On Chudnovsky-Ramanujan TypeFormulae

by

Gleb Glebov

B.Sc. (Mathematics), Southern Methodist University, 2013

Thesis Submitted in Partial Fulfillment of theRequirements for the Degree of

Master of Science

in theDepartment of Mathematics

Faculty of Science

c© Gleb Glebov 2016SIMON FRASER UNIVERSITY

Summer 2016

All rights reserved.However, in accordance with the Copyright Act of Canada, this work may bereproduced without authorization under the conditions for “Fair Dealing.”

Therefore, limited reproduction of this work for the purposes of private study,research, criticism, review and news reporting is likely to be in accordance with

the law, particularly if cited appropriately.

Approval

Name: Gleb Glebov

Degree: Master of Science (Mathematics)

Title: On Chudnovsky-Ramanujan Type Formulae

Examining Committee: Chair: Petr LisonekProfessor

Imin ChenSenior SupervisorAssociate Professor

Stephen ChoiSupervisorProfessor

Nils BruinInternal ExaminerProfessor

Date Defended: June 14, 2016

ii

Abstract

In his well-known 1914 paper, Ramanujan gave a number of rapidly converging series for1/π which involve modular functions of higher level. D. V. and G. V. Chudnovsky havederived an analogous series representing 1/π using the modular function J of level 1, whichresults in highly convergent series for 1/π, often used in practice. In 2013, 12.1×1012 digitsof π were calculated by A. J. Yee and S. Kondo using the Chudnovsky series for π. Thepurpose of this work is to explain the method of D. V. and G. V. Chudnovsky and show howit can be generalised to derive formulae for transcendental constants starting from a one-parameter family of elliptic curves. We find all such closed-form Chudnovsky-Ramanujantype formulae for the family of elliptic curves parameterised by J of level 1, where J is theabsolute j-invariant.

Keywords: elliptic integrals; Dedekind eta function; Eisenstein series; elliptic functions;hypergeometric function; hypergeometric series; modular functions; Weierstrass ellipticfunctions

iii

Acknowledgements

First and foremost, I would like to thank my Lord and Saviour Jesus Christ for there is noneother name under heaven given among men, whereby we must be saved. Amen. In addition,I want to express my thanks to my thesis advisor, Imin Chen, for his encouragement andthoughtful guidance. I also thank the other members of my thesis committee: StephenChoi and Nils Bruin, for their time and extensive comments. Lastly, I thank the NSERCDiscovery Grant Program for supporting this research.

iv

Table of Contents

Approval ii

Abstract iii

Acknowledgements iv

Table of Contents v

List of Tables vii

List of Figures viii

1 Introduction 1

2 Preliminaries 42.1 Topological definitions and theorems . . . . . . . . . . . . . . . . . . . . . . 42.2 Branches of multi-valued functions . . . . . . . . . . . . . . . . . . . . . . . 52.3 Eisenstein series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.4 The Bernoulli numbers and the Fourier expansion of the Eisenstein series . 112.5 Weierstrass’ functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.6 Elliptic curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.7 Uniformisation theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.8 Complex multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.9 The gamma function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.10 The hypergeometric function . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3 Periods and families of elliptic curves 233.1 Families of elliptic curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.2 Periods Ω1, Ω2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4 Picard-Fuchs differential equation 284.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284.2 Derivation of the Picard-Fuchs differential equation . . . . . . . . . . . . . . 31

v

5 Period relations 345.1 Hypergeometric representations of Ω1 . . . . . . . . . . . . . . . . . . . . . 345.2 Quasi-period relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425.3 Complex multiplication period relations . . . . . . . . . . . . . . . . . . . . 43

6 J =∞ case 456.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

7 J = 1 case 507.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

8 J = 0 case 558.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

vi

List of Tables

Table 2.1 Special values of j(τ) and J(τ) at τ =√−N . . . . . . . . . . . . . . 19

Table 2.2 Special values of j(τ) and J(τ) at τ = −1+√−N

2 . . . . . . . . . . . . 20

Table 5.1 Summary: Periods and Quasi-periods . . . . . . . . . . . . . . . . . . 34Table 5.2 Special values of s2(τ) at τ =

√−N . . . . . . . . . . . . . . . . . . . 43

Table 5.3 Special values of s2(τ) at τ = −1+√−N

2 . . . . . . . . . . . . . . . . . 43

vii

List of Figures

Figure 2.1 Two copies of P1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Figure 2.2 Torus with loops α1 and α2 . . . . . . . . . . . . . . . . . . . . . . 8

Figure 3.1 Loops α1 and α2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26Figure 3.2 Two copies of P1 with branch cuts . . . . . . . . . . . . . . . . . . . 26

Figure 5.1 C1/J,∞ = τ : |1/J | < 1, x2 + y2 > 1 . . . . . . . . . . . . . . . . . 35Figure 5.2 C(J−1)/J,i = τ : |(J − 1)/J | < 1, x2 + y2 > 1 . . . . . . . . . . . . 35Figure 5.3 CJ/(J−1),ρ = τ : |J/(J − 1)| < 1, x2 + y2 > 1, x < 0 . . . . . . . . 36

viii

Chapter 1

Introduction

The formula12∞∑n=0

(−1)n 545140134n+ 13591409(6403203)n+1/2

(6n)!(3n)!n!3 = 1

π

has been discovered by D. V. and G. V. Chudnovsky [8] and used in practice for a recordbreaking computation of the digits of π. Ramanujan [19, pp. 23-39] was the first to giveexamples of such formulae by considering modular functions of higher level. The formulaabove arises by considering the modular function J of level 1 around J =∞ and evaluationat the class number of value −1+

√−163

2 . Here, J = j/123 is the absolute Klein invariant andj is the classical j-function.

Without specialised knowledge of classical functions and identities, the derivation ofsuch formulae can seem mysterious. In the case of D. V. and G. V. Chudnovsky [8], theformula above is derived by first proving a precursor formula

a

π√d

√J√

J − 1= F 2 1− s2(τ)

6 − J d

dJF 2.

Here, F = 2F1(1/12, 5/12; 1; 1/J), J = J(τ), and

τ = −b+√−d

2a

with −d = b2 − 4ac, where a, b, c are integers such that a > 0. This formula is valid forimaginary quadratic τ in a certain simply-connected domain of C. Evaluating at a classnumber one singular value τ and then using Clausen’s identity to simplify the series yieldsthe formula for 1/π above.

The purpose of this work is to explain the method of D. V. and G. V. Chudnovsky indetail. We present it in the form which links it to the modern theory of elliptic curvesand modular curves, and generalises it to work for monodromy around any elliptic point inaddition to the cusps. In doing so, we give a fairly transparent but broad framework which

1

in principle suggests a systematic way to tabulate such formulae for genus zero congruencegroups commensurable with SL2(Z). A feature of the method is that it no longer requiresspecialised knowledge of classical functions and identities to carry out, but rather relies onan explicit form of the Picard-Fuchs differential equation and Kummer’s method [16] ofdetermining its solutions [3].

Even in the case of J of level 1, the method gives new formulae (no longer for 1/π, butfor related constants), corresponding to monodromy around J = 1 and J = 0. We provethe following Chudnovsky-Ramanujan type formulae.

Around J = 1, we have

τ + i

2πα2√

3

√J√

1− J

(aτ + i√−d− 1

)= F 2 1− s2(τ)

6 − J d

dJF 2,

where F = 2F1(1/12, 5/12; 1/2; (J − 1)/J), J = J(τ), α = 2iη(i)2, and τ as above.Around J = 0, we have

− τ − ρ2πα2

√3

J1/3

(1− J)1/3

(aτ − ρ√−d− 1

)= F 2

[J

6(1− J) + s2(τ)6

]+ J

d

dJF 2,

where F = 2F1(1/12, 7/12; 2/3; J/(J − 1)), J = J(τ), α = iη(ρ)2√3, and τ as above.Below is the complete list of Ramanujan-type series:

3∞∑n=0

(−1)n 63n+ 8(153)n+1/2

(6n)!(3n)!n!3 = 1

π,

4∞∑n=0

(−1)n 154n+ 15(323)n+1/2

(6n)!(3n)!n!3 = 1

π,

12∞∑n=0

(−1)n 342n+ 25(963)n+1/2

(6n)!(3n)!n!3 = 1

π,

36∞∑n=0

(−1)n 506n+ 31(3 · 1603)n+1/2

(6n)!(3n)!n!3 = 1

π,

12∞∑n=0

(−1)n 16254n+ 789(9603)n+1/2

(6n)!(3n)!n!3 = 1

π,

12∞∑n=0

(−1)n 261702n+ 10177(52803)n+1/2

(6n)!(3n)!n!3 = 1

π,

12∞∑n=0

(−1)n 545140134n+ 13591409(6403203)n+1/2

(6n)!(3n)!n!3 = 1

π,

2

and

8∞∑n=0

28n+ 3(203)n+1/2

(6n)!(3n)!n!3 = 1

π,

72∞∑n=0

11n+ 1(2 · 303)n+1/2

(6n)!(3n)!n!3 = 1

π,

24√

2∞∑n=0

63n+ 5(663)n+1/2

(6n)!(3n)!n!3 = 1

π,

162∞∑n=0

133n+ 8(2553)n+1/2

(6n)!(3n)!n!3 = 1

π.

Ramanujan [19, pp. 23-39] discovered two of them:

72∞∑n=0

11n+ 1(2 · 303)n+1/2

(6n)!(3n)!n!3 = 1

π,

162∞∑n=0

133n+ 8(2553)n+1/2

(6n)!(3n)!n!3 = 1

π.

D. V. and G. V. Chudnovsky [8] proved that

∞∑n=0

(1− s2(τ)6 + n

) (6n)!(3n)!n!3 j(τ)−n =

√−j(τ)π

1√d(123 − j(τ))

.

Here,

s2(τ) = E4(τ)E6(τ)

(E2(τ)− 3

πIm(τ)

), τ = 1 +

√−d

2 .

The Ramanujan-type series listed above can be obtained by substituting τ into the formulaabove corresponding to imaginary quadratic orders of class number one. For a tabulationof Ramanujan-type series obtained from imaginary quadratic orders of low class number,see [7].

In addition to giving a complete derivation of the formula above, we also generalise itand establish some auxiliary formulae of a similar type. We completely exhaust the list ofclosed-form identities for the family of elliptic curves parameterised by J of level 1. Moreformulae can be established by various hypergeometric transformations.

3

Chapter 2

Preliminaries

2.1 Topological definitions and theorems

Definition. A path in a topological space X is a continuous map π : [0, 1] → X. A pathis a loop if π(0) = π(1) = x0 ∈ X. A loop is simple if π(s) = π(t) implies s = t ors, t = 0, 1.

Definition. Suppose γ0 : [0, 1] → G and γ1 : [0, 1] → G are two paths (from z0 to z1)in G ⊆ C. We say γ0 and γ1 are homotopic as paths (from z0 to z1) in G if there is acontinuous function H : [0, 1]× [0, 1]→ G such that

(i) H(0, t) = γ0(t) for 0 ≤ t ≤ 1,

(ii) H(1, t) = γ1(t) for 0 ≤ t ≤ 1,

(iii) H(s, 0) = z0 for 0 ≤ s ≤ 1,

(iv) H(s, 1) = z1 for 0 ≤ s ≤ 1.

Definition. Suppose γ0 : [0, 1]→ G and γ1 : [0, 1]→ G are two loops in G ⊆ C. We say γ0

and γ1 are homotopic as loops in G if there is a continuous function H : [0, 1]× [0, 1]→ G

such that

(i) H(0, t) = γ0(t) for 0 ≤ t ≤ 1,

(ii) H(1, t) = γ1(t) for 0 ≤ t ≤ 1,

(iii) H(s, 0) = H(s, 1) for 0 ≤ s ≤ 1.

The following theorems from the theory of complex integration are quite useful and willbe used later.

4

Theorem 2.1.1. Suppose f(z) is a holomorphic function on an open set G ⊆ C, and letγ0 and γ1 be paths in G. If γ0 and γ1 are homotopic as paths (from z0 to z1) in G, then∫

γ0f(z)dz =

∫γ1f(z)dz.

Proof. See Theorem 2.3.12 and Supplement A to §2.3 in [17].

Theorem 2.1.2. Suppose f(z) is a holomorphic function on an open set G ⊆ C, and letγ0 and γ1 be loops in G. If γ0 and γ1 are homotopic as loops in G, then∫

γ0f(z)dz =

∫γ1f(z)dz.

Proof. See Theorem 2.3.12 and Supplement A to §2.3 in [17].

2.2 Branches of multi-valued functions

Definition. Let K be a field, and let K be a fixed algebraic closure of K. Affine (orCartesian) n-space over a field K, denoted by An or An(K), is the set of n-tuples

(x1, x2, . . . , xn) : xi ∈ K.

The set of K-rational points in An is the set

An(K) = (x1, x2, . . . , xn) ∈ An : xi ∈ K.

Historically, projective space arose through the process of adding “points at infinity” toaffine space [20, Chapter I, §2].

Definition. Let K be a field, and let K be a fixed algebraic closure of K. Projectiven-space over a field K, denoted by Pn or Pn(K), is the set of all (n+ 1)-tuples

(x0, x1, . . . , xn) ∈ An+1

such that at least one xi is nonzero, modulo the equivalence relation:

(x0, x1, . . . , xn) ∼ (y0, y1, . . . , yn)

if there exists λ ∈ K∗ such that xi = λyi for all i. An equivalence class

(λx0, λx1, . . . , λxn) : λ ∈ K∗

5

is denoted by [x0 : x1 : · · · : xn], and x0, x1, . . . , xn are called homogeneous coordinates forthe corresponding point in Pn. The set of K-rational points in Pn is the set

Pn(K) = [x0 : x1 : · · · : xn] ∈ Pn : xi ∈ K.

We see from the definition above that a point in P1 = P1(C), the projective line overC, is given by an equivalence class of homogeneous coordinates of the form [x1 : x2] withx1, x2 6= 0. Here, [x1 : x2] and [x′1 : x′2] are equivalent if [x′1 : x′2] = [λx1 : λx2] for λ 6= 0.The projective line P1 can be viewed as C∪∞, where z ∈ C is identified with [z : 1] ∈ P1

and ∞ = [1 : 0]. The projective line P1 is topologically a sphere – the Riemann sphere.Let z ∈ P1. Then the function w(z) defined by w2 − z = 0 is not single-valued in P1.

To see this first set z = reiθ with θ = arg z and r = |z| > 0. Starting at some point r0eiθ0

with r0 6= 0, and continuing w(z) along a loop that winds once around the origin so thatθ increases by 2π, w(z) comes to the value √r0e

i(θ0+2π)/2 = −√r0eiθ0/2. Continuing along

the same path again leads to √r0eiθ0/2, the original value of w(z).

Let us introduce a branch cut by cutting P1 along (0,∞). If we restrict ourselves soas never to continue w(z) over this cut, we get two single-valued branches of w(z), namelyw =

√reiθ/2 with 0 ≤ θ < 2π, and w =

√reiθ/2 with 2π ≤ θ < 4π. Now we have two copies

of P1 cut along (0,∞). The cut on each “sheet” has two edges; label the edge of the firstquadrant with a + and the edge of the fourth quadrant with a −. Then “glue” the + edgeof the cut on the first sheet to the − edge of the cut on the second sheet, and glue the −edge of the cut on the first sheet to the + edge of the cut on the second sheet. Thus, as wecross the cut, we pass from the first sheet to the second sheet. Topologically, this surface isa sphere.

We associate to a point z on the first sheet the fixed value of√z given by

√reiθ/2 with

0 ≤ θ < 2π, and designate this point on the first sheet by (z,√z). Then, starting from

w =√z, if we continue the function w(z) defined by w2−z = 0 around a simple loop about

the origin, we cross the cut and pass to the second sheet, and when we return to the pointon the second sheet having coordinate z, w becomes −

√z. We designate the point z on

the second sheet by (z,−√z), which distinguishes it from (z,

√z) on the first sheet. We see

that there are two values of w for each z except z = 0 and z =∞, which are called branchpoints of w =

√z. We say

√z ramifies at 0 and ∞.

If w2 = a0z + a1 instead of w2 − z = 0, then we cut P1 from z = −a1/a0 to z = ∞instead of from z = 0 to z = ∞. If w2 = a0z

2 + a1z + a2 with a21 − 4a0a2 6= 0 and

a0 6= 0, then we first write w2 = a0(z − r1)(z − r2) where r1 and r2 are distinct roots ofa0z

2 + a1z + a2. The roots r1 and r2 are the branch points of this function, and we obtaintwo single-valued branches of w =

√a0(z − r1)(z − r2) by cutting P1 along a curve joining

r1 and r2. Gluing two copies of P1, with this cut, along the cut, we obtain a surface onwhich w(z) is single-valued. Topologically, this surface is also a sphere.

6

However, if w2 = a0(z − r1)(z − r2)(z − r3) where r1, r2, r3 are distinct, we obtain adifferent surface. To each z there correspond two values of w (one is the negative of theother). We go from one to the other by continuing w(z) over any loop winding once aroundone of the roots r1, r2, r3. For w = √a0

√z − r1

√z − r2

√z − r3, the factor

√z − r, where

r = r1, r2, r3, changes sign when arg(z − r) changes by 2π. If we cut P1 from r1 to r2, wecannot wind around either r1 or r2 alone without crossing the cut. However, we can choosea path which winds around both r1 and r2. But now both arg(z−r1) and arg(z−r2) changeby 2π, both

√z − r1 and

√z − r2 change sign, and there is no change in w. Then we cut

P1 from r3 to ∞ (see Figure 2.1). This prevents us from winding around all three roots r1,r2, r3. Thus either branch of w(z) is single-valued in the cut P1. If we now take two copiesof the cut P1 and glue them crosswise over the cuts as before, we obtain a surface on whichw2 = a0(z − r1)(z − r2)(z − r3) is single-valued. Topologically, this surface is a torus.

To see this we glue two copies of P1 cut between r1 and r2, and between r3 and ∞along the cuts (see Figure 2.1). Each + edge of a cut is to be glued to the − edge of thecorresponding cut on the other sphere (copy of P1). We rotate the spheres until the slitsface each other, and glue them together. Notice that now the + edges of the cuts on onesphere are opposite the − edges of the cuts on the other sphere. Thus we may glue thespheres along the slits to form a surface which is topologically equivalent to a torus (seeFigure 2.2).

+−

+−r1 r2 r3 ∞

+−

+−r1 r2 r3 ∞

α1α2

Figure 2.1: Two copies of P1

7

α1

α2

Figure 2.2: Torus with loops α1 and α2

Note that on a sphere any loop is homotopic to a point whereas on a torus it is notalways the case (see Figure 2.2). It is easy to conceive of a loop α1 on one sheet and anotherloop α2 on the other sheet which, when viewed as loops on a torus, are not homotopic to apoint. Observe that around either α1 or α2, the function w =

√a0(z − r1)(z − r2)(z − r3)

is single-valued. And, in fact, it is known that integrals of the form∫ z

z0R(ζ, w(ζ))dζ,

where R is a rational function of z and w, can have nonzero values around α1 and α2.

Remark. A torus is an example of a topological object known as a Riemann surface. Recallthat a manifold is a topological space that is locally Euclidean: in other words, around everypoint, there is a neighbourhood that is topologically homeomorphic to the open unit ballin Rn. A Riemann surface is a one-dimensional (over C) connected analytic manifold.Riemann surfaces arise naturally by considering multi-valued functions. We introducedRiemann surfaces informally as was done in Springer [21, pp. 3-10]. We refer the reader to[12, pp. 9-11] for a more technical definition and some examples of Riemann surfaces.

2.3 Eisenstein series

Definition. For s ∈ C, the zeta function ζ(s) is defined by

ζ(s) =∞∑n=1

1ns.

Proposition 2.3.1. The zeta function ζ(s) converges absolutely if Re(s) > 1.

8

Proof. Let log z be the principal value of the logarithm with the branch cut along (−∞, 0].Let σ, t ∈ R and write s = σ + it. Then

∞∑n=1

1|ns|

=∞∑n=1

1|nσ+it|

=∞∑n=1

1|nσ||nit|

=∞∑n=1

1nσ|eit logn|

=∞∑n=1

1nσ.

Next, we see that for σ > 1, ∫ ∞1

x−σdx =[ 1

1− σx1−σ

]∞1.

Moreover, limx→∞ x−σ = 0 and x−σ is decreasing provided σ > 1, which means ζ(s)

converges absolutely for σ > 1 by the Cauchy integral test.

Definition. Let C be considered as a group under addition. Let ω1, ω2 ∈ C such thatτ = ω2/ω1 is not a real number. That is, let ω1, ω2 ∈ C be linearly independent vectorsover R. Then

Λ = [ω1, ω2] = Zω1 + Zω2 = mω1 + nω2 : m,n ∈ Z

is a subgroup of C, called a lattice.

Remark. Since Λ is an additive subgroup, we can form the quotient group

C/Λ = Λ + u : u ∈ C,

where the cosets Λ + u and Λ + v are equal if and only if u− v ∈ Λ. Notice that any z ∈ Cis in the equivalence class of one and only one z = sω1 + tω2 with s, t ∈ [0, 1). That is, eachcoset in C/Λ has a unique coset representative z in

P = sω1 + tω2 : s, t ∈ [0, 1),

which is called the fundamental parallelogram.

Definition. The fundamental parallelogram corresponding to a basis (ω1, ω2) for Λ is theparallelogram P defined by the vertices 0, ω1, ω2, ω1 + ω2 in C. In other words,

P = sω1 + tω2 : s, t ∈ [0, 1).

Remark. The quotient group C/Λ is topologically a torus because it can be topologicallyidentified with the fundamental parallelogram P with its two pairs of opposite sides gluedtogether. When one pair of sides is glued together, a cylinder is obtained. And if the endsof this cylinder are then glued together, we get a torus.

9

Definition. For all integers k > 1, the Eisenstein series G2k(Λ) with associated lattice Λare defined by

G2k(Λ) =∑ω∈Λ′

1ω2k , Λ′ = Λ \ 0.

Moreover, we define

g2(Λ) = 60G4(Λ) and g3(Λ) = 140G6(Λ).

Definition. For all integers k > 1, the normalised Eisenstein series E2k(Λ) with associatedlattice Λ are defined by

E2k(Λ) = G2k(Λ)2ζ(2k) ,

where ζ is the zeta function.

Observe that any function of the lattice Λ = [ω1, ω2] can be viewed as a function of ω1

and ω2. For instance, G2k(Λ) = G2k(ω1, ω2). Bearing this in mind, note that for any λ 6= 0we have G2k(λω1, λω2) = λ−2kG2k(Λ). Thus, if λ = 1/ω1 and τ = ω2/ω1, we have

G2k(1, τ) = G2k(τ) = ω2k1 G2k(Λ), (2.1)

that is,G2k(τ) =

∑(m,n) 6=(0,0)

1(m+ nτ)2k and E2k(τ) = G2k(τ)

2ζ(2k) ,

where the summation is over all integers m, n that do not vanish simultaneously.

Lemma 2.3.2. The series ∑ω∈Λ′

1|ω|s

, Λ′ = Λ \ 0

converges if s > 2.

Proof. Let P be the fundamental parallelogram of Λ and suppose U is the union of fourtranslates of P that surround the origin. In other words, let

U = P ∪ (−ω1 + P ) ∪ (−ω2 + P ) ∪ (−ω1 − ω2 + P ).

Notice that the boundary ∂U of U is compact and does not contain the origin, and thereexists d > 0 such that |u| ≥ d for all u ∈ ∂U . Take m and n such that |m| ≥ |n|. If |n| > 0,then

|ω| = |mω1 + nω2| ≥ |m|∣∣∣∣ω1 + n

mω2

∣∣∣∣ ≥ d|m|as ω1 + (n/m)ω2 ∈ ∂U . If ω = mω1 + nω2 and s > 2, then the terms of

∑ω 6=0 |ω|−s with

n = 0 contribute |ω1|−s∑m 6=0 |m|−s = 2|ω1|−s

∑∞m=1m

−s < ∞. Likewise, the terms with

10

m = 0 contribute 2|ω2|−s∑∞n=1 n

−s <∞. By the inequality above we see that

∑|m|≥|n|>0

1|ω|s

≤ 1ds

∑|m|≥|n|>0

1|m|s

= 1ds

∞∑k=1

∑|n|+|m|=k

1ks

= 1ds

∞∑k=1

1ks

∑|n|+|m|=k

1

= 4ds

∞∑k=1

1ks−1

<∞

because there are four cases: m,n > 0; or m,n < 0; or m > 0 and n < 0; or m < 0 andn > 0. Similarly, ∑

|n|≥|m|>0

1|ω|s

<∞.

So the terms with |m| ≥ |n| > 0 and |n| ≥ |m| > 0 make a finite contribution. This provesthe lemma.

Theorem 2.3.3. The Eisenstein series G2k(Λ) converges absolutely if k > 1.

Proof. This follows directly from the lemma.

2.4 The Bernoulli numbers and the Fourier expansion of theEisenstein series

Definition. The Bernoulli numbers Bk are defined by the coefficients in the power seriesexpansion of z/(ez − 1), namely,

z

ez − 1 =∞∑k=0

Bkk! z

k = 1− z

2 +∞∑k=2

Bkk! z

k.

Since z/(ez − 1) + z/2 = z(ez + 1)/2(ez − 1) is even, B2k+1 = 0 for k ≥ 1. Sayz/(ez − 1) = A0 + A1z + A2z

2 + A3z3 + · · · where An are yet to be determined, and use

ez − 1 = z + z2/2! + z3/3! + · · · to write

1 = (A0 +A1z +A2z2 +A3z

3 + · · · )(

1 + z

2! + z2

3! + z3

4! + · · ·).

11

Equivalently,

1 = A0 +(A02! +A1

)z +

(A03! + A1

2! +A2

)z2 +

(A04! + A1

3! + A22! +A3

)z3 + · · ·

=∞∑n=0

n∑k=0

Ak(n+ 1− k)!z

n.

Multiplying∑nk=0Ak/(n+ 1− k)! by (n+ 1)!k!/(n+ 1)!k! = 1 produces

n∑k=0

Ak(n+ 1− k)! = 1

(n+ 1)!

n∑k=0

(n+ 1k

)Bk, Bk = Akk!.

SetCn =

n∑k=0

(n+ 1k

)Bk.

Then ∞∑n=0

n∑k=0

Ak(n+ 1− k)!z

n =∞∑n=0

Cn(n+ 1)!z

n = 1,

and we see that C0 = 1 whereas C1 = C2 = C3 = · · · = 0, that is,

n∑k=0

(n+ 1k

)Bk =

1 if n = 0

0 if n > 0. (2.2)

An easy calculation shows that B0 = 1. Moreover, if we use (2.2) recursively we obtain theaforementioned Bernoulli numbers:

B0 = 1, B1 = −12 , B2 = 1

6 , B3 = 0, B4 = − 130 , B5 = 0, B6 = 1

42 , B7 = 0, . . .

In other words, the Bernoulli numbers can also be defined by (2.2).

Proposition 2.4.1. For z ∈ C with Im(z) > 0, we have

π cotπz = −πi− 2πi∞∑n=1

qn, q = e2πiz. (2.3)

Proof. Using Euler’s formula one can write

cosπz = eπiz + e−πiz

2 and sin πz = eπiz − e−πiz

2i .

If q = e2πiz and Im(z) > 0, then |q| < 1. Hence,

cotπz = iq + 1q − 1 = −i(1 + q)

∞∑n=0

qn = −i− 2i∞∑n=1

qn,

12

and the result follows.

Corollary. For all integers k ≥ 1 and z ∈ C with Im(z) > 0, we have

∞∑n=−∞

1(z + n)2k = (2πi)2k

(2k − 1)!

∞∑n=1

n2k−1qn, q = e2πiz. (2.4)

Moreover, both series converge absolutely.

Proof. The infinite partial fraction expansion of the cotangent function,

π cotπz =∞∑

n=−∞

1z + n

= 1z

+∞∑n=1

( 1z + n

+ 1z − n

), z ∈ C \ Z, (2.5)

is well-known [1, p. 11]. We claim that (2.5) is uniformly convergent on any compact setK ⊆ C \ Z. If |z| = R > 0, then for large n, R < n2/2R, so from the reverse triangleinequality we have ∣∣∣∣ 1

z + n+ 1z − n

∣∣∣∣ = 2|z − n2/z|

≤ 2||z| − n2/|z||

≤ 4Rn2 .

Hence, (2.5) is uniformly convergent on K by the Weierstrass M -test. Thus, (2.5) istermwise differentiable. Similarly, the derivatives of the summands of (2.5) are boundedby 4R/n2, so the derivatives of (2.5) are also uniformly convergent on compact subsets ofC \ Z. Moreover, since the series on the right-hand side of (2.3) is a complex geometricseries, it converges uniformly in |q| < 1 by the Weierstrass M -test. Let k ≥ 1 be an integer.Differentiating (2.3) and (2.5) 2k − 1 times and equating the results we obtain

(−1)2k−1(2k − 1)!∞∑

n=−∞

1(z + n)2k = −(2πi)2k

∞∑n=1

n2k−1qn,

which reduces to (2.4).

Definition. We define the upper half-plane H by

H = z = x+ iy : x, y ∈ R, y > 0.

Theorem 2.4.2. The (normalised) Eisenstein series E2k(τ) has the Fourier expansion

E2k(τ) = 1− 4kB2k

∞∑n=1

n2k−1 qn

1− qn , q = e2πiτ ,

and converges absolutely for all integers k ≥ 1 and τ ∈ H.

13

Proof. Recall that G2k(τ) converges absolutely if k = 2, 3, 4, . . . and Im(τ) > 0. So, ifk = 2, 3, 4, . . . and Im(τ) > 0, we may write

∑(m,n)6=(0,0)

1(m+ nτ)2k =

∑m 6=0

1m2k +

∞∑m=−∞

∑n6=0

1(m+ nτ)2k

= 2ζ(2k) + 2∞∑n=1

∞∑m=−∞

1(m+ nτ)2k .

Utilising (2.4) we get

∑(m,n)6=(0,0)

1(m+ nτ)2k = 2ζ(2k) + 2(2πi)2k

(2k − 1)!

∞∑n=1

∞∑m=1

m2k−1qnm.

Upon interchanging the order of summation and invoking the definition of G2k(τ) we arriveat

G2k(τ) = 2ζ(2k) + 2(2πi)2k

(2k − 1)!

∞∑m=1

m2k−1 qm

1− qm .

Changing m into n and using the definition of E2k(τ) yields

E2k(τ) = 1 + (2πi)2k

ζ(2k)(2k − 1)!

∞∑n=1

n2k−1 qn

1− qn ,

which, in view of the classical formula [1, p. 12],

ζ(2k) = (−1)k+1(2π)2kB2k2(2k)! ,

becomesE2k(τ) = 1− 4k

B2k

∞∑n=1

n2k−1 qn

1− qn .

Lastly, since ζ(2k) and (2.4) converge absolutely if k = 1, 2, 3, . . . and Im(z) > 0, so doesthe series above.

2.5 Weierstrass’ functions

Definition. For u ∈ C, the Weierstrass function ℘Λ(u) is defined by

℘Λ(u) = 1u2 +

∑ω∈Λ′

[ 1(u− ω)2 −

1ω2

],

where Λ = [ω1, ω2] is a lattice and Λ′ = Λ \ 0.

14

Proposition 2.5.1. For u ∈ C, the Weierstrass function ℘Λ(u) is absolutely convergent ifu ∈ C \ Λ and uniformly convergent on every compact subset of C \ Λ.

Proof. Notice that ∣∣∣∣ 1(u− ω)2 −

1ω2

∣∣∣∣ =∣∣∣∣ u(2ω − u)ω2(u− ω)2

∣∣∣∣ = |u| |2ω − u||ω|2|u− ω|2

.

Let K be a compact set. Define M = max|u| : u ∈ K and let 2M < |ω|. Then

2|ω|+M < 2|ω|+ |ω|2 = 5|ω|2 .

Note that |ω| = |ω − u + u| ≤ |u − ω| + |u|, so |ω| −M ≤ |ω| − |u| ≤ |u − ω| as |u| ≤ M .Moreover, 2M < |ω| implies |ω|/2 < |ω| −M . Hence,

|u| |2ω − u||ω|2|u− ω|2

≤ 5M2|ω|(|ω| −M)2 <

10M|ω|3

.

By Lemma 2.3.2, ∑ω∈Λ′

1|ω|3

<∞,

so by the Weierstrass M -test, ℘Λ(u) is absolutely convergent if u ∈ C \ Λ and uniformlyconvergent on every compact subset of C \ Λ.

Proposition 2.5.2. The Weierstrass function ℘Λ(u) is doubly periodic.

Proof. Since ℘Λ(u) converges uniformly on any compact subsetK of C\Λ by the propositionabove, it can be differentiated termwise in K. Therefore:

℘′Λ(u) = − 2u3 −

∑ω∈Λ′

2(u− ω)3 .

It is easy to show that ℘′Λ(u) is doubly periodic because replacing u by u + ω, whereω = ω1, ω2, does not change ℘′Λ(u) since it is doubly infinite. Now, the double periodicityof ℘′Λ(u) implies that ℘′Λ(u + ω) − ℘′Λ(u) = 0. Thus, ℘Λ(u + ω) − ℘Λ(u) = A = const. Itis readily seen that ℘Λ(u) is even, so upon setting u = −ω/2 we find that A = 0. Hence℘Λ(u) is doubly periodic.

Definition. For u ∈ C, the Weierstrass function ζΛ(u) is defined by

ζΛ(u) = 1u

+∑ω∈Λ′

[ 1u− ω

+ 1ω

+ u

ω2

],

where Λ = [ω1, ω2] is a lattice and Λ′ = Λ \ 0. That is, ζ ′Λ(u) = −℘Λ(u).

15

Definition. The quasi-period η(ω) is defined by the relation η(ω) = ζΛ(u + ω) − ζΛ(u),where ω = ω1, ω2.

The quantitates ω1, ω2 and η1, η2 are related by Legendre’s period relation:

ω2η1 − ω1η2 = 2πi.

Theorem 2.5.3. (Legendre’s Relation) If ω1 and ω2 are the fundamental periods andτ = ω2/ω1 ∈ H, then ω2η1 − ω1η2 = 2πi.

Proof. Let P be the fundamental parallelogram. Translate P by α so that the translatedparallelogram Q encloses u = 0. We see that the new vertices are at α, α + ω1, α + ω2,α + ω1 + ω2. Note that inside of Q the only pole of ζΛ(u) is at u = 0 with residue 1.Integrating ζΛ(u) around Q we get ∫

∂QζΛ(u) du = 2πi

by Cauchy’s theorem. Now,∫∂QζΛ(u) du =

∫ α+ω1

α+∫ α+ω1+ω2

α+ω1+∫ α+ω2

α+ω1+ω2+∫ α

α+ω2

ζΛ(u) du.

Changing u in the second and third integral into u+ ω1 and u+ ω2, respectively, yields∫∂QζΛ(u) du =

∫ α+ω1

αζΛ(u)− ζΛ(u+ ω2) du−

∫ α+ω2

αζΛ(u)− ζΛ(u+ ω1) du

= −η2

∫ α+ω1

αdu+ η1

∫ α+ω2

αdu

= −η2ω1 + η1ω2,

because η(ω) = ζΛ(u+ ω)− ζΛ(u). Therefore, ω2η1 − ω1η2 = 2πi.

2.6 Elliptic curves

The following definition of an elliptic curve will be sufficient for the purposes of this work.

Definition. Given a field K of characteristic different from 2 or 3, an elliptic curve E overK is an algebraic curve defined by the equation

E : y2z = 4x3 − g2xz2 − g3z

3, g2, g3 ∈ K, g32 − 27g2

3 6= 0

with [x : y : z] ∈ P2. This algebraic curve in P2(K), where K is an algebraic closure of K,includes a point at infinity [0 : 1 : 0].

16

Remark. If we associate an affine point (x, y) with the projective point [x : y : 1], we canview E as an algebraic curve in an affine space with z 6= 0 satisfying

E : y2 = 4x3 − g2x− g3, g2, g3 ∈ K, g32 − 27g2

3 6= 0.

In what follows, we will work with this affine patch to describe the elliptic curve E, whichis often called the Weierstrass equation.

Definition. Let K be a field of characteristic different from 2 or 3. Given an elliptic curveE over K, we define the set of K-rational points E(K) to be the set

E(K) = [x : y : z] : y2z = 4x3 − g2xz2 − g3z

3,

where x, y, z ∈ K are not all zero.

Let E be an elliptic curve over C given by

E : y2 = 4x3 − g2x− g3 = 4(x− e1)(x− e2)(x− e3), g2, g3 ∈ C, g32 − 27g2

3 6= 0.

Using Viéta’s formulae it is seen that

e1 + e2 + e3 = 0, e1e2 + e2e3 + e3e1 = −g24 , e1e2e3 = g3

4 .

It follows from these three relations that the (normalised) discriminant ∆ is given by

∆ = 16∏i<j

(ei − ej)2 = g32 − 27g2

3.

The quantity known as the j-invariant of E is defined by

j = 123 g32

∆ .

We also define the absolute Klein invariant J by

J = j

123 .

2.7 Uniformisation theorem

The quantities g2, g3 are related to g2(Λ), g3(Λ). It is known that there is a unique latticeΛ ⊂ C such that g2 = g2(Λ) and g3 = g3(Λ). This is a classical result [14, p. 296]. Inthe modern literature this is known as the uniformisation theorem for elliptic curves. Itsprecise statement is as follows:

17

Theorem 2.7.1. (Uniformisation Theorem) Let E be an elliptic curve over C given bythe Weierstrass equation

E : y2 = 4x3 − g2x− g3, g2, g3 ∈ C, g32 − 27g2

3 6= 0.

Then there is a unique lattice Λ ⊂ C such that

g2 = g2(Λ) and g3 = g3(Λ).

Proof. See [10, p. 285].

Corollary. Let E be an elliptic curve over C given by the Weierstrass equation

E : y2 = 4x3 − g2x− g3, g2, g3 ∈ C, g32 − 27g2

3 6= 0.

Then there is a unique lattice Λ ⊂ C such that j = j(Λ) = j(1, τ), where τ = ω2/ω1 ∈ H.

Proof. Since j = 123g32/(g3

2 − 27g23), it follows from the uniformisation theorem that there

is a unique lattice Λ ⊂ C such that j = j(Λ). If Λ = [ω1, ω2], then j(Λ) can be viewedas a function of ω1 and ω2, that is, j(Λ) = j(ω1, ω2). Therefore, we see from (2.1) thatj(ω1, ω2) = j(1, τ).

Remark. We will denote j(1, τ) by j(τ). Likewise, g2(1, τ), g3(1, τ), and ∆(1, τ) will bedenoted by g2(τ), g3(τ), and ∆(τ), respectively. Also, note that we may identify J withJ(τ) in view of the corollary.

Proposition 2.7.2. If τ ∈ H, then

J(τ) = E4(τ)3

E4(τ)3 − E6(τ)2 .

Proof. Using the definitions of J(τ), g2(τ), g3(τ), and E2k(τ), we see that

J(τ) = g2(τ)3

g2(τ)3 − 27g3(τ)2

= (60G4(τ))3

(60G4(τ))3 − 27(140G6(τ))2

= (4π4E4(τ)/3)3

(4π4E4(τ)/3)3 − 27(8π6E6(τ)/27)2

= E4(τ)3

E4(τ)3 − E6(τ)2 .

The derivative of J(τ) is related to E4(τ) and E6(τ) in the following way:

18

Theorem 2.7.3. Let J = J(τ) and J ′ = dJ/dτ . Then

J ′

J= −2πiE6(τ)

E4(τ) .

Proof. Let q = e2πiτ . Observe that dq/dτ = 2πiq, and

J ′

J= 2E4(τ)E′6(τ)− 3E6(τ)E′4(τ)

E4(τ)3 − E6(τ)2E6(τ)E4(τ)

in view of the proposition above. Using Ramanujan’s differential equations [19, p. 142] forE4(τ) and E6(τ), namely,

qd

dqE4(τ) = E2(τ)E4(τ)− E6(τ)

3 and qd

dqE6(τ) = E2(τ)E6(τ)− E4(τ)2

2 ,

we find thatJ ′

J= −2πiE6(τ)

E4(τ) .

2.8 Complex multiplication

Definition. We define the group SL2(Z) to be

SL2(Z) =(

a b

c d

) ∣∣∣∣ a, b, c, d ∈ Z, ad− bc = 1.

Definition. We define the standard fundamental domain F for SL2(Z) to be

F = τ ∈ H : |Re(τ)| < 1/2, |τ | > 1.

Remark. This has the property that every point in H is SL2(Z)-equivalent to a point inthe closure of F . No two points in the interior of F are SL2(Z)-equivalent.

It is known that j(τ) is rational at τ =√−N if N = 1, 2, 3, 4, 7 [10, pp. 237-238]:

N j(τ) J(τ)1 123 12 203 53/33

3 2 · 303 53/22

4 663 113/23

7 2553 53173/26

Table 2.1: Special values of j(τ) and J(τ) at τ =√−N

19

Moreover, j(τ) is rational at τ = −1+√−N

2 ifN = 3, 7, 11, 19, 27, 43, 67, 163 [10, pp. 237-238]:

N j(τ) J(τ)3 0 07 −153 −53/43

11 −323 −83/33

19 −963 −83

27 −3 · 1603 −403/32

43 −9603 −803

67 −52803 −4403

163 −6403203 −533603

Table 2.2: Special values of j(τ) and J(τ) at τ = −1+√−N

2

These results follow from the theory of complex multiplication of elliptic curves. A surveycan be found in [10].

2.9 The gamma function

Definition. For s ∈ C, the gamma function Γ(s) is defined by

Γ(s) =∫ ∞

0e−xxs−1dx.

Proposition 2.9.1. The gamma function Γ(s) converges absolutely if Re(s) > 0. Moreover,s = 0,−1,−2, . . . are the poles of Γ(s).

Proof. Let σ = Re(s). Note that |e−xxs−1| = e−xxσ−1. If 0 ≤ x ≤ 1, then e−xxσ−1 ≤ xσ−1,so∫ 1

0 e−xxσ−1dx ≤

∫ 10 x

σ−1dx = 1/σ provided σ 6= 0. Moreover, for x ≥ 1 sufficiently large,say x ≥ N , xσ−1 ≤ ex/2 or e−xxσ−1 ≤ e−x/2, so

∫∞N e−xxσ−1dx ≤

∫∞N e−x/2dx = 2e−N/2.

So Γ(s) converges if σ = Re(s) > 0. Integration by parts yields the following functionalequation:

Γ(s+ 1) = sΓ(s) or Γ(s) = Γ(s+ 1)s

, s 6= 0.

Reiterating this functional equation k − 1 times yields

Γ(s) = Γ(s+ k)s(s+ 1) · · · (s+ k − 1) , s 6= 0,−1,−2, . . . , 1− k.

This shows that s = 0,−1,−2, . . . are the poles of Γ(s).

Definition. The Pochhammer symbol is defined by

(α)n = α(α+ 1)(α+ 2) · · · (α+ n− 1),

20

where n is a positive integer and (α)0 = 1.

Remark. It is easily shown that (α)n = Γ(α+ n)/Γ(α).

2.10 The hypergeometric function

Definition. For z ∈ C, the hypergeometric function is defined by

2F1(a, b; c; z) =∞∑n=0

(a)n(b)n(c)n

zn

n! ,

where a, b, c are arbitrary real or complex parameters.

Definition. For z ∈ C, the generalised hypergeometric function is defined by

3F2(a1, a2, a3; b1, b2; z) =∞∑n=0

(a1)n(a2)n(a3)n(b1)n(b2)n

zn

n! ,

where a1, a2, a3, b1, b2 are arbitrary real or complex parameters.

Remark. For the sake of brevity, we shall write

2F1(a, b; c; z) = F (a, b; c; z).

Moreover, since (c)n = c(c+ 1)(c+ 2) · · · (c+ n− 1), we see that c 6= 0,−1,−2, . . .

Proposition 2.10.1. Both the hypergeometric function and the generalised hypergeometricfunction converge absolutely if |z| < 1.

Proof. If un = (a)n(b)n/n!(c)n and c 6= 0,−1,−2, . . ., then∣∣∣∣un+1un

∣∣∣∣ =∣∣∣∣(n+ a)(n+ b)(n+ 1)(n+ c)z

∣∣∣∣→ |z|as n → ∞. So F (a, b; c; z) converges absolutely if |z| < 1 by the ratio test. Likewise, thegeneralised hypergeometric function converges absolutely if |z| < 1.

Now, since F (a, b; c; z) is a power series, it can be differentiated termwise within itsradius of convergence, where it converges absolutely. Hence it is seen that

d

dzF (a, b; c; z) = ab

cF (a+ 1, b+ 1; c+ 1; z). (2.6)

Let a, b, c ∈ C and consider the differential equation

z(1− z)d2y

dz2 + [c− (a+ b+ 1)z]dydz− aby = 0.

21

In order to find its solutions we use the method of Frobenius. Define

P (z) = c− (a+ b+ 1)zz(1− z) and Q(z) = − ab

z(1− z) ,

and notice that limz→0 zP (z) = c and limz→0 z2Q(z) = 0, so since both limits exist, z = 0

is a regular singular point. Thus, our ansatz is y =∑∞n=0 anz

n+r with a0 6= 0. Suppose thisseries converges uniformly in D ⊆ C. Upon differentiating our ansatz we obtain

y′ =∞∑n=0

an(n+ r)zn+r−1 and y′′ =∞∑n=0

an(n+ r)(n+ r − 1)zn+r−2.

Substituting y, y′, and y′′ into the differential equation and using uniqueness of power series[17, Theorem 3.2.5] leads to the characteristic equation r(r − 1 + c) = 0, with the rootsr1 = 0 and r2 = 1− c, and the recursion relation

((n+r)(n+r−1)+c(n+r))an−((n+r−1)(n+r−2)+(1+a+b)(n+r−1)+ab)an−1 = 0.

Therefore:an = (n+ r − 1)(n+ r + a+ b− 1) + ab

(n+ r)(n+ r + c− 1) an−1.

That is,

a1 = (r + a)(r + b)(r + 1)(r + c)a0,

a2 = (r + a+ 1)(r + b+ 1)(r + 2)(r + c+ 1) a1 = (r + a)2(r + b)2

(r + 1)2(r + c)2a0,

a3 = (r + a+ 2)(r + b+ 2)(r + 3)(r + c+ 2) a2 = (r + a)3(r + b)3

(r + 1)3(r + c)3a0,

and so forth. Thus, by induction

an = (r + a)n(r + b)n(r + 1)n(r + c)n

a0, n = 0, 1, 2, . . .

Hence,

y1 = a0

∞∑n=0

(r1 + a)n(r1 + b)n(r1 + 1)n(r1 + c)n

zn+r1 = a0

∞∑n=0

(a)n(b)n(c)n

zn,

y2 = a0

∞∑n=0

(r2 + a)n(r2 + b)n(r2 + 1)n(r2 + c)n

zn+r2 = a0

∞∑n=0

(a+ 1− c)n(b+ 1− c)n(c)n

zn+1−c.

Since a0 6= 0 is a constant and no initial conditions are given, we may put a0 = 1. Thus, ifc 6= 0,−1,−2, . . . we see that F (a, b; c; z) is a solution.

22

Chapter 3

Periods and families of ellipticcurves

3.1 Families of elliptic curves

Suppose E and E are elliptic curves over a field K defined by the Weierstrass equations:

E : y2 = 4x3 − g2x− g3 and E : y2 = 4x3 − g2x− g3.

We say that E and E are isomorphic over K if there exists a nonzero u ∈ K such that

g2 = u4g2 and g3 = u6g3.

This is denoted byE ∼= E,

and is induced by the map(x, y) 7→ (u2x, u3y).

Two lattices Λ and Λ are called homothetic if there exists a nonzero α ∈ C such that

Λ = αΛ.

Theorem 3.1.1. Suppose E and E be elliptic curves over C with associated lattices Λ andΛ, respectively. Then the following are equivalent:

(i) E ∼= E.

(ii) Λ and Λ are homothetic.

(iii) j(E) = j(E).

Proof. See [10, p. 286].

23

Now we compare three families of elliptic curves together with their (normalised) dis-criminants and associated lattices. Consider

E : y2 = 4x3 − g2x− g3, ∆(E) = ∆ = g32 − 27g2

3, Λ(E) = Zω1 + Zω2.

If u = ω1, then E is isomorphic to

Eτ : y2 = 4x3 − g2(τ)x− g3(τ), ∆(Eτ ) = ∆(τ) = g2(τ)3 − 27g3(τ)2, Λτ = Z + Zτ.

If u = ∆−1/12, then E is isomorphic to

E : y2 = 4x3 − γ2x− γ3, ∆(E) = 1, Λ(E) = Zω1 + Zω2,

where

γ2 = ∆−1/3g2 = J1/3 and γ3 = ∆−1/2g3 =

√J − 1

27 .

If u =√g2/g3, then E is isomorphic to

EJ : y2 = 4x3 + g(x+ 1), ∆(EJ) = ∆(J) = 39J2

16(1− J)2 , Λ(EJ) = ZΩ1 + ZΩ2,

whereg = −g

32g2

3= 27J

1− J .

Alternatively, E ∼= EJ if

u = ∆1/12√g2g3

= J1/6( 27J − 1

)1/4.

Also, notice that Eτ is isomorphic to EJ with u =√g2(τ)/g3(τ).

3.2 Periods Ω1, Ω2

Let E : y2 = 4x3 − g2x − g3 be an elliptic curve over C. The uniformisation theorem(Theorem 2.7.1) states that there exists a unique lattice Λ ⊆ C such that g2 = g2(Λ) andg3 = g3(Λ). It is useful, however, to discuss and state a more precise formulation of thistheorem.

The set of points E(C) is a complex Lie group which topologically is a surface over R.More specifically, it is a torus (see Theorem 3.2.1 below and Figure 3.1). LetH1(E(C),Z) de-note the first homology group of the topological space E(C). It is known thatH1(E(C),Z) ∼=Z2, where simple loops α and β in E(C) can be taken as a Z-basis [15, Example 2.3] asillustrated elsewhere [20, Chapter VI, §1, Figure 6.5].

24

Theorem 3.2.1. Let E : y2 = 4x3 − g2x− g3 be an elliptic curve over C.

(i) Let α1 and α2 be simple loops in E(C) giving a Z-basis for H1(E(C),Z). Then

ω1 =∫α1

dx

yand ω2 =

∫α2

dx

y

are R-linearly independent.

(ii) Let Λ = Zω1 + Zω2 ⊆ C be the lattice generated by ω1 and ω2. Then the map

E(C)→ C/Λ

P 7→∫ P

0

dx

y(mod Λ)

is a complex analytic isomorphism of complex Lie groups with the inverse map givenby

C/Λ→ E(C)

z 7→ [℘Λ(z) : ℘′Λ(z) : 1].

Proof. See [20, Chapter VI, §5, Proposition 5.2].

Let X(2)(C) denote the modular curve of level 2 corresponding to the Legendre family[20, Chapter III, §1] of elliptic curves

Eλ : y2 = z(z − 1)(z − λ),

whereJ = 4

27(1− λ+ λ2)3

λ2(1− λ)2 .

It is shown elsewhere [20, Chapter III, §1] that every quadruple (E, e1, e2, e3), where e1, e2, e3

are three points of order 2 on the elliptic curve E, is isomorphic to (Eλ, 0, 1, λ) for someλ 6= 0, 1. Let Y (2)(C) = X(2)(C) \ J−1(0, 1,∞). The invariant J gives an (unramified)S3-covering J : Y (2)(C)→ Y (1)(C), where Y (1)(C) ∼= P1 \ 0, 1,∞; see [20, Appendix C].Let U be a simply-connected open subset of Y (1)(C) and let J0 ∈ U . Let γ be a path in Usuch that γ(0) = J0 and γ(1) = J . By the path-lifting lemma [15, Proposition 1.30], thereexists a unique path γ in Y (2)(C) lifting γ such that γ(0) is a fixed choice of an elementin the fibre over γ(0). Hence, for J ∈ U , starting from an initial fixed choice of labellingof the (distinct) roots e1(J0), e2(J0), e3(J0) of the cubic 4z3 + g(J0)(z + 1), we have awell-defined labelling of the roots e1(J), e2(J), e3(J) of the cubic 4z3 + g(J)(z + 1), whichvary continuously for J ∈ U .

25

Let α1(J) be a simple loop which encircles e2(J) and e3(J), and does not pass throughe1(J); and let α2(J) be a simple loop which encircles e1(J) and e3(J), and does not passthrough e2(J); see Figure 3.2. Then α1(J), α2(J) form a Z-basis for the first homologyH1(EJ ,Z) [20, Chapter VI, §1], which vary continuously for J in Y (1)(C).

α1

α2

Figure 3.1: Loops α1 and α2

+−

+−e2 e3 e1 ∞

+−

+−e2 e3 e1 ∞

α1α2

Figure 3.2: Two copies of P1 with branch cuts

Let α(J) = α1(J), α2(J). We define Ω = Ω1,Ω2 and H = H1,H2 by

Ω = Ω(J) =∫α(J)

dz√4z3 + g(z + 1)

,

H = H(J) = −∫α(J)

z√4z3 + g(z + 1)

dz.

The fact that α1(J), α2(J) form a Z-basis for H1(EJ ,Z) is established by viewing EJ asa double cover of P1 branched at e1(J), e2(J), e3(J), and∞. This in turn can be viewed astwo copies of P1 with branch cuts from e1(J) to ∞, and from e2(J) to e3(J), respectively,glued along the branch cuts with each + edge identified with the − edge on the other copyof P1 as in §2.2; see Figure 3.2.

26

Let Γ1(J) be a path from e1(J) to ∞, and let Γ2(J) be a path from e2(J) to e3(J). Letα(J) = α1(J), α2(J), and let Γ(J) = Γ1(J),Γ2(J). Deforming the loop α(J) into Γ(J) wefind that α(J) is homotopic to 2Γ(J), and so by Theorem 2.1.1 we have

Ω = Ω(J) = 2∫

Γ(J)

dz√4z3 + g(z + 1)

,

H = H(J) = −2∫

Γ(J)

z√4z3 + g(z + 1)

dz.

This is an alternate definition of Ω and H, which appears in the classical literature.Let z0 = α2(J0)/α1(J0) and q0 ∈ SL2(Z) be such that q0(z0) lies in the closure of the

standard fundamental domain F . Then(Γ2(J)Γ1(J)

)= q0

(α2(J)α1(J)

)

for all J ∈ U .It is known that Ω1, Ω2 are linearly independent solutions of the Picard-Fuchs differential

equation [13, p. 34]d2ΩdJ2 + 1

J

dΩdJ

+ 31J − 4144J2(1− J)2 Ω = 0. (3.1)

We prove this formally in the next chapter.

27

Chapter 4

Picard-Fuchs differential equation

In this chapter we give a self-contained derivation of the Picard-Fuchs differential equationas well as some related results that we need. We establish some auxiliary results first.

4.1 Preliminaries

We begin by proving a few results, which we will employ later. First, we have the followinggeneralisation of the result which is sometimes referred to as Leibniz’s rule:

Theorem 4.1.1. (Leibniz’s Rule) Let D be a compact subset of an open convex set Uin C, and let γ be a path in C. Let f(x, t) be a holomorphic function on U . If f(x, t) andfx(x, t) are continuous on D× γ([0, 1]), then for all x ∈ D, the function ϕ(x) =

∫γ f(x, t)dt

is differentiable on D, and its derivative is given by ϕ′(x) =∫γ fx(x, t)dt.

Proof. Pick x0 ∈ D. Consider

ϕ(x)− ϕ(x0)x− x0

=∫γ

f(x, t)− f(x0, t)x− x0

dt.

By hypothesis for each t ∈ γ, f(x, t) is a differentiable function of x. By the (complex)mean value theorem [11], for each t ∈ γ, there exist (possibly) distinct points ξ1 = ξ1(t) andξ2 = ξ2(t) on the line segment from x to x0, which depends on t, such that

Ref(x, t)− f(x0, t)

x− x0

= Refx(ξ1, t),

Imf(x, t)− f(x0, t)

x− x0

= Imfx(ξ2, t),

or

Ref(x, t)− f(x0, t)

x− x0

+ iIm

f(x, t)− f(x0, t)

x− x0

= Refx(ξ1, t)+ iImfx(ξ2, t).

28

We see that fx(ξ, t) is a continuous function of t for ξ = ξ1, ξ2 because fx(x, t) is continuousby hypothesis. Furthermore, recall that if a complex valued function is continuous, its realand imaginary parts are continuous. Besides, the quotient

f(x, t)− f(x0, t)x− x0

is continuous since x 6= x0. Consequently, we can integrate both sides of the equality abovewith respect to t and get

ϕ(x)− ϕ(x0)x− x0

=∫γ

(Refx(ξ1, t)+ iImfx(ξ2, t)

)dt,

where ϕ(x) =∫γ f(x, t)dt. Because fx(x, t) is continuous on the compact set C = D ×

γ([0, 1]), it is uniformly continuous on C by the uniform continuity theorem. By definitionof uniform continuity this means given ε > 0, there exists δ > 0 such that in C we have|fx(x, t)− fx(x0, t)| < ε if |x− x0| < δ. Since ξ = ξ1, ξ2 is on the line segment from x to x0,it can be parametrised as ξ = x0 + s(x− x0) with s ∈ [0, 1], which means

|ξ − x0| = |x0 + s(x− x0)− x0| = |s(x− x0)| ≤ |x− x0| < δ.

So, if ξ = ξ1, ξ2, then |x− x0| < δ implies |ξ − x0| < δ, and so for any ε1, ε2 > 0, we have

|Refx(ξ1, t) − Refx(x0, t)| < ε1 and |Imfx(ξ2, t) − Imfx(x0, t)| < ε2.

Therefore Refx(ξ1, t) + iImfx(ξ2, t) and Refx(x0, t) + iImfx(x0, t) = fx(x0, t) arearbitrarily close, that is,

|Refx(ξ1, t)+ iImfx(ξ2, t) − fx(x0, t)| < ε0

for any ε0 > 0 provided |x−x0| < δ. So, if |γ| denotes the length of γ, then by the estimationlemma, we have ∣∣∣∣ϕ(x)− ϕ(x0)

x− x0−∫γfx(x0, t)dt

∣∣∣∣ < ε0|γ| = η

when |x − x0| < δ. Since ε0 and γ are arbitrary, ε0|γ| = η is arbitrary. Therefore, for anyη > 0, there is δ > 0 such that if |x− x0| < δ, then∣∣∣∣ϕ(x)− ϕ(x0)

x− x0−∫γfx(x0, t)dt

∣∣∣∣ < η,

that is,limx→x0

ϕ(x)− ϕ(x0)x− x0

= ϕ′(x0) =∫γfx(x0, t)dt

for x0 ∈ D. This completes the proof because x0 ∈ D is arbitrary.

29

Lemma 4.1.2. Let D be an open subset of C. Let

Pg(z) = A(z − α1(g))(z − α2(g))(z − α3(g)) ∈ C[z]

be a cubic polynomial with the constant leading coefficient A and roots α1(g), α2(g), α3(g)which are continuous functions of g ∈ D. Let γg : [0, 1] → C be a continuous family ofloops such that γg([0, 1]) does not contain the roots of Pg(z) for every g ∈ D. Let C ⊆ D

be compact with g0 ∈ C. Then there exists δ(C, g0) > 0 such that∫γg

dz√Pg(z)

=∫γg0

dz√Pg(z)

for all g ∈ C with |g − g0| < δ(C, g0).

Proof. Let α(g) be one of the roots. Because γg0([0, 1]) is compact, there exists δg0 > 0such that |p − α(g0)| ≥ δg0 for all p ∈ γg0([0, 1]). Since F (g, t) = γg(t) : D × [0, 1] → C iscontinuous and C × [0, 1] is compact, F (g, t) is uniformly continuous on C × [0, 1]. Hence,there exists δ > 0 such that |g′ − g′′| < δ implies |γg′(t) − γg′′(t)| < δg0/4 for g′, g′′ ∈ C

and all t ∈ [0, 1]. Because α(g) is continuous, there exists δ′ > 0 such that |g − g0| < δ′

implies |α(g)− α(g0)| < δg0/4. Note that the integrand Pg(z)−1/2 is holomorphic on Gg =C\α1(g), α2(g), α3(g) for g ∈ D. Now let g, g′ ∈ C lie in the open disk centred at g0 withradius δ(C, g0) = minδ, δ′. Now,

|γg′(t)− α(g)| = |γg′(t)− γg0(t) + γg0(t)− α(g)|

≥ |γg0(t)− α(g)| − |γg′(t)− γg0(t)|

≥ |γg0(t)− α(g0)| − |α(g)− α(g0)| − |γg′(t)− γg0(t)|

≥ δg0

2

for every t ∈ [0, 1]. So γg′([0, 1]) lies in Gg. Let σ = σ(s) : [0, 1] → C be a path from g0 tog lying in the open disk centred at g0 with radius δ(C, g0). The function H(s, t) = γσ(t) :[0, 1]× [0, 1]→ C is continuous, so it gives a homotopy of loops from γg0 to γg in C. But foreach g′ = σ(s), γg′([0, 1]) lies in Gg by the inequality above. Thus, H(s, t) is a homotopyof loops from γg0 to γg lying in Gg. Hence, by Theorem 2.1.2,

∫γg

dz√Pg(z)

=∫γg0

dz√Pg(z)

.

Theorem 4.1.3. Let D be an open subset of C. Let

Pg(z) = A(z − α1(g))(z − α2(g))(z − α3(g)) ∈ C[z]

30

be a cubic polynomial with the constant leading coefficient A and roots α1(g), α2(g), α3(g)which are continuous functions of g ∈ D. Let γg : [0, 1] → C be a continuous family ofloops such that γg([0, 1]) does not contain the roots of Pg(z) for every g ∈ D. Let C ⊆ D

be compact with g0 ∈ C. Then there exists δ(C, g0) > 0 such that

d

dg

∫γg

dz√Pg(z)

=∫γg

∂

∂g

dz√Pg(z)

for all g ∈ C with |g − g0| < δ(C, g0).

Proof. Since C ⊆ D is compact and g0 ∈ C, in view of the lemma there exists δ(C, g0) > 0such that ∫

γg

dz√Pg(z)

=∫γg0

dz√Pg(z)

for all g ∈ C with |g − g0| < δ(C, g0). By Theorem 4.1.1, we have

d

dg

∫γg0

dz√Pg(z)

=∫γg0

∂

∂g

dz√Pg(z)

.

Because |g − g0| < δ(C, g0), it follows from the lemma that

d

dg

∫γg

dz√Pg(z)

=∫γg

∂

∂g

dz√Pg(z)

.

4.2 Derivation of the Picard-Fuchs differential equation

Recall from §3.2 thatΩ = 2

∫Γg

dz

yand H = −2

∫Γg

zdz

y,

where y =√

4z3 + g(z + 1). Therefore

dΩdg

= −∫

Γg

dz

y3 −∫

Γg

z

y3dz, (4.1)

dHdg

=∫

Γg

z

y3dz +∫

Γg

z2

y3dz (4.2)

by Theorem 4.1.3. Now,

∫Γg

d

(zα

y

)=∫

Γg

αzα−1

ydz −

∫Γg

zα(12z2 + g)2y3 dz = 0, α = 0, 1, 2

31

by the fundamental theorem of complex integration. If α = 0, we get

12∫

Γg

z2

y3dz + g

∫Γg

dz

y3 = 0.

However, setting α = 1, one obtains

∫Γg

dz

y−∫

Γg

z(12z2 + g)2y3 dz = 0,

which can be recast as

2∫

Γg

4z3 + g(z + 1)2y3 dz −

∫Γg

z(12z2 + g)2y3 dz =

∫Γg

g(z + 2)− 4z3

2y3 dz = 0.

Therefore:

Ω2 =

∫Γg

4z3 + g(z + 1)y3 dz +

∫Γg

g(z + 2)− 4z3

y3 dz

= 2g∫

Γg

z

y3dz + 3g∫

Γg

dz

y3 .

Lastly, if α = 2, then ∫Γg

2zydz −

∫Γg

z2(12z2 + g)2y3 dz = 0,

or4∫

Γg

z4z3 + g(z + 1)

2y3 dz −∫

Γg

z2(12z2 + g)2y3 dz =

∫Γg

z4z3 + g(3z + 4)

2y3 dz = 0,

so

H2 = −

∫Γg

z4z3 + g(z + 1)

y3 dz +∫

Γg

z4z3 + g(3z + 4)

y3 dz

= 2g∫

Γg

z2

y3dz + 3g∫

Γg

z

y3dz.

DefineI0 =

∫Γg

dz

y3 , I1 =∫

Γg

z

y3 , I2 =∫

Γg

z2

y3dz.

So far we have shown that

12I2 + gI0 = 0, (4.3)

2gI1 + 3gI0 = Ω2 , (4.4)

2gI2 + 3gI1 = H2 . (4.5)

32

Multiplying (4.4) by 3 and (4.5) by 2, and subtracting the former from the later we obtain

4gI2 − 9gI0 = 2H− 3Ω2 . (4.6)

Multiply (4.3) by 9, add the result to (4.6), and solve for I2 to get

I2 = 2H− 3Ω8(g + 27) .

Using this and (4.3) yieldsI0 = 9Ω− 6H

2g(g + 27) .

Similarly, putting this and (4.4) together we obtain

I1 = gΩ + 18H4g(g + 27) .

Combining (4.1) and (4.2), and the above expressions for I0, I1, I2, we arrive at

4g(g + 27)dΩdg

+ (18 + g)Ω + 6H = 0,

4g(g + 27)dHdg

+ gΩ2 − (18 + g)H = 0.

As g = 27J/(1− J) and dg/dJ = 27/(1− J)2, using the chain rule we deduce that

36J(1− J)dΩdJ

= −3(2 + J)Ω− 2(1− J)H, (4.7)

24J(1− J)dHdJ

= −3JΩ + 2(2 + J)H. (4.8)

Differentiating (4.7) we see that

36J(J − 1)d2ΩdJ2 + (69J − 42)dΩ

dJ+ 2(J − 1)dH

dJ− 3Ω + 2H = 0.

Using (4.7) and (4.8) we obtain expressions for H and dH/dJ in terms of Ω and dΩ/dJalone. Substituting these expressions into the relation above we arrive at the Picard-Fuchsdifferential equation:

d2ΩdJ2 + 1

J

dΩdJ

+ 31J − 4144J2(1− J)2 Ω = 0.

33

Chapter 5

Period relations

5.1 Hypergeometric representations of Ω1

Recall from §2.5 that the quasi-period η(ω) is defined by

η(ω) = ζΛ(z + ω)− ζΛ(z)

for ω ∈ Λ. We see from the definition of ζΛ(u) that ζαΛ(αz) = α−1ζΛ(z), so η(αω) =α−1η(ω). Define

ηk = η(ωk), ηk = η(ωk), Hk = η(Ωk).

From the relations in §3.1 we have

ηk∆−1/12 = ηk,

ηk

√g2g3

= Hk.

We summarise our notation for the periods and quasi-periods in the table below.

Elliptic curve Periods Quasi-periodsE (ω1, ω2) (η1, η2)Eτ (1, τ) (η1ω1, η2ω1)E (ω1, ω2) = (ω1∆1/12, ω2∆1/12) (η1, η2) = (η1∆−1/12, η2∆−1/12)EJ (Ω1,Ω2) = (ω1

√g3/g2, ω2

√g3/g2) (H1,H2) = (η1

√g2/g3, η2

√g2/g3)

Table 5.1: Summary: Periods and Quasi-periods

Lemma 5.1.1. Let ν(J) be one of 1/J , (J − 1)/J , J/(J − 1). Let F be the standardfundamental domain for SL2(Z). Define Cν = τ ∈ F : |ν(J)| < 1. Then Cν is an openset in H which is a union of at most two simply-connected components.

34

Proof. Let H∗ = H ∪ P1(Q). It is known that SL2(Z) acts continuously on the topologicalspace H∗ and the quotient space SL2(Z)\H∗ has the natural structure of a Riemann surface[18, §1.8]. The function J gives a complex analytic isomorphism of Riemann surfacesJ : SL2(Z) \ H∗ → P1, a fortiori, a homeomorphism of topological spaces. This still holdsfor ν(J) as it is a Möbius transformation. Furthermore, the restriction of ν(J) to Cν givesa homeomorphism between Cν and ν(J) ∈ P1 : |ν(J)| < 1 \ ν(∂F). Here, ∂F denotesthe boundary of F . The set ν(J) ∈ P1 : |ν(J)| < 1 \ ν(∂F) is a union of at most twosimply-connected components, which implies the same for Cν .

Let Cν(J),z denote the connected component of Cν(J) with z lying in the closure ofCν(J). The connected components C1/J,∞, C(J−1)/J,i, CJ/(J−1),ρ are depicted below, wherex = Re(τ) and y = Im(τ). Note that the regions are open, unbounded, and extend verticallytowards ∞.

Figure 5.1: C1/J,∞ = τ : |1/J | < 1, x2 + y2 > 1

Figure 5.2: C(J−1)/J,i = τ : |(J − 1)/J | < 1, x2 + y2 > 1

35

Figure 5.3: CJ/(J−1),ρ = τ : |J/(J − 1)| < 1, x2 + y2 > 1, x < 0

Definition. For τ ∈ C, the Dedekind eta function η(τ) is defined by

η(τ) = q1/24∞∏n=1

(1− qn), q = e2πiτ .

Proposition 5.1.2. The Dedekind eta function η(τ) converges absolutely if τ ∈ H.

Proof. Let log z be the principal value of the logarithm with the branch cut along (−∞, 0].Note that |q| < 1 if τ ∈ H. By the reverse triangle inequality we have 1 − |qn| ≤ |1 − qn|,but |q| < 1, so |1− qn| > 0, and so 1− qn does not lie on the branch cut. Now,

m∏n=1

(1− qn) = exp m∑n=1

log(1− qn),

andlimm→∞

exp m∑n=1

log(1− qn)

= exp ∞∑n=1

log(1− qn)

because∑∞n=1 log(1 − qn) is finite if |q| < 1 by virtue of the ratio test and l’Hôpital’s

rule. Naturally this means∑∞n=1 log(1 − qn) converges absolutely if |q| < 1, and thereby∏∞

n=1(1− qn) converges absolutely whenever |q| < 1.

Proposition 5.1.3. If τ ∈ H, then η(τ + 1) = eπi/12η(τ).

Proof. See [2, p. 47].

Proposition 5.1.4. If τ ∈ H, then η(−1/τ) =√−iτη(τ).

Proof. See [2, p. 48].

Proposition 5.1.5. If τ ∈ H, then ∆(τ) = (2π)12η(τ)24.

36

Proof. See [2, p. 51].

Theorem 5.1.6, 5.1.7, 5.1.8 were established by Archinard [3]. However, Theorem 5.1.6was also derived by D. V. and G. V. Chudnovsky [9, p. 610], but it appears to have beenfirst derived by Fricke and Klein [13].

Theorem 5.1.6 (J =∞ case). Suppose τ = ω2/ω1 is in the connected component C1/J,∞

of the open set |J | > 1. Then

ω1(J) = ω1∆1/12 = 2π√2√

3J−1/12F

( 112 ,

512; 1; 1

J

),

= 2π√2√

3J−7/12√J − 1F

( 712 ,

1112; 1; 1

J

),

= 2π√2√

3(J − 1)−1/12F

( 112 ,

712; 1; 1

1− J

),

= 2π√2√

3J1/3(J − 1)−5/12F

( 512 ,

1112; 1; 1

1− J

),

where J = J(τ).

Proof. Recall that EJ is defined by

EJ : y2 = 4x3 + g(x+ 1),

whereg = −g

32g2

3= 27J

1− J , J 6= 0, 1,∞

with Λ(EJ) = ZΩ1(J) + ZΩ2(J). Using the method of Frobenius, it is shown that

ΩJ = (J − 1)1/4

J1/4 F

( 112 ,

512; 1; 1

J

)

is a solution of (3.1). Also, EJ is isomorphic to

Eτ : y2 = 4x3 − g2(τ)x− g3(τ)

with Λτ = Z + Zτ . Note that Λτ is Λ(EJ) multiplied by√g3(τ)/g2(τ). Recall that

ω1

√g3g2

=√g3(τ)g2(τ) and ∆−1/12

√g3g2

= J−1/6(J − 1

27

)1/4.

Therefore: √g3(τ)g2(τ) = ∆(τ)1/12J−1/6

(J − 1

27

)1/4.

37

Using the classical identity ∆(τ) = (2π)12η(τ)24 we find that√g3(τ)g2(τ) = 2πη(τ)2J−1/6

(J − 1

27

)1/4.

Also, recall that Ω1(J), Ω2(J) are linearly independent solutions of (3.1). So there areconstants a, b such that ΩJ = aΩ1(J) + bΩ2(J) is a solution. But τ = Ω2(J)/Ω1(J), soΩJ = (a + bτ)Ω1(J) = (a + bτ)ω1

√g3/g2 = (a + bτ)

√g3(τ)/g2(τ). Now, if τ is in the

connected component C1/J,∞, there are constants A, B such that

(J − 1)1/4

J1/4 F

( 112 ,

512; 1; 1

J

)= (A+Bτ)η(τ)2 (J − 1)1/4

J1/6 ,

that is, there are constants A, B such that

F

( 112 ,

512; 1; 1

J

)= (A+Bτ)η(τ)2J1/12.

We see that∞ is a fixed point of the transformation τ → τ + 1 in C1/J,∞. Using η(τ + 1) =eπi/12η(τ), we find that in C1/J,∞, we have (A + Bτ)η(τ)2 = (A + B(τ + 1))e−πi/6η(τ)2.Upon solving for B and letting τ →∞ we infer that B = 0, so

F

( 112 ,

512; 1; 1

J

)= Aη(τ)2J1/12

for τ in C1/J,∞. It is well known that [2, pp. 21-22]

j = 1q

+ 744 + 196884q +O(q2), q = e2πiτ .

Sinceη(τ) = q1/24

∞∏n=1

(1− qn), q = e2πiτ ,

it is found that η(τ)2J1/12 → 12−1/4 as τ →∞ because j = 123J , and thus A =√

2√

3. As∆(τ) = ω12

1 ∆ and ∆(τ) = (2π)12η(τ)24, we get

ω1(J) = 2π√2√

3J−1/12F

( 112 ,

512; 1; 1

J

).

Applying Euler’s and Pfaff’s transformations we obtain all the proposed formulae.

38

Theorem 5.1.7 (J = 1 case). Suppose τ = ω2/ω1 is in the connected component C(J−1)/J,i

of the open set |(J − 1)/J | < 1. Then

ω1(J) = ω1∆1/12 = 2πατ + i

J−1/12F

( 112 ,

512; 1

2; J − 1J

),

= 2πατ + i

F

( 112 ,

112; 1

2; 1− J),

= 2πατ + i

J1/3F

( 512 ,

512; 1

2; 1− J),

where J = J(τ) and α = 2iη(i)2.

Proof. Let

g = −g32g2

3= 27J

1− J , J 6= 0, 1,∞.

Consider the elliptic curveEJ : y2 = 4x3 + g(x+ 1)

with Λ(EJ) = ZΩ1(J) + ZΩ2(J). Using the method of Frobenius, we find that

ΩJ = (J − 1)1/4

J1/4 F

( 112 ,

512; 1

2; J − 1J

)

is a solution of (3.1). Now, EJ is isomorphic to

Eτ : y2 = 4x3 − g2(τ)x− g3(τ)

with Λτ = Z + Zτ , and Λτ is obtained upon multiplying Λ(EJ) by√g3(τ)/g2(τ). Recall

that

ω1

√g3g2

=√g3(τ)g2(τ) and ∆−1/12

√g3g2

= J−1/6(J − 1

27

)1/4.

Therefore: √g3(τ)g2(τ) = ∆(τ)1/12J−1/6

(J − 1

27

)1/4.

Furthermore, ∆(τ) = (2π)12η(τ)24, so√g3(τ)g2(τ) = 2πη(τ)2J−1/6

(J − 1

27

)1/4.

Further, recall that Ω1(J), Ω2(J) are linearly independent solutions of (3.1). This meansthere are constants a, b such that ΩJ = aΩ1(J)+bΩ2(J) is a solution. But τ = Ω2(J)/Ω1(J),so ΩJ = (a + bτ)Ω1(J) = (a + bτ)ω1

√g3/g2 = (a + bτ)

√g3(τ)/g2(τ). Therefore, for τ in

39

the connected component C(J−1)/J,i, there are constants A, B such that

(J − 1)1/4

J1/4 F

( 112 ,

512; 1

2; J − 1J

)= (A+Bτ)η(τ)2 (J − 1)1/4

J1/6 ,

orF

( 112 ,

512; 1

2; J − 1J

)= (A+Bτ)η(τ)2J1/12.

Now, in C(J−1)/J,i, i is a fixed point of the transformation τ → −1/τ . Replacing τ by −1/τand employing η(−1/τ)2 = τη(τ)2/i, we see that in C(J−1)/J,i we must have (A+Bτ)η(τ)2 =(A−B/τ)η(−1/τ)2 = (A−B/τ)τη(τ)2/i. Solving for A and taking the limit as τ → i weconclude that A = Bi, so

F

( 112 ,

512; 1

2; J − 1J

)= B(τ + i)η(τ)2J1/12

for τ in C(J−1)/J,i. Taking the limit as τ → i, we get 1/B = 2iη(i)2. Hence,

F

( 112 ,

512; 1

2; J − 1J

)= τ + i

2iη(i)2 η(τ)2J1/12.

Using ∆(τ) = ω121 ∆ and the classical identity ∆(τ) = (2π)12η(τ)24, yields

ω1(J) = 2πατ + i

J−1/12F

( 112 ,

512; 1

2; J − 1J

).

Using Euler’s and Pfaff’s transformations gives the other two identities.

Theorem 5.1.8 (J = 0 case). Suppose τ = ω2/ω1 is in the connected component CJ/(J−1),ρ

of the open set |J/(J − 1)| < 1. Then

ω1(J) = ω1∆1/12 = 2πατ − ρ

(1− J)−1/12F

( 112 ,

712; 2

3; J

J − 1

),

= 2πατ − ρ

F

( 112 ,

112; 2

3; J),

= 2πατ − ρ

√1− JF

( 712 ,

712; 2

3; J),

where J = J(τ) and α = iη(ρ)2√3.

Proof. Recall that if

g = −g32g2

3= 27J

1− J , J 6= 0, 1,∞

then we define EJ byEJ : y2 = 4x3 + g(x+ 1),

40

with Λ(EJ) = ZΩ1(J) + ZΩ2(J). Using the method of Frobenius, it is shown that

ΩJ = (J − 1)1/6

J1/6 F

( 112 ,

712; 2

3; J

J − 1

)

is a solution of (3.1). The elliptic curve EJ is isomorphic to

Eτ : y2 = 4x3 − g2(τ)x− g3(τ)

with Λτ = Z + Zτ , where Λτ is Λ(EJ) multiplied by√g3(τ)/g2(τ). Recall that

ω1

√g3g2

=√g3(τ)g2(τ) and ∆−1/12

√g3g2

= J−1/6(J − 1

27

)1/4.

Therefore: √g3(τ)g2(τ) = ∆(τ)1/12J−1/6

(J − 1

27

)1/4.

Using ∆(τ) = (2π)12η(τ)24 shows that√g3(τ)g2(τ) = 2πη(τ)2J−1/6

(J − 1

27

)1/4.

Recall that Ω1(J), Ω2(J) are linearly independent solutions of (3.1). Consequently, thereare constants a, b such that ΩJ = aΩ1(J)+bΩ2(J) is a solution. Because τ = Ω2(J)/Ω1(J),we get ΩJ = (a+ bτ)Ω1(J) = (a+ bτ)ω1

√g3/g2 = (a+ bτ)

√g3(τ)/g2(τ). Hence, if τ is in

the connected component CJ/(J−1),ρ, there are constants c, d such that

(J − 1)1/6

J1/6 F

( 112 ,

712; 2

3; J

J − 1

)= (c+ dτ)η(τ)2 (J − 1)1/4

J1/6 ,

or, there are constants A, B such that

F

( 112 ,

712; 2

3; J

J − 1

)= (A+Bτ)η(τ)2(1− J)1/12.

Observe that ρ is a fixed point of the transformation τ → −1/(τ + 1) in CJ/(J−1),ρ. Fromthe functional equations

η(−1/τ) =√−iτη(τ) and η(τ + 1) = eπi/12η(τ),

we infer that in CJ/(J−1),ρ, we have

(A+Bτ)η(τ)2 = −ie−πi/6(τ + 1)(A− B

τ + 1

)η(τ)2.

41

Note that ieπi/6 = ρ. Solving for A and letting τ → ρ gives A = −ρB. Therefore, if τ is inCJ/(J−1),ρ, then

F

( 112 ,

712; 2

3; J

J − 1

)= B(τ − ρ)η(τ)2(1− J)1/12.

Now, as τ → ρ we obtain B = 1/(ρ − ρ)η(ρ)2 = 1/η(ρ)2i√

3. Lastly, since ∆(τ) = ω121 ∆

and ∆(τ) = (2π)12η(τ)24, we arrive at

ω1(J) = 2πατ − ρ

(1− J)−1/12F

( 112 ,

712; 2

3; J

J − 1

).

Lastly, employing Euler’s and Pfaff’s transformations leads to all the stated identities.

5.2 Quasi-period relations

The following result appears in [8] and [9, p. 610], but we provide a proof.

Theorem 5.2.1. If ω = ω1, ω2 and η = η1, η2, then

η = −2√

3J2/3√J − 1dωdJ.

Proof. Let ω = ω1, ω2, η = η1, η2, and Ω = Ω1,Ω2, H = H1,H2. Using (4.7) we obtain

H = 32

(J + 2)Ω− 12J(J − 1)dΩdJ

J − 1 = 32J + 2J − 1Ω− 18J dΩ

dJ. (5.1)

Now, Ω = ω√g3/g2, H = η

√g2/g3 and ω = ω∆1/12, η = η∆−1/12. Recall that

∆−1/12√g3g2

= J−1/6( 27J − 1

)−1/4and ∆1/12

√g2g3

= J1/6( 27J − 1

)1/4.

Thus it is plain that

Ω = ωJ−1/6( 27J − 1

)−1/4and H = ηJ1/6

( 27J − 1

)1/4,

anddΩdJ

= 31/4(J + 2)36J7/6(J − 1)3/4 ω + 31/4(J − 1)1/4

3J1/6dω

dJ.

42

Substituting these into (5.1) gives

η

( 27J − 1

)1/4= 3

2J + 2J − 1 ωJ

−1/3( 27J − 1

)−1/4

− 18J5/6[ 31/4(J + 2)

36J7/6(J − 1)3/4 ω + 31/4(J − 1)1/4

3J1/6dω

dJ

]= −18J5/6 31/4(J − 1)1/4

3J1/6dω

dJ.

Solving for η we obtainη = −2

√3J2/3√J − 1dω

dJ.

5.3 Complex multiplication period relations

Definition. We define s2(τ) by

s2(τ) = E4(τ)E6(τ)

(E2(τ)− 3

πIm(τ)

).

It is known that s2(τ) is rational at τ =√−N if N = 2, 3, 4, 7 [9, Lemma 4.1]:

N s2(τ)2 5/143 5/114 11/217 85/133

Table 5.2: Special values of s2(τ) at τ =√−N

Moreover, s2(τ) is rational at τ = −1+√−N

2 if N = 7, 11, 19, 27, 43, 67, 163 [9, Lemma 4.1]:

N s2(τ)7 5/2111 32/7719 32/5727 160/25343 640/90367 33440/43617163 77265280/90856689

Table 5.3: Special values of s2(τ) at τ = −1+√−N

2

43

Theorem 5.3.1. We have

2iω1η1Im(τ)− ω21

[2iIm(τ)3g3

2g2s2(τ)

]= 2πi.

Proof. From

g2(τ) = 4π4E4(τ)3 and g3(τ) = 8π6E6(τ)

27 ,

and the fact that g2(τ) = ω41g2 and g3(τ) = ω6

1g3, we obtain

3g32g2

s2(τ) = π2

3ω21

[E2(τ)− 3

πIm(τ)

]. (5.2)

Obviously,

2ω1η1Im(τ)−[2ω1η1 −

2πIm(τ)

]Im(τ) = 2π,

or, what is the same thing,

2ω1η1Im(τ)− 2π2Im(τ)3

[3ω1η1π2 − 3

πIm(τ)

]= 2π.

It is shown elsewhere [14, p. 298] that

E2(τ) = 3ω1η1π2 .

Hence2ω1η1Im(τ)− 2π2Im(τ)

3

[E2(τ)− 3

πIm(τ)

]= 2π.

Using (5.2), and multiplying through by i, completes the proof.

Suppose τ satisfiesaτ2 + bτ + c = 0, (5.3)

where a, b, c are integers such that a > 0, that is,

τ = −b+√−d

2a ,

where −d = b2 − 4ac. If τ is as in (5.3), then

ω1η1√−d

a− ω2

1

[√−da

3g32g2

s2(τ)]

= 2πi.

However, it will deem convenient to rewrite this as

ω12π

[η1 − ω1

3g32g2

s2(τ)]

= a√d. (5.4)

44

Chapter 6

J =∞ case

In this section we derive the Chudnovsky series for 1/π.In what follows, we will often not specify branches of the n-th root functions during

intermediate calculations. For instance, we use the following algebraic identity, which isvalid up to a 6-th root of unity dependent on the quantities g2, g3, J , or on τ if we regardthese quantities as functions of τ .

Lemma 6.0.1. Up to a 6-th root of unity, we have

3g32g2

√J√

J − 1= (J∆)1/6√

12. (6.1)

Proof. Note that up to a 6-th root of unity we have

3g32g2

(j∆)−1/6√J√

J − 1= 3g3

2g2(123g3

2)−1/6

√123g3

2/∆123g3

2/∆− 123

= 3g32g2

12g2√(123g3

2/∆− 123)∆

= 3g3

4√

3g32 − 3∆

.

But ∆ = g32 − 27g2

3, so3g3

4√

3g32 − 3∆

= 3g3

4√

81g23

,

which simplifies to 1/12.

Remark. In Theorem 6.0.3, 6.0.4, 7.0.2, 7.0.3, 8.0.1, 8.0.2 we have exact equalities with theprincipal branch of the n-th root function involved. This was established by first showingthat we have an identity f(τ) = ε(τ)g(τ) on some connected subset U of C, where f(τ) andg(τ) are continuous on U , ε(τ) is an n-th root of unity, and n does not depend on τ ∈ U .Thus, f(τ)/g(τ) is (or can be extended to) a continuous function on a connected set U to a

45

finite subset with the induced discrete topology, and is hence constant. Thus, if there is oneτ0 ∈ U such that f(τ0)/g(τ0) = 1, then in fact ε(τ) = 1 on all of U . This was establishedfor some suitable τ0. In Theorem 6.0.3 and 6.0.4 we used τ0 =

√−2; in Theorem 7.0.2 and

7.0.3 we used τ0 =√−2; and in Theorem 8.0.1 and 8.0.2 we used τ0 = −1+

√−7

2 .

Lemma 6.0.2. We have (16

)n

(56

)n

(12

)n

= 12−3n (6n)!(3n)! . (6.2)

Proof. Notice that (p

q

)n

= q−nn∏k=1

(qk + p− q), p, q ∈ N

follows directly from the definition of (a)n. Therefore,(16

)n

(56

)n

(36

)n

= 6−3nn∏k=1

(6k − 5)(6k − 3)(6k − 1)

= 3 · 5 · 7 · · · (6n− 1)63n

= 6−3n (6n)!2 · 4 · 6 · · · 6n

= 12−3n (6n)!(3n)! ,

and the proof is complete.

Theorem 6.0.3. Let√z be the principal value of the square root function with the branch

cut along (−∞, 0). If τ is as in (5.3) and lies in C1/J,∞, then we have

a

π√d

√J√

J − 1= F 2 1− s2(τ)

6 − J d

dJF 2,

where F = F (1/12, 5/12; 1; 1/J).

Proof. If we denote F (1/12, 5/12; 1; 1/J) by F , then by Theorem 5.1.6

ω1 = 2π121/4J

−1/12F and Ω1 = 2π121/4 (J∆)−1/12F,

so

dω1dJ

= 2π121/4

(J−1/12dF

dJ+ F

d

dJJ1/12

)= 2π

121/4

(J−1/12dF

dJ− F

12J13/12

)= − 2π

121/4J−1/12

(F

12J −dF

dJ

).

46

Further, from Theorem 5.2.1,

η1 = −2√

3J2/3√J − 1dω1dJ

∆1/12

= 4π√

3121/4 J

7/12√J − 1(F

12J −dF

dJ

)∆1/12

= 4π√

3121/4 J

−5/12√J − 1(F

12 − JdF

dJ

)∆1/12.

Substituting the above expressions for ω1 and η1 into (5.4) gives

(J∆)−1/12

121/4 F

[ 2√

3121/4J

−5/12√J − 1(F

12 − JdF

dJ

)∆1/12

− (J∆)−1/12

124 F3g32g2

s2(τ)]

= a

2π√d,

or, what is the same thing,√J − 1√J

F

[F

12 − JdF

dJ

− (J∆)−1/6√

12F

3g32g2

√J√

J − 1s2(τ)

]= a

2π√d.

Using (6.1) leads to the desired result.

Theorem 6.0.4. Let√z be the principal value of the square root function with the branch

cut along (−∞, 0). If τ is as in (5.3) and lies in C1/J,∞, then we have

a

π√d

√j√

j − 123 =∞∑n=0

(1− s2(τ)6 + n

) (6n)!(3n)!n!3 j

−n.

Proof. Using Clausen’s formula [1, p. 116]

2F1(a, b; c; z)2 = 3F2(2a, 2b, a+ b; 2a+ 2b, c; z), c = a+ b+ 12 ,

we arrive at

2F1

( 112 ,

512; 1; 1

J

)2= 3F2

(16 ,

56 ,

12; 1, 1; 1

J

).

On account of (6.2) we have

F

( 112 ,

512; 1; z

)2=∞∑n=0

(6n)!(3n)!n!3

zn

123n ,

47

hence

zd

dzF

( 112 ,

512; 1; z

)2=∞∑n=0

(6n)!(3n)!n!3

n

123n zn.

Changing z into 1/z yields this

−z ddzF

( 112 ,

512; 1; 1

z

)2=∞∑n=0

(6n)!(3n)!n!3

n

123n z−n.

Upon setting z = J = j/123, and utilising Theorem 6.0.3, we obtain the desired result.

6.1 Examples

The values τ =√−N , N = 2, 3, 4, 7, for which J(τ) is rational, satisfy (5.3) and lie in

C1/J,∞. Hence, the formula above holds whenever τ =√−N , N = 2, 3, 4, 7. We state all

the possible identities.τ =√−2:

5√

528π =

∞∑n=0

( 328 + n

) (6n)!(3n)!n!3 20−3n.

τ =√−3:

5√

1566π =

∞∑n=0

( 111 + n

) (6n)!(3n)!n!3 2−n30−3n.

τ =√−4:

11√

33252π =

∞∑n=0

( 563 + n

) (6n)!(3n)!n!3 66−3n.

τ =√−7:

85√

2557182π =

∞∑n=0

( 8133 + n

) (6n)!(3n)!n!3 225−3n.

The values τ = −1+√−N

2 , N = 7, 11, 19, 27, 43, 67, 163, for which J(τ) is rational,satisfy (5.3) and lie in C1/J,∞. So the theorem above holds for τ = −1+

√−N

2 , N =7, 11, 19, 27, 43, 67, 163. We give all the possible formulae.

τ = −1+√−7

2 :5√

1563π =

∞∑n=0

(−1)n( 8

63 + n

) (6n)!(3n)!n!3 15−3n.

τ = −1+√−11

2 :16√

277π =

∞∑n=0

(−1)n( 15

154 + n

) (6n)!(3n)!n!3 32−3n.

48

τ = −1+√−19

2 :16√

6171π =

∞∑n=0

(−1)n( 25

342 + n

) (6n)!(3n)!n!3 96−3n.

τ = −1+√−27

2 :

80√

302277π =

∞∑n=0

(−1)n( 31

506 + n

) (6n)!(3n)!n!3 3−n160−3n.

τ = −1+√−43

2 :

320√

158127π =

∞∑n=0

(−1)n( 263

5418 + n

) (6n)!(3n)!n!3 960−3n.

τ = −1+√−67

2 :

880√

330130851π =

∞∑n=0

(−1)n( 10177

261702 + n

) (6n)!(3n)!n!3 5280−3n.

τ = −1+√−163

2 :

213440√

10005272570067π =

∞∑n=0

(−1)n( 13591409

545140134 + n

) (6n)!(3n)!n!3 640320−3n.

49

Chapter 7

J = 1 case

We will now derive formulae that naturally arise from the hypergeometric series for Ω1

around J = 1.We begin with the following proposition.

Proposition 7.0.1. Let√z be the principal value of the square root function with the

branch cut along (−∞, 0). Then

E4(τ)E6(τ) = 2π2

9ω21J1/3

√27√

J − 1, ω1 = ω1∆1/12.

Proof. Since

g2(τ) = 4π4E4(τ)3 and g3(τ) = 8π6E6(τ)

27 ,

we getE4(τ)E6(τ) = 2π2

9g2(τ)g3(τ) .

But g2(τ) = ω41g2 and g3(τ) = ω6

1g3, so

E4(τ)E6(τ) = 2π2

9ω21

g2g3.

Lastly, it follows from (6.1) that

g2g3

= J1/3

∆1/6

√27√

J − 1,

so the proposed identity follows at once since ω1 = ω1∆1/12.

50

Theorem 7.0.2. Let√z be the principal value of the square root function with the branch

cut along (−∞, 0). If τ is as in (5.3) and lies in C(J−1)/J,i, then we have

i

π

[a(τ + i)2

2iα2√

3d

√J√

J − 1− F 2

τ + i

E4(τ)E6(τ)

]= F 2 1− s2(τ)

6 − J d

dJF 2,

where F = F (1/12, 5/12; 1/2; (J − 1)/J) and α = 2iη(i)2.

Proof. Let F = F (1/12, 5/12; 1/2; (J − 1)/J). Recall from Theorem 5.1.7 that

ω1 = J−1/12m

τ0F,

where m = 2πα and τ0 = τ + i. If J ′ 6= 0, then locally τ ′ = 1/J ′ by the inverse functiontheorem. Hence,

dω1dJ

= J−1/12m

τ0

dF

dJ+ F

d

dJ

(m

τ0J−1/12

)= J−1/12m

τ0

[dF

dJ−( 1

12J + 1τ0J ′

)F

].

Because the zeroes of a non-constant analytic function on C are isolated, the above holdsin a deleted neighbourhood of C(J−1)/J,i. From Theorem 5.2.1,

η1 = −2√

3J7/12√J − 1mτ0

[dF

dJ−( 1

12J + 1τ0J ′

)F

]∆1/12.

Using (5.4) we find that

a√d

= − mF

2πτ0(J∆)1/12

[2√

3J7/12√J − 1mτ0

[dF

dJ− F

( 112J + 1

τ0J ′

)]∆1/12

+ mF

τ0(J∆)1/123g32g2

s2(τ)]

or

2πaτ20

m2√d

= −J−1/12F

[2√

3J7/12√J − 1[dF

dJ− F

( 112J + 1

τ0J ′

)]+ F

J1/12∆1/63g32g2

s2(τ)].

Consequently,

2πaτ20

m2√d

= −J−1/12F

[2√

3J7/12√J − 1[dF

dJ− F

( 112J + 1

τ0J ′

)]+ F

J1/12

√J − 1

2√

3J1/3 s2(τ)]

51

because it is easily seen from (6.1) that

3g32g2

∆−1/6 =√J − 1

2√

3J1/3 .

Equivalently,2πaτ2

0m2√

3d

√J√

J − 1= F 2

[16 + 2J

τ0J ′− s2(τ)

6

]− 2JF dF

dJ.

However,2JF dF

dJ= J

d

dJF 2,

so2πaτ2

0m2√

3d

√J√

J − 1− F 2 2J

τ0J ′= F 2 1− s2(τ)

6 − J d

dJF 2.

Recall from Theorem 2.7.3 thatJ ′

J= −2πiE6(τ)

E4(τ) .

Using this we obtain

2πaτ20

m2√

3d

√J√

J − 1− i

π

F 2

τ0

E4(τ)E6(τ) = F 2 1− s2(τ)

6 − J d

dJF 2.

Since m = 2πα we have

i

π

[aτ2

02iα2√

3d

√J√

J − 1− F 2

τ0

E4(τ)E6(τ)

]= F 2 1− s2(τ)

6 − J d

dJF 2.

Theorem 7.0.3. Let√z be the principal value of the square root function with the branch

cut along (−∞, 0). If τ is as in (5.3) and lies in C(J−1)/J,i, then we have

τ + i

2πα2√

3

√J√

1− J

(aτ + i√−d− 1

)= F 2 1− s2(τ)

6 − J d

dJF 2,

where F = F (1/12, 5/12; 1/2; (J − 1)/J) and α = 2iη(i)2.

Proof. Let F = F (1/12, 5/12; 1/2; (J − 1)/J). Using Proposition 7.0.1 and Theorem 7.0.2we obtain

i

π

[a(τ + i)2

2iα2√

3d

√J√

J − 1− F 2

τ + i

2π2

9ω21J1/3

√27√

J − 1

]= F 2 1− s2(τ)

6 − J d

dJF 2.

We see from Theorem 5.1.7 that

ω21 = 4π2α2

(τ + i)2J1/6F2,

52

soi

π

[a(τ + i)2

2iα2√

3d

√J√

J − 1− τ + i

18α2

√27J√J − 1

]= F 2 1− s2(τ)

6 − J d

dJF 2

ori(τ + i)

π

√J√

J − 1

(a(τ + i)2iα2√

3d−√

36α2

)= F 2 1− s2(τ)

6 − J d

dJF 2,

which simplifies to

τ + i

2πα2√

3

√J√

1− J

(aτ + i√−d− 1

)= F 2 1− s2(τ)

6 − J d

dJF 2.

7.1 Examples

From (2.6), we have

d

dJF

(a, b; c; J − 1

J

)2= 2abcJ2F

(a, b; c; J − 1

J

)F

(a+ 1, b+ 1; c+ 1; J − 1

J

).

The values τ =√−N , N = 2, 3, 4, 7, for which J(τ) is rational, satisfy (5.3) and lie in

C(J−1)/J,i. Thus, the above theorem holds for τ =√−N , N = 2, 3, 4, 7. We give all possible

identities.τ =√−2:

5(1 +√

2)(−2 +√

2)√

2√

3√

51344πη(i)4 =

328F

( 112 ,

512; 1

2; 98125

)2

− 3100F

( 112 ,

512; 1

2; 98125

)F

(1312 ,

1712; 3

2; 98125

).

τ =√−3:

5(1 +√

3)(−3 +√

3)√

3√

51584πη(i)4 =

111F

( 112 ,

512; 1

2; 121125

)2

− 1225F

( 112 ,

512; 1

2; 121125

)F

(1312 ,

1712; 3

2; 121125

).

53

τ =√−4:

− 11√

11672πη(i)4 =

563F

( 112 ,

512; 1

2; 3372

113

)2

− 1032113F

( 112 ,

512; 1

2; 3372

113

)F

(1312 ,

1712; 3

2; 3372

113

).

τ =√−7:

85(1 +√

7)(−7 +√

7)√

7√

8524337219πη(i)4 =

8133F

( 112 ,

512; 1

2; 35 · 7 · 192

53173

)2

− 163252173F

( 112 ,

512; 1

2; 35 · 7 · 192

53173

)F

(1312 ,

1712; 3

2; 35 · 7 · 192

53173

).

54

Chapter 8

J = 0 case

We will now derive formulae which directly follow from the hypergeometric series for Ω1

around J = 0. The outline of this chapter is similar to that of the two preceding chapters.

Theorem 8.0.1. Let√z be the principal value of the square root function with the branch

cut along (−∞, 0). If τ is as in (5.3) and lies in CJ/(J−1),ρ, then we have

i

π

[a(τ − ρ)2

2α2√

3dJ1/3

(1− J)1/3 + F 2

τ − ρE4(τ)E6(τ)

]= F 2

[J

6(1− J) + s2(τ)6

]+ J

d

dJF 2,

where F = F (1/12, 7/12; 2/3; J/(J − 1)) and α = iη(ρ)2√3.

Proof. Let F = F (1/12, 7/12; 2/3; J/(J − 1)). Recall from Theorem 5.1.8 that

ω1 = m

τ0(1− J)−1/12F,

where m = 2πα and τ0 = τ − ρ. If J ′ 6= 0, then locally τ ′ = 1/J ′ by the inverse functiontheorem. Thus,

dω1dJ

= (1− J)−1/12m

τ0

dF

dJ+ F

d

dJ

(m

τ0(1− J)−1/12

)= (1− J)−1/12m

τ0

[dF

dJ+( 1

12(1− J) −1τ0J ′

)F

].

Because the zeroes of a non-constant analytic function on C are isolated, the above holdsin a deleted neighbourhood of CJ/(J−1),ρ. Using Theorem 5.2.1 we get

η1 = −2√

3J2/3√J − 1(1− J)1/12

m

τ0

[dF

dJ+( 1

12(1− J) −1τ0J ′

)F

]∆1/12.

55

By virtue of (5.4) we obtain

a√d

= − mF

2πτ0((1− J)∆)1/12

[2√

3J2/3√J − 1(1− J)1/12

m

τ0

[dF

dJ+( 1

12(1− J) −1τ0J ′

)F

]∆1/12

+ mF

τ0((1− J)∆)1/123g32g2

s2(τ)]

or

2πaτ20

m2√d

= − F

(1− J)1/6

[2√

3J2/3√J − 1[dF

dJ+( 1

12(1− J) −1τ0J ′

)F

]+ F

∆1/63g32g2

s2(τ)].

Using (6.1) we get3g32g2

∆−1/6 =√J − 1

2√

3J1/3 ,

which yields

2πaτ20

m2√d

= − F

(1− J)1/6

[2√

3J2/3√J − 1[dF

dJ+( 1

12(1− J) −1τ0J ′

)F

]+√J − 1

2√

3J1/3 s2(τ)F],

or2πaτ2

0m2√

3dJ1/3√J − 1

(1− J)1/6 = −2JF[dF

dJ+( 1

12(1− J) −1τ0J ′

)F

]− s2(τ)

6 F 2.

Therefore,

− 2πaτ20

m2√

3dJ1/3√J − 1

(1− J)1/6 + 2Jτ0J ′

F 2 = F 2[

J

6(1− J) + s2(τ)6

]+ J

d

dJF 2

as2JF dF

dJ= J

d

dJF 2.

Recall from Theorem 2.7.3 thatJ ′

J= −2πiE6(τ)

E4(τ) .

Hence,

− 2πaτ20

m2√

3dJ1/3√J − 1

(1− J)1/6 + i

π

F 2

τ0

E4(τ)E6(τ) = F 2

[J

6(1− J) + s2(τ)6

]+ J

d

dJF 2,

that is,

i

π

[aτ2

02α2√

3dJ1/3

(1− J)1/3 + F 2

τ0

E4(τ)E6(τ)

]= F 2

[J

6(1− J) + s2(τ)6

]+ J

d

dJF 2.

56

Theorem 8.0.2. Let√z be the principal value of the square root function with the branch

cut along (−∞, 0). If τ is as in (5.3) and lies in CJ/(J−1),ρ, then we have

− τ − ρ2πα2

√3

J1/3

(1− J)1/3

(aτ − ρ√−d− 1

)= F 2

[J

6(1− J) + s2(τ)6

]+ J

d

dJF 2,

where F = F (1/12, 7/12; 2/3; J/(J − 1)) and α = iη(ρ)2√3.

Proof. Let F = F (1/12, 7/12; 2/3; J/(J − 1)). Using Proposition 7.0.1 and Theorem 8.0.1we obtain

i

π

[a(τ − ρ)2

2α2√

3dJ1/3

(1− J)1/3 + F 2

τ − ρ2π2

9ω21J1/3

√27√

J − 1

]= F 2

[J

6(1− J) + s2(τ)6

]+ J

d

dJF 2.

We see from Theorem 5.1.8 that

ω21 = 4π2α2

(τ − ρ)2(1− J)1/6F2,

soi

π

(a(τ − ρ)2

2α2√

3d− τ − ρ

6α2√−3)

J1/3

(1− J)1/3 = F 2[

J

6(1− J) + s2(τ)6

]+ J

d

dJF 2,

which simplifies to

− τ − ρ2πα2

√3

J1/3

(1− J)1/3

(aτ − ρ√−d− 1

)= F 2

[J

6(1− J) + s2(τ)6

]+ J

d

dJF 2.

8.1 Examples

From (2.6), we get

d

dJF

(a, b; c; J

J − 1

)2= − 2ab

c(J − 1)2F

(a, b; c; J

J − 1

)F

(a+ 1, b+ 1; c+ 1; J

J − 1

).

The values τ = −1+√−N

2 , N = 7, 11, 19, 27, 43, 67, 163, for which J(τ) is rational,satisfy (5.3) and lie in CJ/(J−1),ρ. So the theorem above holds for τ = −1+

√−N

2 , N =7, 11, 19, 27, 43, 67, 163. We state all possible identities.

57

τ = −1+√−7

2 :

3−√−3

πη(ρ)4 71/6 5756 =

− 40567F

( 112 ,

712; 2

3; 125189

)2

+ 50015309F

( 112 ,

712; 2

3; 125189

)× F

(1312 ,

1912; 5

3; 125189

).

τ = −1+√−11

2 :

3−√−3

πη(ρ)4 111/671/3 4693 =

− 48539F

( 112 ,

712; 2

3; 512539

)2

+ 28841503F

( 112 ,

712; 2

3; 512539

)× F

(1312 ,

1912; 5

3; 512539

).

τ = −1+√−19

2 :

3−√−3

πη(ρ)4 191/6 8513 =

− 1121539F

( 112 ,

712; 2

3; 512513

)2

+ 224789507F

( 112 ,

712; 2

3; 512513

)× F

(1312 ,

1912; 5

3; 512513

).

τ = −1+√−27

2 :

3−√−3

πη(ρ)4 271/62531/3 206831 =

− 392064009F

( 112 ,

712; 2

3; 6400064009

)2

+ 840004097152081F

( 112 ,

712; 2

3; 6400064009

)× F

(1312 ,

1912; 5

3; 6400064009

).

58

τ = −1+√−43

2 :

3−√−3

πη(ρ)4 431/6211/3 20024381 =

− 745601536003F

( 112 ,

712; 2

3; 512000512001

)2

+ 32000112347867429F

( 112 ,

712; 2

3; 512000512000

)× F

(1312 ,

1912; 5

3; 512000512001

).

τ = −1+√−67

2 :

3−√−3

πη(ρ)4 671/62171/3 1760392553 =

− 9937840255552003F

( 112 ,

712; 2

3; 8518400085184001

)2

+ 53240003109848868443429F

( 112 ,

712; 2

3; 8518400085184001

)× F

(1312 ,

1912; 5

3; 8518400085184001

).

τ = −1+√−163

2 :

3−√−3

πη(ρ)4 1631/61858011/3 533600817710201 =

− 11363838226240455794119168003F

( 112 ,

712; 2

3; 151931373056000151931373056001

)2

+ 94957108160009892775193720748560806619429F

( 112 ,

712; 2

3; 151931373056000151931373056001

)× F

(1312 ,

1912; 5

3; 151931373056000151931373056001

).

59

Bibliography

[1] G. E. Andrews, R. Askey, and R. Roy, Special Functions, Cambridge University Press,Cambridge, 1999.

[2] T. Apostol, Modular Functions and Dirichlet Series in Number Theory, 2nd ed.,Springer-Verlag, New York, 1976.

[3] N. Archinard, Exceptional sets of hypergeometric series, J. Number Theory 101 (2003)244-269.

[4] B. C. Berndt, Number Theory in the Spirit of Ramanujan, Amer. Math. Soc., Provi-dence, RI, 2006.

[5] J. M. Borwein and P. B. Borwein, More Ramanujan-type series for 1/π, RamanujanRevisited, G. E. Andrews, R. A. Askey, B. C. Berndt, K. G. Ramanathan, and R. A.Rankin, eds., Academic Press, Boston, 1988, 359-374.

[6] ——, Some observations on computer aided analysis, Notices Amer. Math. Soc. 39(1992) 825-829.

[7] ——, Class number three Ramanujan type series for 1/π, J. Comput. Appl. Math. 46(1993) 281-290.

[8] D. V. Chudnovsky and G. V. Chudnovsky, Approximation and complex multiplicationaccording to Ramanujan, Ramanujan Revisited, G. E. Andrews, R. A. Askey, B. C.Berndt, K. G. Ramanathan, and R. A. Rankin, eds., Academic Press, Boston, 1988,375-472.

[9] ——, Use of computer algebra for Diophantine and differential equations, Computeralgebra, Lecture Notes in Pure and Appl. Math. 113, D. V. Chudnovsky and R. D.Jenks, eds., Dekker, New York, 1989, 1-81.

[10] D. Cox, Primes of the Form x2 + ny2 : Fermat, Class Field Theory, and ComplexMultiplication, 2nd ed., Wiley, New York, 1989.

[11] J.-Cl. Evard and F. Jafari, A Complex Rolle’s Theorem, Amer. Math. Monthly 99(1992) 858-861.

60

[12] H. M. Farkas and I. Kra, Riemann Surfaces, 2nd ed., Springer-Verlag, New York, 1992.

[13] R. Fricke and F. Klein, Vorlesungen über die Theorie der elliptischen Modulfunctionen,Teubner, Leipzig, 1890.

[14] A. G. Greenhill, The Applications of Elliptic Functions, Macmillan and Co., New York,1892.

[15] A. Hatcher, Algebraic Topology, Cambridge University Press, Cambridge, 2002.

[16] E. E. Kummer, Über die hypergeometrische Reihe, J. Reine Angew. Math. 15 (1836)39-83, 127-172.

[17] J. E. Marsden and M. J. Hoffman, Basic Complex Analysis, 2nd ed., W. H. Freemanand Co., New York, 1987.

[18] T. Miyake, Modular Forms, Springer, New York, 2006.

[19] S. Ramanujan, Collected Papers, Cambridge University Press, Cambridge, 1927.

[20] J. H. Silverman, The Arithmetic of Elliptic Curves, Springer-Verlag, New York, 1985.

[21] G. Springer, Introduction to Riemann Surfaces, Addison-Wesley, Reading, MA, 1957.

61