Modular equations of a continued fraction of order six

Yoonjin Lee; Yoon Kyung Park

doi:10.1515/math-2019-0003

Article Open Access

Modular equations of a continued fraction of order six

Yoonjin Lee and Yoon Kyung Park

Published/Copyright: April 9, 2019

Published by

Become an author with De Gruyter Brill

Submit Manuscript Author Information Explore this Subject

$Open Mathematics$

From the journal Open Mathematics Volume 17 Issue 1

Abstract

We study a continued fraction X(τ) of order six by using the modular function theory. We first prove the modularity of X(τ), and then we obtain the modular equation of X(τ) of level n for any positive integer n; this includes the result of Vasuki et al. for n = 2, 3, 5, 7 and 11. As examples, we present the explicit modular equation of level p for all primes p less than 19. We also prove that the ray class field modulo 6 over an imaginary quadratic field K can be obtained by the value X² (τ). Furthermore, we show that the value 1/X(τ) is an algebraic integer, and we present an explicit procedure for evaluating the values of X(τ) for infinitely many τ’s in K.

Keywords: Ramanujan continued fraction; modular function; modular equation; ray class fields

MSC 2010: 11Y65; 11F03; 11R37; 11R04; 14H55

1 Introduction

A continued fraction X(τ) of order six is defined as the quotient of mock theta functions

X(τ):=q14∑n=0∞(−q;q2)nqn2+2n(q4;q4)n/∑n=0∞qn2(q2;q2)n

where τ ∈ ℌ, ℌ is the complex upper half plane, q := e^{2 π i τ} and (a;q)n=∏j=1n(1−aqj−1). This was studied by Vasuki et al.[10], and they expressed X(τ) by an infinite product

X(τ)=q14∏n=1∞(1−q6n−5)(1−q6n−1)(1−q6n−4)(1−q6n−2).

Other types of Ramanujan continued fractions have been studied before. One of these is a continued fraction of order twelve U(τ), defined by

U(τ):=q(1−q)1−q3+q3(1−q2)(1−q4)(1−q3)(1+q6)+q3(1−q8)(1−q10)(1−q3)(1+q12)+⋯.

Mahadeva Naika et al. [8] found the infinite product form of U(τ):

U(τ)=q∏n=1∞(1−q12n−1)(1−q12n−11)(1−q12n−5)(1−q12n−7)

and the modular equations of levels 3 and 5 by theory of hypergeometric series. Moreover, they evaluated U(τ) at τ = i m for several rational numbers m. For examples,

Ui2=63−94−163−94+1,U−32=−108+108234−12+12234−108+108234+12+12234

and

U−52=1265+7215−1623−2794−11265+7215−1623−2794+1.

This follows from the fact that U(τ) is a generator of the function field on Γ₁(12) and a singular value U(τ) is an algebraic unit for any imaginary quadratic quantity τ. This value U(τ) is also contained in some ray class field over an imaginary quadratic field; this is proved by the authors [7].

Vasuki et al. studied X(τ) using the identities of theta functions [10]. They obtained the modular equations of X(τ) of levels 2, 3, 5, 7 and 11 [10, Theorem 2.1-2.5]. However, there is no result on evaluating X(τ) at an imaginary quadratic number τ and generating a class field over an imaginary quadratic field.

The goal of this paper is to study a continued fraction X(τ) of order six by using the modular function theory. We first prove the modularity of X(τ) in Theorem 3.1, and then we find the modular equations of X(τ) of level n for any positive integer n in Theorem 3.2. Table 1 shows the exact modular equations of X(τ) of levels 2, 3, 5, 7, 11, 13, 17 and 19; this result includes the result of Vasuki et al. in [10, Theorems 2.1-2.5] for n = 2, 3, 5, 7 and 11. We show that the ray class field modulo 6 over an imaginary quadratic field K can be obtained by the value X² (τ) (Theorem 4.1 and Corollary 4.5). We also show that the value 1/X(τ) is an algebraic integer contained in a certain number field (Theorem 4.2). Furthermore, we find an explicit relation between X(τ) and C(τ), where C(τ), called Ramanujan’s cubic continued fraction, will be defined in (4.1). We can evaluate the values X(θ) at infinitely many θ’s by using such a relation (Theorem 4.3). Examples of (2) and (3) of Theorem 4.3 are presented in Examples 4.8 and 4.10. We use MAPLE program to find the modular equations and examples.

Table 1

The modular equation F_p(x, y) of X(τ) of levels 2, 3, 5, 7, 11, 13, 17 and 19

p	The modular equation F_p(x, y) of X(τ) of level p
2	(x⁴ − y²)(1 − y²) + 4x⁴y²

3	(x³ − y) − 3xy²(xy − 1)

5	(x⁵ − y)(x + y⁵) + 5 xy(x³y − xy³ − 2x⁴y⁴ + x⁴ + y⁴)

7	(x⁷ − y) (x + y⁷)
7	+7xy(x⁴ + 2 x⁵y + (−x² + 3 x⁶) y² + (1 − 3 x⁴) y⁴ − 2 xy⁵ + (−4 x⁶ + 3 x²) y⁶)

11	(x¹¹ − y) (x − y¹¹)
	+11xy(x⁴ − 2 x⁸ + 6 x⁹y + (−6 x¹⁰ − 15 x⁶ + x²) y² + 45 x⁷y³ + (1 − 45 x⁸ − 28 x⁴) y⁴
	+84x⁵y⁵ + (27 x¹⁰ − 84 x⁶ − 15 x²) y⁶ + 45 x³y⁷ + (−2 + 27 x⁸ − 45 x⁴) y⁸ + 6 xy⁹
	+ (−6 x² − 22 x¹⁰ + 27 x⁶) y¹⁰)

13	(x¹³ − y)(x − y¹³)
	+ 13xy(x⁴ − 3 x⁸ + x¹² + (−x³ − 2 x¹¹ + 10 x⁷) y + (2 x² − 20 x⁶ − 10 x¹⁰) y²
	+ (−x + 10 x⁵ + 68 x⁹) y³ + (−27 x¹² − 13 x⁴ − 115 x⁸ + 1) y⁴ + (10 x³ + 90 x¹¹ + 131 x⁷) y⁵
	+ (-20 x² − 64 x⁶ − 180 x¹⁰) y⁶ + (131 x⁵ + 90 x⁹ + 10X) y⁷ + (−115 x⁴ − 117 x⁸ + 81 x¹² − 3) y⁸
	+ (−81 x¹¹ + 90 x⁷ + 68 x³) y⁹ + (−10 x² + 162 x¹⁰ − 180 x⁶) y¹⁰ + (−2 x + 90 x⁵ − 81 x⁹) y¹¹
	+ (−27 x⁴ + 81 x⁸ + 1 − 56 x¹²) y¹²)

17	(x¹⁷ − y) (x + y¹⁷)
	+ 17 xy(−2 x¹⁶ − 5 x⁸ + x⁴ + 7 x¹² + (16 x¹⁵ + 21 x⁷ − 28 x¹¹ − 2 x³) y
	+ (14 x⁶ + 34 x¹⁰ − 2 x² − 14 x¹⁴) y² + (2 x − 35 x¹³ + 168 x⁹ − 35 x⁵) y³
	+ (160 x⁸ + 30 x¹² − 30 x⁴ + 1 + 63 x¹⁶) y⁴ + (−378 x⁷ + 959 x¹¹ − 252 x¹⁵ + 35 x³) y⁵
	+ (−210 x⁶ + 14 x² + 306 x¹⁰ + 306 x¹⁴) y⁶ + (1512 x¹³ − 1583 x⁹ − 21 x + 378 x⁵) y⁷
	+ (−405 x¹⁶ + 160 x⁴ − 5 − 1190 x⁸ + 1440 x¹²) y⁸ + (1701 x¹⁵ + 1583 x⁷ − 3402 x¹¹ − 168 x³) y⁹
	+ (306 x⁶ + 1134 x¹⁴ + 34 x² − 1890 x¹⁰) y¹⁰ + (3402 x⁹ − 959 x⁵ + 28 x − 2835 x¹³) y¹¹
	+ (1440 x⁸ + 729 x¹⁶ − 2430 x¹² + 30 x⁴ + 7) y¹² + (35 x³ − 1458 x¹⁵ + 2835 x¹¹ − 1512 x⁷) y¹³
	+ (306 x⁶ + 1134 x¹⁰ − 1458 x¹⁴ − 14 x²) y¹⁴ + (252 x⁵ + 1458 x¹³ − 1701 x⁹ − 16x) y¹⁵
	+ (729 x¹² − 386 x¹⁶ − 2+63 x⁴ − 405 x⁸) y¹⁶)

19	(x¹⁹ − y) (x + y¹⁹)
	− 19xy (6 x¹⁶ − x⁴ + 6 x⁸ − 12 x¹² + (5 x⁵ − 52 x⁹ − 26 x¹⁷ + 65 x¹³) y
	+ (65 x¹⁰ + 18 x¹⁸ + 4 x² − 42 x⁶ − 87 x¹⁴) y² + (−135 x¹⁵ − 20 x¹¹ + 2 x⁷) y³
	+ (−261 x¹⁶ − 200 x⁸ + 33 x⁴ − 1+21 x¹²) y⁴ + (520 x⁹ + 585 x¹⁷ − 5 x − 1916 x¹³) y⁵
	+ (591 x⁶ − 1512 x¹⁰ + 63 x¹⁴ − 42 x² − 324 x¹⁸) y⁶ + (−180 x¹⁵ − 1144 x¹¹ − 2 x³) y⁷
	+ (6 + 2091 x⁸ + 1755 x¹⁶ − 4536 x¹² − 200 x⁴) y⁸ + (−520 x⁵ − 4212 x¹⁷ + 52 x + 4680 x¹³) y⁹
	+ (−5400 x¹⁴ + 1458 x¹⁸ + 6273 x¹⁰ − 1512 x⁶ + 65 x²) y¹⁰ + (162 x¹⁵ + 20 x³ + 1144 x⁷) y¹¹
	+ (15957 x¹² − 12 − 10206 x¹⁶ − 4536 x⁸ + 21 x⁴) y¹² + (−4680 x⁹ + 1916 x⁵ − 65 x + 3645 x¹⁷) y¹³
	+ (63 x⁶ − 87 x² − 2187 x¹⁸ − 5400 x¹⁰ + 8019 x¹⁴) y¹⁴ + (−162 x¹¹ + 180 x⁷ + 135 x³) y¹⁵
	+ (−10206 x¹² + 6 + 8748 x¹⁶ − 261 x⁴ + 1755 x⁸) y¹⁶ + (−585 x⁵ + 4212 x⁹ − 3645 x¹³ + 26 x) y¹⁷
	+ (1458 x¹⁰ − 324 x⁶ − 2187 x¹⁴ + 18 x² + 1036 x¹⁸) y¹⁸)

2 Preliminaries

We begin with some definitions and properties regarding the theory of modular functions. Let Γ(1) = SL₂ (ℤ) be the full modular group. For any positive integer N, the congruence subgroups Γ(N), Γ₁ (N),Γ¹(N), Γ₀ (N) and Γ⁰(N) of Γ(1) consist of matrices abcd congruent to 1001,1∗01,10∗1,∗∗0∗ and ∗0∗∗ modulo N, respectively. Let ℌ^∗ = ℌ ∪ ℚ ∪ {∞}, where ℌ is the complex upper half plane {τ ∈ ℂ : Im (τ) > 0}.

A congruence subgroup Γ acts on ℌ^∗ by γ (τ) = (a τ + b) / (c τ + d) for γ ∈ Γ. Then the quotient space Γ ∖ ℌ^∗ is a compact Riemann surface with an appropriate complex structure. One can identify γ with its action on ℌ^∗. We call an element s∈ ℚ ∪ {∞} a cusp. If there exists γ ∈ Γ such that γ(s₁) = s₂, then we say that two cusps s₁ and s₂ are equivalent under Γ. We also call the equivalence class of such s a cusp. We note that there exist at most finitely many inequivalent cusps of Γ. For any cusp s of Γ, there exists an element ρ of SL₂(ℤ) such that ρ(s) = ∞. Let h be the smallest positive integer satisfying

ρ−11h01ρ∈±1⋅Γ, (2.1)

and we call h the width of s. One can easily check that the width h of s depends only on the equivalence class of s under Γ, and it is independent of the choice of ρ.

We call a ℂ-valued function f(τ) a modular function on a congruence subgroup Γ if f(τ) satisfies the following three conditions:

f(τ) is meromorphic on ℌ.
f(τ) is invariant under Γ, in other words, (f ∘ γ)(τ) = f(τ) for all γ ∈ Γ.
f(τ) is meromorphic at all cusps of Γ.

More precisely, the last condition means the following: for a cusp s of Γ, let h be the width of s and ρ be an element of SL₂(ℤ) such that ρ(s) = ∞ by (2.1). We thus have

(f∘ρ−1)(τ+h)=f∘ρ−11h01ρ(ρ−1τ)=f∘ρ−1(τ),

and so f ∘ ρ⁻¹ has a Laurent series expansion in q_h = e^{2 π i τ/h} and (f∘ρ−1)(τ)=∑n≥n0anqhn with a_n₀ ≠ 0 for some integer n₀. We call n₀ the order of f(τ) at the cusp s and write n₀ = ord_s f(τ). We say that f(τ) has a zero (respectively, a pole) at s if ord_sf(τ) is positive (respectively, negative). Moreover, a modular function f(τ) is holomorphic on ℌ^∗ if f(τ) is holomorphic on ℌ and ord_sf(τ) ≥ 0 for all cusp s. We may identify a modular function on Γ with a meromorphic function on the compact Riemann surface Γ∖ℌ^∗. Any holomorphic modular function on Γ is constant.

Let A₀ (Γ) be the field of all modular functions on Γ, and let A₀ (Γ)_ℚ be the subfield of A₀(Γ) consisting of all modular functions f(τ) whose Fourier coefficients belong to ℚ. One can identify A₀(Γ) with the field ℂ (Γ ∖ ℌ^∗) of all meromorphic functions on the compact Riemann surface Γ∖ℌ^∗. Note that if f(τ) ∈ A₀ (Γ) is nonconstant, then the extension degree [A₀ (Γ) : ℂ (f(τ))] is finite and it is the total degree of poles of f(τ). In this paper we consider the modular functions with neither zeros nor poles on ℌ, and so the total degree of poles of f(τ) is − ∑_s ord_sf(τ), where the summation runs over all the inequivalent cusps s at which f(τ) has poles.

The Klein form is a main tool for obtaining our main theorems. The Weierstrass σ-function is defined by

σ(z;L):=z∏ω∈L−01−zωezω+12(zω)2,

where L is a lattice in ℂ and z ∈ ℂ. The Weierstrass ζ-function is also defined by the logarithmic derivative of σ(z; L) as follows:

ζ(z;L):=σ′(z;L)σ(z;L)=1z+∑ω∈L−01z−ω+1ω+zω2.

We notice that σ(z; L) is holomorphic with only simple zeros at every lattice point ω ∈ L and ζ(z; L) is meromorphic with only simple poles at every lattice point ω ∈ L. Moreover, it is easily checked that σ (λ z ; λ L) = λ σ(z; L) and ζ(λ z; λ L) = λ⁻¹ ζ(z; L) for any λ ∈ ℂ^×. The Weierstrass ℘-function ℘(z; L) is defined by

℘(z;L):=−ζ′(z;L)=1z2+∑ω∈L−01(z−ω)2−1ω2.

Since ℘ (z + ω; L) = ℘(z; L) and ddz[ζ(z+ω;L)]=0 for any ω ∈ L, ζ(z + ω; L) – ζ(z; L) depends only on a lattice point ω ∈ L and not on z ∈ ℂ. Hence we may let

η(ω;L):=ζ(z+ω;L)−ζ(z;L)

for ω ∈ L. Fixing the basis {ω₁, ω₂} for L = ℤω₁ + ℤ ω₂, define the Weierstrass η-function by

η(z;L):=a1η(ω1;L)+a2η(ω2;L)

for z = a₁ ω₁ + a₂ ω₂ ∈ ℂ and a₁, a₂ ∈ ℝ. By using the fact that η(z; L) does not depend on the choice of the basis {ω₁, ω₂}, η(z; L) is well-defined and η(rz;L) = r η(z; L) for any r ∈ ℝ.

Now we define the Klein form by

K(z;L):=e−η(z;L)z/2σ(z;L)

and

Ka(τ)=K(a1τ+a2;Zτ+Z),

where 𝔞 = (a₁, a₂) ∈ ℝ². We note that K_𝔞 (τ) is holomorphic and nonvanishing on ℌ if 𝔞 ∈ ℝ² − ℤ². Furthermore, K_𝔞 (τ) is homogeneous of degree 1, that is, K (λ z; λ L) = λ K (z; L).

Let abcd ∈ SL₂ (ℤ) and 𝔞 = (a₁, a₂) ∈ ℝ². The Klein form K_𝔞 satisfies the following properties (K0)–(K5):

K_{− 𝔞} (τ) = − K_𝔞 (τ).
K_𝔞 (γ τ) = (c τ + d)⁻¹ K_{𝔞 γ} (τ).
For 𝔟 = (b₁, b₂) ∈ ℤ², we have that
Ka+b(τ)=ε(a,b)Ka(τ),

where ε (𝔞, 𝔟) = (−1)^{b₁ b₂ + b₁ + b₂} e^{π i (b₂ a₁ – b₁ a₂)}.
For 𝔞 = (r/N, s/N) = (1/N) ℤ² − ℤ² and γ ∈ Γ(N) with an integer N > 1, we obtain that Ka(γτ)=εa(γ)(cτ+d)−1Ka(τ),

where ε_𝔞 (γ) = − (−1)^{((a−1)r + cs + N) (br + (d−1) s + N) / N²} e^{π i (br² + (d−a) rs − cs²)/N²}.
Let τ ∈ ℌ, z = a₁ τ + a₂ with 𝔞 = (a₁, a₂) ∈ ℚ² − ℤ², and let q = e^{2 π i τ} and q_z = e^{2 πiz} = e^2πia₂ e^{2πia₁ τ}. Then
Ka(τ)=−12πieπia2(a1−1)qa1(a1−1)/2(1−qz)∏n=1∞(1−qnqz)(1−qnqz−1)(1−qn)2

and ord_q K_𝔞 (τ) = 〈a₁〉 (〈a₁〉 − 1)/2, where 〈a₁〉 is a rational number such that 0 ≤ 〈a₁〉 < 1 with a₁ − 〈a₁〉 ∈ ℤ.
Let f(τ)=∏aKam(a)(τ) be a finite product of Klein forms with m(𝔞) ∈ ℤ and 𝔞 = (r/N, s/N) = (1/N) ℤ² − ℤ² for an integer N > 1, and let k = −∑_𝔞 m(𝔞). Then f(τ) is a modular form of weight k on Γ(N) if and only if
∑am(a)r2≡∑am(a)s2≡∑am(a)rs≡0(modN) if N is odd,∑am(a)r2≡∑am(a)s2≡0(mod2N),∑am(a)rs≡0(modN) if N is even.

We can refer to [6] for more details on Klein forms.

Consider the modular function on Γ = Γ₀ (N₁) ∩ Γ⁰ (N₂)∩ Γ₁ (N₃) ∩ Γ¹(N₄) ∩ Γ(N₅). Let N = lcm(N₃, N₄, N₅), m = lcm(N₁, N₃, N₅) lcm(N₂, N₄, N₅)/N and β=lcm(N2,N4,N5)001. Then β⁻¹ Γ β = Γ₁ (N) ∩ Γ₀ (mN), which contains 1101. For f ∈ A₀(Γ) and γ ∈ Γ, we have f ∘ β ∈ A₀(β⁻¹ Γ β) = A₀ (Γ₁ (N) ∩ Γ₀(mN)) because

(f∘β)∘(β−1γβ)=f∘γβ=f∘β.

Since the width of ∞ is 1 on Γ₁(N) ∩ Γ₀ (mN), (f ∘ β)(τ) has the Fourier expansion ∑_{n ∈ ℤ} a_n qⁿ. It means that f ∘ β is easier to treat than f.

The last thing that we need to discuss is some information on cusps of congruence subgroup Γ₁ (N) ∩ Γ₀(mN). For positive integers N and m, let Γ = Γ₁(N) ∩ Γ₀(mN). When Γ∖Γ(1)/Γ(1)_∞ = {Γ γ₁ Γ(1)_∞, …, Γγ_g Γ(1)_∞, the set {γ₁ (∞), …, γ_g (∞)} is a set of all inequivalent cusps of Γ such that γ_i (∞) and γ_j(∞) are not equivalent under Γ for any distinct i and j. Consider the set

M:=(c¯,d¯)∈Z/mNZ2:(c¯,d¯)=1¯.

In fact, (c, d) = 1 if and only if (c, d, mN) = 1. Moreover, we define the subgroup Δ of (ℤ/mN ℤ)^×

Δ:=±(1+Nk)¯∈Z/mNZ×:k=0,…,m−1.

If there exists s ∈ Δ and n ∈ ℤ / mNℤ satisfying

c2¯=s¯⋅c1¯ and d2¯=s¯⋅d1¯+n¯⋅c1¯,

we define an equivalence relation ∼ on M by (c₁, d₁) ∼ (c₂, d₂) for (c₁, d₁), (c₂, d₂) ∈ M. Hence, a map ϕ : Γ ∖ Γ(1) / Γ(1)_∞ → M / ∼ by ϕ(Γabcd)=[(c¯,d¯)] is well-defined and bijective. Now we have the following lemma.

Lemma 2.1

Suppose that a, c, a′, c′ ∈ ℤ and (a, c) = (a′, c′) = 1. Consider ± 1 / 0 as ∞. Then with the notation Δ as before, a/c and a′/c′ are equivalent under Γ₁ (N) ∩ Γ₀(mN) if and only if there exist s ∈ Δ ⊂ (ℤ / mN ℤ)^× and n ∈ ℤ such that a′c′≡s¯−1a+ncs¯c(modmN).

For a positive divisor x of mN, let π_x : (ℤ / mN ℤ)^× → (ℤ / xℤ)^× be the natural surjective homomorphism. For a positive divisor c of mN, let sc,1′¯,…,sc,nc′¯∈(Z/(mN/c)Z)× be all the distinct coset representatives of π_mN/c(Δ) in (ℤ / (mN/c)ℤ)^×, where n_c = ϕ(mN/c) / |π_mN/c(Δ)| and ϕ is the Euler’s ϕ-function. Then we take s_c,i ∈ (ℤ / mN ℤ)^× such that πmN/c(sc,i¯)=sc,i′¯ for any i = 1, …, n_c, and let

Sc:=sc,1¯,…,sc,nc¯∈(Z/mNZ)×.

For a positive divisor c of mN, let ac,1′¯,…,ac,mc′¯∈(Z/cZ)× be all distinct coset representatives of π_c (Δ ∩ ker(π_mN/c)) ⊂ (ℤ / c ℤ)^×, where m_c = ϕ(c) / |π_c (Δ ∩ ker(π_mN/c))| = ϕ(c) ⋅ |π_mN/c(Δ)| / |π_c₀(Δ)| and c₀ = mN/(c, mN/c). We take a_c,j ∈ (ℤ / mN ℤ)^× such that πc(ac,j¯)=ac,j′¯ for any j = 1, …, m_c. Moreover, one can choose a representative a_c,j of a_c,j such that 0 < a_c,1, …, a_{c, m_c} < mN and (a_c,j, mN) = 1, and let

Ac:=ac,1,…,ac,mc.

We then have a set of inequivalent cusps by using the sets S_c and A_c for 0 < c ∣ mN.

Lemma 2.2

With the notation as above, let

S:=(c¯⋅sc,i¯,ac,j¯)∈(Z/mNZ)2:0<c∣mN,sc,j¯∈Sc,ac,j∈Ac.

For given (c ⋅ s_c,i, a_c,j) ∈ S, we can take x, y ∈ ℤ such that (x, y) = 1, x = c ⋅ s_c,i and y = a_c,j because (c ⋅ s_c,i, a_c,j, mN) = 1. Then the set of y/x with such x and y is a set of all the inequivalent cusps of Γ₁(N) ∩ Γ₀ (mN) and the number of such cusps is

|S|=∑0<c|mNnc⋅mc=∑0<c|mNϕ(c)ϕ(mN/c)|πc0(Δ)|,

where c₀ = mN/(c,mN/c).

The following lemma gives us the width of each cusp of Γ₁(N) ∩ Γ₀(mN).

Lemma 2.3

Consider ±1/0 as ∞. Let a/c be a cusp of Γ = Γ₁(N) ∩ Γ₀(mN), where a, c ∈ ℤ and (a, c) = 1. Then the width h of a cusp a/c in Γ ∖ ℌ^∗ is given by

h=m/(c2/4,m)ifN=4,(m,2)=1and(c,4)=2,mN/(c,N)⋅(m,c2/(c,N))otherwise.

The proofs of Lemmas 2.1, 2.2 and 2.3 are given in [3, Lemma 1-3].

3 A continued fraction X(τ) of order six

In this section we prove the existence of a modular equation of X(τ) (Theorem 3.2) for any level by showing the modularity of X(τ) (Theorem 3.1). The followings are parts of our main results.

Theorem 3.1

The function field generated by X²(τ) over ℂ is the field of modular functions on Γ₁(6) ∩ Γ⁰(2).
X(τ) is a modular function on 〈Γ1(12)∩Γ0(4),1061,−7−665〉.

Theorem 3.2

For any positive integer n, one can find a modular equation F_n (x, y) of X(τ) of level n.

We note that

X(τ)=ζ245∏j=05K(1/6,j/6)(τ)K(2/6,j/6)(τ)

as the finite product of Klein forms by (K4) with ζ_N = e^{2 πi/N}.

Proof of Theorem 3.1

Note that
C(f(τ))=A0(Γ)⟺C(f(Nτ))=A0N001−1ΓN001,

where Γ is a congruence subgroup of genus zero, f(τ) is a modular function on Γ, and N is a positive integer. It is thus sufficient to prove that
C(X2(2τ))=A0(Γ0(12)). (3.1)

By (K5), X²(τ) is a modular function on Γ(6). Since X² (τ + 2) = X² (τ), X²(τ) is a modular function on 〈Γ(6),1201〉 = Γ₁(6) ∩ Γ⁰(2). Furthermore, X² (2 τ) is a modular function on 2001−1⋅Γ1(6)∩Γ0(2)⋅2001=Γ1(6)∩Γ0(12). In other words, X² (2 τ) ∈ A₀ (Γ₁ (6) ∩ Γ₀(12)) = A₀ (Γ₀ (12)) by using the fact that (Γ₁(6) ∩ Γ₀ (12)) ⋅ {± I} = Γ₀(12).

There are six inequivalent cusps on Γ₀(12), and we observe the behaviour of X² (2 τ) at each cusp from the following table.
cusp s∞01/21/31/41/6ordsX2(2τ)1000−10

We note that the degree d = [A₀ (Γ₀ (12)) : ℂ (X² (2 τ)) ] of the field extensions is the total degree of poles. Let S_Γ₀(12) = {∞, 0, 1/2, 1/3, 1/4, 1/6} be the set of inequivalent cusps on Γ₀(12). Since
d=−∑s∈SΓ0(12)ordsX2(2τ)<0ordsX2(2τ)=−ord1/4X2(2τ)=1,

we conclude that A₀ (Γ₀ (12)) = ℂ (X² (2 τ)).
By (K5), X(τ) is a modular function on Γ(12). From (K1) and (K2), one can check that
X∘1401(τ)=X(τ+4)=X(τ),X∘1061(τ)=X(τ)

and
X∘−7−665(τ)=X(τ).

Hence X(τ) is a modular function on Γ′, where Γ′:=〈Γ(12),1401,1061,−7−665〉=〈Γ1(12)∩Γ0(4),1061,−7−665〉. □

Remark 3.3

Since

X∘20012001−1γ2001=X∘2001

for γ ∈ Γ′, we have that X(2 τ) is invariant under the action of Γ″=2001−1Γ′2001=〈Γ1(12)∩Γ0(2),−7−3125〉. We denote Γ = Γ / {± I} (respectively, Γ = Γ) if – I ∈ Γ (respectively, if − I ∉ Γ) for any Γ ∈ SL₂ (ℤ). Since [Γ₀ (12) : Γ₁(12) ∩ Γ⁰(2)] = 4 and Γ₁ (12) ∩ Γ⁰(2) ⊂ Γ″ ⊂ Γ₀(12), we obtain that [Γ₀(12) : Γ″] = 2.

Assume that there exists a congruence subgroup Γ_X such that A₀(Γ_X) = ℂ (X(τ)), i.e., the genus of the field of modular functions is zero. Then the congruence subgroup ΓX′ corresponding to X(2 τ) is also of genus zero. Moreover, Γ″⊂ΓX′ and [Γ0(12)¯:ΓX′¯]=2 by Theorem 3.1 (1). Unfortunately, it dose not guarantee that the genus of Γ″ is zero. That is the reason why we need to use X²(τ) or X² (2 τ) instead of X(τ).

For convenience, we fix that

f(τ)=1X2(2τ).

The following lemma shows the existence of an affine plain model, so-called modular equation.

Lemma 3.4

Let n be a positive integer. Then we have

Q(f(τ),f(nτ))=A0(Γ0(12n))Q.

Proof

By Theorem 3.1 (2), ℚ(f (τ)) = A₀ (Γ₀ (12))_ℚ. Clearly, f(τ) ∈ A₀ (Γ₀ (12n))_ℚ since Γ₀ (12) ⊃ Γ₀ (12n). For β=n001, we get Γ₀(12) ∩ β⁻¹ Γ₀(12) β = Γ₀ (12n) and f(n τ) = (f ∘ β) (τ) ∈ A₀ (β⁻¹ Γ₀ (12) β) _ℚ ⊂ A₀ (Γ₀ (12n))_ℚ.

It is thus sufficient to show that ℚ (f(τ), f(n τ)) ⊃ A₀ (Γ₀ (12n))_ℚ. We take M_i ∈ Γ₀(12) and write

Γ0(12)=∪iΓ0(12n)⋅Mi (3.2)

as a disjoint union.

We claim that all functions f ∘ β ∘ M_i are distinct. Suppose that

f∘β∘Mi=f∘β∘Mi′

for i and i′ with i ≠ i′. Then f ∘ β M_i′ Mi−1 β⁻¹ = f and β M_i′ Mi−1 β⁻¹ ∈ ℚ^× ⋅ Γ₀(12). We get M_i′ Mi−1 ∈ Γ₀ (12n) because M_i′ Mi−1 ∈ β⁻¹ Γ₀(12) β and M_i, Mi′ ∈ Γ₀ (12); but, this is a contradiction to (3.2). Consequently, f(τ) and (f ∘ β)(τ) = f (n τ) generate A₀ (Γ₀ (12n)). □

By the table in the proof of Theorem 3.1 (2), f(τ) has a simple pole at ∞ and a simple zero at 1/4 as a modular function on Γ₀(12). Next, we investigate the behaviour of f at all cusps s ∈ ℚ ∪ {∞}.

Lemma 3.5

Let a, c, a′, c′ ∈ ℤ and f(τ) = 1 / X² (2 τ). Then we obtain the following assertions:

f(τ) has a pole at a/c ∈ ℚ ∪ {∞} with (a, c) = 1 if and only if (a, c) = 1 and c ≡ 0 (mod 12).
f(n τ) has a pole at a′/ c′ ∈ ℚ ∪ {∞} if and only if there exist a, c ∈ ℤ such that a/c = n a′/c′, (a, c) = 1 and c ≡ 0 (mod 12).
f(τ) has a zero at a/c ∈ ℚ ∪ {∞} if and only if (a, c) = 1 and c ≡ ± 4 (mod 12).
f(n τ) has a pole at a′/ c′ ∈ ℚ ∪ {∞} if and only if there exist a, c ∈ ℤ such that a/c = n a′/c′, (a, c) = 1 and c ≡ ± 4 (mod 12).

Proof

We know that f(τ) has the only simple pole at a/c ∈ ℚ ∪ {∞} if and only if there exists s ∈ Δ = (ℤ / 12 ℤ)^× with (ac)≡(s¯0) (mod 12) by Lemma 2.1. This proves (1) and (2).

To prove (3) and (4), it is enough to find the condition for a/c ∈ ℚ ∪ {∞} to be equivalent to 1/4 under Γ₀(12). By using Lemma 2.1 again, (ac)≡s¯−1+4n4s¯ (mod 12) for s ∈ (ℤ / 12 ℤ)^× and n ∈ ℤ. Hence, c should be congruent to ± 4 (mod 12) and (a, c) = 1. □

We denote d₁ (respectively, d_n) by the total degree of poles of f(τ) (respectively, f(n τ)). Then there exists a polynomial 𝓕_n (x, y) such that

Fn(x,y)=∑0≤i≤dn0≤j≤d1Ci,jxiyj∈Q[X,Y]

and 𝓕_n (f (τ), f(n τ)) = 0. Ishida-Ishii [5] proved the following lemma using the theory of algebraic functions. This is useful when we check which coefficients C_i,j of 𝓕_n (x, y) are zero.

Lemma 3.6

For any congruence subgroup Γ, let f₁ (τ) and f₂ (τ) be nonconstant such that ℂ (f₁ (τ)), f₂ (τ)) = A₀ (Γ) with the total degree D_j of poles of f_j(τ) for j = 1, 2, and let

F(x,y)=∑0≤i≤D20≤j≤D1Ci,jxiyj∈C[x,y]

be such that F(f₁ (τ), f₂ (τ)) = 0. Let S_Γ be a set of all the inequivalent cusps of Γ, and let

Sj,0=s∈SΓ:fj(τ)haszerosats

and

Sj,∞=s∈SΓ:fj(τ)haspolesats

for j = 1, 2. Let

a=−∑s∈S1,∞∩S2,0ordsf1(τ),b=∑s∈S1,0∩S2,0ordsf1(τ).

We assume that a (respectively, b) is 0 if S_{1, ∞} ∩ S_2,0 (respectively, S_1,0 ∩ S_2,0) is empty. Then we obtain the following assertions:

C_{D₂, a} ≠ 0. In addition, if S_{1, ∞} ⊂ S_{2, ∞} ∪ S_2,0, then C_{D₂, j} = 0 for any j ≠ a.
C_{0, b} ≠ 0. In addition, if S_{1, 0} ⊂ S_{2, ∞} ∪ S_2,0, then C_{0, j} = 0 for any j ≠ b.
C_{i, D₁} = 0 for 0 ≤ i < |S_1,0 ∩ S_{2, ∞}|, D₂ − |S_{1, ∞} ∩ S_{2, ∞}| < i ≤ D₂.
C_{i, 0} = 0 for 0 ≤ i < |S_1,0 ∩ S_{2, 0}|, D₂ − |S_{1, ∞} ∩ S_{2, 0}| < i ≤ D₂.

If we interchange the roles of f₁ (τ) and f₂ (τ), then we may have more properties similar to (1)-(4). Suppose that there exist r ∈ ℝ and N, n₁, n₂ ∈ ℤ with N > 0 such that
fj(τ+r)=e2πinj/Nfj(τ)

for j = 1, 2. Then we get the following assertion:
If n₁ i + n₂ j ≢ n₁ D₂ + n₂ a (mod N), then C_{i, j} = 0. Here note that n₂ b ≡ n₁ D₂ + n₂ a (mod N).

Proof

See [5, Lemma 3 and 6]. □

Proof of Theorem 3.2

If ℂ (f₁ (τ), f₂ (τ)) is the field of all modular functions on some congruence subgroup for which f₁ (τ) and f₂ (τ) are nonconstant, then the extension degree [ℂ (f₁ (τ), f₂ (τ)) : ℂ (f_j (τ))] is equal to the total degree d_j of poles of f_j (j = 1,2). Hence we get the polynomial Φ(x, y) ∈ ℂ [x, y] such that Φ(f₁ (τ), y) (respectively, Φ(x, f₂(τ))) is a minimal polynomial of f₂ (τ) (respectively, f₁ (τ)) over ℂ (f₁ (τ)) (respectively, ℂ (f₂ (τ))). Let f₁(τ) = f(τ) and f₂ (τ) = f(n τ). By Lemma 3.6, for any positive integer n we obtain a polynomial 𝓕_n (x, y) ∈ ℚ [x, y] such that 𝓕_n (f(τ), f(n τ)) = 0 with deg_x 𝓕_n (x, y) = d₂ and deg_y 𝓕_n (x, y) = d₁ since the q-expansion of f(τ) has only rational coefficients. Hence we get 𝓕_n (1/X² (τ), 1/X² (n τ)) = 0 by substituting f(τ) = 1/X² (2 τ). Let

F~n(x,y)=x2d2y2d1Fn1x2,1x2

be a polynomial with F͠_n (X(τ), X(n τ)) = 0. By factorizing this, we can choose only one irreducible factor F_n (x, y) ∈ ℤ[x, y] satisfying F_n (X(τ), X(n τ)) = 0. In detail, by replacing x (respectively, y) by the q-expansion of X(τ) (respectively, X(nτ)) in each irreducible factor of F̃_n (x, y), one of the irreducible factors of F̃_n (x, y) can be written as O(q^N) for a suitable integer N; then we denote such an irreducible factor by F_n (x, y). Thus, F_n (x, y) is the modular equation of X(τ) of level n for a positive integer n. □

Using Lemma 3.6, we obtain the modular equations of f(τ) of levels 2 and 3 as follows.

Theorem 3.7

The modular equations of X(τ) of levels 2 and 3 are found as follows:

(Modular equation of level 2)
X4(τ)=X2(2τ)1−X2(2τ)1+3X2(2τ),
(Modular equation of level 3)
X3(τ)−X(3τ)+3X(τ)X2(3τ)−3X2(τ)X3(3τ)=0.

Proof

By Lemma 3.4, ℚ (f (τ), f (2 τ)) = A₀ (Γ₀ (24))_ℚ. We may write the set S_{Γ₀ (24)} of inequivalent cusps as
SΓ0(24)=∞,0,12,13,14,16,18,112.

By Lemma 3.5 and the property (K4) of the Klein form, with considering the width at each cusp,
ord1/4f(τ)=ord1/8f(τ)=1 and ord1/12f(τ)=ord∞f(τ)=−1

and
ord1/8f(2τ)=2 and ord∞f(2τ)=−2.

In other words, both of the total degrees of poles of f(τ) and f(2 τ) are 2, and there exists a polynomial 𝓕₂ (x, y) satisfying 𝓕₂ (f (τ), f (2 τ)) = 0. Let
F2(X,Y)=∑0≤i,j≤2Ci,jxiyj.

Since the set of poles of f(τ) and the set of zeros of f(2 τ) are disjoint, we may assume that C_2,0 = 1. By substituting x = f(τ) and y = f (2 τ) to 𝓕₂ (x, y),
F2(x,y)=x2−x2y+y2+3y

because all coefficients of q-expansion of 𝓕₂ (f (τ), f (2 τ)) should be zero. Consider the polynomial F͠₂ (x, y) := x⁴ y⁴ 𝓕₂ (1/x², 1/y²) = 3 x⁴ y² + x⁴ + y⁴ − y². Since F͠₂ (x, y) is irreducible, we get that
X4(τ)=X2(2τ)(1−X2(2τ))1+3X2(2τ).
Note that
Q(f(τ),f(3τ))=A0(Γ0(36))Q

and
SΓ0(36)=∞,0,12,13,23,14,16,56,19,112,512,118,

where S_Γ₀(36) is the set of all inequivalent cusps of Γ₀ (36). For s ∈ S_Γ₀(36),
ordsf(τ)=−1 if s∈1/12,5/12,∞,≥0otherwise

and
ordsf(3τ)=−3 if s=∞,≥0otherwise.

So we may write the modular equation of f (τ) of level 3 as
F3(x,y)=∑0≤i,j≤3Ci,jxiyj.

Moreover, the only zero of f (3 τ) is 1/12, and there is no cusp s at which both f(τ) and f(3 τ) vanish simultaneously. This means that we may assume that C_0,3 = 1 by Lemma 3.6 (2). By substituting q-expansions of f(τ) and f (3 τ) to X and Y in 𝓕₃ (x, y), respectively, we get
F3(x,y)=y3−9x+6xy2+3x2y−x3y2

and
F~3(x,y):=x6y6×F31x2,1y2

=(x3+y+3xy2+3x2y3)(x3−y+3xy2−3x2x3). (3.3)

We note that x³ + y + 3 x y² + 3 x² y³ = 2 q^3/4 + ⋯ by substituting x = X(τ) and y = X(3 τ), so we take the other factor of (3.3) as the modular equation of X(τ); thus,
F3(x,y)=x3−y+3xy2−3x2y3

and
X3(τ)−X(3τ)+3X(τ)X2(3τ)−3X2(τ)X3(3τ)=0.
□

The following theorem is useful for finding the modular equations of f(τ) and X(τ).

Theorem 3.8

With the notation as above, let p be a prime ≥ 5. Then the modular equation 𝓕_p (x, y) = ∑_{0 ≤ i, j ≤ p+1} C_i,j xⁱ y^j ∈ ℚ[x, y] of f(τ) = 1/X² (2 τ) satisfies the following conditions:

C_{p+1, 0} ≠ 0, C_{0, p+1} ≠ 0,
C_{j, p+1} = 0, C_{p+1, j} = 0 (j = 1, …, p+1),
C_j,0 = 0, C_0,j = 0 (j = 0, …, p).

Proof

Assume that p ≥ 5 is a prime. We may take the set S_{Γ₀ (12p)} of inequivalent cusps on Γ₀(12p) as follows:

SΓ0(12p)=∞,0,12,13,14,16,112,1p,12p,13p,14p,16p.

Consider ∞ and 0 as 1/12p and 1/1 respectively. By Lemma 2.3, the width h_1/c of the cusp 1/c is 12p/ (12p, c²) for a divisor c of 12p. Hence

ord1/12f(τ)=−p,ord∞f(τ)=−1, ord1/12f(pτ)=−1, and ord∞f(pτ)=−p.

So we obtain the modular equation 𝓕_p (x, y) of f(τ) of level p:

Fp(x,y)=∑0≤i,j≤p+1Ci,jxiyj.

Let f₁ (τ) = f(τ) and f₂ (τ) = f(p τ). Using the notation in Lemma 3.6,

a=0 and b=∑s∈S1,0∩S2,0ordf1(τ)=p+1

because S_{1, 0} = S_2,0 = {1/4, 1/4p}. Therefore, we get (1) because

Cp+1,0=Cp+1,a≠0 and C0,p+1=C0,b≠0.

Moreover,

Cp+1,j=0 for j≠0,C0,j=0 for j≠p+1.

By switching the roles of f(τ) and f(p τ), let f₁(τ) = f(p τ), f₂ (τ) = f(τ) and

Fp′(x,y)=∑0≤i,j≤p+1Ci,j′xiyj

satisfying Fp′ (f(p τ), f(τ)) = 0. By using Lemma 3.6 again, we get

Cj,p+1=Cp+1,j′=0 for j≠0,Cj,0=C0,j′=0 for j≠p+1.

□

In [10], Vasuki et al. computed the modular equations of X(τ) of levels 2, 3, 5, 7 and 11, where they used different methods for each level. They did not present the general method for getting the modular equation of higher levels. We can get the modular equation of any integer level. The following table, which can be computed by Theorem 3.2, shows the modular equations of levels 2, 3, 5, 7, 11, 13, 17 and 19.

Since f(τ) is a generator of a field of modular functions on congruence subgroup with genus zero, its modular equation has similar properties to the modular equation of the classical elliptic modular function j(τ). We state those properties in Theorem 3.9 after setting some notation.

Consider Γ := Γ₀(12). For any integer a with (a, 6) = 1, we choose σ_a ∈ Γ(1) such that σa≡a−100a (mod 12). Clearly, σ_a ∈ Γ₀(12). For instance, we may choose σ_a as

σ±1=±1001 and σ±5=±5121229.

For every integer n with (n, 6) = 1, we can write

Γ100nΓ=⋃0<a|n⋃0≤b<n/a(a,b,n/a)=1Γσaab0na

as the disjoint union and |Γ∖Γ(100n)Γ|=n∏p|n(1+1/p) [9, Proposition 3.36].

Define the polynomial

Φn(x,τ):=∏0<a|n∏0≤b<n/a(a,b,n/a)=1x−(f∘αa,b)(τ)

of degree n ∏_p|n (1+1/p) with αa,b:=σaab0n/a. Each coefficient of the Φ_n(x, y) (in x) is an elementary symmetric function of f ∘ α_a,b and invariant under Γ. It means that they are in ℂ (f(τ)) and Φ_n (x, τ) ∈ ℂ (f (τ))[x] because A₀ (Γ) = ℂ (f (τ)). By abuse of notation we may write Φ_n (x, f(τ)) instead of Φ_n (x, τ). By using α_{a, b} = αn,0=σnn001,(f∘αn,0)(τ)=f(nτ). Since f ∘ α_n,0 is a zero of Φ_n (x, f(τ)), Φ_n (f (n τ), f(τ)) = 0. Let f(n τ) = f₁ (τ), f(τ) = f₂ (τ) and the set S_{j, ∞} (respectively, S_{j, 0}) be the set of cusps which are poles (respectively, zeros) of f_j (τ). With the same notation as Lemma 3.6, write −a for the order ord_s f₁ (τ) of f₁ at the cusp s in S_{1, ∞} ∩ S_2,0. When we multiply Φ_n (x, f(τ)) by a suitable power of f(τ), we get a polynomial F_n (x, f(τ)) ∈ ℂ [x, f(τ)] satisfying F_n (f(n τ), f(τ)) = 0. Since f(τ) = 1/X² (2 τ) and (n, 6) = 1, a might be zero and we regard Φ_n (x, f(τ)) as a polynomial of x and f(τ). Hence we have the following theorem.

Theorem 3.9

With the same notation as above, for any positive integer n such that (n, 6) = 1, let f(τ) = 1/X²(2 τ) and Φ_n(x, y) be a polynomial with Φ_n(f(τ), f(n τ)) = 0. Then we have the following assertions:

Φ_n (x, y) ∈ ℤ [x, y] and deg_x Φ_n (x, y) = deg_y Φ_n (x, y) = n ∏_p|n (1+1/p).
Φ_n (x, y) is irreducible both as a polynomial in x over ℂ (y) and as a polynomial in y over ℂ (x).
Φ_n (x, y) = Φ_n (y, x).
If n is not a square, Φ_n (x, x) is a polynomial of degree > 1 whose leading coefficient is ±1.
(Kronecker’s congruence) Let p be an odd prime. Then
Φp(x,y)≡(xp−y)(x−yp)(modpZ[x,y]).

Proof

Since f(τ) = q⁻¹ + 2q + q³ + ⋯, f(τ) has a Fourier expansion

f(τ)=q−1+∑m=1∞cf(m)qm (cf(m)∈Z).

Let ψ_k be the automorphism of ℚ(ζ_n) over ℚ defined by ψ_k (ζ_n) = ζnk for an integer k with (k, n) = 1. Since

f∘ab0n/a(τ)=fa2τ+abn=ζn−abq−a2/n+∑m=1∞cf(m)ζnabmqa2m/n,

ψ_k induces an automorphism ψ̂_k of ℚ(ζ_n) ((q^1/n)) over ℚ(ζ_n) such that

ψ^kf∘ab0n/a(τ)=ζn−abkq−a2/n+∑m=1∞cf(m)ζnabkmqa2m/n.

When we take an integer b′ such that 0 ≤ b′ < n/a and b′ ≡ bk (mod n/a), we get ab′ ≡ abk (mod n) and

ψ^k(f∘αa,b)=ψ^kf∘σaab0n/a=ψ^kf∘ab0n/a=f∘ab′0n/a=f∘σaab′0n/a=f∘αa,b′.

This implies that ψ̂_k(Φ_n(x, f(τ)) = Φ_n(x, f(τ)).

We note that the degree of Φ_n(x, f(τ)) in x is n ∏_p|n (1 + 1/p), and let d := n ∏_p|n (1 + 1/p). By using that (f ∘ α_{1, 0})(τ) = f(τ/n), Φ_n(f(τ/n), f(τ)) = 0 and [ℂ(f(τ/n), f(τ)) : ℂ (f(τ))] ≤ d.

We have Γα_a,b ⊂ Γ 100n Γ, and so there exist γ, γ′, γ_a,b ∈ Γ such that γ 100n γ_a,b = γ′ α_a,b. Consider an embedding ξ_a,b of ℂ(f(τ/n), f(τ)) to the field of all meromorphic functions on ℌ containing ℂ(f(τ/n), f(τ)) over ℂ(f(τ)) defined by

ξa,b(h)=h∘γa,b.

So ξ_a,b(f)(τ) = f(τ) and ξ_a,b(f(τ/n)) = (f ∘ α_a,b)(τ) because

ξa,b(f(τ/n))=ξa,bf∘100n(τ)=f∘100nγa,b(τ)=f∘αa,b(τ).

In other words, α_a,b ≠ α_a′,b′ means that f ∘ α_a,b ≠ f ∘ α_a′,b′; hence, there exist distinct d embeddings ξ_a,b of ℂ(f(τ/n), f(τ)) over ℂ(f(τ)) and [ℂ(f(τ/n), f(τ)) : ℂ(f(τ))] = d. Therefore, Φ_n(x, f(τ)) is irreducible over ℂ(f(τ)).

Let F(x, y) be the polynomial determined in Lemma 3.6. Let f₁(τ) = f(τ), f₂(τ) = f(nτ), d_j be the total degree of f_j(τ) for j = 1, 2 and

a=−∑s∈S1,∞∩S2,0ordsf(τ).

Then the polynomial F(x, y) is written as

F(x,y)=Cd2,axn+∑0≤i≤d20≤j≤d1Ci,jxiyj.

We have the following relation:

Fn(x,f(τ))=Cd2,a⋅f(τ)aΦn(x,f(τ))

because F(x, f(τ)) (respectively, F(f(τ/n), y)) is the minimal polynomial of f(τ/n) (respectively, f(τ)) over ℂ(f(τ)) (respectively, ℂ(f(τ/n)). As we mentioned before Theorem 3.9, a = 0. Moreover, both F(x, y) and Φ_n(x, y) are in ℤ[x, y]; thus (1) and (2) are proved.

(3) Note that (f ∘ α_n,0)(τ) = f(nτ). From that Φ_n(f(nτ), f(τ)) = 0, Φ_n(f(τ), f(τ/n)) = 0 and f(τ/n) is a root of Φ_n(f(τ), x) = 0. Since Φ_n(x, f(τ)) ∈ ℤ[x, f(τ)] and Φ_n(x, f(τ)) is irreducible, we may assume that there is a polynomial G(x, f(τ)) such that

Φn(f(τ),x)=G(x,f(τ))Φn(x,f(τ)).

By exchanging x with f(τ) and multiplying G (x, f(τ)),

Φn(f(τ),x)=G(x,f(τ))Φn(x,f(τ))=G(x,f(τ))G(f(τ),x)Φn(f(τ),x);

so, we get G := G (x, y) = ±1.

We claim that G = 1. If G = –1, then Φ_n(f(τ), x) = –Φ_n(x, f(τ)) = 0. By substituting x = f(τ), we have 2Φ_n(f(τ), f(τ)) = 0, so f(τ) is a root of Φ_n(x, f(τ)) = 0. In other words, x – f(τ) divides Φ_n(x, f(τ)) as a polynomial. However, Φ_n(x, f(τ)) is irreducible over ℂ(f(τ)) by (2), which is a contradiction. Hence, G = 1, and so Φ_n(f(τ), x) = Φ_n(x, f(τ)).

(4) We observe that each factor of Φ_n(f(τ), f(τ)) is

f(τ)−(f∘αa,b)(τ)=q−1−ζn−abq−a2/n+O(q1/n). (3.4)

Assume that n is not a square. Let c_a,b be the coefficient of the lowest term of (3.4). Then c_a,b is 1 or −ζn−ab. Hence, the leading coefficient of Φ_n(f(τ), f(τ)) is

∏0<a|n∏0≤b<n/a(a,b,n/a)=1ca,b,

which is a unit. It is an integer by (1), thus (4) is proved.

(5) Assume that p is an odd prime. We write

g(τ)≡h(τ)(modα)

for g(τ), h(τ) ∈ ℤ[ ζ_p]((q^1/p)) satisfying

g(τ)−h(τ)∈αZ[ζp]((q1/n)).

Consider the Fourier expansion of (f ∘ α_a,b)(τ) (mod 1 – ζ_p as follows:

(f∘α1,b)(τ)=ζp−bq−1/p+∑m=1∞cf(m)ζpbmqm/p≡q−1/p+∑m=1∞cf(m)qm/p(mod1−ζp)=(f∘α1,0)(τ),

(f∘αp,0)(τ)=q−p+∑m=1∞cf(m)qpm≡q−p+∑m=1∞cf(m)pqpm(modp)≡f(τ)p(modp).

On the other hand,

((f∘α1,0)(τ))p=q−1/p+∑m=1∞cf(m)qm/pp ≡q−1+∑m=1∞cf(m)qm(mod1−ζp) =f(τ).

Hence we get the following congruence equation

Φp(x,f(τ))=∏0≤b<px−(f∘α1,b)(τ)x−(f∘αp,0)(τ)≡x−(f∘α1,0)(τ)p(x−f(τ)p)(mod1−ζp)≡xp−(f∘α1,0)(τ)p(x−f(τ)p)(mod1−ζp)≡xp−f(τ)(x−f(τ)p)(mod1−ζp).

This means that

∑iβiXi:=Φp(x,f(τ))−(xp−f(τ))(x−f(τ)p)

belongs to (1 – ζ_p) ℤ[x, f(τ)], where β_i ∈ (1 – ζ_p) ℤ[f(τ)]. However, since we already know that Φ_p(x, y) – (x^p – y)(x – y^p) is contained in ℤ[x, y], we have ∑_i β_ixⁱ ∈ ℤ[x, f(τ)], that is, β_i ∈ ℤ[f(τ)]. Hence, each β_i is a polynomial in f(τ) whose coefficient is divisible by (1 – ζ_p); thus, β_i ∈ pℤ[f(τ)]. Therefore, the modular equation of f(τ) satisfies that

Φp(x,y)−(xp−y)(x−yp)∈pZ[x,y].

□

Remark 3.10

As we mentioned in Remark 3.3, we cannot conclude that X(τ) can generate the field of modular functions for certain congruence subgroup. But, in Table 1, we observe that the modular equations F_n(x, y) of X(τ) satisfy the following congruence:

Fp(x,y)≡(xp−y)(x+yp) if p=5,7,17,19,(xp−y)(x−yp) if p=11,13.

Hence we conjecture that

Fp(x,y)≡(xp−y)(x−yp) if p≡±1(mod12),(xp−y)(x+yp) if p≡±5(mod12).

4 Ray class fields and evaluation of X(τ)

In this section, we prove our remaining results. We first show that X(τ) generates the ray class field modulo 6 over K. We then prove that as a value, 1/X(τ) is integral and X(rτ) can also be expressed in terms of radicals for any positive rational number r if X(τ) can be written in terms of radicals. We also present some examples at the end of this section.

Theorem 4.1

Let K be an imaginary quadratic field with discriminant d_K and τ ∈ K ∩ ℌ be a root of the primitive equation ax² + bx + c = 0 such that b² – 4ac = d_K and (a, 6) = 1, where a, b, c ∈ ℤ. Then K(X²(τ)) is the ray class field modulo 6 over K.

Theorem 4.2

Let K be an imaginary quadratic field and τ ∈ K ∩ ℌ. Then the value 1/X(τ) is an algebraic integer.

Theorem 4.3

Suppose that X(τ) can be evaluated in terms of radicals. Then X(rτ) can also be expressed in terms of radicals for any positive rational number r.

Let K be an imaginary quadratic field and d_K its discriminant. For a positive integer N, denote K_(N) by the ray class field modulo N over K. In this section we first show that X²(τ) generates the ray class field K₍₆₎ modulo 6 over K, where τ ∈ K ∩ ℌ is a root of the primitive equation ax² + bx + c = 0 such that b² – 4ac = d_K.

The following lemma gives us the ray class field generated by X²(τ), and it is used for the proof of Theorem 4.1.

Lemma 4.4

Let K be an imaginary quadratic field with discriminant d_K, and let τ ∈ K ∩ ℌ be a root of the primitive equation ax² + bx + c = 0 in ℤ[x] such that b² – 4ac = d_K. Let Γ′ be any congruence subgroup such that Γ(N) ⊂ Γ′ ⊂ Γ₁(N). Suppose that (N, a) = 1. Then the field generated over K by all the values h(τ) with h ∈ A₀(Γ′)_ℚ is defined and finite at τ; this field is the ray class field modulo N over K.

Proof

See [2, Corollary 5.2].□

Proof of Theorem 4.1

If Γ′ is the congruence subgroup such that ℚ(X²(τ)) = A₀(Γ′)_ℚ, then Γ′ = Γ₁(6) ∩ Γ⁰(2) and Γ(6) ⊂ Γ′ ⊂ Γ₁(6) by Theorem 3.1 (2). For an imaginary quadratic field K with discriminant d_K, consider τ ∈ K ∩ ℌ satisfying aτ² + bτ + c = 0, where b² – 4ac = d_K, (a, 6) = 1 and a, b, c ∈ ℤ. Since X is defined and finite at this τ, K(X²(τ)) is the ray class field modulo 6 over K by Lemma 4.4.□

Corollary 4.5

Let K be an imaginary quadratic field. If ℤ[τ] is the integral closure of ℤ in K, then K (X²(τ)) is the ray class field modulo 6 over K.

Proof

Assume that ℤ[τ] is the ring of integers in K. If aτ² + bτ + c = 0, where a, b, c ∈ ℤ and (a, b, c) = 1, then a should be 1. Hence, K(X²(τ)) is the ray class field modulo 6 over K.□

The Hauptmodul is the normalized generator of a genus zero function field with q-series q⁻¹ + O(q). Now, we focus on proving that 1/X(τ) is an algebraic integer.

Lemma 4.6

The Hauptmodul of A₀(Γ₁(6)) is 1/X⁴(τ) – 4.

Proof

Let g(τ) = 1/X⁴(τ). Since g(τ) is written as the product of Klein forms:

g(τ)=ζ6∏j=05K(2/6,j/6)4K(1/6,j/6)4(τ)=q−1∏n=1∞(1−q6n−4)4(1−q6n−2)4(1−q6n−5)4(1−q6n−1)4,

g(τ) is a modular function on Γ(6) by (K5). Moreover, g(τ + 1) = g(τ); this implies that g(τ) ∈ A₀(Γ₁(6)). Note that there are four inequivalent cusps ∞, 0, 1/2 and 1/3 with widths 1, 6, 3 and 2, respectively. By (K4), the orders of g(τ) at ∞, 0, 1/2 and 1/3 are –1, 0, 1 and 0, respectively. Hence, g(τ) generates the field A₀(Γ₁(6)), and it has q-expansion q⁻¹ + 4 + 6q + 4q² – 3q³ + O(q⁴). It implies that g(τ) – 4 = 1/X⁴(τ) – 4 is the Hauptmodul of Γ₁(6).□

In [4], the cubic continued fraction C(τ) is defined as follows:

C(τ)=q1/31+q+q21+q2+q41+⋯. (4.1)

Then we get the following lemma.

Lemma 4.7

Let X(τ) and C(τ) be defined as before. Then

X4(τ)=C3(τ)1+C3(τ).

Proof

In [4, Lemma 15], the Hauptmodul of Γ₁(6) is 1/C³(τ) – 3. By uniqueness of Hauptmodul, 1/C³(τ) = 1/X⁴(τ) – 1, so the result follows.□

Proof of Theorem 4.2

By [4, Theorem 16], 1/C(τ) is an algebraic integer for τ ∈ K ∩ ℌ. Hence we have that 1/X⁴(τ) and 1/X(τ) are algebraic integers because 1/C³(τ) = 1/X⁴(τ) – 1.□

Proof of Theorem 4.3

For given r ∈ ℚ_{> 0}, we write r as r = a/b for relatively prime integers a and b. For a prime factor p of r, let

τ0=pτ if p∣a,τ/p if p∣b,

and take the modular equation F_p(x, y) of X(τ) of level p in Table 1. Let

P(t):=Fp(X(τ),t) if p∣a,Fp(t,X(τ)) if p∣b.

By solving the equation P(t) = 0, we can get finitely many solutions s₁, …, s_l which are written in terms of radicals. For a sufficiently large N and q₀ = e^2πiτ₀, we check the absolute values

absj:=q01/4∏n=1N(1−q06n−5)(1−q06n−1)(1−q06n−4)(1−q06n−2)−sj

for j = 1, …, l. Then there is only one j′ with extremely small abs_j′, and so we have X(τ₀) = s_j′. We may repeat these steps until we get X(rτ).□

Example 4.8

(Evaluation of X(i)). One can obtain that

X(i)=124−33 (4.2)

by going through the following procedure:

By [1, Theorem 3 (4.1)], C((1 + i)/2) = (1 – 3 )/2.
Since X⁴(τ) = C³(τ)/(1 + C³(τ)), X⁴((1 + i)/2) = (3 – 2 3 )/9.
By comparing the approximation of X((1 + i)/2) with is⋅(3−23)/94 (s = 0, 1, 2, 3), we can see that

X1+i2=−i3−2394=−i3−2343.
Using the fact that the modular equation of level 2 is

x4+3x4y2−y2+y4, (4.3)

four zeros of (4.3) are

s1:=124+33,s2:=s1,s3:=i124−33,s4:=−s3

when x := X((1 + i)/2).
Hence we find X(1 + i) = s₄.
Actually, X(i) should be ϵⁱ ⋅ X(1 + i)(s = 0, 1, 2, 3). By comparing the approximation of X(i) with ϵⁱ ⋅ s₄ again, we get

X(i)=124−33.

In detail, for q₀ = e^–2π, let

absj:=q01/4∏n=1100(1−q06n−5)(1−q06n−1)(1−q06n−4)(1−q06n−2)−ϵjs4.

Then abs₁ = 0.5 ⋅ 10⁻⁹.

Remark 4.9

One can use the value C(i)=63−3−14 for some steps. Then we get the value

X(i)=2742−3−1343+342+3,

which is equal to (4.2).

Example 4.10

Using the same method as Example 4.8, we obtain the following evaluations.

X(−2)=1−33+323−32+33,
X(−5)=3ξ5−13+3ξ5,

where ξ5=−31+815+145−183.

In [1, Theorems 3 and 5], we can find the values

C(−2/2)=(6−2)/2,C((1+−5)/2)=(5−3)(5−3)/2,

and these are useful for finding the values X(−2) and X(−5) .

Acknowledgement

The first-named author is supported by the National Research Foundation of Korea(NRF) grant funded by the Korea government(MEST)(NRF-2017R1A2B2004574). The second-named author is supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (NRF-2017R1D1A1B03029519).

References

[1] Chan H.H., On Ramanujan’s cubic continued fraction, Acta Arith., 1995, 73, no.4, 343—355.10.4064/aa-73-4-343-355Search in Google Scholar

[2] Cho B., Koo J.K., Construction of class fields over imaginary quadratic fields and applications, Quart. J. Math. 2010, 61, 199–126.10.1093/qmath/han035Search in Google Scholar

[3] Cho B., Koo J.K., Park Y.K., Arithmetic of the Ramanujan-Göllnitz-Gordon continued fraction, J. Number Theory, 2009, 129, 922–947.10.1016/j.jnt.2008.09.018Search in Google Scholar

[4] Cho B., Koo J.K., Park Y.K., On Ramanujan’s cubic continued fraction as a modular function, Tohoku Math. J., 2010, 62 (4), 579–603.10.2748/tmj/1294170348Search in Google Scholar

[5] Ishida N., Ishii N., The equations for modular function fields of principal congruence subgroups of prime level, Manuscripta Math., 1996, 90, 271–285.10.1007/BF02568306Search in Google Scholar

[6] Kubert D., Lang S., Modular Units, Springer-Verlag, New York-Berlin, 1981.10.1007/978-1-4757-1741-9Search in Google Scholar

[7] Lee Y., Park Y.K.. A continued fraction of order twelve as a modular function, Math. Comp., 2018, 87, No. 312, 2011–2036.10.1090/mcom/3259Search in Google Scholar

[8] Mahadeva Naika M.S., Dharmendra B.N., Shivashankar K., A continued fraction of order twelve, Cent. Eur. J. Math., 2008, 6, 393–404.10.2478/s11533-008-0031-ySearch in Google Scholar

[9] Shimura G., Introduction to the arithmetic theory of automorphic functions, in: Kanô Memorial Lectures, no. 1, in:Publications of the Mathematical Society of Japan, vol. 11, Iwanami Shoten Publishers/Princeton University Press, Tokyo/Princeton, N. J., 1971, xiv+267.Search in Google Scholar

[10] Vasuki K.R., Bhaskar N., Sharath G., On a continued fraction of order six, Ann. Univ. Ferrara Sez. VII Sci. Mat., 2010, 56, No. 1, 77–89.10.1007/s11565-010-0090-4Search in Google Scholar

Received: 2018-08-15

Accepted: 2019-01-03

Published Online: 2019-04-09

This work is licensed under the Creative Commons Attribution 4.0 Public License.

Articles in the same Issue

https://doi.org/10.1515/math-2019-0003

Keywords for this article

Ramanujan continued fraction; modular function; modular equation; ray class fields

Creative Commons

BY 4.0