algebraic geometry. Dedekind sums exhibit many beautiful properties, the most famous 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

5 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 continue 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 perfect ways, Ayla and Asha, thank you for your medicine and being the profound teachers that you are. v

6 TABLE OF CONTENTS 1 Introduction More History, Definitions and Notations Bernoulli Polynomials and More Dedekind and Bernoulli meet The Multivariate Dedekind Bernoulli sum A New Proof of Hall Wilson Zagier s Reciprocity Theorem The 4-variable reciprocity theorem Bibliography vi

7 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 B 1 ( h b ) ( ) ha B 1, (1.1) b where {x} 1/2 (x Z), B 1 (x) = 0 (x Z), (1.2) 1

8 2 and we call x x = {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. Although Dedekind sums are studied in number theory, they have appeared in many areas of mathematics such as analytical number theory [6], combinatorial geometry [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 ) 1 ab 4. (1.3) Since Dedekind, many mathematicians such as Rademacher [11], Carlitz [5], and Apostol [1] have introduced Dedekind-like sums that generalize B 1 (u) to periodized Bernoulli polynomials, B k (u) (see Definition 2.4). In suit with Dedekind, reciprocity relations have been proven for the respective Dedekind-like sums.

9 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, Wilson 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.

10 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 B k (u) is defined through the generating function ze uz e z 1 = k 0 B k (u) z k. (2.1) k! 4

11 5 Definition 2.2 (Bernoulli numbers). The Bernoulli numbers are B k := B k (0) and have the generating function z e z 1 = k 0 B k k! zk. (2.2) 12.12]. It has been proven that Bernoulli functions are in fact polynomials [2, Theorem Theorem 2.1 (k-th Bernoulli polynomial). The functions B k (u) are polynomials in u given by B k (u) = k n=0 ( ) k B n u k n. (2.3) n An important property of k-th Bernoulli polynomials is the Fourier expansion formula [2, Theorem 12.19] k Z\{0} e 2πiu k m = (2πi)m m! B m (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 B m (u) with period 1 that coincides with B m (u) on [0, 1), except we set B 1 (u) = 0 for u Z [7].

12 6 Lemm.2 (Raabe s formula [7]). For a N, x R, f mod a ( B m x + f ) = m Bm (ax). a Proof. Begin by embedding the right-hand side into a generating function: m Bm (ax) Y m 1 m! m=0 = m 0 = eax Y a e Y a 1 = exy e Y a 1. B m (ax) (Y/a)m 1 m! Now, embed the left-hand side into a generating function: m=0 f mod a ( B m x + f ) Y m 1 a m! = a 1 f=0 = exy e (x+ f a )Y e Y 1 e Y 1 + e(x+ 1 a )Y 2 a )Y a 1 a )Y e Y 1 + e(x+ e Y e(x+ e Y 1 = exy (1 + e 1 a Y + e 2 a Y + + e a 1 a Y ) e Y 1

13 7 (applying the geometric series identity n 1 k=0 rk = 1 rn 1 r where r = e Y a ) e Y 1 1 e Y 1 e Y a = exy = exy 1 e Y a = exy e Y a 1. a a 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 ( f ) B m x + = a 1 m Bm (ax). a Definition 2.3. Let β(α, Y ) = m=0 B m (α) Y m 1, (2.5) m! where α R.

14 8 Lemm.3. β(α, Y ) = 1 ey +1 2 e Y 1 for α Z, e {α}y for α Z. e Y 1 (2.6) Proof. Recall that B m (α) coincides with B m (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 Z. β(α, Y ) = = m=0 m=0 = e{α}y e Y 1. B m (α) Y m 1 m! B m ({α}) Y m 1 m! 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 B 0 (α) = 1 and B m (α) = 0 for odd m 3, as well as

15 9 the first few Bernoulli numbers [2]. B 0 (0)Y 1 + B 1 (0) + B 2 (0) Y 2! + B 3(0) Y 2 3! + B 4(0) Y 3 4! + = 1 e Y = 1 e Y = 1 e Y = 2 + ey 1 2(e Y 1) = 1 2 e Y + 1 (e Y 1). Recall that B 1 (α) = 0 for all α Z, and so we have the following identity: B 0 (0)Y 1 + B 1 (0) + B 2 (0) Y 2! + B 3 (0) Y 2 3! + B 4 (0) Y 3 4! + = 1 2 e Y + 1 (e Y 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], Mikolás [10] and Hall, Wilson and Zagier [7]. The latter introduced the generalized Dedekind Rademacher sum, the function we wish to generalize to n variables.

16 10 Definition 2.4. The generalized Dedekind Rademacher sum is S m,n a b c = x y z h mod c ( B m a h + z ) x c ( B n b h + z ) y, (2.7) c 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 a b c Ω x y z = X Y Z m,n 0 1 m!n! S m,n a b c (X/a) m 1 (Y/b) n 1, (2.8) x y z where Z = X Y. Hall,Wilson and Zagier proved the following reciprocity theorem for the generalized Dedekind Rademacher sum. Theorem 2.4. [7] Let a, b, c Z >0 and pairwise relatively prime, x, y, z R, and

17 11 X, Y, Z three variables such that X + Y + Z = 0. Then a b c b c a c a b Ω x y z + Ω y z x + Ω z x y X Y Z Y Z X Z X Y 1/4 if (x, y, z) (a, b, c)r + Z 3, = 0 otherwise. (2.9) To clarify, we call the result a b c b c a c a b Ω x y z + Ω y z x + Ω z x y = 0 X Y Z Y Z X Z X Y 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.

18 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 (p 1, p 2,..., p n ) and (,,..., a n ), and real numbers (x 1, x 2,..., x n ). For 1 k n, set A k = (,,..., â k,..., a n ), X k = (x 1, x 2,..., ˆx k,..., x n ) and P k = (p 1, p 2,..., ˆp k,..., p n ), where â k means we omit the entry. Then the multivariate Dedekind Bernoulli 12

19 13 sum is S Pk A k X k x k = h mod n i=1 B pi (a i h + x k ) x i. (3.1) We see that if (,, a 3 ) = (a, b, c), (x 1, x 2, x 3 ) = (x, y, z), and P 3 = (m, n) then A 3 = (a, b), X 3 = (x, y) and thus we recover Hall, Wilson and Zagier s generalized Dedekind Rademacher sum, S m,n A 3 X 3 c = z h mod c ( B m a h + z ) x c ( B n b h + z ) y. c Moreover, by letting (x 1, x 2,..., x n ) = 0 and P k = (p 1, p 2,..., ˆp k,..., p n ), 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 S Pk A k P k p k = h mod n i=1 ( ) ai h B pi. (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

20 14 (y 1, y 2,..., y n ) where y n = y 1 y 2 y n 1, let Y k = (y 1, y 2,..., ŷ k,..., y n ). Then Ω A k X k Y k x k y k =... 1 p p 1 p k p 1!p 2! p k 1!p k+1! p n! S P k A k n X k x k n i=1 ( yi a i ) pi 1. The series of summands p 1... p k p n is understood as summations over all non-negative integers p 1,..., ˆp k,..., p n. Our main result is the following reciprocity law involving multivariate Dedekind Bernoulli sums. Theorem 3.1. Let (x 1, x 2,..., x n ) R and (p 1, p 2,..., p n ), (,,..., a n ) Z >0 where (a u, a v ) = 1 for all 1 u < v n. For 1 k n, let A k = (,,..., â k,..., a n ), X k = (x 1, x 2,..., ˆx k,..., x n ) and P k = (p 1, p 2,..., ˆp k,..., p n ). For nonzero variables y 1, y 2,..., y n, let Y k = (y 1, y 2,..., ŷ k,..., y n ) such that y 1 +

21 15 y y n = 0, then n Ω k=1 A k X k Y k x k y k = 0 if xu hu a u xv hv a v Z whenever 1 u < v n and h u, h v Z. We now embark on a journey of transforming Ω A k X k Y k x k y k 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

22 16 sum can be rewritten as follows, {a b} {a c} {c b} = a b 1 (a c 2 ) (c b 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}.

23 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 = =, 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.

24 18 We will now begin manipulating Ω multivariate Dedekind Bernoulli sums. S Pk A k X k x k n i=1 m i i = A k X k Y k h mod n i=1 x k y k using the following identity for B pi (a i h + x k x i ) m i i (applying 2.2, Raabe s formula) = H n i=1 B pi ( xk + h k x ) i + h i, a i where H = h 1 mod h 2 mod h n mod a n includes the original summand h

25 19 we now call h k. Let r i = x i+h i a i, then Ω A k X k Y k x k y k =... 1 p p 1 p k p 1!p 2! p k 1!p k+1! p n! S P k A k n = H = H... p k p 1 1 p p 1!p 2! p k 1!p k+1! p n! n X k n i=1 x k n i=1 ( ) pi 1 yi a i B pi (r k r i ) y p i 1 i n β (r k r i, y i ), (3.3) i=1 where Definition 2.3 is applied in the final equality. It is now clear that (3.3) depends on the differences r k r i for 1 k < i n and all β (r k r i, y i ) 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.

26 20 Consider the overall sum of Ω A k X k Y k x k y k as k ranges from 1 to n. n Ω k=1 A k X k Y k x k y k = n k=1 H n 1 = k=1 H n β (r k r i, y i ) i=1 n β (r k r i, y i ) + H i=1 n 1 β (r n r j, y j ) j=1 (since {r k r i } 0 for each 1 k < i n, we apply Lemm.3) n 1 = k=1 H n 1 = = k=1 H n 1 k=1 n i=1 n i=1 e {r k r i }y i e y i 1 + H n 1 j=1 e {r k r i }y i e y i 1 eyk 1 e y k 1 + H H ni=1 + H n 1 e {rn r j}y j e y j 1 n 1 j=1 e {r k r i }yi e y k n 1 ni=1 k=1 H e {rn r j}y j e y j 1 eyn 1 e yn 1 e {r k r i }y i n 1 j=1 e{rn r j}y j j=1 e{rn r j}yj e yn H n. f=1 (ey f 1)

27 21 We will drop the denominator and continue to focus only on the numerator: n 1 k=1 H + H e {r k r n}y n+y k n 1 n 1 n 1 = k=1 n 1 e {r k r i }y i i=1 j=1 e {rn rj}yj e yn H H n 1 k=1 + H H n 1 k=1 H n 1 e {rn r j}y j j=1 n 1 e {r k r n}( y 1 y n 1 )+y k e {r k r i }y i n 1 e {r k r n}( y 1 y n 1 ) i=1 i=1 e {r k r i }y i j=1 e {rn rj}yj e ( y1 yn 1) H n 1 e {r k r n}y n e {r k r i }y i i=1 n 1 e {rn r j}y j. j=1 Recall that 1 {r i r j } = { r j r i } since r i r j Z. Thus the numerator equals n 1 k=1 H n 1 k=1 + H n 1 e {rn r k}y k e ({r k r i } {r k r n})y i H n 1 i=1 e {r k r n}y k n 1 e ({r k r i } {r k r n})y i (3.4) i=1 j=1 e {rj rn}yj H n 1 e {rn r j}y j. j=1 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),

28 22 we can see that only three types of exponents appear, {r n r i }, {r i r n }, and {r i r j } {r i r n }. A critical charasteristic of the latter difference is that, by applying Lemma 3.2 or Lemma 3.3, {r j r i } {r j r n } = {r n r i } or {r i r n }, respectively. This means the exponents can be condensed to just the first two forms, {r n r i } and {r i r n }. Moreover, we can represent these numbers by their sign since the sign of {r j r i } {r j r n } determines if it is equal to {r n r i } or {r i r n }. This lends us to the following representations: {r n r i } = +, {r i r n } = and C ij = sign ({r j r i } {r j r n }). Thus every term of n we will state explicitly. k=1 Ω A k X k Y k x k y k can be represented by a sign vector, which The exponent corresponding to k = 1 in n 1 k=1 H n 1 e {rn r k}y k e ({r k r i } {r k r n})y i i=1 is {r n r 1 }y 1 + ({r 1 r 2 } {r 1 r n }) y ({r 1 r n 1 } {r 1 r n }) y n 1

29 23 and is represented as the sign vector (+, C 12,..., C 1,n 1 ). Similarly, the (k = 1)-exponent in is n 1 k=1 H n 1 e {r k r n}y k e ({r k r i } {r k r n})y i i=1 {r 1 r n }y 1 + ({r 1 r 2 } {r 1 r n }) y ({r 1 r n 1 } {r 1 r n }) y n 1, and the sign vector representation is (, C 12,..., C 1,n 1 ). Finally, the terms n 1 H j=1 e {rj rn}yj and H n 1 e {rn r j}y j j=1

30 24 are represented as the respective sign vectors (,,..., ) and (+, +,..., +). Note that if the sign of, say, {r 1 r n } and {r 4 r n } are the same, it is not necessarily true that {r 1 r n } = {r 4 r n }. We will address this futher in the following argument. To prove that the terms of n k=1 Ω A k X k Y k x k y k do in fact cancel, we proceed to construct two matrices M n and M n that consist of the sign vector representations of the terms of n k=1 Ω A k X k Y k x k to proving M n = M n after row swapping. y k and will show that n k=1 Ω A k X k Y k x k y k = 0 equates Let M n be the matrix of all sign vectors representing the exponents of the positive terms of n k=1 Ω A k X k Y k x k y k 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 for each k is placed in the kth row of matrix M n. H e{rn r k}y k n 1 i=1 e ({r k r i } {r k r n})y i Similarly, the sign vector representation for every exponent from the negative terms n 1 k=1 H e {r k r n}y k n 1 i=1 e ({r k r i } {r k r n})y i for each k is placed in the kth

31 25 row of the matrix M n. Finally, place the sign vector representing n 1 H j=1 e {r j r n}y j in the last row of M n and the sign vector representing n 1 H j=1 e{rn r j}y j in the last row of M n. Notice that the placement of entry C ki depends on the indices k and i of each term in n k=1 Ω A k X k Y k x k y k in the same row and column in both matrices. and given the symmetry among the terms, C ki lives This latter fact implies that if sign ({r n r i }) = sign ({r n r j }) then {r n r i } = {r n r j } only when i = j. Thus we have constructed the following matrices, + C 12 C 13 C 1,n 1 C 21 + C 23 C 2,n 1 C M n = 31 C 32 + C 3,n C n 1,1 C n 1,2 C n 1,3 +

32 26 and C 12 C 13 C 1,n 1 C 21 C 23 C 2,n 1 C M n = 31 C 32 C 3,n C n 1,1 C n 1,2 C n 1, Last, we will show that M n = M n after row swapping implies n k=1 Ω vanishes. Assume M n = M n up to row swapping. Then for each sign row vector A k X k Y k x k y k (C s1,..., C s,s 1, +, C s,s+1,..., C s,n 1 ) M n there exists (C t1,..., C t,t 1,, C t,t+1,..., C t,n 1 ) M n such that (C s1,..., C s,s 1, +, C s,s+1,..., C s,n 1 ) = (C t1,..., C t,t 1,, C t,t+1,..., C t,n 1 ).

33 27 Also, we have for some row f in M n (C f1,..., C f,f 1, +, C f,f+1,..., C f,n 1 ) = (+,..., +) and for some row g in M n (,..., ) = (C g1,..., C g,g 1, +, C g,g+1,..., C g,n 1 ). For each identity, we will show that the sign row vectors correspond to canceling terms of n k=1 Ω A k X k Y k x k y k M n, this corresponds to the exponent. Beginning with (C s1,..., C s,s 1, +, C s,s+1,..., C s,n 1 ) ({r s r 1 } {r s r n }) y ({r s r s 1 } {r s r n }) y s 1 + {r n r s }y s + ({r s r s+1 } {r s r n }) y s ({r s r n 1 } {r s r n }) y n 1, and more specifically, represents the positive term e 0 ({r s r 1 } {r s r n }) y ({r s r s 1 } {r s r n }) y s 1 + {r n r s }y s + ({r s r s+1 } {r s r n }) y s ({r s r n 1 } {r s r n }) y n 1 1 C A.

34 28 Similarly, (C t1, C t2,..., C t,t 1,, C t,t+1,..., C t,n 1 ) M n represents the exponent ({r t r 1 } {r t r n }) y ({r t r t 1 } {r t r n }) y t 1 {r t r t }y t + ({r t r t+1 } {r t r n }) y t ({r t r n 1 } {r t r n }) y n 1, and so also the negative term e 0 ({r t r 1 } {r t r n }) y ({r t r t 1 } {r t r n }) y t 1 {r t r t }y t + ({r t r t+1 } {r t r n }) y t ({r t r n 1 } {r t r n }) y n 1 1 C A. Then (C s1,..., C s,s 1, +, C s,s+1,..., C s,n 1 ) = (C t1,..., C t,t 1,, C t,t+1,..., C t,n 1 ),

35 29 implies sign ({r s r 1 } {r s r n }) y sign ({r s r s 1 } {r s r n }) y s 1 + {r n r s }y s + sign ({r s r s+1 } {r s r n }) y s sign ({r s r n 1 } {r s r n }) y n 1 = sign ({r t r 1 } {r t r n }) y sign ({r t r t 1 } {r t r n }) y t 1 {r t r t }y t + sign ({r t r t+1 } {r t r n }) y t sign ({r t r n 1 } {r t r n }) y n 1. If sign ({r s r i } {r s r n }) = sign ({r t r i } {r t r n }) then, by Lemmas 3.2 and 3.3, ({r s r i } {r s r n }) = ({r t r i } {r t r n }). Thus, ({r s r 1 } {r s r n }) y ({r s r s 1 } {r s r n }) y s 1 + {r n r s }y s + ({r s r s+1 } {r s r n }) y s ({r s r n 1 } {r s r n }) y n 1 = ({r t r 1 } {r t r n }) y ({r t r t 1 } {r t r n }) y t 1 {r t r t }y t + ({r t r t+1 } {r t r n }) y t ({r t r n 1 } {r t r n }) y n 1

36 30 and so e 0 1 ({r s r 1 } {r s r n }) y ({r s r s 1 } {r s r n }) y s 1 + {r n r s }y s e = 0. + ({r s r s+1 } {r s r n }) y s ({r s r n 1 } {r s r n }) y n ({r t r 1 } {r t r n }) y ({r t r t 1 } {r t r n }) y t 1 {r t r t }y t + ({r t r t+1 } {r t r n }) y t ({r t r n 1 } {r t r n }) y n 1 C A C A As for the identities (C f1, C f2,..., C f,f 1, +, C f,f+1,..., C f,n 1 ) = (+, +,..., +) and (,,..., ) = (C g1, C g2,..., C g,g 1, +, C g,g+1,..., C g,n 1 ), a similar argument follows. (C f1, C f2,..., C f,f 1, +, C f,f+1,..., C f,n 1 ) represents the exponent ({r f r 1 } {r f r n }) y ({r f r f 1 } {r f r n }) y f 1 + {r n r f }y f + ({r f r f+1 } {r f r n }) y f ({r f r n 1 } {r f r n }) y n 1

37 31 and thus the term e 0 ({r f r 1 } {r f r n }) y ({r f r f 1 } {r f r n }) y f 1 + {r n r f }y f + ({r f r f+1 } {r f r n }) y f ({r f r n 1 } {r f r n }) y n 1 1 C A. The identity (C f1, C f2,..., C f,f 1, +, C f,f+1,..., C f,n 1 ) = (+, +,..., +) means {r f r i } {r f r n } = {r n r i }. Thus e 0 ({r f r 1 } {r f r n }) y ({r f r f 1 } {r f r n }) y f 1 + {r n r f }y f = e 0 = H + ({r f r f+1 } {r f r n }) y f ({r f r n 1 } {r f r n }) y n 1 {r n r 1 }y {r n r f 1 }y f 1 {r f r n }y f +{r n r f+1 }y f {r n r n 1 }y n 1 n 1 e {rn r i}y i. i=1 The exponent of the negative term H vector (+, +,..., +) which cancels with the above. 1 C A 1 C A n 1 j=1 e{rn r j}y j is represented by the sign

38 32 (C g1, C g2,..., C g,g 1, +, C g,g+1,..., C g,n 1 ) represents the exponent ({r g r 1 } {r g r n }) y ({r g r g 1 } {r g r n }) y g 1 {r g r n }y g + ({r g r g+1 } {r g r n }) y g ({r g r n 1 } {r g r n }) y n 1 and thus the negative term e 0 ({r g r 1 } {r g r n }) y ({r g r g 1 } {r g r n }) y g 1 {r g r n }y g + ({r g r g+1 } {r g r n }) y g ({r g r n 1 } {r g r n }) y n 1 1 C A. The identity (,,..., ) = (C g1, C g2,..., C g,g 1, +, C g,g+1,..., C g,n 1 ) means {r g r i } {r g r n } = {r i r n }. Thus e 0 1 ({r g r 1 } {r g r n }) y ({r g r g 1 } {r g r n }) y g 1 {r g r n }y g = e = H + ({r g r g+1 } {r g r n }) y g ({r g r n 1 } {r g r n }) y n {r 1 r n }y 1 {r g 1 r n }y g 1 {r g r n }y g {r g+1 r n }y g+1 {r n 1 r n }y n 1 n 1 e {r i r n}y i. i=1 C A C A

39 33 The exponent of the positive term H vector (,,..., ), which cancels with the above. n 1 j=1 e {r j r n}y j is represented by the sign Thus the matching of rows of the matrices M n and M n correspond to the matching of like terms of opposite signs and we have n just revealed the potential of our main result, k=1 Ω A k X k Y k x k y k = 0. We have M n = M n, up to row swapping, implies n Ω k=1 A k X k Y k x k y k = 0. We must now show that under certain conditions, we can indeed find M n = M n up to row swapping. By proving the following properties of the matrices M n and M n, we are able to quickly show M n = M n up to row swapping and thus our reciprocity theorem follows. Lemma 3.4. M n and M n are of the form such that M n 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 M n (excluding the last row

40 34 of s) represent the terms in the exponent of the sum n 1 k=1 H e {rn r k}y k n 1 e ({r k r i } {r k r n})y i ; i=1 when k = i, the exponent is {r n r k } = +. 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 {r k r n}y k n 1 e ({r k r i } {r k r n})y i ; i=1 when k = i, this exponent is {r k r n } =. Lemma 3.5. As above, let C ij = sign ({r i r j } {r i r n }) where i, j, n Z >0 and 1 i < j n 1. Then C ij = + if and only if C ji =. Proof. Assume C ij = +. Then {r i r j } {r i r n } 0 and by Lemma 3.2 {r i r j } {r i r n } = {r n r j }. We want to show C ji = which is to show {r j r i } {r j r n } < 0. {r j r i } {r j r n } = {r j r i } {r j r n } = {r i r j } + {r n r j } = {r i r n } C ji =.

41 35 Assume C ij =. Then {r i r j } {r i r n } < 0 and by Lemma 3.3 {r i r j } {r i r n } = {r j r n }. We want to show C ji = + which is to show {r j r i } {r j r n } > 0. {r j r i } {r j r n } = {r j r i } + {r i r j } {r i r n } = 1 {r i r n } = {r n r i } C ji = +, recalling that {r j r i } + {r i r j } = 1. By Lemma 3.5, it follows that the sign of one difference of fractional parts {r i r j } {r i r n } is dependent on the sign of another difference of fractional parts {r j r i } {r j r n }. The lemma implies C ij determines C ji. Thus, we will rename C ji = C ij, meaning if C ij = + then C ji = and if C ij = then C ji = +. Thus, we can update the sign vector matrices utilizing this new information as

42 36 + C 12 C 13 C 1,n 1 C 12 + C 23 C 2,n 1 C M n = 13 C 23 + C 3,n C 1,n 1 C 2,n 1 C 3,n 1 + and C 12 C 13 C 1,n 1 C 12 C 23 C 2,n 1 C M n = 13 C 23 C 3,n ,. C 1,n 1 C 2,n 1 C 3,n and we can state the following property. Lemma 3.6. M n and M n exhibit antisymmetry about the diagonal. Lemma 3.7. As above, let C ij = sign ({r i r j } {r i r n }) where i, j, n Z >0 and 1 i < j n 1. If C ij = + and C ik = then C jk =.

43 37 Proof. Assume C ij = + and C ik =. Then {r i r j } {r i r n } > 0 and {r i r k } {r i r n } < 0, and by Lemmas 3.2 and 3.3 {r i r j } {r i r n } = {r n r j } (3.5) and {r i r k } {r i r n } = {r k r i }. (3.6) Then the difference (3.5)-(3.6) is positive and we get {r i r j } {r i r k } = {r n r j } + {r k r i }. The final identity is positive, which means the left hand side is positive. Then by Lemma 3.2 {r i r j } {r i r k } = {r k r j }, and we have {r k r j } = {r n r j } + {r k r i }.

44 38 We want to show C jk = which reduces to showing {r j r k } {r j r n } < 0. {r j r k } {r j r n } = {r j r k } {r j r n } = {r k r j } + {r n r j } = {r k r i } C jk =. Lemma 3.8. There exists a unique row with k + s, for each 0 k n 1, in the matrix M n. Proof. We begin by showing that every row of the matrix M n is unique. Assume on the contrary that row m and row l of M n are equal. Then we can view the rows as follows: row m: C 1m C 2m + C ml C m,n 1 row l: C 1l C 2l C ml + C l,n 1. Then C ml = + and C ml = +, but by Lemma 3.5, C ml = + implies C ml =, a contradiction. Therefore, the rows of the matrix M n 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 M n contain exactly i + s, are not equal

45 39 (and thus 1 i < n 1) and look the same as above. We will examine the elements C ml of row m and C ml of row l. Let C ml = +. Since the mth row does not contain only + s, there exists a in column, say, w. Then by Lemma 3.7, the entry C wl 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 C ml = + 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. M n and M n share all entries C ij for 1 i < j n 1. Then the k + s in the kth row of M n 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 M n. We can now prove our main Theorem 3.1. Proof. Our goal is to show n k=1 Ω A k X k Y k x k y k. Assume xu h u a u xv hv a v Z whenever

46 40 1 u < v n and h u, h v Z which allows us to implement the following matrix argument. We ve seen that n k=1 Ω A k X k Y k x k y k gives rise to the following matrices + C 12 C 13 C 1,n 1 C 12 + C 23 C 2,n 1 C M n = 13 C 23 + C 3,n C 1,n C 2,n 1 C 3,n 1 + and C 12 C 13 C 1,n 1 C 12 C 23 C 2,n 1 C M n = 13 C 23 C 3,n ,.. C 1,n 1 C 2,n 1 C 3,n

47 41 where M n represents the exponents of all positive terms of n k=1 Ω A k X k Y k x k y k and M n represents all negative terms. As was shown above, to prove that the matrices M n andm n contain the same rows is equivalent to showing every positive term of n k=1 Ω A k X k Y k x k y k has a canceling negative term. Thus our problem reduces to showing M n = M n after row swapping. We ve proved in Lemma 3.4 that M n 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 M n 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 M n as they appear in matrix M n such that M n = M n, which, by our previous argument, implies n k=1 Ω A k X k Y k x k y k = 0.

48 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 generalized Dedekind Rademacher sum ultimately uses cotangent identities to show a b c b c a c a b Ω x y z + Ω y z x + Ω z x y X Y Z Y Z X Z X Y 1/4 if (x, y, z) (a, b, c)r + Z 3, = 0 otherwise. 42

49 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 (,, a 3 ) Z >0 and (x 1, x 2, x 3 ) R. For non-zero variables y 1, y 2, y 3 such that y 1 + y 2 + y 3 = 0 we want to show that a 3 a 3 a 3 Ω x 1 x 2 x 3 + Ω x 2 x 3 x 1 + Ω x 3 x 1 x 2 y 1 y 2 y 3 y 2 y 3 y 1 y 3 y 1 y 2 1/4 if (x 1, x 2, x 3 ) (,, a 3 )R + Z 3, = 0 otherwise. Recall identity (3.3), n Ω k=1 A k X k Y k x k y k = n k=1 H n β (r k r i, y i ), i=1 where β(α, Y ) = 1 ey +1 2 e Y 1 for α Z, e {α}y for α Z, e Y 1

50 44 and H = h 1 mod h 2 mod h n mod a n. Then we have a 3 a 3 a 3 Ω x 1 x 2 x 3 + Ω x 2 x 3 x 1 + Ω x 3 x 1 x 2 y 1 y 2 y 3 y 2 y 3 y 1 y 3 y 1 y = β (r k r i, y i ). We must examine the following cases. k=1 H i=1 (i) (x 1, x 2, x 3 ) (,, a 3 )R + Z 3 ; (ii) (x i, x j ) (a i, a j )R + Z 2 for some 1 i < j 3 but not (i); (iii) None of the above. We will begin with case (i). Let (x 1, x 2, x 3 ) (,, a 3 )R+Z 3. Then x i = λa i +z i for each i where λ R and z i Z. Thus h i + x i a i h j + x j a j = h i + λa i + z i a i = h i + z i a i h j + λa j + z j a j h j + z j a j. (4.1) Since h i + z i only permutes the modular values of h i, we can make a change in

51 45 indices (letting h i z i = h i ) so that (4.1) becomes h i a i h j a j and a 3 a 3 a 3 Ω x 1 x 2 x 3 + Ω x 2 x 3 x 1 + Ω x 3 x 1 x 2 y 1 y 2 y 3 y 2 y 3 y 1 y 3 y 1 y = β (r k r i, y i ) = = k=1 H 3 k=1 H 3 k=1 H i=1 3 β i=1 3 β i=1 ( hk h ) i, y i a i ( hk h ) i, y i, a i since we sum h i over a complete residue system moda i. Since (a i, a j ) = 1, then h i a i can split up the sum 3 k=1 h j a j 3i=1 H β Z occurs only when h i = h j = 0. Thus, we ( ) h k al h i a i, y i into two parts, a term when all

52 46 h i = 0 and a sum over the remaining h i Z where (h 1, h 2, h 3 ) (0, 0, 0) 3 k=1 H 3 β i=1 ( hk h ) i, y i a l a i =β (0, y 2 ) β (0, y 3 ) + β (0, y 1 ) β (0, y 3 ) + β (0, y 1 ) β (0, y 2 ) ( ) ( ) h + β 1 h 2 h, y 2 β 1 h 3 a 3, y 3 ( ) ( ) h +β 2 a H\{0,0,0} 2 h 1 h, y 1 β 2 h 3 a 3, y 3 ( ) ( ). h +β 3 a 3 h 1 h, y 1 β 3 a 3 h 2, y 2 Let us first address the term when all h i = 0. We use the following cotangent identities cot (y) = i e2πy + 1 e 2πy 1, (4.2) cot (α) cot (β) 1 cot (α) + cot (β) =. cot (α + β) (4.3) Combined with the definition of cot y, β(0, y) becomes β(0, y) = 1 i cot y 2i. (4.4)

53 47 Let y k = y k 2i. Then β (0, y 2 ) β (0, y 3 ) + β (0, y 1 ) β (0, y 3 ) + β (0, y 1 ) β (0, y 2 ) = 1 e y e y e y e y e y e y e y 2 1 e y e y 1 1 e y e y 1 1 e y 2 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 y3 + cot y1) + cot y1 cot y3) = 1 ( ( ) ) cot y cot y2 1 cot y3 1 + cot y 4 cot y1 + y3 1 cot y3. By assumption, y 1 + y 2 + y 3 = 0, so it follows that y 1 + y 3 = y 2 and cot (y 1 + y 3) = cot ( y 2) = cot y 2. Thus, ( ( ) cot y 1 cot y3 1 ) 1 4 cot y2 + cot y cot y1 + y3 1 cot y3 = 1 ( ( ) ) cot y cot y2 1 cot y3 1 + cot y 4 cot y2 1 cot y3 = 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

54 48 sum before. It is just 3 k=1 H\{0,0,0} i=1 3 β (r k r i, y i ) when (x 1, x 2, x 3 ) = 0. Since r k r i Z, we can apply the same matrix argument from the proof of Theorem 3.1 and the terms cancel. Assume case (ii). Let (x i, x j ) (a i, a j )R + Z 2 for some 1 i < j 3 but not (i). Without loss of generality, we assume (x 1, x 2 ) (, )R+Z 2. Then, as before, x 1 = λ + z 1 and x 2 = λ + z 2 for λ R and z 1, z 2 Z so that h 1 + x 1 h 2 + x 2 = h 1 + λ + z 1 h 2 + λ + z 2 = h 1 + z 1 h 2 + z 2. Since z 1 and z 2 permute the summands over h 1 and h 2, we introduce a change of variables and let h 1 = h 1 + z 1 and h 2 = h 2 + z 2. We can rewrite the differences involving r 3 as h i + x i h ) 3 + x 3 ( hi = + λ r 3 a i a 3 a i ( hi = h ) 3 + x 3 λa 3 a i a 3 ) ( hi = r 3. a i

55 49 Then a 3 a 3 a 3 Ω x 1 x 2 x 3 + Ω x 2 x 3 x 1 + Ω x 3 x 1 x 2 y 1 y 2 y 3 y 2 y 3 y 1 y 3 y 1 y 2 = 3 k=1 H = 3 β (r k r i, y i ) i=1 h 1 mod h 2 mod h 3 mod a 3 = H β +β +β ( h 1 h 2 ( h 2 h 1 ( ) ( ) β h1 h 2, y 2 β h1 r 3, y 3 ( ) ( ) +β h2 a 2 h 1, y 1 β h2 r 3, y 3 ( ) ( ) +β r 3 h 1 λ, y 1 β r 3 h 2, y 2 ) ( ) h, y 2 β 1 r 3, y 3 ) ( ) h, y 1 β 2 r 3, y 3 ( ). r 3 h 2, y 2 ( r 3 h 1, y 1 ) β We will again split up the final expression into two parts, one part includes all terms where h 1 = h 2 = 0 and the other part are all summands where h 1 and h 2 are not

56 50 both zero First, we show ( ) ( ) h β 1 h 2 h, y 2 β 1 r 3, y 3 ( ) ( ) h +β 2 a H 2 h 1 h, y 1 β 2 r 3, y 3 ( ) ( ) +β r 3 h 1, y 1 β r 3 h 2, y 2 = β (0, y 2 ) β ( r 3, y 3 ) +β (0, y 1 ) β ( r 3, y 3 ) h 3 mod a 3 +β ( r 3, y 1 ) β ( r 3, y 2 ) ( ) ( ) h β 1 h 2 h, y 2 β 1 r 3, y 3 ( ) ( ) + h +β 2 a H 2 h 1 h, y 1 β 2 r 3, y 3 ( ) ( ). (h 1,h 2 ) (0,0) +β r 3 h 1, y 1 β r 3 h 2, y 2 β +β H +β ( h 1 h 2 ( h 2, y 2 ) β ) h 1, y 1 β ( ) r 3 h 1, y 1 β ) r 3, y 3 ) r 3, y 3 ( ) = 0. r 3 h 2, y 2 ( h 1 ( h 2

57 51 ( ) ( ) h β 1 h 2 h, y 2 β 1 r 3, y 3 ( ) ( ) h +β 2 a H 2 h 1 h, y 1 β 2 r 3, y 3 ( ) ( ) +β r 3 h 1, y 1 β r 3 h 2, y 2 = ( 1 e y e{ r3}y3 + 1 e y 1 ) + 1 e{ r3}y3 e{ r3}y1 e{ r3}y e y 2 1 e y e y 1 1 e y 3 1 e y 1 1 e y 2 1 h 3 mod a 3 After multiplying each term by the common denominator and combining terms, we get the numerator h 3 mod a 3 e y 1+y 2 +{ r 3 }y 3 e y 2+{ r 3 }y 3 + e y 1+{ r 3 }y 3 e { r 3}y 3 +e y 1+y 2 +{ r 3 }y 3 e y 1+{ r 3 }y 3 + e y 2+{ r 3 }y 3 e { r 3}y 3 +2e { r 3}y 1 +{ r 3 }y 2 +y 3 2e { r 3}y 1 +{ r 3 }y 2 (writing all exponents in terms of y 1 and y 2 ) = h 3 mod a 3 e { r 3}y 1 +{ r 3 }y 2 e { r 3}y 1 +{ r 3 }y 2 + e { r 3}y 1 { r 3 }y 2 e { r 3}y 1 { r 3 }y 2 + e { r 3}y 1 +{ r 3 }y 2 e { r 3}y 1 { r 3 }y 2 = 0. +e { r 3}y 1 +{ r 3 }y 2 e { r 3}y 1 { r 3 }y 2 +2e { r 3}y 1 { r 3 }y 2 2e { r 3}y 1 +{ r 3 }y 2

58 52 We must now show Since ± H (h 1,h 2 ) (0,0) β +β +β ( h 1 h 2 ( h 2, y 2 ) β ) h 1, y 1 β ( ) r 3 h 1, y 1 β ) r 3, y 3 ) r 3, y 3 ( ) = 0. r 3 h 2, y 2 ( h 1 ( h 2 ( ) ( ) h 1 h 2 h, ± i a i r 3 Z, we see that this is just another special case of our proof of Theorem 3.1 when (x 1, x 2, x 3 ) = (0, 0, x 3 λa 3 ). Thus, these terms vanish. Last, we apply our Theorem 3.1 to the final case.

59 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,, a 3, a 4 Z, x 1, x 2, x 3, x 4 R, and y 1, y 2, y 3, y 4 be four nonzero variables such that y 1 + y 2 + y 3 + y 4 = 0. Then 4 Ω k=1 A k X k Y k x k y k = 1 8i ( cot ( y1 ) + cot 2i ( y2 ) + cot 2i ( y3 ) + cot 2i ( y4 )) 2i 53

60 54 if (x 1, x 2, x 3, x 4 ) (,, a 3, a 4 ) R+Z 4. The sum vanishes for all other cases, except possibly when (x i, x j, x k ) (a i, a j, ) R + Z 3 and ( hi + x i a i h l + x l, h j + x j a l a j h l + x l, h k + x k a l h ) l + x l Z 3 a l for all h i, h j, h k, h l 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 Ω k=1 A k X k Y k x k y k = n k=1 H n β (r k r i, y i ), i=1

61 55 where β(α, Y ) = 1 ey +1 2 e Y 1 for α Z, e {α}y e Y 1 for α Z and H = h 1 mod h 2 mod h n mod a n. Our approach to proving Theorem 5.1 will mimic that of our new proof of Theorem 2.4 outlined previously. Thus, we must check the following cases (i) (x 1, x 2, x 3, x 4 ) (,, a 3, a 4 ) R + Z 4 ; (ii) (x i, x j, x k ) (a i, a j, )R + Z 3 for some 1 i < j < k 4 but not (i); (iii) (x i, x j ) (a i, a j )R + Z 2 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 (x 1, x 2, x 3, x 4 ) (,, a 3, a 4 ) R + Z 4. As in our simplified proof of Theorem 2.4, our concern is if h i a i h j a j Z. Since this occurs only when h i = h j = 0 after introducing a change in variables, we can split the sum

62 56 4 k=1 Ω A k X k Y k x k y k into the terms where h i = 0 for all i = 1, 2, 3, 4 and the remaining terms when not all h i = 0 4 Ω k=1 A k X k Y k x k y k = 4 k=1 H 4 β (r k r i, y i ) i=1 =β(0, y 2 )β(0, y 3 )β(0, y 4 ) + β(0, y 1 )β(0, y 3 )β(0, y 4 ) + β(0, y 1 )β(0, y 2 )β(0, y 4 ) + β(0, y 1 )β(0, y 2 )β(0, y 3 ) β k=1 H\{0,0,0,0} i=1 ( hk h i a i, y i ). We know that since in the last sum h k h i a i of Theorem 3.1 and 4 4 β k=1 H\{0,0,0,0} i=1 Z, we can apply the idea of the proof ( hk h ) i, y i = 0. a i The interesting part consists of the remaining terms. We will now show that the remaining terms in fact simplify to a compact finite sum.

63 57 Let y k = y k 2i. Then 4 Ω k=1 A k X k Y k x k y k =β(0, y 2 )β(0, y 3 )β(0, y 4 ) + β(0, y 1 )β(0, y 3 )β(0, y 4 ) + β(0, y 1 )β(0, y 2 )β(0, y 4 ) + β(0, y 1 )β(0, y 2 )β(0, y 3 ) = 1 8i (cot (y 2) cot (y3) cot (y4) + cot (y1) cot (y3) cot (y4) + cot (y1) cot (y2) cot (y4) + cot (y1) cot (y2) cot (y3)) = 1 8i (cot (y 3) cot (y4) (cot (y2) + cot (y1)) + cot (y1) cot (y2) (cot (y3) + cot (y4))) (applying the cotangent identity to the sums cot(y 1)+cot(y 2) and cot(y 3)+cot(y 4)) ( = 1 cot (y3) cot (y4) ( 8i + cot (y1) cot (y2) cot(y2) cot(y1) 1 cot(y1 +Y 2 ) cot(y 3) cot(y 4) 1 cot(y 3 +y 4) ) )

64 58 (since y 1 + y 2 + y 3 + y 4 = 0, then cot (y 3 + y 4) = cot ( y 1 y 2) = cot (y 1 + y 2)) = 1 8i cot (y 3 ) cot (y4) (cot (y2) cot (y1) 1) cot (y1) cot (y2) (cot (y3) cot (y4) 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) cot (y 1 + y2) (cot (y2) + cot (y1)) + 1 = 1 cot (y 8i 3 + y4) (cot (y3) + cot (y4)) 1 cot (y1 + y 2) = 1 8i = 1 8i cot (y 1 + y2) (cot (y2) + cot (y1)) cot (y1 + y2) (cot (y3) + cot (y4)) 1 ( cot (y 2) + cot (y 1) + cot (y 1 + y 2) 1 cot (y 1 + y 2) + cot (y 3) + cot (y 4) = 1 8i (cot (y 1) + cot (y 2) + cot (y 3) + cot (y 4)). ) 1 cot (y1 + y2) Next, we assume case (iii). Let (x i, x j ) (a i, a j )R + Z 2 for some 1 i < j 4 and not case (i) and not case (ii). Without loss of generality, assume (x 1, x 2 )

65 59 (, )R + Z 2. Then, again, we can divide the sum 4 k=1 Ω A k X k Y k x k y k into terms where h 1 = h 2 = 0 and the remaining terms when not both h 1 and h 2 are zero 4 Ω k=1 A k X k Y k x k y k = + h 3 mod a 3 h 4 mod a 4 H (h 1,h 2 ) (0,0) β (0, y 2 ) β ( r 3, y 3 ) β ( r 4, y 4 ) +β (0, y 1 ) β ( r 3, y 3 ) β ( r 4, y 4 ) +β ( r 3, y 1 ) β ( r 3, y 2 ) β ( r 3 r 4, y 4 ) +β ( r 4, y 1 ) β ( r 4, y 2 ) β ( r 4 r 3, y 3 ) ( ) ( ) ( ) h β 1 h 2 h, y 2 β 1 h r 3, y 3 β 1 r 4, y 4 ( ) ( ) ( ) h +β 2 h 1 h, y 1 β 2 h r 3, y 3 β 2 r 4, y 4 ( ) ( ). +β r 3 h 1 a 1, y 1 β r 3 h 2, y 2 β ( r 3 r 4, y 4 ) ( ) ( ) +β r 4 h 1, y 1 β r 4 h 2, y 2 β ( r 4 r 3, y 3 ) 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.

66 60 h 3 mod a 3 h 4 mod a 4 = h 3 mod a 3 h 4 mod a 4 = h 3 mod a 3 h 4 mod a 4 β (0, y 2 ) β ( r 3, y 3 ) β ( r 4, y 4 ) +β (0, y 1 ) β ( r 3, y 3 ) β ( r 4, y 4 ) +β ( r 3, y 1 ) β ( r 3, y 2 ) β ( r 3 r 4, y 4 ) +β ( r 4, y 1 ) β ( r 4, y 2 ) β ( r 4 r 3, y 3 ) 1 ey 2 +1 e{ r 3 }y 3 e{ r 4 }y 4 2 e y 2 1 e y 3 1 e y ey e{ r 3 }y 3 e{ r 4 }y 4 e y 1 1 e y 3 1 e y e{ r 3 }y 1 e{ r 3 }y 2 e{ r 3 r 4 }y 4 e y 1 1 e y 2 1 e y e{ r 4 }y 1 e{ r 4 }y 2 e{ r 4 r 3 }y 3 e y 1 1 e y 2 1 e y ey 2 +1 e{ r 3 }y 3 e{ r 4 }y 4 2 e y 2 1 e y 3 1 e y ey 1 +1 e{ r 3 }y 3 e{ r 4 }y 4 2 e y 1 1 e y 3 1 e y 4 1 ( ) e y 1 1 e y 1 1 ) ( e y 2 1 e y e{ r 3 }y 1 e y 1 1 e{ r 3 }y 2 e y 2 1 e{ r 3 r 4 }y 4 e y 4 1 ( 2 2 ( ey 3 1 e y 3 1 )) + e{ r 4 }y 1 e y 1 1 e{ r 4 r}y 2 e y 2 1 e{ r 4 r 3 }y 3 e y 3 1 ( 2 2 ( ey 4 1 e y 4 1 )) e y 1+y 2 +{ r 3 }y 3 +{ r 4 }y 4 e y 2+{ r 3 }y 3 +{ r 4 }y 4 + e y 1+{ r 3 }y 3 +{ r 4 }y 4 e { r 3}y 3 +{ r 4 }y 4 + e y 1+y 2 +{ r 3 }y 3 +{ r 4 }y 4 e y 1+{ r 3 }y 3 +{ r 4 }y 4 +e y 2+{ r 3 }y 3 +{ r 4 }y 4 e { r 3}y 3 +{ r 4 }y 4 +2e { r 3}y 1 +{ r 3 }y 2 +y 3 +{ r 3 r 4 }y 4 2e { r 3}y 1 +{ r 3 }y 2 +{ r 3 r 4 }y 4 = h 3 mod a 3 h 4 mod a 4 +2e { r 4}y 1 +{ r 4 }y 2 +{ r 4 r 3 }y 3 +y 4 2e { r 4}y 1 +{ r 4 }y 2 +{ r 4 r 3 }y 3 2 (e y 1 1) (e y 2 1) (e y 3 1) (e y 4 1).

67 61 We write the final equality in terms of y 1, y 2 and y 3 and get the numerator e { r 4}y 1 +{ r 4 }y 2 +({ r 3 } { r 4 })y 3 e { r 4}y 1 +{ r 4 }y 2 +({ r 3 } { r 4 })y 3 + e { r 4}y 1 { r 4 }y 2 +({ r 3 } { r 4 })y 3 e { r 4}y 1 { r 4 }y 2 +({ r 3 } { r 4 })y 3 + e { r 4}y 1 +{ r 4 }y 2 +({ r 3 } { r 4 })y 3 e { r 4}y 1 { r 4 }y 2 +({ r 3 } { r 4 })y 3 + e { r 4}y 1 +{ r 4 }y 2 +({ r 3 } { r 4 })y 3 e { r 4}y 1 { r 4 }y 2 +({ r 3 } { r 4 })y 3 + 2e ({ r 3} { r 3 r 4 })y 1 +({ r 3 } { r 3 r 4 })y 2 +{ r 4 r 3 }y 3 2e ({ r 3} { r 3 r 4 })y 1 +({ r 3 } { r 3 r 4 })y 2 { r 3 r 4 }y 4 + 2e { r 4}y 1 { r 4 }y 2 { r 3 r 4 }y 3 2e { r 4}y 1 +{ r 4 }y 2 +{ r 4 r 3 }y 3. Let every term in an exponent be represented by its sign. Then sign{ r 4 } = +, sign ( { r 4 }) =, sign ({ r 3 } { r 4 }) = C 03, sign ({ r 3 } { r 3 r 4 }) = C 30, and let {r 3 r 4 } = {r 4 r 3 } = +.

68 62 We can construct the following matrices + + C 03 + C C 03 + C M 4 = 03 C 30 C 30 + C 30 C C 03 C 03 + C 03 C and M 4 = 03, C 30 C 30 C 30 C where M 4 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 negative terms in the numerator. We can apply Lemmas 3.2 and 3.3 to the differences { r 3 } { r 4 } and { r 3 } { r 3 r 4 }, where a = 0 and b = 0, respectively. Then we can make a statement about the entries C 03 and C 30. Lemma 5.2. C 03 = + if and only if C 30 =. Proof. Assume C 30 = +. This means the difference { r 3 } { r 4 } is positive and,

69 63 by Lemma 3.2, { r 3 } { r 4 } = { r 4 r 3 }. Then { r 3 } { r 3 r 4 } = { r 3 } { r 3 r 4 } = { r 3 } + { r 4 r 3 } = { r 4 }. Thus, { r 3 } { r 3 r 4 } is negative and so C 30 =. Thus, C 30 = C 03 and the matrices become + + C 03 + C 03 + C 03 C C 03 + C 03 + C M 4 = 03 C, M 4 = 03. C 03 C 03 + C 03 C 03 C 30 C 30 + C 03 C

70 64 It is left to show that M 4 = M 4 after row swapping. The matrices M 4 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 C 03 = + and the matrices are equal up to row swapping. The same is true for C 03 =. 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 (x i, x j, x k ) (a i, a j, )R + Z 3 for some 1 i < j < k 4 and not case (i). Without loss of generality, let (x 1, x 2, x 3 ) (,, a 3 )R + Z 3. Then as before, we divide the sum into the terms when h 1 = h 2 = h 3 = 0 and the remaining

71 65 terms when not all h 1, h 2 and h 3 are zero 4 Ω k=1 A k X k Y k x k y k = + β (0, y 2 ) β (0, y 3 ) β ( r 4, y 4 ) +β (0, y 1 ) β (0, y 3 ) β ( r 4, y 4 ) +β (0, y 1 ) β (0, y 2 ) β ( r 4, y 4 ) +β ( r 4, y 1 ) β ( r 4, y 2 ) β ( r 4 r 3, y 3 ) ( ) ( ) ( ) h β 1 h 2 h, y 2 β 1 h 3 h a 3, y 3 β 1 r 4, y 4 ( ) ( ) ( ) h +β 2 h 1 h, y 1 β 2 h 3 h a 3, y 3 β 2 r 4, y 4 ( ) ( ) ( ). h H +β 3 a 3 h 1 h, y 1 β 3 a 3 h 2 h, y 2 β 3 a 3 r 4, y 4 ( ) ( ) ( ) +β r 4 h 1, y 1 β r 4 h 2, y 2 β r 4 h 3 a 3, y 3 h 4 mod a 4 (h 1,h 2,h 3 ) (0,0,0) 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

72 66 goes awry in the first summand. h 4 mod a 4 β (0, y 2 ) β (0, y 3 ) β ( r 4, y 4 ) +β (0, y 1 ) β (0, y 3 ) β ( r 4, y 4 ) +β (0, y 1 ) β (0, y 2 ) β ( r 4, y 4 ) +β ( r 4, y 1 ) β ( r 4, y 2 ) β ( r 4 r 3, y 3 ) 1 ey ey 3 +1 e{ r 4 }y 4 2 e = y e y 3 1 e y ey ey 3 +1 e{ r 4 }y 4 2 e y e y 3 1 e y 4 1 h 4 mod a ey ey 2 +1 e{ r 4 }y 4 e y e y 2 1 e y e{ r 4 }y 1 e{ r 4 }y 2 e{ r 4 }y 3 e y 1 1 e y 2 1 e y ey ey 3 +1 e{ r 4 }y 4 ( ) e y e = y e y 3 1 e y 4 1 e y ey ey 3 +1 e{ r 4 }y 4 ( ) e y e y e y 3 1 e y 4 1 e y 2 1 h 4 mod a ey ey 2 +1 e{ r 4 }y 4 ( ). e y 3 1 e y e y 2 1 e y 4 1 e y e{ r 4 }y 1 e{ r 4 }y 2 e{ r 4 }y 3 ( 4 ( ey 4 1 )) e y 1 1 e y 2 1 e y e y 4 1

73 67 Expanding terms we get the numerator h 4 mod a 4 e y 1+y 2 +y 3 +{ r 4 }y 4 e y 2+y 3 +{ r 4 }y 4 +e y 1+y 2 +{ r 4 }y 4 e y 1+{ r 4 }y 4 +e y 1+y 3 +{ r 4 }y 4 e y 3+{ r 4 }y 4 +e y 1+{ r 4 }y 4 e { r 4}y 4 e y 1+y 2 +y 3 +{ r 4 }y 4 e y 1+y 3 +{ r 4 }y 4 +e y 1+y 2 +{ r 4 }y 4 e y 1+{ r 4 }y 4 +e y 2+y 3 +{ r 4 }y 4 e y 3+{ r 4 }y 4 +e y 2+{ r 4 }y 4 e { r 4}y 4 e y 1+y 2 +y 3 +{ r 4 }y 4 e y 1+y 2 +{ r 4 }y 4 +e y 1+y 3 +{ r 4 }y 4 e y 1+{ r 4 }y 4 +e y 2+y 3 +{ r 4 }y 4 e y 2+{ r 4 }y 4 +e y 3+{ r 4 }y 4 e { r 4}y 4 +4e { r 4}y 1 +{ r 4 }y 2 +{ r 4 }y 3 +y 4 4e { r 4}y 1 +{ r 4 }y 2 +{ r 4 }y 3