A Geometric Approach to Carlitz-Dedekind Sums

Transcription

1 A Geometric Approach to Carlitz-Dedekind 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 y Asia R Matthews San Francisco, California May2007

2 Copyright y Asia R Matthews 2007

3 CERTIFICATION OF APPROVAL I certify that I have read A Geometric Approach to Carlitz-Dedekind Sums y Asia R Matthews and that in my opinion this work meets the criteria for approving a thesis sumitted 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. Serkan Hosten Professor of Mathematics Dr. Federico Ardila Professor of Mathematics

4 A Geometric Approach to Carlitz-Dedekind Sums Asia R Matthews San Francisco State University 2007 A Carlitz polynomial is a polynomial generalization of the Dedekind sum, which in turn is an arithmetic sum playing a central role in various mathematical areas, such as theta functions, group actions on manifolds, and integer-point enumeration in polytopes. The most important property of any Dedekind-like sum is reciprocity which Carlitz proved algeraically for his polynomials. In this paper we give a geometric proof of Carlitz reciprocity and derive Dedekind reciprocity from our result. This approach gives rise to alternate geometric pictures from which we get a new version of Carlitz reciprocity and some new theorems. Finally, using Brion s decomposition theorem for lattice points in polyhedra, we discover two new theorems relating Carlitz sums to the generating function of two and three-dimensional simplices. In three dimensions we rederive the Mordell-Pommersheim theorem, which marks the first appearance of Dedekind sums in Ehrhart polynomials. I certify that the Astract is a correct representation of the content of this thesis. Chair, Thesis Committee Date

5 ACKNOWLEDGMENTS My est friend Rueen: thank you for understanding that I had to do it, without really understanding what I was doing. Matthias, my advisor in Mathematics as well as Life Matters: thank you for leading the way. My thesis committee. My professors and teachers along the way who have made Math my passion. My Girlfriends, all. My Baa. My family (including all close, far, extended, half, lawful, unlawful, and undiscovered) for eing my family. You are all ingredients in this tasty recipe. v

6 TABLE OF CONTENTS 1 Introduction 1 2 Generating Functions and Geometry A Two-Dimensional Example of Generating Functions An Example in 3 Dimensions 16 3 Carlitz Reciprocity Dimensions n Dimensions 21 4 Two Rays in Two Dimensions 27 5 Perpendicular Rays in the Plane Reciprocity From Any Quadrant in the Plane Reflected Cones 35 6 Brion Decompositions The Triangle The Tetrahedron Dedekind-Rademacher-Carlitz Sums The Relation to Dedekind Sums 53 Biliography 60 vi

7 LIST OF FIGURES 2.1 A ray in the first quadrant The fundamental parallelogram Π The fundamental parallelepiped Π A slice of the fundamental parallelepiped Two-ray decomposition of the first quadrant Perpendicular rays in the plane The cones K 3 and K A simple polytope: a triangle The tetrahedron P. 47 vii

8 Chapter 1 Introduction Reciprocity is a very useful mathematical tool that, in many cases, reduces a complex argument to a trivial one. We consider two important reciprocity laws in this paper; Dedekind reciprocity and Carlitz reciprocity. In the 1880 s Richard Dedekind introduced what is now known as a Dedekind sum. Definition 1.1 (Dedekind sum). For positive integers a and, where 1 s (a, ) := (( ka )) (( )) k, 1

9 2 x x 1/2 if x R \ Z, ((x)) = 0 if x Z. (1.1) Here, the greatest integer function x denotes the greatest integer less than or equal to x. The sum s(a, ) was introduced to study another function that Dedekind was interested in, the Dedekind eta function defined on the upper half-plane of the complex numers as η(τ) = e πıτ 12 Π n=1 (1 e 2πınτ ). Dedekind sums are used primarily in the fields of analytic numer theory and discrete geometry and are also found in cominatorics, topology, algeraic numer theory, algorithmic complexity, and continued fractions. The Dedekind sum has no closed form and can take a priori quite a long time to compute for large values of. This is a property of the function that has the potential to make it devastatingly difficult to work with. Fortunately, Dedekind proved the following reciprocity theorem that eliminated this circumstance and which has had profound implications in any field in which Dedekind sums appear. Theorem 1.1 (Dedekind reciprocity). Let a and e relatively prime positive integers. Then s (a, ) + s (, a) = ( a + 1 a + ). a The simple act of switching the parameters in the Dedekind sum and then adding the two sums together gives a very simple and symmetric closed-form result which

10 3 reduces the computations to a much more practical length of time. In fact, this is one of the reasons that the theorem is so noteworthy. A priori, it takes aout steps to compute a single Dedekind sum, s (a, ), which can take quite a long time if = 1000, for example. On the other hand, using the reciprocity theorem, it takes aout log 2 steps: for = 1000, aout ten steps! This is due to the property of Dedekind sums that s(a, ) = s(a mod, ) where a mod is the remainder of a when divided y. The process is similar to the Euclidean algorithm and takes the same time to compute. For example, using this property of the modulo we have s(381, 125) = s(6, 125). Then using reciprocity we reduce this to s(125, 6) = s(5, 6). Leonard Carlitz studied various generalizations of the classical Dedekind sums. He is said to e one of the most prolific mathematicians of all time ecause he pulished over 700 research papers. Carlitz polynomials are of the following form. Definition 1.2 (Carlitz polynomial). For indeterminates u and v, and positive integers a and, define c (u, v; a, ) := u k 1 v k a. These polynomials arose while Carlitz was studying Dedekind sums through Bernoulli polynomials. Carlitz showed [6] that these polynomials are related y a reciprocity law similar to that of Dedekind sums. Theorem 1.2 (Carlitz reciprocity). If u and v are indeterminates, and a and are

11 4 relatively prime positive integers, then (u 1) c(u, v; a, ) + (v 1) c(v, u;, a) = u a 1 v 1 1. (1.2) The theorem was proved algeraically y Carlitz in 1975, ut first appeared in this form in a paper y Berndt and Dieter [4] in It is interesting to note that the two reciprocity laws that we have mentioned are so closely related that it is possile to algeraically otain Dedekind reciprocity from Carlitz reciprocity. Before proving this result, we note a few useful identities. Lemma 1.3. Let {x} := x x denote the fractional-part function. If a and are relatively prime positive integers, then 1 k = ( 1), (1.3) 2 1 k 2 ( 1)(2 1) =, (1.4) 6 1 ka (a 1)( 1) =, (1.5) 2 1 { } ka = 1 2, (1.6) 1 { } 2 ka ( 1)(2 1) =. (1.7) 6 Proof. The first two identities are trivial. The sum in equation (1.5) can e inter-

12 5 preted as the numer of integer points etween the lines y = 0 and y = a x for x = 1, 2,..., 1. Draw the rectangle with vertices (0, 0), (a, 0), (0, ), and (a, ). There are exactly (a 1)( 1) integer points in the interior of this rectangle. If we draw a line etween the origin and the point (a, ), we decompose the rectangle into the triangle in which we are interested, and a similar triangle. Because a and are relatively prime, there are no integer points along the line etween the points (0, 0) and (a, ). Therefore, there are (a 1)( 1) 2 integer points in each triangle and we have proved equation (1.5). To show (1.6) we sustitute the fractional-part function into equation (1.5) as follows: (a 1)( 1) 2 1 ka = = 1 ( ka { }) ka = a( 1) 2 1 { } ka. Then, using the identity in (1.3) we have proved (1.6). Note that (1.6) implies that 1 { ka } = ( 1) = 1 k 2 = 1 { k }. Because the function 1 { ka } is periodic with period and (a, ) = 1, the sum over a complete period is just a permutation of 1 { k }. We use this fact to prove equation (1.7) y noting that 1 { ka } 2 = 1 { k } 2 = 1 ( k ) 2 = k2. This along with equation (1.4) gives the result. Lemma 1.4. If a and are relatively prime positive integers, then s(a, ) = 1 1 k { } ka 1 4, (1.8)

13 6 and, equivalently, s(a, ) = 1 1 k ka a( 1)(2 1) (1.9) Proof. Using Equation (1.1), we replace the sawtooth function in the Dedekind sum with the fractional-part function as follows: 1 (( )) (( )) ka k 1 ({ } ka s (a, ) = = ({ } ka = 1 ) ( k 2 1 ) 2 = 1 1 { } ka k 1 1 { } ka 1 1 k = 1 1 { } ka k 1 4. ) ({ } k 1 ) 2 The third equality is otained y oserving that for k ranging from 1 to 1, the fractional-part function { } k in the second sum is equal to k. After expanding, we 1 use a few of the identities from Lemma 1.3 in the final equality. Converting the fractional-part function in (1.8) to the floor function, we have (1.9). 1 4 Proof that Theorem 1.2 implies Theorem 1.1. By applying the operators u u twice

14 7 and then v v once to (u 1) u k 1 v 1 k a + (v 1) and then setting u = v = 1, we otain k k 1 k 2 (k 1) 2 + k a a ka v k 1 u ka = u a 1 v (k 1) 2 ka = (a 1) 2 ( 1). This reduces to 2 k k 1 + a 2 ka = (a 1)( 1) 2 + (a 1) 2 ( 1), which is a relation of greatest integer functions. Now we convert the greatest-integer functions to fractional-part functions using identities from Lemma 1.3: 2 ( k k a { }) k 1 + a which we expand to get 2 k { } k +2 a a 1 k ( ka { } ka = 2 a { }) 2 ka = k 2 + a (a 1)( 1) (a 1) 2 ( 1), k 2 1 (a 1)( 1) (2a 1). 2

15 8 Multiplying the equation y 6 and simplify the right-hand side yields 12 2 k (a2 + 1) a 1 k 2 3(a 1)( 1) (2a 1) = 2 2 (a 1)(2a 1) + a 2 ( 1)(2 1) + ( 1)(2 1) 3(a 1)( 1) (2a 1) = 3a 2 + 3a 2 + a a + 1. and thus we have the conversion of Carlitz reciprocity to fractional-part functions, namely, 1 12a k { } ka + 12 k { } k = 3a 2 + 3a 2 + a a + 1. (1.10) a Now we turn our attention to Dedekind reciprocity. Using Lemma 1.4, we convert the greatest-integer functions in the left-hand side of Equation (1.1) to fractionalpart functions: 1 1 k 1 12a k { } ka 1 4 } { ka + 1 { } k k a 1 = 1 a a ( a a + ) a { } k + 12 k = 3a 2 + 3a 2 + a a + 1. a

16 9 Multiplying the first equation y 12a and rearranging the terms to isolate the fractional-part functions on the left-hand side of the equation gives the same result as (1.10). In this paper we will also consider a more generalized version of the Carlitz polynomial. Definition 1.3. The generalization of the Carlitz polynomial, where u 1, u 2,..., u n are indeterminates and a 1, a 2,..., a n are positive integers, is defined as the polynomial c (u 1, u 2,..., u n ; a 1, a 2,..., a n ) := a 1 1 j k j k ka2 ka3 u1 k 1 a u 1 a 2 u 1 3 u j k kan a 1 n. Berndt and Dieter proved a general polynomial reciprocity theorem [4, Theorem 5.1] for Carlitz polynomials in n indeterminates. We use the slightly more specific form given in [1, Theorem 1.2] which states the following. Theorem 1.5 (Berndt Dieter). If a 1, a 2,..., a n are pairwise relatively prime positive integers, then (u 1 1) c (u 1, u 2,..., u n ; a 1, a 2,..., a n ) + (u 2 1) c (u 2, u 3,..., u n, u 1 ; a 2, a 3,..., a n, a 1 ) + + (u n 1) c (u n, u 1,..., u n 1 ; a n, a 1,..., a n 1 ) = u a u a u an 1 n.

17 10 In the following chapters we will give a novel proof of Theorem 1.2 and of its higher-dimensional analog, Theorem 1.5. Our approach will lead us to some new reciprocity theorems and relations for Carlitz polynomials. Finally, we derive a relation etween Carlitz polynomials and certain familiar geometric figures y realizing Carlitz sums as generating functions of polyhedra.

18 Chapter 2 Generating Functions and Geometry 2.1 A Two-Dimensional Example of Generating Functions In this paper we take a geometric approach to Carlitz reciprocity. It was recently oserved that Carlitz polynomials show up in the generating functions of certain geometric ojects. A geometric proof of Carlitz reciprocity was introduced y Beck in [1] in Here we present a novel way of realizing Carlitz polynomials in a geometric setting. Looking at the elements that make up a Carlitz polynomial c(u, v; a, ), it is natural to associate the variales u and v with the x and y axes, and the parameters a and with the point (a, ) in the first quadrant. From there we draw a ray from the origin through the point (a, ) and we have decomposed the first quadrant into 11

19 12 two cones. Figure 2.1: A ray in the first quadrant. These pointed cones are defined as the intersection of finitely many half-spaces that intersect in exactly one point, the vertex. The pointed cones defined aove are K 1 = {λ 1 (0, 1) + λ 2 (a, ) : λ 1, λ 2 0} R 2, K 2 = {λ 1 (1, 0) + λ 2 (a, ) : λ 1 0, λ 2 > 0} R 2, such that K 1 is closed and K 2 is half-open. Proposition 2.1. Define Q (2) 1 := {(x, y) R 2 : x, y 0}, the first quadrant in the plane. Then K 1 K 2 = Q (2) 1 and K 1 K 2 =. Proof. It is clear that K 1 K 2 = Q (2) 1. Suppose y contradiction that (u, v) K 1 K 2. Note that, (u, v) K 1 implies that (u, v) = (λa, λ 2 + λ) for λ, λ 2 0 and

20 13 (u, v) K 2 implies that (u, v) = (λ 1 + λa, λ) for λ 0, λ 1 > 0. Then λa = λ 1 + λa and hence λ 1 = 0 which is a contradiction. The process now requires that we list the integer points in each of these cones. The method that we employ uses generating functions to encode integer points in a set S R d. Instead of vectors, each coordinate point is listed as the multidegree of a monomial. For example, the point (a, ) is encoded in the monomial u a v in the indeterminates u and v. This clever method is helpful for it raises the discrete set of points to a continuous function, therey making it easier to work with. We define a polyhedron as the intersection of finitely many half-spaces, and call it a rational polyhedron if the half-spaces defining the polyhedron can e written in terms of rational parameters. Definition 2.1. If S is a rational polyhedron, σ S R d is called the integer-point transform of S, where σ S (z) = σ S (z 1, z 2, z d ) := m S Z d z m. We use the notation m := (m 1,..., m d ) and z m := z m 1 1 z m d d. The two cones that we created in the first quadrant are rational cones ecause the two rays that ound each cone go through rational points. We tile the cone with copies of the fundamental parallelogram. By tiling, we are simplifying the infinite list of integer points in each cone to just translates of the lattice points in

21 14 the fundamental parallelogram of each cone. Consistent with our definition of the cones K 1 and K 2, we define the fundamental parallelograms Π 1 = {λ 2 e 2 + λ(a, ) : 0 λ 2, λ < 1}, Π 2 = {λ 1 e 1 + λ(a, ) : 0 < λ 1 1, 0 λ < 1}. Figure 2.2: The fundamental parallelogram Π 1. In order to illustrate the process of writing the integer-point transform of each cone, consider the fundamental parallelogram Π 1. We egin y listing the translates of the vertices of Π 1. These are the non-negative integer cominations of the generators (0, 1) and (a, ) and so we list them using the generating function:

22 15 (m 1,m 2 )=j(0,1)+k(a,) j,k 0 u m 1 v m 2 = j 0 (u 0 v 1 ) j k 0(u a v ) k = 1 (1 v)(1 u a v ). (2.1) Now we list the integer points in the interior of Π 1. Because the parallelogram is half-open, it contains the point (0, 0). Recall that the lattice point (a, ) is the first integer point from the origin along the ray (a, ). Because this property remains fixed for any integer translation of the ray in the plane, the point (a, + 1) is the first integer point along the ray extending from (0, 1) through (a, + 1). This means that there are no integer points on the oundary lines of the parallelogram, except at the vertices. Now, for each point along the horizontal axis etween 0 and a, the vertical height of the fundamental parallelogram is one unit. Therefore, at each of these values along the horizontal axis, there will e exactly one interior lattice point in the parallelogram. We use the greatest integer function to denote this point. The vertical integer point will lie etween the lines y = x and y = x + 1, for these a a lines descrie two sides of the parallelogram. Therefore, the vertical integer point will have the value y = a x + 1 = a x + 1 for each x and we have that the set of interior lattice points in Π 1 is the set { (0, 0), ( x, a x + 1 ) : 1 x a 1, x Z }. Putting this information together with (2.1) we have

23 16 ( ) σ K1 (u, v) = u 0 v 0 + u k v k a +1 = 1 + a 1 uk v k a +1 (1 v)(1 u a v ) = u m 1 v m 2 (m 1,m 2 )=j(0,1)+k(a,) j,k uv c (u, v; a, ) (v 1) (u a v 1). We otain the integer-point transform for K 2 in the same way, careful to adjust our sums for the half-open cone: Π 2 does not contain the point (0, 0). Instead, the point that will tile the vertices of the paralellogram is the point (1, 0) which is on the closed side of the cone. Thus, σ K2 (u, v) = u + 1 vk u ka +1 (1 u)(1 u v a ) = u + uv c (v, u;, a) (u 1) (u v a 1). (2.2) 2.2 An Example in 3 Dimensions This construction is more complex for dimensions > 2, and is also much more difficult to depict, so we illustrate it in three dimensions efore generalizing. Let Q (3) 1 denote the first (non-negative) orthant in 3-space. In three dimensions, for coprime positive integers a,, c, draw an infinite ray from the origin through the point (a,, c). Now we decompose Q (3) 1 into three (half-open) vertex cones,

24 17 K 1 = {λ 2 e 2 + λ 3 e 3 + λ(a,, c) : λ 2, λ 3, λ 0}, K 2 = {λ 1 e 1 + λ 3 e 3 + λ(a,, c) : λ 1, λ 3, λ 0}, K 3 = {λ 1 e 1 + λ 2 e 2 + λ(a,, c) : λ 1, λ 2, λ 0}. The fundamental parallelepiped Π 3 = {λ 1 e 1 + λ 2 e 2 + λ (a,, c) : 0 λ 1, λ 2 < 1, 0 λ < 1} corresponding to the cone K 3 is ounded y the three rays that extend from the origin through the points (1, 0, 0), (0, 1, 0), and (a,, c). Figure 2.3: The fundamental parallelepiped Π 3.

25 18 As in the two-dimensional case, the integer translates of the vertices of Π 3 gives the denominator for the integer-point transform of K 3 : u m 1 v m 2 w m 3 = (m 1,m 2,m 3 )=i(1,0,0)+j(0,1,0)+k(a,,c) i,j,k 0 1 (1 u) (1 v) (1 u a v w c ). We multiply this y the emedded list of interior points of Π 3 in order to generate the infinite list of lattice points of K 3. To get the interior points of Π 3, consider a slice of the parallelepiped at w = k for the integer 0 < k < c as in Figure 2.3. Figure 2.4: A slice of the fundamental parallelepiped. Note that the parallelogram at this slice of Π 3 has area 1. Also note that it contains no lattice points on its edges or vertices. Π 3 therefore contains exactly one lattice point at w = k. It is immediately clear that this point has coordinates

26 19 ( ka c + 1, k ) c + 1, k. Hence, the interior lattice points of Π3 can e listed as the set of these points for each integer 1 k c 1. The final step of this process is to identify the edges of the parallelepiped that are open and those that are closed in order to list the one vertex point that will tile the vertices of Π 3. For the cone K 3, the closed face is the one along the u 1 - u 2 plane, though the edges along each of those axes are open. Therefore, the point (1, 1, 0) is contained in Π 3 and will tile the vertices of the parallelepipeds. Thus, σ K3 (u, v, w) = uv + w 1 k 1 u ka c +1 v k c +1 c k (1 u) (1 v) (1 u a v w c ) = uv c (w, u, v; c, a, ) (1 u) (1 v) (1 u a v w c ).

27 Chapter 3 Carlitz Reciprocity Dimensions We now have an expression that lists the lattice points in each of our cones, so that in fact we have a list of the lattice points in the first quadrant Q (1) 2 of the plane. Note that we can also write the integer-point transform of the first quadrant directly for it is also a rational cone. The integer-point transform of Q (2) 1 is σ (2) Q (u, v) = 1 1 (1 u)(1 v). Proof of Theorem 1.2. By Proposition 2.1 we have that K 1 K 2 = Q (2) 1 and K 1 K 2 =. This means that we have created two different expressions σ K1 + σ K2 and σ Q (2) 1 that list the integer points in the first quadrant. Naturally we set these two 20

28 21 expressions equal to each other to otain 1 + uv c (u, v; a, ) (v 1) (u a v 1) + u + uv c (v, u;, a) (u 1) (u a v 1) = 1 (u 1)(v 1). To simplify, multiply the equation y (u 1)(v 1)(u a v 1) and we have: (u 1) + uv(u 1) c (u, v; a, ) + u(v 1) + uv(v 1) c (v, u;, a) = u a v 1 uv [(u 1) c(u, v; a, ) + (v 1) c(v, u;, a)] = u a v 1 (u 1) u(v 1) (u 1) c(u, v; a, ) + (v 1) c(v, u;, a) = u a 1 v 1 1. Thus we have constructed a novel proof of Theorem 1.2, the polynomial reciprocity theorem in two dimensions due to Carlitz, using a geometric picture. 3.2 n Dimensions It is now possile to generalize this geometric approach to Carlitz reciprocity. We will prove Theorem 1.5 in a geometric manner similar to the two-dimensional case. Namely, we construct a single ray in n-dimensional space, and then we decompose the non-negative orthant into n cones, K 1, K 2,..., K n. Again we avoid the issue of over-counting along the common faces of the cones y defining our cones to e half-open. Let a := (a 1, a 2,..., a n ) R n such that (a i, a j ) = 1 whenever i j.

29 22 Then we define K 1 = {λ 2 e 2 + λ 3 e λ n e n + λa : λ i 0 i = 2, 3,..., n, λ 0}, K 2 = {λ 1 e 1 + λ 3 e λ n e n + λa : λ i 0 i > 2, λ 1 > 0, λ 0},. K j = λ 1 e λ j 1 e j 1 + λ j+1 e j λ n e n + λa : λ i 0 i > j, λ 1 > 0 i < j, λ 0,. K n = {λ 1 e 1 + λ 2 e λ n 1 e n 1 + λa : λ i > 0 i = 1, 2,..., n, λ 0}. Proposition 3.1. Let Q (n) 1 := {z = (z 1, z 2,..., z n ) R n : z i 0 i = 1, 2,..., n}, the first orthant in n-space. Then n i=1 K i = Q (n) 1 and n i=1 K n =. Proof. It is clear that n i=1 K i = Q (n) 1. Suppose y contradiction that z n i=1 K n. Note that for j k, where j, k = 1, 2,..., n, if z K j then z = (λ 1 + λa 1, λ 2 + λa 2,... λ j 1 + λa j 1, λa j, λ j+1 + λa j+1,..., λ n + λa n ), where λ i 0 i > j and λ 1 > 0 i < j. Also, z K k implies that z = (γ 1 + γa 1, γ 2 + γa 2,... γ k 1 + γa k 1, γa k, γ k+1 + γa k+1,..., γ n + γa n ),

30 23 where γ i 0 i > k and γ i > 0 i < k. Suppose, without loss of generality, that j < k. Then λa j = γ j + γa j and γa k = λ k + λa k. Note that λ γ since otherwise γa j = γ j +γa j and λa k = λ k +λa k imply that γ j = 0 and λ k = 0. Therefore, a j = γ j λ γ and a k = λ k λ γ. If λ > γ,then a k < 0 which is a contradiction. Now suppose that λ < γ. Then a j < 0. Thus K j and K k are disjoint for all j, k = 1, 2,..., n and we have shown that the set {K i : i = 1, 2,..., n} is a disjoint decomposition of the first orthant. We want to construct the integer-point transform for each cone K 1, K 2,..., K n, as well as for Q (n) 1. Because we are now in n dimensions, we identify the fundamental parallelepiped for each cone y the n rays that descrie it. The construction of this generating function proceeds as in the two-dimensional case where the process is roken down to three steps: 1. List the vertices of the translates of the fundamental parallelepiped Π j and encode these in the generating function. 2. Encode the list of the interior points in Π j. 3. Determine the vertex of the half-open parallelepiped that will tile all of the vertices and add this point to the list of interior points. Proof of Theorem 1.5. For each parallelepiped Π j, j = 1,..., n, we first compute the integer translates of the vertices: the non-negative integer comination of all unit

31 24 vectors e 1, e 2,..., e n, and the vector a = (a 1, a 2,..., a n ) and not the unit vector e j. If we let m K j Z n e defined through m = k 1 e 1 + k 2 e k j 1 e j 1 + k j+1 e j k n e n + ka, k i 0, and let the monomial u x := u x 1 1 u x 2 2 u xn n for all x := (x 1, x 2,..., x n ) Z n. Then m K j Z n u m = = 1 (1 u 1 ) (1 u j 1 ) (1 u j+1 ) (1 u n ) (1 u a 1 1 u a 2 2 u an n ) 1 (1 u a ) 1 k n k j (1 u k ). Note that the interior points at each slice of Π j where u j = k, k j, have coordinates ( ka1 + 1,..., a j kaj 1 a j + 1, k, kaj+1 a j kan + 1,..., a j ) + 1. Finally, we must determine the vertex which will tile all vertices of translates of Π j. Because Π j is closed on all edges of the unit vectors e 1, e 2,..., e j 1, the point (1, 1,..., 1, 0,..., 0), with the first 0 is in the j-th entry, will tile the vertices.

32 25 Putting all of this together we otain the integer-point transform of K j, ka 1 a j +1 kaj 1 +1 a j kaj+1 +1 a j σ Kj (u) = u 1u 2 u j 1 + a j 1 u 1 uj 1 u k j uj+1 u (1 u a ) 1 k n (1 u k ) k j kan +1 a j n (3.1) = u 1u 2 u j 1 + u a c (u j, u j+1,..., u n, u 1,..., u j 1 ; a j, a j+1,..., a n, a 1,..., a j 1 ) (1 u a ). 1 k n (1 u k ) k j (3.2) We are now ready to estalish a geometric proof of Theorem 1.5. We egin with the equation σ K1 (u) + σ K2 (u) + + σ Kn (u) = σ KQ (u). (n) 1 Sustituting the expression from (3.1) gives: n u 1 u 2 u j 1 + u a c (u j, u j+1,..., u n, u 1,..., u j 1 ; a j, a j+1,..., a n, a 1,..., a j 1 ) (1 u a ) 1 k n (1 u k ) k j 1 = n i=1 (1 u i). j=1 Now, y multiplying the equation y (1 u a ) and isolating the Carlitz polynomials

33 26 on the left-hand side of the equation we simplify to otain: n u 1 u 2 u n (u j 1) c (u j, u j+1,..., u n, u 1,..., u j 1 ; a j, a j+1,..., a n, a 1,..., a j 1 ) j=1 = u a 1 (u 1 1) u 1 (u 2 1) u 1 u 2 (u 3 1) u 1 u n 1 (u n 1) = u a u 1 u 2 u n. Dividing y u 1 u 2 u n yields Theorem 1.5.

34 Chapter 4 Two Rays in Two Dimensions The geometry used in the previous section is very simple: one ray in space. We now turn to a slightly more complex geometric picture: two rays in space. To simplify further, we consider only the case of two rays in the first quadrant of the plane. Let a,, c, d e positive integers and draw infinite rays through the points (a, ) and (c, d) in the first quadrant. This construction decomposes the first quadrant into three rational cones. The integer-point transform of each of the two exterior cones, K 1 and K 3, is easily computed in the same manner as in Chapter 2. However, the cone in the middle, K 2 is ounded on either side y a non-unit vector, and therefore the interior points of the fundamental parallelogram of this cone are not trivial to list. To avoid this prolem, we add the constraint that the fundamental parallelogram should have no interior 27

35 28 Figure 4.1: Two-ray decomposition of the first quadrant. lattice points. This is equivalent to saying that it has area 1. The fundamental parallelogram Π 2 is descried y the rays (a, ) and (c, d) and therefore has area a c d = ad c. Thus we have the constraint ad c = 1. This construction leads to the following new theorem. Theorem 4.1. If a,, c, d are such that ad c = 1, then (u 1) ( u a v 1 ) c (u, v; c, d) + (v 1) ( u c v d 1 ) c (v, u;, a) = u a+c 1 v +d 1 u a v u c v d + u a 1 v + u c v d 1 u a 1 v 1 u c 1 v d

36 29 Proof. Again, we prove the theorem geometrically. To simplify our computations, define K 1 = {λ 2 e 2 + λ cd (c, d) : λ 2 > 0, λ cd 0} R 2, K 2 = {λ a (a, ) + λ cd (c, d) : λ a, λ cd 0} R 2, K 3 = {λ 1 e 1 + λ a (a, ) : λ 1 > 0, λ a 0} R 2, so that K 2 is closed and K 1 and K 3 are half-open. Thus K 1 K 2 K 3 = Q (2) 1 and K 1, K 2, K 3 are pairwise disjoint. With the method introduced in Chapter 2, the integer-point transform of each cone ecomes the following: and σ K1 (u, v) = v + c 1 uk v kd c +1 v + uv c (u, v; c, d) = (1 v) (1 u c v d ) (v 1) (u c v d 1), 1 σ K2 (u, v) = (1 u a v ) (1 u c v d ) = 1 (u a v 1) (u c v d 1), σ K3 (u, v) = u + 1 vk u ka +1 (1 u) (1 u a v ) = u + uv c (v, u;, a) (u 1) (u a v 1). Because the fundamental parallelogram Π 2 has no interior integer points, the generating function σ K2 is relatively simple. Then the sum of the integer-point transforms of each of the cones is equal to the integer-point transform of Q (2) 1 so we have σ K1 (u, v) + σ K2 (u, v) + σ K3 (u, v) = σ (2) Q (u, v). 1

37 30 Replacing these with the expressions aove we have v + uv c (u, v; c, d) (v 1) (u c v d 1) + 1 u + uv c (v, u;, a) + (1 u a v ) (1 u c v d ) (u 1) (u a v 1) = 1 (u 1)(v 1). By clearing the denominator and isolating the Carlitz sums on the left, (u 1) ( u a v 1 ) c (u, v; c, d) + (v 1) ( u c v d 1 ) c (v, u;, a) = u a+c 1 v +d 1 u a v u c v d + u a 1 v + u c v d 1 u a 1 v 1 u c 1 v d Thus we have proved Theorem 4.1. Recall that we are ale to otain classical Dedekind reciprocity from Carlitz reciprocity. It is interesting to consider what would happen if we performed the same computations on Theorem 4.1. In the classical case, we applied the operators u u twice and then v v once and then set u = v = 1. If we perform these exact operations, we otain a trivial result, ut we lose the greatest integer functions and therefore can not otain, through conversion to the fractional-part function, the Dedekind sums that we are seeking. However, applying the operators u u and v v twice each, and then setting u = v = 1 yields a nontrivial result. Corollary 4.2 (Rademacher). If a,, c, d are positive integers such that ad c = 1, then s(a, ) + s(d, c) = ( a + a c + d + d ). c

38 31 Proof. Let a,, c, d e relatively prime positive integers such that ad c = 1. Derivate Theorem 4.1 using the operators u u and v v twice each, and then set u = v = 1 to otain: c 1 2 kd c 1 kd c 1 kd c 1 c 1 2a + 4 k + 2(2a 1) k + (2a 1) 2 c c c 1 2 ka 1 ka 1 ka d + 4c k + 2c(2d 1) + 2c 2 k + (2d 1) = (a + c 1) 2 ( + d 1) 2 a 2 2 c 2 d 2 + (a 1) c 2 (d 1) 2 (a 1) 2 ( 1) 2 (c 1) 2 (d 1) c 2 Then sustituting the identities from Propositions 1.3 and 1.4 yields the simplification: 12(ad + c) (s(a, ) + s(d, c)) = 3 3a 3 3c 3d + 3a 2 d 3ac + 3ad 3 2 c + 3acd 3c 2 + 3ad 2 3cd c 2 3a 2 d 2 + 4a 2 c 4a 2 d + 4c 2 d 4acd 2 + 6a 6ad + a c + d + ad2 c = (ad + c) + d + a2 d ( a + a c + d + d c 6 + ac + 6cd ). We use the assumption that ad c = 1 to otain the second equality, and the theorem follows.

39 32 Corollary 4.2 was given y Hans Rademacher in 1956 [9]. It was derived algeraically from classical Dedekind reciprocity using the following properties of Dedekind sums for the relation ad c = 1: s(d, ) = s(a, ) and s(, d) = s( c, d) = s(c, d). Taking the sum of the two reciprocity identities s(d, c) + s(c, d) + s(d, ) + s(, d) = ( c d + d c + 1 ) 1 cd ( 12 d + d + 1 ) d and sustituting the aove properties gives Corollary 4.2.

40 Chapter 5 Perpendicular Rays in the Plane We now turn our focus ack to a single ray in the plane, the ray through the points (0, 0) and (a, ). Extending the ray to the line through these points, and drawing the perpendicular line through the origin, we decompose each quadrant of the plane into two cones. If we lael each of these cones counter-clockwise y convention, eginning with the positive x-axis, then each cone has its vertex at the origin and K 1 is the cone defined y the rays (1, 0) and (a, ), K 2 is defined y (a, ) and (0, 1), and so on. 5.1 Reciprocity From Any Quadrant in the Plane We showed in Section 2.1 that Carlitz reciprocity can e derived from the geometric picture of a ray in the first quadrant of the plane. In fact, we can do the same for 33

41 34 Figure 5.1: Perpendicular rays in the plane. any quadrant. In the second quadrant, letting K 7 e closed and K 8 half-open and using the integer-point transform of each cone we have: (u 1) c (u, 1v ) ( ) ( ) ( ) a ;, a + v 1 c, u; a, = u 1 1, (5.1) v v which is Carlitz reciprocity in the variales u and 1/v.

42 Reflected Cones We would like to have a relation etween the cones σ K3 (u, v) and σ K7 (u, v). The (half-open) cones that make up a half-space have the property that the sum of their integer-point transfoms is zero [3, Theorem 9.2]. For example, suppose that K 1 is closed, K 2 and K 8 are open, and K 3 is half-open: closed on the side that it shares with K 2. Then K 3, K 2, K 1 and K 8 make up an (open) half-space and σ K3 (u, v) + σ K2 (u, v) + σ K1 (u, v) + σ K8 (u, v) = 0. Figure 5.2: The cones K 3 and K 7. Similarly, if we define K 1 to e closed, K 2 and K 8 open as aove, and K 7 to e half-open where it is closed on the side that it shares with K 8, then K 2, K 1, K 8 and

43 36 K 7 make up an (open) half-space so that σ K2 (u, v) + σ K1 (u, v) + σ K8 (u, v) + σ K7 (u, v) = 0. Putting these two equations together we have σ K3 (u, v) = σ K7 (u, v), (5.2) where K 3 = {λ 1 (0, 1) + λ 2 (, a) : λ 1 > 0, λ 2 0} R 2, K 7 = {λ 1 (0, 1) + λ 2 (, a) : λ 1 0, λ 2 > 0} R 2. This construction gives rise to an interesting version of Carlitz reciprocity. Theorem 5.1. Let a and e relatively prime positive integers. Then for the indeterminates u and v, uv 1 (v 1) (u v a 1) c ( v 1, u; a, ) + u 1 v(u 1) ( u v a 1 ) c ( u 1, v;, a ) = u ( u v a 1 ) + v ( u v a 1 ), (5.3)

44 37 or equivalently, uv a 1 (v 1) c ( v 1, u; a, ) + u 1 v(u 1) c ( u 1, v;, a ) = uv a + u v. (5.4) Before we prove the theorem, we first state a useful identity. Proposition 5.2. If a and are relatively prime positive integers and k Z 0, such that 1 k 1, then ka = ka 1 1. Proof. First note that for any integers a and, a a since a a and a a. Now, let a, Z 0 and let 1 k 1. Clearly ka 1 < ka, thus ka < ka 1. Therefore, ka ka 1 ka 1, with equality etween the first two expressions if and only if ka 1, we have that ka + 1 = ka 1. Z. Since ka and ka 1 differ y no more than Proof of Theorem 5.1. Let us define the cones K 3 and K 7 as the (half-open) cones as they appear in (5.2). Then the integer-point transform of each cone is written: σ K3 (u, v) = v + 1 u k v ka +1 (1 v)(1 u v a ) ka uk v σ K7 (u, v) = u v a + 1 (1 v 1 ) (1 u v a ) = v + 1 v c ( 1, v;, a) u u (v 1) (u v a 1), = u v a + uv 1 c ( u, 1 ;, a) v. (v 1 1) (u v a 1) Putting this into (5.2), we have v + 1 v c ( ( 1, v;, a) u u (v 1) (u v a 1) = u v a + uv 1 c u, 1;, a) v, (v 1 1) (u v a 1)

45 38 and thus, u ( 1 u v a) c ( u, v 1 ;, a ) = u 1 v ( u v a 1 ) c ( u 1, v;, a ). (5.5) Now we multiply this y (u 1) so that we can use Carlitz reciprocity to replace c (u, v 1 ;, a): u ( 1 u v a) ( u 1 v (a 1) 1 + v 1 (v 1) c ( u 1, v; a, )) = u 1 v(u 1) ( u v a 1 ) c ( u 1, v; ; a ), so that uv 1 (v 1) ( u v a 1 ) c ( u 1, v; a, ) + u 1 v(u 1) ( u v a 1 ) c ( u 1, v; ; a ) = u ( u v a 1 ) ( u 1 v (a 1) 1 ), and finally, uv 1 (v 1) (u v a 1) c ( v 1, u; a, ) + u 1 v(u 1) ( u v a 1 ) c ( u 1, v;, a ) = u ( u v a 1 ) + v ( u v a 1 ). It is interesting to note that Equation (5.5) can e derived algeraically y a

46 39 simple change of variales. Proposition 5.3. The Carlitz polynomials c (u, v 1 ;, a) and c (u 1, v;, a) are related in the following way: u ( 1 u v a) c ( u, v 1 ;, a ) = u 1 v ( u v a 1 ) c ( u 1, v;, a ). Proof. Beginning with the right hand side of the equation, u 1 v ( u v a 1 ) c ( u 1, v;, a ) 1 = u k+ v 1 ka +1 a replace k with k in oth sums and the expression ecomes u k v ka +1, 1 u k ka 1 v +1 u k ka v +1+a. Here note that ka ka ka = ka 1 y Proposition 5.2. Then, ecause ka 1 < / Z and there does not exist any integer x such that ka 1 = ka and hence ka + 1 = ka. Thus, u 1 v ( u v a 1 ) c ( u 1, v;, a ) 1 = u k ka 1 v < x < ka, we have u k ka v +a.

47 40 This is equal to the left-hand side of the equation u ( 1 u v a) c ( u, v 1 ;, a ) 1 = u k ka 1 v +a + u k ka v. Note, however, that the construction of the equation itself is not intuitive. We emphasize that the set-up of the equation itself is a result of the geometry.

48 Chapter 6 Brion Decompositions 6.1 The Triangle Until now, we have only considered pointed cones with the vertex at the origin. Let us now turn to a different oject. A convex polytope is the ounded intersection of finitely many half-spaces, and is called rational if all of its vertices have rational coordinates. A well-known theorem relating to convex polytopes is Brion s theorem [5] which gives an expression for the numer of integer points in a rational polytope. Definition 6.1. If v is a vertex of a closed polytope P, then a vertex cone K v is the smallest cone with apex v that contains P. Theorem 6.1 (Brion s theorem). Suppose P is a rational convex polytope, and let K v e a vertex cone with vertex v for each vertex v of P. Then we have the following 41

49 42 identity of rational functions: σ P (z) = v σ Kv (z), where the sum is over all vertices of P. We construct a triangle in two dimensions using the positive quadrant and the half-space y x for relatively prime a and as illustrated in Figure 6.1. a Figure 6.1: A simple polytope: a triangle. This polytope P is the convex hull of the points (0, 0), (a, 0), and (0, ). If we call the (closed) vertex cones K 1, K 2 and K 3 with vertices at each of these three

50 43 points, respectively, then Brion s theorem says that σ P (z) = σ K1 (z) + σ K2 (z) + σ K3 (z). This construction allows us to give a novel expression for the Carlitz sum as the integer-point transform of a triangle. Theorem 6.2. Let a and e relatively prime positive integers and u and v e indeterminates. If P is the triangle with vertices (0, 0), (a, 0) and (0, ), then the Carlitz polynomial c ( v, 1 ;, a) and the integer-point transform of P are related in u the following manner: (u 1) σ P (u, v) = u a v c ( v, u 1 ;, a ) + u ( u a + v ) v+1 1 v 1. Proof. Formally, we define our cones K 1 = {j(1, 0) + k(0, 1) : j, k 0}, K 2 = {(a, 0) + j( 1, 0) + k( a, ) : j, k 0}, K 3 = {(, 0) + j(a, ) + k(0, 1) : j, k 0}. Now let us compute the generating functions. The integer-point transform of the

51 44 first cone, K 1, is simple; we have done it efore: σ K1 (u, v) = 1 (1 u)(1 v). To compute σ K3 (u, v), shift the cone K 3 down units so that the vertex is at the origin and call this cone K 3. K 3 is now a familiar oject as well. We computed the integer-point transform of a very similar oject (permute a and ) in Section 5.2. We have σ K 3 (u, v) = 1 + a 1 uk v (1 v 1 ) (1 u a v ) = 1 + uv 1 c (u, v 1 ; a, ) (1 v 1 ) (1 u a v ). k a To shift the cone ack up, multiply σ K 3 (u, v) y v to otain σ K3 (u, v) = v+1 + uv c (u, v 1 ; a, ) (v 1) (u a v 1). Applying the same process to cone K 2 we have the integer-point transform σ K2 (u, v) = ua+1 + u a v c (v, u 1 ;, a) (u 1) (u a v 1). Using the integer-point transform of each of these three cones in Brion s theorem

52 45 we otain σ P (u, v) = 1 (u 1)(v 1) ua+1 + u a v c (v, u 1 ;, a) v+1 + uv c (u, v 1 ; a, ). (u 1) (u a v 1) (v 1) (u a v 1) We egin simplifying y re-writing the generating functions in a slightly different form: σ P (u, v) = 1 (u 1)(v 1) + u2a+1 + u 2a v c (v, u 1 ;, a) v2+1 + uv 2 c (u, v 1 ; a, ). (u 1) (u a v ) (v 1) (u a v ) (6.1) Clearing the denominator gives (u 1)(v 1) ( u a v ) σ P (u, v) = u a v + u 2a+1 (v 1) v 2+1 (u 1) + u 2a v(v 1) c ( v, u 1 ;, a ) uv 2 (u 1) c ( u, v 1 ; a, ). Note that all ut the factor v in the final term of this equation can e replaced using reciprocity from a variation of (5.4) from Theorem 5.1. By replacing the variale u with 1 and v with 1, Theorem 5.1 ecomes v u uv (u 1) c ( u, v 1 ; a, ) u a v(v 1) c ( v, u 1 ;, a ) = u a v uv,

53 46 and sustituting this into (6.1) we have (u 1)(v 1) ( u a v ) σ P (u, v) = u 2a v(v 1) c ( v, u 1 ;, a ) uv +1 (v 1) c ( v, u 1 ;, a ) + u a v + u 2a+1 (v 1) v 2+1 (u 1) u a v +1 + uv 2. (6.2) Notice that we now have only one Carlitz sum in the aove equation! The right-hand side ecomes u a v(v 1) ( u a v ) c ( v, u 1 ;, a ) +u a v +u 2a+1 (v 1) uv 2 (v 1) v +1 ( u a v ), so that (u 1)(v 1) ( u a v ) σ P (u, v) = u a v(v 1) ( u a v ) c ( v, u 1 ;, a ) + u a v + u 2a+1 (v 1) uv 2 (v 1) v +1 ( u a v ). (6.3) Dividing oth sides of this equation y (v 1) ( u a v ), we have Theorem 6.2.

54 The Tetrahedron Dedekind-Rademacher-Carlitz Sums It is natural to extend the application of Brion s theorem to a convex polytope in higher dimensions. We consider a simple polytope in the first octant Q (3) 1 : the tetrahedron P defined P = {(x, y, z) R 3 ; x, y, z 0, x a + y + z c 1 }, (6.4) with vertices (0, 0, 0), (a, 0, 0), (0,, 0), (0, 0, c) where a, and c are positive integers. Figure 6.2: The tetrahedron P.

55 48 The tth dilate of a suset S R d is the set {(tx 1, tx 2,..., tx d ) : (x 1, x 2,..., x d ) S}. Let us consider the tth dilate of P for t Z >0. This gives us the dilated tetrahedron tp with vertices (0, 0, 0), (ta, 0, 0), (0, t, 0), and (0, 0, tc). Note that t = 1 gives the tetrahedron in Figure 6.2. Using the motivation of the triangle in two dimensions and the tetrahedron tp, we will use Brion s theorem to find an expression for the integer-point transform of tp. Definition 6.2 (Dedekind-Rademacher-Carlitz (DRC) sum). If a,, and c are positive integers, and u, v, w are indeterminates, then define c 1 1 c(u, v, w; a,, c) := u ja + ka c v j w k. k=0 j=0 We discover that the integer-point transform of tp is an expression of DRC sums. Theorem 6.3. Let tp e the dilated tetrahedron with vertices (0, 0, 0), (ta, 0, 0), (0, t, 0), (0, 0, tc) for t Z >0 and a, and c relatively prime positive integers. Then

56 49 for indeterminates u, v, w, (u 1)(v 1)(w 1) ( u a v ) (u a w c ) ( v w c) σ P (u, v, w) = u (t+2)a (v 1)(w 1) ( v w c) [ (u 1) + c ( u 1, v, w; a,, c )] v (t+2) (u 1)(w 1) (u a w c ) [ (v 1) + c ( v 1, u, w;, a, c )] + w (t+2)c (u 1)(w 1) ( u a v ) [ (w 1) + c ( w 1, u, v; c, a, )] ( u a v ) (u a w c ) ( v w c). Proof. Let P e the tetrahedron defined aove with t = 1 and define the vertex cones K 0 = {λ 1 (1, 0, 0) + λ 2 (0, 1, 0) + λ 3 (0, 0, 1) : λ 1, λ 2, λ 3 0}, K 1 = {(a λ 1, 0, 0) + λ 2 (0,, 0) + λ 3 (0, 0, c) : λ 1, λ 2, λ 3 0}, K 2 = {λ 1 (a, 0, 0) + (0, λ 2, 0) + λ 3 (0, 0, c) : λ 1, λ 2, λ 3 0}, K 3 = {λ 1 (a, 0, 0) + λ 2 (0,, 0) + (0, 0, c λ 3 ) : λ 1, λ 2, λ 3 0}. Brion s theorem states that the sum of the integer-point transforms of these cones is equal to the integer-point transform of P. We approach the definition of these generating functions as we did in the two-dimensional case y shifting the vertex of each cone K i to the origin. If we call this shifted cone K i, then the fundamental parallelepiped Π i for each i = 0, 1, 2, 3 is defined through

57 50 Π 0 = {λ 1 (1, 0, 0) + λ 2 (0, 1, 0) + λ 3 (0, 0, 1) : 0 λ 1, λ 2, λ 3 < 1}, Π 1 = {λ 1 ( 1, 0, 0) + λ 2 ( a,, 0) + λ 3 ( a, 0, c) : 0 λ 1, λ 2, λ 3 < 1}, Π 2 = {λ 1 (a,, 0) + λ 2 (0, 1, 0) + λ 3 (0,, c) : 0 λ 1, λ 2, λ 3 < 1}, Π 3 = {λ 1 (a, 0, c) + λ 2 (0,, c) + λ 3 (0, 0, 1) : 0 λ 1, λ 2, λ 3 < 1}. By construction, there exists exactly one integer point in the interior of Π 1 at each integer j along the y-axis and k along the z-axis. In other words, for j, k Z 0, (x, j, k) Π 1 is written as (x, j, k) = ( λ 1 λ 2 a λ 3 a, λ 2, λ c ) and hence λ 2 = j and λ 3 = k. Because 0 λ c i < 1, we have that j = 0, 1,..., a 1 and k = 0, 1,..., 1, and hence (x, j, k) = ( ja ) ka c, j, k. Now, writing the integerpoint transform of K 1 is as simple as multiplying the integer-point transform of K 1 y u a. That is, σ K1 (u, v, w) = u a σ K 1 (u, v, w). We would like to know how this relates to the dilated tetrahedron tp. Let tk i denote the vertex cone related to tp (as we related K i to P ) for i = 0, 1, 2, 3. Then tk 1 = (ta, 0, 0) + K 1, tk 2 = (0, t, 0) + K 2, tk 3 = (0, 0, tc) + K 3, and tk 0 = K 0. So, the cones K i and tk i, i = 1, 2, 3 are similar cones! That is, they have the same generators ut different vertices. Therefore, for t = 1, 2, 3, writing the

58 51 integer-point transform of tk 1 is as trivial as multiplying the integer-point transform of K 1 y u ta. Similarly for tk 2 and tk 3, we have σ tk2 (u, v, w) = v t σ K 2 (u, v, w) and σ tk3 (u, v, w) = w tc σ K 3 (u, v, w). Note that ja ka c = ja Therefore, the integer-point transform of tk 1 ecomes + ka c 1 for 1 j 1 and 1 k c 1. c 1 1 ja σ tk1 (u, v, w) = u ta k=0 j=0 u ka c v j w k (1 u 1 ) (1 u a v ) (1 u a w c ) = u ta 1 + c 1 1 ja k=0 j=0 u + ka c 1 v j w k u 1 u 2a 1 (u 1) (u a v ) (u a w c ) = u(t+2)a [(u 1) + c (u 1, v, w; a,, c)] (u 1) (u a v ) (u a w c ). Similarly, c 1 a 1 j σ tk2 (u, v, w) = v t k=0 j=0 uj v a k a w k (1 v 1 ) (1 u a v ) (1 v w c ) = v(t+2) [(v 1) + c (v 1, u, w;, a, c)] (v 1) (u a v ) (v w c ),

59 52 and 1 a 1 σ tk3 (u, v, w) = w tc k=0 j=0 uj v k jc w a kc (1 w 1 ) (1 u a w c ) (1 v w c ) = w(t+2)c [(w 1) + c (w 1, u, v; c, a, )] (w 1) (u a w c ) (v w c ). The integer-point transform for K 0 doesn t change when we dilate the tetrahedron so we have that the generators of K 0 are the three unit vectors and σ K0 (u, v, w) = u m 1 v m 2 w m 3 (m 1,m 2,m 3 )=i(1,0,0)+j(0,1,0)+k(0,0,1) i,j,k 0 1 = (u 1)(v 1)(w 1). By Brion s theorem, we have that σ P (u, v, w) = σ K0 (u, v, w) + σ tk1 (u, v, w) + σ tk2 (u, v, w) + σ tk3 (u, v, w) and hence σ P (u, v, w) = u (t+2)a [(u 1) + c (u 1, v, w; a,, c)] (u 1) (u a v ) (u a w c ) + w(t+2)c [(w 1) + c (w 1, u, v; c, a, )] (w 1) (u a w c ) (v w c ) v(t+2) [(v 1) + c (v 1, u, w;, a, c)] (v 1) (u a v ) (v w c ) 1 (u 1)(v 1)(w 1).

60 53 Clearing the denominator gives the theorem The Relation to Dedekind Sums It was perhaps unclear in the previous section the reason for calling the polynomial c(u, v, w; a,, c) the Dedekind-Rademacher-Carlitz sum. We hope that the following will help to clarify this. We egin y stating a well-known theorem [7, 8]; this was the first instant that an explicit formula for a lattice-point count in three-dimensional polytopes was given. The lattice-point enumerator for the t th dilate of P R d is denoted L P (t) and is equivalent to # ( tp Z d), the discrete volume of P. Theorem 6.4 (Mordell, Pommersheim). Let P e given y (6.4) and let a, and c e pairwise relatively prime. Then L P (t) = ac a + ac + c t3 + t 2 + ( s(c, a) s(ca, ) s(a, c)) t ( a + + c + 1 ( c 4 12 a + ca + a c + 1 )) t + 1. ac We will prove that Theorem 6.3 implies Theorem 6.4. Before we prove this interesting result, we give some necessary identities.

61 54 Proposition 6.5. If a, and c are relatively prime positive integers, then 1 k=0 j=0 1 j k=0 j=0 1 k=0 j=0 { jc a + kc } = a 1, (6.5) 2 { jc a + kc } a(c 1) = c s(a, c) +, (6.6) 4 { jc a + kc } 2 = (a 1)(2a 1) 6a. (6.7) Proof. A Dedekind Rademacher sum is like a Dedekind sum with a shift in the sawtooth argument. We egin y noting the following invaluale property of one such function: (( x + kc )) = ((ax)), (6.8) a k=0 which is given as an exercise in [3]. Using this, we can simplify the single sum k=0 { jc a + kc } = k=0 [(( jc a + kc )) + 1 ] = 2 (( )) jac + a { } jac 2 = + a 1. 2 (6.9) We use this result to prove the first two identities as follows. (6.5): Comining (6.9) with Identity (1.6) we have that 1 k=0 j=0 { jc a + kc } = 1 j=0 ({ } jac + a 1 ) = a

62 55 (6.6): Now, we use (6.9) to show 1 j k=0 j=0 = a { jc a + kc } j=1 = a [(( )) j + 1 a 2 j=1 j a ] [(( jc a 1 { } jac { j = a a k=0 j=1 )) + ] = a s(c, a) + 2 } ({ } jc + 1 ) a 2 a(a 1) 4, where the final expression was otained y expanding the product and using (6.8) with x = 0. (6.7): We use Identity (1.7) and the second Bernoulli polynomial, B 2 (x) := x 2 x to otain the final equation. The Bernoulli function B k (x) := B k ({x}) has the property [2] that a 1 B k=0 k (x) = a 1 Bk (ax). Therefore, we re-write { jc + } kc 2 a in terms of Bernoulli polynomials and replace this in the doule sum as follows: 1 k=0 j=0 { jc a + kc } 2 = 1 ( kac = a 1 B2 k=0 1 ( k=0 j=0 ) + a 1 2 ( jc B 2 a + kc ) { jc + a + kc } 1 ) 6 a 6 = (a 1)(2a 1) 6a. Proof that Theorem 6.3 implies Theorem 6.4. Let tp e the dilated tetrahedron

63 56 given in Theorem 6.3. If we divide oth sides of the equation in the theorem y (u 1)(v 1)(w 1) ( u a v ) (u a w c ) ( v w c), then we have a rational expression for σ tp (u, v, w) where the numerator is u (t+2)a (v 1)(w 1) ( v w c) [ (u 1) + c ( u 1, v, w; a,, c )] v (t+2) (u 1)(w 1) (u a w c ) [ (v 1) + c ( v 1, u, w;, a, c )] + w (t+2)c (u 1)(w 1) ( u a v ) [ (w 1) + c ( w 1, u, v; c, a, )] ( u a v ) (u a w c ) ( v w c), and the denominator is (u 1)(v 1)(w 1) ( u a v ) (u a w c ) ( v w c). The lattice-point enumerator of tp is the numer of integer points in tp. We otain this from the integer-point transform of tp y setting u = v = w = 1. Because this operation reduces oth numerator and denominator to zero, we use L Hospital s rule to reduce. Because we have three indeterminates, the numer of times we use L Hospital and the indeterminates with respect to which we derivate is not immediately clear. However, oth numerator and denominator are symmetric, reducing the prolem to the numer of derivations. It turns out that taking the

64 57 partial derivative with respect to u once, v twice, and w three times does the trick. Beginning with the denominator, we derivate as descried and set u = v = w = 1. The result is very simple: 12c 2. Turning to the numerator, after differentiating and setting u = v = w = 1 we otain a complex expression in t. Fortunately there is a way to further simplify the expression! Because the constant term of L P is 1 [3, Corollary 3.15], we are only concerned with the non-constant terms of the expression: 2a 2 c 3 t 3 6a 2 c 3 t 2 6c 2 t 2 4a 2 c 3 t 18c 2 t c 1 j + 6ct a + k c 1 j 6c 2 t c a + k c j=0 k=0 c 1 12ct j=0 k= c 2 t j=0 k=0 1 6ct j=0 k=0 j k a + k c j=0 k=0 1 6ct j=0 k=0 jc a + kc + 6c 2 t 2 jc a + kc 1 j=0 k=0 ( jc 1 + a + kc jc a + kc ). jc a + kc As efore, we replace all greatest-integer functions with fractional-part functions to

65 58 otain 2a 2 c 3 t c 3 t 2 3ac 3 t 2 6c 2 t 2 2 c 3 a t 6ac3 t 3c 3 t ac 3 t c 1 { j + 6a 2 c 2 t + 6ac 2 t 18c 2 t 5a 2 ct 6ct a + k } c k=0 j=0 c 1 { j + 6c 2 t a + k } c 1 { j + 12ct k c a + k } c k=0 j= ct k=0 j=0 1 6c 2 t 2 k=0 j= c 2 t k k=0 j=0 { jc a + kc } k=0 j=0 1 24c 2 t { jc a + kc } + 12c2 a { jc a + kc } k=0 j=0 j 1 a 1 t k=0 j=0 1 6ct k=0 j=0 { jc a + kc } { jc a + kc } { jc a + kc Now we replace the fractional-part functions in this expression with identities from Proposition 6.5. Simplifying the expression gives: } 2. 2a 2 c 3 t c 3 t 2 3ac 3 t 2 3c 2 t 2 2 c 3 a t 3c3 t ac 3 t 3 2 c 2 t 3ac 2 t 9c 2 t a 2 ct c a t + 12c2 (s(a, c) + s(ac, ) + s(c, a)) t. Recall that this is only the non-constant part of the numerator of the lattice-point enumerator of P. We divide y the denominator 12c 2 and after rearranging we