Representation by Quaternary Quadratic Forms whose Coefficients are 1, 2, 7 or 14

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1 Representation by Quaternary Quadratic Forms whose Coefficients are 1, 2, 7 or 14 by Jamilah Alanazi, B. Math (King Faisal University) A thesis submitted to the Faculty of Graduate and Postdoctoral Affairs in partial fulfillment of the requirements for the degree of Master of Science School of Mathematics and Statistics Ottawa-Carleton Institute for Mathematics and Statistics Carleton University Ottawa, Ontario, Canada c Copyright 2015, Jamilah Alanazi

2 Abstract We determine explicit formulae for the number of representations of a positive integer n by the quaternary quadratic forms a 1 x a 2 x a 3 x a 4 x 2 4, where a 1, a 2, a 3, a 4 {1, 2, 7, 14} which satisfy the simplifying assumptions a 1 a 2 a 3 a 4 and gcd(a 1, a 2, a 3, a 4 ) = 1. We use a modular form approach. We then extend our work to determine explicit formulae for the number of representations of n by the octonary quadratic forms x 2 1+x 2 2+x 2 3+x 2 4+7(x 2 5+x 2 6+x 2 7+x 2 8), x 2 1+x 2 2+7(x 2 3+x 2 4+x 2 5+x 2 6+x 2 7+x 2 8) and x x x x x x (x x 2 8). i

3 Dedication To my mother who was always supporting me and my husband Mohammed who never gave up. ii

4 Acknowledgements I thank God for everything that I have been blessed with in my lifespan. I would especially like to convey my special thanks to my supervisor Dr. Ayşe Alaca who has always been there for me, shown acceptance and a welcoming attitude towards me. I thank Dr. Şaban Alaca who have offered so much assistance during my program, as well as Carleton University and all the professors whose courses I have studied over the years. I also, thank Saudi Arabia for the financial support needed to do my Master s degree in Canada. iii

5 Contents Abstract i Dedication ii Acknowledgements iii 1 Introduction 1 2 Basic Concepts Modular Forms Eisenstein Series Dimension Formulae Representations by Quaternary Quadratic Forms with Coefficients 1, 2, 7 and Preliminaries The space M 2 (Γ 0 (56), χ 0 ) The space M 2 (Γ 0 (56), χ 3 ) The space M 2 (Γ 0 (56), χ 5 ) The space M 2 (Γ 0 (56), χ 6 ) iv

6 4 Representations by Octonary Quadratic Forms and Future Work Representations by Octonary Quadratic Forms with Coefficients 1 and Conclusion and Future Work v

7 Chapter 1 Introduction Let N, N 0, Z, Q, R and C denote the sets of positive integers, nonnegative integers, integers, rational numbers, real numbers and complex numbers respectively. Let a 1, a 2, a 3, a 4 N, and n N 0. Let N(a 1, a 2, a 3, a 4 ; n) denote the number of representations of n by the quaternary quadratic form a 1 x a 2 x a 3 x a 4 x 2 4, that is N(a 1, a 2, a 3, a 4 ; n) := card{(x 1, x 2, x 3, x 4 ) Z 4 n = a 1 x a 2 x a 3 x a 4 x 2 4}. It is clear that N(a 1, a 2, a 3, a 4 ; 0) = 1. Since N(a 1, a 2, a 3, a 4 ; n) is invariant under a permutation of a 1, a 2, a 3, a 4, we may suppose that a 1 a 2 a 3 a 4. (1.0.1) Note that if gcd(a 1, a 2, a 3, a 4 )= d, then N(a 1, a 2, a 3, a 4 ; n) = N(a 1 /d, a 2 /d, a 3 /d, a 4 /d; n/d). So we may also suppose that gcd(a 1, a 2, a 3, a 4 ) = 1. (1.0.2) 1

8 CHAPTER 1. INTRODUCTION 2 Our first objective in this thesis is to determine explicit formulae for N(a 1, a 2, a 3, a 4 ; n), where a 1, a 2, a 3, a 4 {1, 2, 7, 14} which satisfy the simplifying assumptions (1.0.1) and (1.0.2). We then extend our work to determine explicit formulae for the number of representations of n by the octonary quadratic forms x 2 1 +x 2 2 +x 2 3 +x (x 2 5 +x 2 6 +x 2 7 +x 2 8), x x (x x x x x x 2 8) and x x x x x x (x x 2 8), which we denote by N(1 4, 7 4 ; n), N(1 2, 7 6 ; n) and N(1 6, 7 2 ; n) respectively. Over the years many people have worked on the problem of representations of integers by quadratic forms. In 1770 Lagrange [16] proved that every positive integer can be written as a sum of four integer squares. Jacobi [12] gave formulae for N(1, 1, 1, 1; n) as 8σ(n) if 4 n, N(1, 1, 1, 1; n) = 8σ(n) 32σ(n/4) = 24σ(n) if 4 n, where σ(n) is the sum of divisors function. See [26]. In 1860, Liouville [18] gave a formula for N(1, 1, 2, 2; n) and in 1861 he gave two more formulae for N(1, 1, 1, 2; n) and N(1, 2, 2, 2; n). Also, Benz [4], Demuth [10], and Pepin [23] gave proofs for these formulae. Williams [29] gave a completely arithmetic proof of the Liouville formulae for N(1, 1, 1, 2; n) and N(1, 2, 2, 2; n). In Chapter 2 we present some basic properties of modular groups and modular forms. In Chapter 3 we determine an explicit formula for N(a 1, a 2, a 3, a 4 ; n) for each of the twenty-two quaternary quadratic forms given by (a 1, a 2, a 3, a 4 ) = (1, 1, 7, 7), (2, 2, 7, 7), (1, 2, 7, 14), (1, 1, 14, 14), (1, 1, 1, 7), (1, 2, 2, 7), (1, 7, 7, 7), (1, 1, 2, 14), (2, 7, 7, 14), (1, 7, 14, 14),

9 CHAPTER 1. INTRODUCTION 3 (1, 1, 2, 7), (2, 2, 2, 7), (2, 7, 7, 7), (1, 1, 1, 14), (1, 2, 2, 14), (1, 7, 7, 14), (2, 7, 14, 14), (1, 14, 14, 14), (1, 2, 7, 7), (1, 1, 7, 14), (2, 2, 7, 14), (1, 2, 14, 14). To the best of our knowledge, these are the only remaining diagonal quaternary quadratic forms with coefficients 1, 2, 7 and 14 for which explicit formulae for N(a 1, a 2, a 3, a 4 ; n) have not been determined so far. In Chapter 4 we determine the number of representations of a positive integer n by the octonary quadratic forms N(1 4, 7 4 ; n), N(1 2, 7 6 ; n) and N(1 6, 7 2 ; n). We conclude our thesis by indicating some further directions for our research.

10 Chapter 2 Basic Concepts In this chapter we present some basic concepts for modular forms. For more information one can see [8], [9], [14], [15], [21], [22], [25], [27], and [28]. 2.1 Modular Forms Definition The modular group SL 2 (Z) is defined as {( a b SL 2 (Z) = c d ) } a, b, c, d Z, ad bc = 1, which acts on the upper half plane H = {z C Im(z) > 0} by the linear fractional transformation ( a b c d ) (z) = az + b, for z H. cz + d 4

11 CHAPTER 2. BASIC CONCEPTS 5 Note that the modular group SL 2 (Z) is generated by two elements ( ) ( ) T = and S = Definition Let N N. We define the principal congruence subgroup Γ(N) by {( a b Γ(N) := c d ) SL 2 (Z) a d 1(mod N), b c 0(mod N)}. A subgroup Γ of SL 2 (Z) is called a congruence subgroup if it contains Γ(N) for some positive integer N. The smallest such N is called the level of Γ. The two important congruence subgroups are {( ) } a b Γ 0 (N) := SL 2 (Z) c 0(mod N), c d {( ) } a b Γ 1 (N) := Γ 0 (N) a d 1(mod N). c d Definition A Dirichlet character (mod N) is a function χ : Z C satisfying (i) χ(ab) = χ(a)χ(b) for any a, b Z, (ii) χ(a) 0 if gcd(a, N) = 1, (iii)χ(a) = 0 if gcd(a, N) > 1, (iv) χ(a) = χ(b) if a b (mod N). Definition The trivial character is the Dirichlet character of modulus 1 and is denoted by χ 0. Definition The conductor of a Dirichlet character χ is the smallest positive integer M dividing its modulus such that there exists a Dirichlet character ψ of

12 CHAPTER 2. BASIC CONCEPTS 6 modulus M with χ(a) = ψ(a) for all a Z with (a, N) = 1. We say that a Dirichlet character modulo N is primitive if its conductor equals its modulus. Definition Let k Z. A weakly modular function of weight k for a congruence subgroup Γ is a meromorphic function f : H C which satisfies ( ) ( az + b a b f = (cz + d) k f(z), for cz + d c d ) Γ and z H. ( ) a b Definition [14] Let γ = Γ and z H. We define the weight k c d operator [γ] k on a function from H to C by ( ) (f [γ] k ) (z) = (cz + d) k az + b f. cz + d Definition [14] Let Γ be a congruence subgroup of level N in SL 2 (Z) and k Z. A function f : H C is called a modular form of weight k for Γ if it satisfies the following conditions (i) f is weakly modular for Γ, (ii) f is holomorphic on H, (iii) f [α] k is holomorphic at for all α SL 2 (Z). Note that by condition (iii), f [α] k has a Fourier expansion of the form f [α] k = a n qn, n q N = e 2πiz/N. n=0 We say that f is a cusp form of weight k for Γ if f [α] k vanishes at, which mean a 0 = 0 for every α SL 2 (Z) in the Fourier expansion of f [α] k. Definition [14] Let N be a positive integer and let χ be a Dirichlet character.

13 CHAPTER 2. BASIC CONCEPTS 7 A function f : H C which is holomorphic on H and f [α] k is holomorphic at for all α is a modular form of weight k for Γ 0 (N) with character χ if ( a b f [γ] k = χ(d)f(z) for all γ = c d ) Γ 0 (N). We write M k (Γ 0 (N), χ) to denote the space of modular forms of weight k and character χ, and S k (Γ 0 (N), χ) to denote the subspace of cusp forms of weight k and character χ. Let k Z. We write E k (Γ 0 (N), χ) to denote the subspace of Eisenstein series. It is known (see for example [28, p.83]) that M k (Γ 0 (N), χ) = E k (Γ 0 (N), χ) S k (Γ 0 (N), χ). (2.1.1) Definition The Dedekind eta function is defined on the upper half plane H by the product formula η(z) = q 1/24 An eta quotient is defined as a finite product of the form (1 q n ), q = e 2πiz. (2.1.2) f(z) = η r (z), (2.1.3) where runs through a finite set of positive integers and the exponents r are non-zero integers. By taking N to be the least common multiple of the s we can write the

14 CHAPTER 2. BASIC CONCEPTS 8 eta quotient (2.1.3) as f(z) = η r (z), (2.1.4) 1 N where some of the exponents r may be 0. When all exponents are non-negative, f(z) is said to be an eta product. For q C with q < 1 we set F (q) = (1 q n ). (2.1.5) Appealing to (2.1.5) we can express the eta function (2.1.2) as η(z) = q 1/24 (1 q n ) = q 1/24 F (q), q = e 2πiz. Definition For q C with q < 1 Ramanujan s theta function ϕ(q) is defined by ϕ(q) = q n2. n= We note that for quaternary quadratic forms a 1 x 2 1+a 2 x 2 2+a 3 x 2 3+a 4 x 2 4 (a 1, a 2, a 3, a 4 N), we have N(a 1, a 2, a 3, a 4 ; n)q n = ϕ(q a 1 )ϕ(q a 2 )ϕ(q a 3 )ϕ(q a 4 ), (2.1.6) n=0 and for octonary quadratic forms a 1 x a 8 x 2 8 (a 1,..., a 8 N), we have N(a 1,..., a 8 ; n)q n = ϕ(q a 1 ) ϕ(q a 8 ). (2.1.7) n=0

15 CHAPTER 2. BASIC CONCEPTS 9 The infinite product representation of ϕ(q) is due to Jacobi ([3]), ϕ(q) = F 5 (q 2 ) F 2 (q)f 2 (q 4 ), (2.1.8) where F (q) is given by (2.1.5). It follows from (2.1.2), (2.1.5) and (2.1.8) that η(z) = η5 (2z) η 2 (z)η 2 (4z). (2.1.9) 2.2 Eisenstein Series Definition Let χ and ψ be Dirichlet characters. For n N we define σ (k 1,χ,ψ) (n) by σ (k 1,χ,ψ) (n) := 1 m n ψ(m)χ(n/m)m k 1. (2.2.1) We set σ (k 1,χ,ψ) (n) = 0 for n / N. If χ and ψ are trivial characters then σ (k 1,χ,ψ) (n) becomes the sum of divisors function σ k 1 (n) = m k 1. 1 m n Definition Let ψ be a Dirichlet character of modulus N. We define the generalized Bernoulli numbers {B k,ψ } k N by the formal series N xe ax ψ(a) e Nx 1 = x k B k,ψ k!. a=1 k=0 Let χ and ψ be primitive Dirichlet characters with conductors L and M, respec-

16 CHAPTER 2. BASIC CONCEPTS 10 tively. We set E k,χ,ψ (q) = c 0 + ( ψ(m)χ(n/m)m )q k 1 n, (2.2.2) n 1 m n where c 0 is written in terms of the generalized Bernoulli numbers defined by 0 if L > 1; c 0 = B k,ψ if L = 1. 2k If χ and ψ are trivial characters, then the Eisenstein series E 2,χ0,χ 0 (q) and E 4,χ0,χ 0 (q) become L(q) := E 2 (q) := E 2,χ0,χ 0 (q) = σ(n)q n, (2.2.3) and E 4 (q) := E 4,χ0,χ 0 (q) = σ 3 (n)q n. (2.2.4) The following theorem can be found in [28]. Theorem Suppose t, k are positive integers. Let χ and ψ be Dirichlet characters with conductors L and M, respectively. The power series E k,χ,ψ (q t ) with LMt N and χψ = ε form a basis for the Eisenstein subspace E k (Γ 0 (N), ε). Except if k = 2, χ = ψ = 1, t > 1 then L(q) tl(q t ) is a modular form of weight 2 in M 2 (Γ 0 (t)). 2.3 Dimension Formulae Let N, k N and χ a Dirichlet character. In this section we state formulae for the dimensions of M k (Γ 0 (N), χ), E k (Γ 0 (N), χ) and S k (Γ 0 (N), χ). First, we state the

17 CHAPTER 2. BASIC CONCEPTS 11 trivial character case. Let µ 0 (N) =N p N(1 + 1/p), 0 if 4 N, µ 0,2 (N) = 4 p N (1 + ( )) otherwise, p 0 if 2 N or 9 N, µ 0,3 (N) = 3 p N (1 + ( )) otherwise, p c 0 (N) = d N φ(gcd(d, N/d)), where φ is Euler totient function and p runs through the prime divisors of N. Also, let g(n) := 1 + µ 0(N) 12 µ 0,2(N) 4 µ 0,3(N) 4 c 0(N). 2 the following proposition which is taken from [28, Section 6.1, p. 93]. Proposition We have dims 2 (Γ 0 (N)) = g(n), and for k 4 even, ( k ) dims k (Γ 0 (N)) = (k 1) (g(n) 1) c(n) k k +µ 0,2 (N) + µ 0,3 (N), 4 3 where is the floor function. The dimension of the Eisenstein subspace is c 0 (N) 1 if k = 2, dime k (Γ 0 (N)) = c 0 (N) if k 2.

18 CHAPTER 2. BASIC CONCEPTS 12 Example Let N = 56. We have c(56) = 8, µ 0 (N) = 96, µ 0,2 (56) = µ 0,3 (56) = 0. Hence g(56) = = 5. Thus by Propostion we have dims 2 (Γ 0 (56)) = 5 and dime 2 (Γ 0 (56)) = 7. Second, we state the non-trivial character case. The formulae are taken from [28, Section 6.3, p ]. Let υ p (N) denote the largest r N 0 such that p r N and let c be the conductor of χ. We set p r 2 + p r 2 1 if 2 v p (c) r and 2 r, λ (p,n,vp(c)) = 2 p r 1 2 if 2 v p (c) r and 2 r, 2 p r vp(c) if 2 v p (c) > r. The rational numbers γ 3 and γ 4 are defined as follows 1/3 if k 2 (mod 3), γ 3 (k) = 0 if k 1 (mod 3), 1/3 if k 0 (mod 3), 1/4 if k 2 (mod 4), γ 4 (k) = 0 if k is odd, 1/4 if k 0 (mod 4).

19 CHAPTER 2. BASIC CONCEPTS 13 Let χ be a Dirichlet character of modulus N for which χ( 1) = ( 1) k. dims k (Γ 0 (N), χ) dimm 2 k (Γ 0 (N), χ) = k 1 12 µ 0(N) 1 2 λ(p, N, v p (c)) p N +γ 4 (k) χ(x) + γ 3 (k) χ(x), (2.3.1) x A 4 (N) x A 3 (N) where A 4 (N) = { x Z/NZ : x = 0 } and A 3 (N) = { x Z/NZ : x 2 + x + 1 = 0 }. To compute dimm k (Γ 0 (N), χ) for k 2, we use the fact that dims k (Γ 0 (N), χ) = 0 for k 0. dimm k (Γ 0 (N), χ) = (dims 2 k (Γ 0 (N), χ) dimm k (Γ 0 (N), χ)) ( 1 k = 12 µ 0(N) 1 2 λ(p, N, v p (c)) + γ 4 (2 k) x A 4 (N) p N χ(x) + γ 3 (2 k) x A 3 (N) ) χ(x), (2.3.2) and dime k (Γ 0 (N), χ) = dimm k (Γ 0 (N), χ) dims k (Γ 0 (N), χ). (2.3.3) ( Example For N = 56, k = 2, χ 3 (m) = ( ) χ 6 (m) =, we have 56 m x A 4 (N) χ(x) = x A 3 (N) 28 m ), χ 5 (m) = χ(x) = 0. ( 8 m ) and Also, we have

20 CHAPTER 2. BASIC CONCEPTS 14 Thus by (2.3.1) (2.3.3) we obtain χ χ 3 χ 5 χ 6 p 56 λ(p, 56, v p(c)) χ dimm 2 (Γ 0 (56), χ) dims 2 (Γ 0 (56), χ) dime 2 (Γ 0 (56), χ) χ χ χ

21 Chapter 3 Representations by Quaternary Quadratic Forms with Coefficients 1, 2, 7 and Preliminaries We recall that, for a 1, a 2, a 3, a 4 N and n N 0, N(a 1, a 2, a 3, a 4 ; n) denotes the number of representations of n by the quaternary form a 1 x a 2 x a 3 x a 4 x 2 4, that is N(a 1, a 2, a 3, a 4 ; n) := card{(x 1, x 2, x 3, x 4 ) Z 4 n = a 1 x a 2 x a 3 x a 4 x 2 4}. We also have the simplifying assumptions a 1 a 2 a 3 a 4, (3.1.1) 15

22 CHAPTER 3. QUATERNARY QUADRATIC FORMS 16 Table M 2 (Γ 0 (56), χ 0 ) M 2 (Γ 0 (56), χ 3 ) M 2 (Γ 0 (56), χ 5 ) M 2 (Γ 0 (56), χ 6 ) (1, 1, 7, 7) (1, 1, 1, 7) (1, 2, 7, 7) (1, 1, 2, 7) (2, 2, 7, 7) (1, 2, 2, 7) (1, 1, 7, 14) (2, 2, 2, 7) (1, 2, 7, 14) (1, 7, 7, 7) (2, 2, 7, 14) (2, 7, 7, 7) (1, 1, 14, 14) (1, 1, 2, 14) (1, 2, 14, 14) (1, 1, 1, 14) (2, 7, 7, 14) (1, 2, 2, 14) (1, 7, 14, 14) (1, 7, 7, 14) (2, 7, 14, 14) (1, 14, 14, 14) and gcd(a 1, a 2, a 3, a 4 ) = 1. (3.1.2) We also recall that χ 0 denotes the trivial character. For m Z we define six characters by χ 1 (m) = ( 7 ), χ 2 (m) = m ( 4 ), χ 3 (m) = m ( 28 ), (3.1.3) m χ 4 (m) = ( 8 ) ( 8, χ 5 (m) =, χ 6 (m) = m m) ( 56 ). (3.1.4) m Under the simplifying assumptions (3.1.1) and (3.1.2) there are twenty-six quaternary quadratic forms a 1 x a 2 x a 3 x a 4 x 2 4 for which ϕ(q a 1 )ϕ(q a 2 )ϕ(q a 3 )ϕ(q a 4 ) M 2 (Γ 0 (56), χ) where χ {χ 0, χ 3, χ 5, χ 6 }. Their coefficients (a 1, a 2, a 3, a 4 ) are listed in Table Formulae for the four quaternary quadratic forms (1, 1, 1, 1), (1, 1, 2, 2), (1, 2, 2, 2), (1, 1, 1, 2) appeared in [1], [29]. In this chapter we determine formulae for the remaining twenty-two quaternary quadratic forms listed in Table

23 CHAPTER 3. QUATERNARY QUADRATIC FORMS 17 We use the following theorem to determine if an eta quotient f(z) = 1 N η r (z) is in M k (Γ 0 (N), χ). See [11], [13, Corollary 2.3, p. 37], [14, Theorem 5.7, p. 99] and [17]. Theorem ( Ligozat) Let N N and let f(z) = which satisfies the following conditions: (L1) r 0 (mod 24), (L2) 1 N 1 N (L3) for each d N, N r 0 (mod 24), 1 N gcd(d, ) 2 r 0. Then f(z) is in M k (Γ 0 (N), χ), where χ is given by ( ( 1) k s ) χ(m) =, m 1 N η r (z) be an eta quotient with weight k = 1 2 r, 1 N and s = r. 1 N In addition to the above conditions if f(z) also satisfies the condition gcd(d, ) 2 r (L4) for each d N, > 0, 1 N then f(z) is in S k (Γ 0 (N), χ).

24 CHAPTER 3. QUATERNARY QUADRATIC FORMS The space M 2 (Γ 0 (56), χ 0 ) In this section we determine formulae for N(a 1, a 2, a 3, a 4 ; n) for the quaternary quadratic forms listed in the first column of Table in terms of σ(n), σ(n/2), σ(n/4), σ(n/7), σ(n/8), σ(n/14), σ(n/28), σ(n/56), and a k (n) (1 k 5) defined by A 1 (q) = A 2 (q) = A 3 (q) = A 4 (q) = A 5 (q) = a 1 (n)q n = η(2z)η(4z)η(14z)η(28z), (3.2.1) a 2 (n)q n = η3 (2z)η 3 (28z) η(4z)η(14z), (3.2.2) a 3 (n)q n = η(z)η3 (4z)η(7z)η 3 (28z) η(2z)η(8z)η(14z)η(56z), (3.2.3) a 4 (n)q n = η3 (2z)η(8z)η 3 (14z)η(56z), (3.2.4) η(z)η(4z)η(7z)η(28z) a 5 (n)q n = η 4 (4z)η 4 (28z) η(2z)η(8z)η(14z)η(56z). (3.2.5) There is no linear relationship among the A k (q), 1 k 5. The first fifty-six values of a k (n), are given in Table Table n a 1 (n) a 2 (n) a 3 (n) a 4 (n) a 5 (n) n a 1 (n) a 2 (n) a 3 (n) a 4 (n) a 5 (n)

25 CHAPTER 3. QUATERNARY QUADRATIC FORMS Theorem Let (a 1, a 2, a 3, a 4 ) be as in the first column of Table Then ϕ(q a 1 )ϕ(q a 2 )ϕ(q a 3 )ϕ(q a 4 ) M 2 (Γ 0 (56), χ 0 ). Proof. Appealing to (2.1.9) for each quadratic form, we then check conditions (L1), (L2) and (L3) of Theorem for each form. We have N = 56. First we consider (1, 1, 7, 7) ϕ 2 (q)ϕ 2 (q 7 ) = η 10 (2z)η 10 (14z) η 4 (z)η 4 (4z)η 4 (7z)η 4 (28z). Table 3.2.2(a) r It can be seen from Table 3.2.2(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.2.2(b) d /

26 CHAPTER 3. QUATERNARY QUADRATIC FORMS 20 From Table 3.2.2(b) the condition (L3) is also satisfied. Thus ϕ 2 (q)ϕ 2 (q 7 ) M 2 (Γ 0 (56), χ 0 ). Second we consider (2, 2, 7, 7) ϕ 2 (q 2 )ϕ 2 (q 7 ) = η 10 (4z)η 10 (14z) η 4 (2z)η 4 (7z)η 4 (8z)η 4 (28z). Table 3.2.3(a) r It can be seen from Table 3.2.3(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.2.3(b) d / From Table 3.2.3(b) the condition (L3) is also satisfied. Thus ϕ 2 (q 2 )ϕ 2 (q 7 ) M 2 (Γ 0 (56), χ 0 ). Third we consider (1, 2, 7, 14) ϕ(q)ϕ(q 2 )ϕ(q 7 )ϕ(q 14 ) = η3 (2z)η 3 (4z)η 3 (14z)η 3 (28z) η 2 (z)η 2 (7z)η 2 (8z)η 2 (56z). Table 3.2.4(a) r It can be seen from Table 3.2.4(a) that conditions (L1) and (L2) are satisfied.

27 CHAPTER 3. QUATERNARY QUADRATIC FORMS gcd(d, ) 2 r Table 3.2.4(b) d /7 96/ From Table 3.2.4(b) the condition (L3) is also satisfied. Thus ϕ(q)ϕ(q 2 )ϕ(q 7 )ϕ(q 14 ) M 2 (Γ 0 (56), χ 0 ). Fourth we consider (1, 1, 14, 14) ϕ 2 (q)ϕ 2 (q 14 ) = η 10 (2z)η 10 (28z) η 4 (z)η 4 (4z)η 4 (14z)η 4 (56z). Table 3.2.5(a) r It can be seen from Table 3.2.5(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.2.5(b) d / From Table 3.2.5(b) the condition (L3) is also satisfied. Thus ϕ 2 (q)ϕ 2 (q 14 ) M 2 (Γ 0 (56), χ 0 ). Theorem A k (q) (1 k 5) given by (3.2.1) (3.2.5) are in S 2 (Γ 0 (56), χ 0 ). Proof. We will check conditions (L1), (L2) and (L4) of Theorem We have N = 56. First we consider A 1 (q) = η(2z)η(4z)η(14z)η(28z).

28 CHAPTER 3. QUATERNARY QUADRATIC FORMS 22 Table 3.2.6(a) r It can be seen from Table 3.2.6(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.2.6(b) d /7 24/7 48/7 6 48/ From Table 3.2.6(b) the condition (L4) is also satisfied. Thus A 1 (q) S 2 (Γ 0 (56), χ 0 ). Then A 2 (q) = η3 (2z)η 3 (28z) η(4z)η(14z). Table 3.2.7(a) r It can be seen from Table 3.2.7(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.2.7(b) d /7 36/7 24/7 3 24/ From Table 3.2.7(b) the condition (L4) is also satisfied. thus A 2 (q) S 2 (Γ 0 (56), χ 0 ). Then A 3 (q) = η(z)η3 (4z)η(7z)η 3 (28z) η(2z)η(8z)η(14z)η(56z).

29 CHAPTER 3. QUATERNARY QUADRATIC FORMS 23 Table 3.2.8(a) r It can be seen from Table 3.2.8(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.2.8(b) d /7 12/7 72/7 9 24/ From Table 3.2.8(b) the condition (L4) is also satisfied. Thus A 3 (q) S 2 (Γ 0 (56), χ 0 ). Then A 4 (q) = η3 (2z)η(8z)η 3 (14z)η(56z). η(z)η(4z)η(7z)η(28z) Table 3.2.9(a) r It can be seen from Table 3.2.9(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.2.9(b) d /7 36/7 24/7 3 72/ From Table 3.2.9(b) the condition (L4) is also satisfied. Thus A 4 (q) S 2 (Γ 0 (56), χ 0 ). Then A 5 (q) = η 4 (4z)η 4 (28z) η(2z)η(8z)η(14z)η(56z).

30 CHAPTER 3. QUATERNARY QUADRATIC FORMS 24 Table (a) r It can be seen from Table (a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table (b) d /7 12/7 96/7 3 48/ From Table (b) the condition (L4) is also satisfied. Thus A 5 (q) S 2 (Γ 0 (56), χ 0 ). Theorem (a) {A 1 (q),..., A 5 (q)} constitute a basis for S 2 (Γ 0 (56), χ 0 ). (b) L(q) tl(q t ) (t = 2, 4, 7, 8, 14, 28, 56) constitute a basis for E 2 (Γ 0 (56), χ 0 ). (c) L(q) tl(q t ) (t = 2, 4, 7, 8, 14, 28, 56) together with A k (q) (1 k 5) constitute a basis for M 2 (Γ 0 (56), χ 0 ). Proof. (a) By Theorem 3.2.2, A k (q) (1 k 5) S 2 (Γ 0 (56), χ 0 ). There is no linear relationship among them. By Example 2.3.2, we have dims 2 (Γ 0 (56), χ 0 ) = 5. Thus A k (q) (1 k 5) constitute a basis for S 2 (Γ 0 (56), χ 0 ). (b) By Example 2.3.2, we have dime 2 (Γ 0 (56), χ 0 ) = 7. By Theorem 2.2.3, L(q) tl(q t ) (t = 2, 4, 7, 8, 14, 28, 56) constitute a basis for E 2 (Γ 0 (56), χ 0 ). (c) It follows from (a), (b) and (2.1.1) that the dimension of M 2 (Γ 0 (56), χ 0 ) is 12 and therefore L(q) tl(q t ) (t = 2, 4, 7, 8, 14, 28, 56) together with A k (q) (1 k 5) constitute a basis for M 2 (Γ 0 (56), χ 0 ). Theorem (a) ϕ 2 (q)ϕ 2 (q 7 ) = 4 3 L(q) 8 3 L(q2 ) L(q4 ) 28 3 L(q7 ) L(q14 ) L(q28 ) A 3(q) 16 3 A 4(q) A 5(q),

31 CHAPTER 3. QUATERNARY QUADRATIC FORMS 25 (b) ϕ 2 (q 2 )ϕ 2 (q 7 ) = 2 3 L(q) 2 3 L(q2 ) 4 3 L(q4 ) 14 3 L(q7 ) L(q8 ) L(q14 ) L(q28 ) L(q56 ) 10 3 A 1(q) + 4A 2 (q) 2 3 A 3(q) 20 3 A 4(q) A 5(q), (c) ϕ(q)ϕ(q 2 )ϕ(q 7 )ϕ(q 14 ) = 2 3 L(q) 2 3 L(q2 ) 4 3 L(q4 ) 14 3 L(q7 ) L(q8 ) L(q14 ) L(q28 ) L(q56 ) A 1(q) A 3(q) A 4(q) A 5(q), (d) ϕ 2 (q)ϕ 2 (q 14 ) = 2 3 L(q) 2 3 L(q2 ) 4 3 L(q4 ) 14 3 L(q7 ) L(q8 ) L(q14 ) L(q28 ) L(q56 ) A 1(q) 4A 2 (q) A 3(q) A 4(q) A 5(q). Proof. Let (a 1, a 2, a 3, a 4 ) be one of the quadratic forms listed in the first column of Table By Theorem and Theorem (c), ϕ(q a 1 )ϕ(q a 2 )ϕ(q a 3 )ϕ(q a 4 ) must be a linear combinations of L(q) tl(q t ) (t = 2, 4, 7, 8, 14, 28, 56) and A k (q) (1 k 5), namely ϕ(q a 1 )ϕ(q a 2 )ϕ(q a 3 )ϕ(q a 4 ) = x 1 (L(q) 2L(q 2 )) + x 2 (L(q) 4L(q 4 )) +x 3 (L(q) 7L(q 7 )) + x 4 (L(q) 8L(q 8 )) + x 5 (L(q) 14L(q 14 )) +x 6 (L(q) 28L(q 28 )) + x 7 (L(q) 56L(q 56 )) + y 1 A 1 (q) + y 2 A 2 (q) +y 3 A 3 (q) + y 4 A 4 (q) + y 5 A 5 (q). We equate the first 60 coefficients of q n on both sides of the equation above to obtain a system of linear equations with the unknowns x 1, x 2, x 3, x 4, x 5, x 6, x 7, y 1, y 2, y 3, y 4, y 5. Then, using MAPLE we solve the system to find the asserted coefficients.

32 CHAPTER 3. QUATERNARY QUADRATIC FORMS 26 We now give an explicit formulae for N(a 1, a 2, a 3, a 4 ; n) for the quadratic forms (a 1, a 2, a 3, a 4 ) in Theorem in terms of σ(n/d) (d = 1, 2, 4, 7, 14, 28, 56) and a k (n) (1 k 5). Theorem Let n N. Then (a) N(1, 1, 7, 7; n) = 4 3 σ(n) 8 3 σ(n/2) σ(n/4) σ(n/7) σ(n/14) σ(n/28) a 3(n) 16 3 a 4(n) a 5(n), (b) N(2, 2, 7, 7; n) = 2 3 σ(n) 2 3 σ(n/2) σ(n/14) σ(n/4) σ(n/7) σ(n/8) σ(n/28) σ(n/56) 3 3 a 1(n) + 4a 2 (n) 2 3 a 3(n) 20 3 a 4(n) a 5(n), (c) N(1, 2, 7, 14; n) = 2 3 σ(n) 2 3 σ(n/2) σ(n/4) σ(n/7) σ(n/8) σ(n/14) + σ(n/28) 3 3 σ(n/56) a 1(n) a 3(n) a 4(n) a 5(n), (d) N(1, 1, 14, 14; n) = 2 3 σ(n) 2 3 σ(n/2) σ(n/4) σ(n/7) σ(n/8) σ(n/14) + σ(n/28) 3 3 σ(n/56) a 1(n) 4a 2 (n) a 3(n) a 4(n) a 5(n). Proof. From (2.1.6), (2.2.3) and Theorem 3.2.4, we obtain (a) N(1, 1, 7, 7; n)q n = ϕ 2 (q)ϕ 2 (q 7 ) n=0 = 1 + ( 4 3 σ(n) 8 3 σ(n/2) σ(n/4) σ(n/7) + 3 ) σ(n/28) a 3(n) 16 3 a 4(n) a 5(n) q n, 3 σ(n/14)

33 CHAPTER 3. QUATERNARY QUADRATIC FORMS 27 (b) (c) (d) N(2, 2, 7, 7; n)q n = ϕ 2 (q 2 )ϕ 2 (q 7 ) n=0 = 1 + ( 2 3 σ(n) 2 3 σ(n/2) σ(n/28) ) 3 a 5(n) n=0 = 1 + q n, σ(n/4) σ(n/7) + σ(n/8) σ(n/14) σ(n/56) 10 3 a 1(n) + 4a 2 (n) 2 3 a 3(n) 20 3 a 4(n) N(1, 2, 7, 14; n)q n = ϕ(q)ϕ(q 2 )ϕ(q 7 )ϕ(q 14 ) ( 2 3 σ(n) 2 3 σ(n/2) σ(n/4) σ(n/7) σ(n/8) σ(n/14) + σ(n/28) σ(n/56) a 1(n) a 3(n) a 4(n) + 4 ) 3 a 5(n) q n, N(1, 1, 14, 14; n)q n = ϕ 2 (q)ϕ 2 (q 14 ) n=0 = 1 + ( 2 3 σ(n) 2 3 σ(n/2) 4 3 σ(n/4) σ(n/7) + σ(n/8) σ(n/14) σ(n/28) 3 3 σ(n/56) a 1(n) 4a 2 (n) a 3(n) a 4(n) + 16 ) 3 a 5(n) q n. Equating the coefficients of q n on both sides of equations (a) (d) yields the results. For (a 1, a 2, a 3, a 4 ) = (1, 1, 7, 7), (1, 1, 14, 14), (2, 2, 7, 7), (1, 2, 7, 14), the values of N(a 1, a 2, a 3, a 4 ; n) for 1 n 20 are given in Table One can verify them by using Table Table n N(1, 1, 7, 7; n) N(2, 2, 7, 7; n) N(1, 2, 7, 14; n)

34 CHAPTER 3. QUATERNARY QUADRATIC FORMS 28 N(1, 1, 14, 14; n) For example, using Table 3.2.1, we obtain N(2, 2, 7, 7; 16) = 2 3 σ(16) 2 3 σ(8) σ(16/14) σ(4) σ(16/7) σ(2) σ(16/28) σ(16/56) 3 3 a 1(16) +4a 2 (16) 2 3 a 3(16) 20 3 a 4(16) a 5(16) = 2 3 (31) 2 3 (15) 4 3 = 20, (7) + (3) 3 3 ( 1) 2 20 ( 1) which agrees with the value of N(2, 2, 7, 7; 16) in Table The space M 2 (Γ 0 (56), χ 3 ) Let χ 0 be the trivial character and χ 1, χ 2, χ 3 as in (3.1.3). We define the Eisenstein series E 2,χ3,χ 0 (q) = σ (χ3,χ 0 )(n)q n, (3.3.1) E 2,χ0,χ 3 (q) = 4 + E 2,χ1,χ 2 (q) = E 2,χ2,χ 1 (q) = σ (χ0,χ 3 )(n)q n, (3.3.2) σ (χ1,χ 2 )(n)q n, (3.3.3) σ (χ2,χ 1 )(n)q n. (3.3.4) We determine N(a 1, a 2, a 3, a 4 ; n) for the six quaternary quadratic forms listed in the second column of Table in terms of σ (χ,ψ) (n), where χ, ψ {χ 0, χ 1, χ 2, χ 3 }, and

35 CHAPTER 3. QUATERNARY QUADRATIC FORMS 29 b k (n) (1 k 4) defined by B 1 (q) = B 2 (q) = B 3 (q) = B 4 (q) = b 1 (n)q n = η2 (2z)η 3 (7z), (3.3.5) η(z) b 2 (n)q n = η3 (8z)η 2 (28z), (3.3.6) η(56z) b 3 (n)q n = η2 (4z)η 3 (56z), (3.3.7) η(8z) b 4 (n)q n = η3 (z)η 2 (14z). (3.3.8) η(7z) There is no linear relationship among the B k (q), 1 k 4. The first fifty-six values of b k (n), 1 k 4, are given in Table Table n b 1 (n) b 2 (n) b 3 (n) b 4 (n) n b 1 (n) b 2 (n) b 3 (n) b 4 (n)

36 CHAPTER 3. QUATERNARY QUADRATIC FORMS Theorem Let (a 1, a 2, a 3, a 4 ) be as in the second column of Table Then ϕ(q a 1 )ϕ(q a 2 )ϕ(q a 3 )ϕ(q a 4 ) M 2 (Γ 0 (56), χ 3 ). Proof. We appeal to (2.1.9) for each of the six quadratic forms and then check the conditions (L1), (L2) and (L3) of Theorem for each quadratic form. We have N = 56. First we consider (1, 1, 1, 7) ϕ 3 (q)ϕ(q 7 ) = η 15 (2z)η 5 (14z) η 6 (z)η 6 (4z)η 2 (7z)η 2 (28z). Table 3.3.2(a) r It can be seen from Table 3.3.2(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.3.2(b) d / From Table 3.3.2(b) the condition (L3) is also satisfied. Thus ϕ 3 (q)ϕ(q 7 ) M 2 (Γ 0 (56), χ 3 ). Now for the form (1, 2, 2, 7) we have ϕ(q)ϕ 2 (q 2 )ϕ(q 7 ) = η(2z)η 8 (4z)η 5 (14z) η 2 (z)η 2 (7z)η 4 (8z)η 2 (28z).

37 CHAPTER 3. QUATERNARY QUADRATIC FORMS 31 Table 3.3.3(a) r It can be seen from Table 3.3.3(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.3.3(b) d / From Table 3.3.3(b) the condition (L3) is also satisfied. Thus ϕ(q)ϕ 2 (q 2 )ϕ(q 7 ) M 2 (Γ 0 (56), χ 3 ). Then for the form (1, 7, 7, 7) we have ϕ(q)ϕ 3 (q 7 ) = η 5 (2z)η 15 (14z) η 2 (z)η 2 (4z)η 6 (7z)η 6 (28z). Table 3.3.4(a) r It can be seen from Table 3.3.4(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.3.4(b) d / From Table 3.3.4(b) the condition (L3) is also satisfied. Thus ϕ(q)ϕ 3 (q 7 ) M 2 (Γ 0 (56), χ 3 ). Now for the form (1, 1, 2, 14) we have ϕ 2 (q)ϕ(q 2 )ϕ(q 14 ) = η 8 (2z)η(4z)η 5 (28z) η 4 (z)η 2 (8z)η 2 (14z)η 2 (56z).

38 CHAPTER 3. QUATERNARY QUADRATIC FORMS 32 Table 3.3.5(a) r It can be seen from Table 3.3.5(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.3.5(b) d / From Table 3.3.5(b) the condition (L3) is also satisfied. Thus ϕ 2 (q)ϕ(q 2 )ϕ(q 14 ) M 2 (Γ 0 (56), χ 3 ). Now for the form (2, 7, 7, 14) we have ϕ(q 2 )ϕ 2 (q 7 )ϕ(q 14 ) = η 5 (4z)η 8 (14z)η(28z) η 2 (2z)η 4 (7z)η 2 (8z)η 2 (56z). Table 3.3.6(a) r It can be seen from Table 3.3.6(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.3.6(b) d /7 96/ From Table 3.3.6(b) the condition (L3) is also satisfied. Thus ϕ(q 2 )ϕ 2 (q 7 )ϕ(q 14 ) M 2 (Γ 0 (56), χ 3 ). Now for the form (1, 7, 14, 14) we have ϕ(q)ϕ(q 7 )ϕ 2 (q 14 ) = η5 (2z)η(14z)η 8 (28z) η 2 (z)η 2 (4z)η 2 (7z)η 4 (56z).

39 CHAPTER 3. QUATERNARY QUADRATIC FORMS 33 Table 3.3.7(a) r It can be seen from Table 3.3.7(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.3.7(b) d /7 24/ From Table 3.3.7(b) the condition (L3) is also satisfied. Thus ϕ(q)ϕ(q 7 )ϕ 2 (q 14 ) M 2 (Γ 0 (56), χ 3 ). Theorem B k (q) (1 k 4) given by (3.3.5) (3.3.8) are in S 2 (Γ 0 (56), χ 3 ). Proof. We will check conditions (L1),(L2) and (L4) of Theorem We have N = 56. First we consider B 1 (q) = η2 (2z)η 3 (7z). η(z) Table 3.3.8(a) r It can be seen from Table 3.3.8(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.3.8(b) d /7 24/7 24/ /

40 CHAPTER 3. QUATERNARY QUADRATIC FORMS 34 From Table 3.3.8(b) the condition (L4) is also satisfied. Thus B 1 (q) S 2 (Γ 0 (56), χ 3 ). Secondly we consider B 2 (q) = η3 (8z)η 2 (28z). η(56z) Table 3.3.9(a) r It can be seen from Table 3.3.9(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.3.9(b) d /7 12/7 48/ From Table 3.3.9(b) the condition (L4) is also satisfied. Thus B 2 (q) S 2 (Γ 0 (56), χ 3 ). Thirdly we consider B 3 (q) = η2 (4z)η 3 (56z). η(8z) Table (a) r It can be seen from Table (a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table (b) d /7 12/7 48/7 3 24/

41 CHAPTER 3. QUATERNARY QUADRATIC FORMS 35 From Table (b) the condition (L4) is also satisfied. Thus B 3 (q) S 2 (Γ 0 (56), χ 3 ). Fourthly we consider B 4 (q) = η3 (z)η 2 (14z). η(7z) Table (a) r It can be seen from Table (a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table (b) d /7 24/7 3 24/ From Table (b) the condition (L4) is also satisfied. Thus B 4 (q) S 2 (Γ 0 (56), χ 3 ). Theorem (a) {B 1 (q), B 2 (q), B 3 (q), B 4 (q)} is a basis for S 2 (Γ 0 (56), χ 3 ). (b) {E 2,χ3,χ 0 (q t ), E 2,χ0,χ 3 (q t ), E 2,χ1,χ 2 (q t ), E 2,χ2,χ 1 (q t ) t = 1, 2} is a basis for E 2 (Γ 0 (56), χ 3 ). (c) {E 2,χ3,χ 0 (q t ), E 2,χ0,χ 3 (q t ), E 2,χ1,χ 2 (q t ), E 2,χ2,χ 1 (q t ) t = 1, 2} together with B k (q) (1 k 4) constitute a basis for M 2 (Γ 0 (56), χ 3 ). Proof. (a) By Theorem 3.3.2, B k (q) (1 k 4) S 2 (Γ 0 (56), χ 3 ). There is no linear relationship among them. By Example 2.3.3, we have dims 2 (Γ 0 (56), χ 3 ) = 4. Therefore, B k (q) (1 k 4) constitute a basis for S 2 (Γ 0 (56), χ 3 ). (b) By Example 2.3.3, we have dime 2 (Γ 0 (56), χ 3 ) = 8. By Theorem 2.2.3, {E 2,χ3,χ 0 (q t ), E 2,χ0,χ 3 (q t ), E 2,χ1,χ 2 (q t ), E 2,χ2,χ 1 (q t ) t = 1, 2} is a basis for E 2 (Γ 0 (56), χ 3 ). (c) By Example 2.3.3, we have dimm 2 (Γ 0 (56), χ 3 ) = 12. Therefore, by (2.1.1)

42 CHAPTER 3. QUATERNARY QUADRATIC FORMS 36 {E 2,χ3,χ 0 (q t ), E 2,χ0,χ 3 (q t ), E 2,χ1,χ 2 (q t ), E 2,χ2,χ 1 (q t ) t = 1, 2} together with B k (q) (1 k 4) constitute a basis for M 2 (Γ 0 (56), χ 3 ). Theorem Let χ 0 be the trivial character and χ 1, χ 2, χ 3 be as in (3.1.3). Then (a) ϕ 3 (q)ϕ(q 7 ) = 7 2 E 2,χ 3,χ 0 (q) 1 4 E 2,χ 0,χ 3 (q) E 2,χ 1,χ 2 (q) 1 2 E 2,χ 2,χ 1 (q) + 3B 2 (q) 21B 3 (q) 3 2 B 4(q), (b) ϕ(q)ϕ 2 (q 2 )ϕ(q 7 ) = 7 4 E 2,χ 3,χ 0 (q) 1 4 E 2,χ 0,χ 3 (q 2 ) E 2,χ 1,χ 2 (q 2 ) 1 4 E 2,χ 2,χ 1 (q) 7 8 B 1(q) B 2(q) 21 2 B 3(q) 1 8 B 4(q), (c) ϕ(q)ϕ 3 (q 7 ) = 1 2 E 2,χ 3,χ 0 (q) 1 4 E 2,χ 0,χ 3 (q) 1 4 E 2,χ 1,χ 2 (q) E 2,χ 2,χ 1 (q) 3 2 B 1(q) + 3B 2 (q) + 3B 3 (q), (d) ϕ 2 (q)ϕ(q 2 )ϕ(q 14 ) = 7 4 E 2,χ 3,χ 0 (q) 1 4 E 2,χ 0,χ 3 (q 2 ) E 2,χ 1,χ 2 (q 2 ) 1 4 E 2,χ 2,χ 1 (q) B 1(q) 1 2 B 2(q) B 3(q) B 4(q), (e) ϕ(q 2 )ϕ 2 (q 7 )ϕ(q 14 ) = 1 4 E 2,χ 3,χ 0 (q) 1 4 E 2,χ 0,χ 3 (q 2 ) 1 4 E 2,χ 1,χ 2 (q 2 ) E 2,χ 2,χ 1 (q) B 1(q) 1 2 B 2(q) 1 2 B 3(q) 3 8 B 4(q), (f) ϕ(q)ϕ(q 7 )ϕ 2 (q 14 ) = 1 4 E 2,χ 3,χ 0 (q) 1 4 E 2,χ 0,χ 3 (q 2 ) 1 4 E 2,χ 1,χ 2 (q 2 ) E 2,χ 2,χ 1 (q) 1 8 B 1(q) B 2(q) B 3(q) B 4(q). Proof. Let (a 1, a 2, a 3, a 4 ) be one of the quadratic forms listed in the second column of Table By Theorem we have ϕ(q a 1 )ϕ(q a 2 )ϕ(q a 3 )ϕ(q a 4 ) M 2 (Γ 0 (56), χ 3 ). Therefore, by Theorem (c), ϕ(q a 1 )ϕ(q a 2 )ϕ(q a 3 )ϕ(q a 4 ) must be a linear combination of {E 2,χ3,χ 0 (q t ), E 2,χ0,χ 3 (q t ), E 2,χ1,χ 2 (q t ), E 2,χ2,χ 1 (q t ) t = 1, 2} and B k (q) (1

43 CHAPTER 3. QUATERNARY QUADRATIC FORMS 37 k 4), namely ϕ(q a 1 )ϕ(q a 2 )ϕ(q a 3 )ϕ(q a 4 ) =x 1 E 2,χ3,χ 0 (q) + x 2 E 2,χ3,χ 0 (q 2 ) + x 3 E 2,χ0,χ 3 (q) + x 4 E 2,χ0,χ 3 (q 2 ) + x 5 E 2,χ1,χ 2 (q) + x 6 E 2,χ1,χ 2 (q 2 ) + x 7 E 2,χ2,χ 1 (q) + x 8 E 2,χ2,χ 1 (q 2 ) + y 1 B 1 (q) + y 2 B 2 (q) + y 3 B 3 (q) + y 4 B 4 (q). We equate the first twenty coefficients of q n on both sides of the equation above to obtain a system of linear equations with the unknowns x 1, x 2, x 3, x 4, x 5, x 6, x 7, x 8 and y 1, y 2, y 3, y 4. Then, using MAPLE we solve the system to find the asserted coefficients. Theorem Let n N. Let σ χi,χ j (n) be as in (2.2.1) for i, j {0, 1, 2, 3}. Then (a) N(1, 1, 1, 7; n) = 7 2 σ χ 3,χ 0 (n) 1 4 σ χ 0,χ 3 (n) σ χ 1,χ 2 (n) 1 2 σ χ 2,χ 1 (n) + 3b 2 (n) 21b 3 (n) 3 2 b 4(n), (b) N(1, 2, 2, 7; n) = 7 4 σ χ 3,χ 0 (n) 1 4 σ χ 0,χ 3 (n/2) σ χ 1,χ 2 (n/2) 1 4 σ χ 2,χ 1 (n) 7 8 b 1(n) b 2(n) 21 2 b 3(n) 1 8 b 4(n), (c) N(1, 7, 7, 7; n) = 1 2 σ χ 3,χ 0 (n) 1 4 σ χ 0,χ 3 (n) 1 4 σ χ 1,χ 2 (n) σ χ 2,χ 1 (n) 3 2 b 1(n) + 3b 2 (n) + 3b 3 (n), (d) N(1, 1, 2, 14; n) = 7 4 σ χ 3,χ 0 (n) 1 4 σ χ 0,χ 3 (n/2) σ χ 1,χ 2 (n/2) 1 4 σ χ 2,χ 1 (n) b 1(n) 1 2 b 2(n) b 3(n) b 4(n), (e) N(2, 7, 7, 14; n) = 1 4 σ χ 3,χ 0 (n) 1 4 σ χ 0,χ 3 (n/2) 1 4 σ χ 1,χ 2 (n/2) σ χ 2,χ 1 (n) b 1(n) 1 2 b 2(n) 1 2 b 3(n) 3 8 b 4(n),

44 CHAPTER 3. QUATERNARY QUADRATIC FORMS 38 (f) N(1, 7, 14, 14; n) = 1 4 σ χ 3,χ 0 (n) 1 4 σ χ 0,χ 3 (n/2) 1 4 σ χ 1,χ 2 (n/2) σ χ 2,χ 1 (n) 1 8 b 1(n) b 2(n) b 3(n) b 4(n). Proof. From (2.1.6), (3.3.1)-(3.3.4) and Theorem 3.3.4, we obtain (a) N(1, 1, 1, 7; n)q n = ϕ 3 (q)ϕ(q 7 ) n=0 = 7 2 E 2,χ 3,χ 0 (q) 1 4 E 2,χ 0,χ 3 (q) E 2,χ 1,χ 2 (q) 1 2 E 2,χ 2,χ 1 (q) + 3B 2 (q) 21B 3 (q) 3 2 B 4(q), ( 7 = σ χ3,χ0(n) 1 4 σ χ0,χ3(n) σ χ1,χ2(n) 1 2 σ χ2,χ1(n) + 3b 2 (n) 21b 3 (n) 3 ) 2 b 4(n) q n, (b) (c) N(1, 2, 2, 7; n)q n = ϕ(q)ϕ 2 (q 2 )ϕ(q 7 ) n=0 = 7 4 E 2,χ 3,χ 0 (q) 1 4 E 2,χ 0,χ 3 (q 2 ) E 2,χ 1,χ 2 (q 2 ) 1 4 E 2,χ 2,χ 1 (q) 7 8 B 1(q) B 2(q) 21 2 B 3(q) 1 8 B 4(q), ( 7 = σ χ3,χ0(n) 1 4 σ χ0,χ3(n/2) σ χ1,χ2(n/2) 1 4 σ χ2,χ1(n) 7 8 b 1(n) b 2(n) 21 2 b 3(n) 1 ) 8 b 4(n) q n, N(1, 7, 7, 7; n)q n = ϕ(q)ϕ 3 (q 7 ) n=0 = 1 2 E 2,χ 3,χ 0 (q) 1 4 E 2,χ 0,χ 3 (q) 1 4 E 2,χ 1,χ 2 (q) E 2,χ 2,χ 1 (q) 3 2 B 1(q) + 3B 2 (q) + 3B 3 (q), ( = σ χ 3,χ 0 (n) 1 4 σ χ 0,χ 3 (n) 1 4 σ χ 1,χ 2 (n) σ χ 2,χ 1 (n) 3 2 b 1(n) ) + 3b 2 (n) + 3b 3 (n) q n,

45 CHAPTER 3. QUATERNARY QUADRATIC FORMS 39 (d) (e) N(1, 1, 2, 14; n)q n = ϕ 2 (q)ϕ(q 2 )ϕ(q 14 ) n=0 = 7 4 E 2,χ 3,χ 0 (q) 1 4 E 2,χ 0,χ 3 (q 2 ) E 2,χ 1,χ 2 (q 2 ) 1 4 E 2,χ 2,χ 1 (q) B 1(q) 1 2 B 2(q) B 3(q) B 4(q), ( 7 = σ χ3,χ0(n) 1 4 σ χ0,χ3(n/2) σ χ1,χ2(n/2) 1 4 σ χ2,χ1(n) b 1(n) 1 2 b 2(n) b 3(n) + 3 ) 8 b 4(n) q n, N(2, 7, 7, 14; n)q n = ϕ(q)ϕ 2 (q 7 )ϕ(q 14 ) n=0 = 1 4 E 2,χ 3,χ 0 (q) 1 4 E 2,χ 0,χ 3 (q 2 ) 1 4 E 2,χ 1,χ 2 (q 2 ) E 2,χ 2,χ 1 (q) B 1(q) 1 2 B 2(q) 1 2 B 3(q) 3 8 B 4(q) 1 2 D 1(q) 7 5 D 3(q) 12 5 D 4(q) D 5(q) D 6(q), ( 1 =1 + 4 σ χ3,χ0(n) 1 4 σ χ0,χ3(n/2) 1 4 σ χ1,χ2(n/2) σ χ2,χ1(n) b 1(n) 1 2 b 2(n) 1 2 b 3(n) 3 ) 8 b 4(n) q n, (f) N(1, 7, 14, 14; n)q n = ϕ(q)ϕ(q 7 )ϕ 2 (q 14 ) n=0 = 1 4 E 2,χ 3,χ 0 (q) 1 4 E 2,χ 0,χ 3 (q 2 ) 1 4 E 2,χ 1,χ 2 (q 2 ) E 2,χ 2,χ 1 (q) 1 8 B 1(q) B 2(q) B 3(q) B 4(q), ( 1 = σ χ3,χ0(n) 1 4 σ χ0,χ3(n/2) 1 4 σ χ1,χ2(n/2) σ χ2,χ1(n) 1 8 b 1(n) b 2(n) b 3(n) + 1 ) 8 b 4(n) q n. Equating the coefficients of q n on both sides of equations (a) (f) yields the results. The values of N(a 1, a 2, a 3, a 4 ; n) for 1 n 20 for the quadratic forms (a 1, a 2, a 3, a 4 ) in Theorem are given in Table One can verify them by using Table

46 CHAPTER 3. QUATERNARY QUADRATIC FORMS 40 Table n N(1, 1, 1, 7; n) N(1, 2, 2, 7; n) N(1, 7, 7, 7; n) N(1, 1, 2, 14; n) N(2, 7, 7, 14; n) N(1, 7, 14, 14; n) For example, by substituting n = 16 in Theorem 3.3.5(b), we obtain N(1, 2, 2, 7; 16) = 7 4 σ χ 3,χ 0 (16) 1 4 σ χ 0,χ 3 (8) σ χ 1,χ 2 (8) 1 4 σ χ 2,χ 1 (16) 7 8 b 1(16) b 2(16) 21 2 b 3(16) 1 8 b 4(16). Then appealing to (2.2.1) and (3.1.3), we obtain σ χ3,χ 0 (16) = 16, σ χ0,χ 3 (8) = 1, σ χ1,χ 2 (8) = 1, σ χ2,χ 1 (16) = 16. From Table 3.3.1, we have b 1 (16) = 1, b 2 (16) = b 3 (16) = 0, b 4 (16) = 11. Thus we have N(1, 2, 2, 7; 16) = 7 4 (16) (16) ( 11) = 26, 8 which agrees with the value of N(1, 2, 2, 7; 16) in Table

47 CHAPTER 3. QUATERNARY QUADRATIC FORMS The space M 2 (Γ 0 (56), χ 5 ) Let χ 0 be the trivial character and χ 5 as in (3.1.4). We define the Eisenstein series E 2,χ5,χ 0 (q) = σ (χ5,χ 0 )(n)q n, (3.4.1) E 2,χ0,χ 5 (q) = σ (χ0,χ 5 )(n)q n. (3.4.2) We determine N(a 1, a 2, a 3, a 4 ; n) for the quaternary quadratic forms listed in the third column of Table in terms of σ (χ,ψ) (n), where χ, ψ {χ 0, χ 5 }, and c k (n) (1 k 6) defined by C 1 (q) = C 2 (q) = C 3 (q) = C 4 (q) = C 5 (q) = C 6 (q) = c 1 (n)q n = η3 (2z)η(7z)η 2 (8z)η(28z), (3.4.3) η(z)η 2 (4z) c 2 (n)q n = η(2z)η2 (7z)η(8z)η 3 (28z), (3.4.4) η 2 (14z)η(56z) c 3 (n)q n = η2 (z)η 3 (4z)η(14z)η(56z), (3.4.5) η 2 (2z)η(8z) c 4 (nq n ) = c 5 (n)q n = c 6 (n)q n = η6 (2z)η(8z)η 4 (28z) η 2 (z)η 3 (4z)η(14z)η(56z), (3.4.6) η4 (2z)η(7z)η 6 (28z) η(z)η(4z)η 3 (14z)η 2 (56z), (3.4.7) η4 (4z)η 6 (14z)η(56z) η(2z)η 2 (7z)η(8z)η 3 (28z). (3.4.8) There is no linear relationship among the C k (q), 1 k 6. The first fifty-six values of c k (n), 1 k 6, are given in Table Table n c 1 (n) c 2 (n) c 3 (n) c 4 (n) c 5 (n) c 6 (n) n c 1 (n) c 2 (n) c 3 (n) c 4 (n) c 5 (n) c 6 (n)

48 CHAPTER 3. QUATERNARY QUADRATIC FORMS Theorem Let (a 1, a 2, a 3, a 4 ) be as in the third column of Table Then ϕ(q a 1 )ϕ(q a 2 )ϕ(q a 3 )ϕ(q a 4 ) M 2 (Γ 0 (56), χ 5 ). Proof. We have N = 56. First we consider quadratic form (1, 2, 7, 7). By (2.1.9) we have ϕ(q)ϕ(q 2 )ϕ 2 (q 7 ) = η3 (2z)η 3 (4z)η 10 (14z) η 2 (z)η 4 (7z)η 2 (8z)η 4 (28z). Table 3.4.2(a) and Table 3.4.2(b). Table 3.4.2(a) r It can be seen from Table 3.4.2(a) that conditions (L1) and (L2) are satisfied.

49 CHAPTER 3. QUATERNARY QUADRATIC FORMS gcd(d, ) 2 r Table 3.4.2(b) d / From Table 3.4.2(b) the condition (L3) is also satisfied. Thus ϕ(q)ϕ(q 2 )ϕ 2 (q 7 ) M 2 (Γ 0 (56), χ 5 ). Secondly we consider the form (1, 1, 7, 14). By (2.1.9) we have ϕ 2 (q)ϕ(q 7 )ϕ(q 14 ) = η10 (2z)η 3 (14z)η 3 (28z) η 4 (z)η 4 (4z)η 2 (7z)η 2 (56z). Table 3.4.3(a) r It can be seen from Table 3.4.3(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.4.3(b) d /7 12/ From Table 3.4.3(b) the condition (L3) is also satisfied. Thus ϕ 2 (q)ϕ(q 7 )ϕ(q 14 ) M 2 (Γ 0 (56), χ 5 ). Thirdly we consider the form (2, 2, 7, 14). We have ϕ 2 (q 2 )ϕ(q 7 )ϕ(q 14 ) = η10 (4z)η 3 (14z)η 3 (28z) η 4 (2z)η 2 (7z)η 4 (8z)η 2 (56z). Table 3.4.4(a) r It can be seen from Table 3.4.4(a) that conditions (L1) and (L2) are satisfied.

50 CHAPTER 3. QUATERNARY QUADRATIC FORMS gcd(d, ) 2 r Table 3.4.4(b) d /7 180/ From Table 3.4.4(b) the condition (L3) is also satisfied. Thus ϕ 2 (q 2 )ϕ(q 7 )ϕ(q 14 ) M 2 (Γ 0 (56), χ 5 ). Fourthly we consider the form (1, 2, 14, 14). We have ϕ(q)ϕ(q 2 )ϕ 2 (q 14 ) = η 3 (2z)η 3 (4z)η 10 (28z) η 2 (z)η 2 (8z)η 4 (14z)η 4 (56z). Table 3.4.5(a) r It can be seen from Table 3.4.5(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.4.5(b) d / From Table 3.4.5(b) the condition (L3) is also satisfied. Thus ϕ(q)ϕ(q 2 )ϕ 2 (q 14 ) M 2 (Γ 0 (56), χ 5 ). Theorem C k (q) (1 k 6) given by (3.4.3) (3.4.8) are in S 2 (Γ 0 (56), χ 5 ). Proof. We will check conditions (L1), (L2) and (L4) of Theorem We have N = 56. First we consider C 1 (q) = η3 (2z)η(7z)η 2 (8z)η(28z). η(z)η 2 (4z)

51 CHAPTER 3. QUATERNARY QUADRATIC FORMS 45 Table 3.4.6(a) r It can be seen from Table 3.4.6(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.4.6(b) d /7 30/7 12/7 9 96/ From Table 3.4.6(b) the condition (L4) is also satisfied. Thus C 1 (q) S 2 (Γ 0 (56), χ 5 ). Secondly we consider C 2 (q) = η(2z)η2 (7z)η(8z)η 3 (28z). η 2 (14z)η(56z) Table 3.4.7(a) r It can be seen from Table 3.4.7(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.4.7(b) d /7 18/7 36/ / From Table 3.4.7(b) the condition (L4) is also satisfied. Thus C 2 (q) S 2 (Γ 0 (56), χ 5 ). Thirdly we consider C 3 (q) = η2 (z)η 3 (4z)η(14z)η(56z). η 2 (2z)η(8z)

52 CHAPTER 3. QUATERNARY QUADRATIC FORMS 46 Table 3.4.8(a) r It can be seen from Table 3.4.8(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.4.8(b) d /7 6/7 60/7 6 24/ From Table 3.4.8(b) the condition (L4) is also satisfied. Thus C 3 (q) S 2 (Γ 0 (56), χ 5 ). Fourthly we consider C 4 (q) = η6 (2z)η(8z)η 4 (28z) η 2 (z)η 3 (4z)η(14z)η(56z). Table 3.4.9(a) r It can be seen from Table 3.4.9(a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table 3.4.9(b) d /7 54/7 12/7 3 48/ From Table 3.4.9(b) the condition (L4) is also satisfied. Thus C 4 (q) S 2 (Γ 0 (56), χ 5 ). Fifthly we consider C 5 (q) = η4 (2z)η(7z)η 6 (28z) η(z)η(4z)η 3 (14z)η 2 (56z).

53 CHAPTER 3. QUATERNARY QUADRATIC FORMS 47 Table (a) r It can be seen from Table (a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table (b) d /7 6 36/7 6 24/ From Table (b) the condition (L4) is also satisfied. Thus C 5 (q) S 2 (Γ 0 (56), χ 5 ). Sixthly we consider C 6 (q) = η4 (4z)η 6 (14z)η(56z) η(2z)η 2 (7z)η(8z)η 3 (28z). Table (a) r It can be seen from Table (a) that conditions (L1) and (L2) are satisfied gcd(d, ) 2 r Table (b) d /7 18/ / From Table (b) the condition (L4) is also satisfied. Thus C 6 (q) S 2 (Γ 0 (56), χ 5 ). Theorem (a) {C 1 (q),..., C 6 (q)} is a basis for S 2 (Γ 0 (56), χ 5 ). (b) {E 2,χ5,χ 0 (q t ), E 2,χ0,χ 5 (q t ) t = 1, 7} is a basis for E 2 (Γ 0 (56), χ 5 ).

54 CHAPTER 3. QUATERNARY QUADRATIC FORMS 48 (c) {E 2,χ5,χ 0 (q t ), E 2,χ0,χ 5 (q t ) t = 1, 7} together with C k (q) (1 k 6) constitute a basis for M 2 (Γ 0 (56), χ 5 ). Proof. (a) By Theorem 3.4.2, C k (q) (1 k 6) S 2 (Γ 0 (56), χ 5 ). There is no linear relationship among them. By Example 2.3.3, we have dims 2 (Γ 0 (56), χ 5 ) = 6. Thus C k (q) (1 k 6) constitute a basis for S 2 (Γ 0 (56), χ 5 ). (b) By Example 2.3.3, we have dime 2 (Γ 0 (56), χ 5 ) = 4. By Theorem 2.2.3, {E 2,χ5,χ 0 (q t ), E 2,χ0,χ 5 (q t ) t = 1, 7} constitute a basis for E 2 (Γ 0 (56), χ 5 ). (c) By Example 2.3.3, we have dimm 2 (Γ 0 (56), χ 5 ) = 10. Therefore, by (2.1.1) {E 2,χ5,χ 0 (q t ), E 2,χ0,χ 5 (q t ) t = 1, 7} together with C k (q) (1 k 6) constitute a basis for M 2 (Γ 0 (56), χ 5 ). Theorem Let χ 0 be the trivial character and χ 5 be as in (3.1.4). Then (a) ϕ(q)ϕ(q 2 )ϕ 2 (q 7 ) = 4 3 E 2,χ 5,χ 0 (q) 28 3 E 2,χ 5,χ 0 (q 7 ) E 2,χ 0,χ 5 (q) 7 3 E 2,χ 0,χ 5 (q 7 ) 4C 1 (q) 2 3 C 2(q) C 3(q) + 3C 4 (q) + C 5 (q) C 6 (q), (b) ϕ 2 (q)ϕ(q 7 )ϕ(q 14 ) = 4 3 E 2,χ 5,χ 0 (q) 28 3 E 2,χ 5,χ 0 (q 7 ) E 2,χ 0,χ 5 (q) 7 3 E 2,χ 0,χ 5 (q 7 ) + 4C 1 (q) C 2(q) C 3(q) 3C 4 (q) C 5 (q) + C 6 (q), (c) ϕ 2 (q 2 )ϕ(q 7 )ϕ(q 14 ) = 2 3 E 2,χ 5,χ 0 (q) 14 3 E 2,χ 5,χ 0 (q 7 ) E 2,χ 0,χ 5 (q) 7 3 E 2,χ 0,χ 5 (q 7 ) 2C 1 (q) 3C 2 (q) C 3 (q) C 4(q) + 2C 5 (q) C 6(q), (d) ϕ(q)ϕ(q 2 )ϕ 2 (q 14 ) = 2 3 E 2,χ 5,χ 0 (q) 14 3 E 2,χ 5,χ 0 (q 7 ) E 2,χ 0,χ 5 (q) 7 3 E 2,χ 0,χ 5 (q 7 ) + 2C 1 (q) + 3C 2 (q) + C 3 (q) 1 3 C 4(q) 2C 5 (q) C 6(q). Proof. Let (a 1, a 2, a 3, a 4 ) be one of the quadratic forms listed in the third column of Table By Theorem 3.4.1, ϕ(q a 1 )ϕ(q a 2 )ϕ(q a 3 )ϕ(q a 4 ) (M 2 (Γ 0 (56), χ 5 ). Therefore by Theorem (c), ϕ(q a 1 )ϕ(q a 2 )ϕ(q a 3 )ϕ(q a 4 ) must be a linear combina-