Modular equations of a continued fraction of order six

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.


Introduction
A continued fraction X(τ) of order six is de ned as the quotient of mock theta functions X(τ) := q ∞ n= (−q; q )n q n + n (q ; q )n ∞ n= q n (q ; q )n where τ ∈ H, H is the complex upper half plane, q := e πiτ and (a; q)n = n j= ( − aq j− ). This was studied by Vasuki et al. [10], and they expressed X(τ) by an in nite product Other types of Ramanujan continued fractions have been studied before. One of these is a continued fraction of order twelve U(τ), de ned by This follows from the fact that U(τ) is a generator of the function eld on Γ ( ) 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 eld over an imaginary quadratic eld; 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 , , , and [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 eld over an imaginary quadratic eld.
The goal of this paper is to study a continued fraction X(τ) of order six by using the modular function theory. We rst prove the modularity of X(τ) in Theorem 3.1, and then we nd 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 , , , , , , and ; this result includes the result of Vasuki et al. in [10, Theorems 2.1-2.5] for n = , , , and . We show that the ray class eld modulo over an imaginary quadratic eld K can be obtained by the value X (τ) (Theorem 4.1 and Corollary 4.5). We also show that the value /X(τ) is an algebraic integer contained in a certain number eld (Theorem 4.2). Furthermore, we nd an explicit relation between X(τ) and C(τ), where C(τ), called Ramanujan's cubic continued fraction, will be de ned in (4.1). We can evaluate the values X(θ) at in nitely many θ's by using such a relation (Theorem 4.3). Examples of (2) and (3)  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) satis es the following three conditions: 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 SL (Z) such that ρ(s) = ∞ by (2.1). We thus have and so f • ρ − has a Laurent series expansion in q h = e πiτ/h and (f • ρ − )(τ) = n≥n an q n h with an ≠ for some integer n . We call n the order of f (τ) at the cusp s and write n = ords f (τ). We say that f (τ) has a zero (respectively, a pole) at s if ords f (τ) is positive (respectively, negative). Moreover, a modular function f (τ) is holomorphic on H * if f (τ) is holomorphic on H and ords f (τ) ≥ 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 A (Γ) be the eld of all modular functions on Γ, and let A (Γ) Q be the sub eld of A (Γ) consisting of all modular functions f (τ) whose Fourier coe cients belong to Q. One can identify A (Γ) with the eld C(Γ\H * ) of all meromorphic functions on the compact Riemann surface Γ\H * . Note that if f (τ) ∈ A (Γ) is nonconstant, then the extension degree [A (Γ) : C(f (τ))] is nite 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 ords f (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 de ned by where L is a lattice in C and z ∈ C. The Weierstrass ζ -function is also de ned by the logarithmic derivative of σ(z; L) as follows: 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(λz; λL) = λσ(z; L) and ζ (λz; λL) = λ − ζ (z; L) for any λ ∈ C × . The Weierstrass ℘-function ℘(z; L) is de ned by Since ℘(z + ω; L) = ℘(z; L) and d dz [ζ (z + ω; L)] = 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 for z = a ω + a ω ∈ C and a , a ∈ R. By using the fact that η(z; L) does not depend on the choice of the basis {ω , ω }, η(z; L) is well-de ned and η(rz; L) = rη(z; L) for any r ∈ R. and Ka(τ) = K(a τ + a ; Zτ + Z), where a = (a , a ) ∈ R . We note that Ka(τ) is holomorphic and nonvanishing on H if a ∈ R − Z . Furthermore, Ka(τ) is homogeneous of degree , that is, K(λz; λL) = λK(z; L). Let γ = a b c d ∈ SL (Z) and a = (a , a ) ∈ R . The Klein form Ka satis es the following properties (K0)-(K5):  We can refer to [6] for more details on Klein forms. .
Since the width of ∞ is on Γ (N) ∩ Γ (mN), (f • β)(τ) has the Fourier expansion n∈Z an q n . 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 Γ (N) ∩ Γ (mN). If there exists s ∈ ∆ and n ∈ Z/mNZ satisfying is well-de ned and bijective. Now we have the following lemma. For a positive divisor c of mN, let a c, , . . . , a c,mc ∈ (Z/cZ) × be all distinct coset representatives of We take a c,j ∈ (Z/mNZ) × such that πc(a c,j ) = a c,j for any j = , . . . , mc. Moreover, one can choose a representative a c,j of a c,j such that < a c, , . . . , ac,m c < mN and (a c,j , mN) = , and let We then have a set of inequivalent cusps by using the sets Sc and Ac for < c | mN.

Lemma 2.2. With the notation as above, let
For given (c · s c,i , a c,j ) ∈ S, we can take x, y ∈ Z such that (x, y) = , x = c · s c,i and y = a c,j because (c·s c,i , a c,j , mN) = . Then the set of y/x with such x and y is a set of all the inequivalent cusps of Γ (N)∩Γ (mN) and the number of such cusps is The following lemma gives us the width of each cusp of Γ (N) ∩ Γ (mN).
Then the width h of a cusp a/c in Γ\H * is given by

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. 1. The function eld generated by X (τ) over C is the eld of modular functions on Γ ( )∩Γ ( ).

Theorem 3.2. For any positive integer n, one can nd a modular equation Fn(x, y) of X(τ) of level n.
We note that as the nite product of Klein forms by (K4) with ζ N = e πi/N .

Proof of Theorem 3.1. (1)
Note that 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 su cient to prove that There are six inequivalent cusps on Γ ( ), and we observe the behaviour of X ( τ) at each cusp from the following table.
(2) By (K5), X(τ) is a modular function on Γ( ). From (K1) and (K2), one can check that Assume that there exists a congruence subgroup Γ X such that A (Γ X ) = C(X(τ)), i.e., the genus of the eld of modular functions is zero. Then the congruence subgroup Γ X corresponding to X( τ) is also of genus zero. Moreover, Γ ⊂ Γ X and [Γ ( ) : Γ X ] = 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 X (τ) or X ( τ) instead of X(τ).
For convenience, we x that The following lemma shows the existence of an a ne plain model, so-called modular equation.
By the table in the proof of Theorem 3.1 (2), f (τ) has a simple pole at ∞ and a simple zero at / as a modular function on Γ ( ). Next, we investigate the behaviour of f at all cusps s ∈ Q ∪ {∞}. We denote d (respectively, dn) by the total degree of poles of f (τ) (respectively, f (nτ)). Then there exists a polynomial Fn(x, y) such that Fn(x, y) = ≤i≤dn ≤j≤d and Fn(f (τ), f (nτ)) = . Ishida-Ishii [5] proved the following lemma using the theory of algebraic functions. This is useful when we check which coe cients C i,j of Fn(x, y) are zero.
We assume that a (respectively, b) is if S ,∞ ∩ S , (respectively, S , ∩ S , ) is empty. Then we obtain the following assertions: If we interchange the roles of f (τ) and f (τ), then we may have more properties similar to (1)- (4). Suppose that there exist r ∈ R and N, n , n ∈ Z with N > such that f j (τ + r) = e πin j /N f j (τ) for j = , . Then we get the following assertion: 5. If n i + n j ≢ n D + n a (mod N), then C i,j = . Here note that n b ≡ n D + n a (mod N).
Using Lemma 3.6, we obtain the modular equations of f (τ) of levels and as follows.

(Modular equation of level )
Proof. In other words, both of the total degrees of poles of f (τ) and f ( τ) are 2, and there exists a polynomial F (x, y) Since the set of poles of f (τ) and the set of zeros of f ( τ) are disjoint, we may assume that C , = . By F (x, y) = x − x y + y + y because all coe cients of q-expansion of F (f (τ), f ( τ)) should be zero. Consider the polynomial F (x, y) := x y F ( /x , /y ) = x y + x + y − y . Since F (x, y) is irreducible, we get that (2) Note that Moreover, the only zero of f ( τ) is / , and there is no cusp s at which both f (τ) and f ( τ) vanish simultaneously. This means that we may assume that C , = by Lemma 3.6 (2). By substituting q-expansions of f (τ) and f ( τ) to X and Y in F (x, y), respectively, we get We note that x + y + xy + x y = q / + · · · by substituting x = X(τ) and y = X( τ), so we take the other factor of (3.3) as the modular equation of X(τ); thus, The following theorem is useful for nding the modular equations of f (τ) and X(τ). Moreover, C p+ ,j = for j ≠ , C ,j = for j ≠ p + .
In [10], Vasuki et al. computed the modular equations of X(τ) of levels , , , and , where they used di erent 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 , , , , , , and . Since f (τ) is a generator of a eld 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.
By exchanging x with f (τ) and multiplying G(x, f (τ)), This means that Hence, each β i is a polynomial in f (τ) whose coe cient is divisible by ( − ζp); thus, β i ∈ pZ[f (τ)]. Therefore, the modular equation of f (τ) satis es that Remark 3.10. As we mentioned in Remark 3.3, we cannot conclude that X(τ) can generate the eld of modular functions for certain congruence subgroup. But, in Table 1, we observe that the modular equations Fn(x, y) of X(τ) satisfy the following congruence: Hence we conjecture that

Ray class elds and evaluation of X(τ)
In this section, we prove our remaining results. We rst show that X(τ) generates the ray class eld modulo over K. We then prove that as a value, /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. Let K be an imaginary quadratic eld and d K its discriminant. For a positive integer N, denote K (N) by the ray class eld modulo N over K. In this section we rst show that X (τ) generates the ray class eld K ( ) modulo over K, where τ ∈ K ∩ H is a root of the primitive equation ax + bx + c = such that b − ac = d K .
The following lemma gives us the ray class eld generated by X (τ), and it is used for the proof of Theorem 4.1.

Lemma 4.4. Let K be an imaginary quadratic eld with discriminant d K , and let τ ∈ K ∩ H be a root of the primitive equation ax
Let Γ be any congruence subgroup such that Γ(N) ⊂ Γ ⊂ Γ (N). Suppose that (N, a) = . Then the eld generated over K by all the values h(τ) with h ∈ A (Γ ) Q is de ned and nite at τ; this eld is the ray class eld modulo N over K.
Corollary 4.5. Let K be an imaginary quadratic eld. If Z[τ] is the integral closure of Z in K, then K(X (τ)) is the ray class eld modulo over K.
Proof. Assume that Z[τ] is the ring of integers in K. If aτ + bτ + c = , where a, b, c ∈ Z and (a, b, c) = , then a should be . Hence, K(X (τ)) is the ray class eld modulo over K.
In [4], the cubic continued fraction C(τ) is de ned as follows: (4.1) Then we get the following lemma.