On the common zeros of quasi-modular forms for Γ+0(N) of level N = 1, 2, 3

Bo-Hae Im; Hojin Kim; Wonwoong Lee

doi:10.1515/math-2024-0065

Article Open Access

On the common zeros of quasi-modular forms for Γ⁺₀(N) of level N = 1, 2, 3

Bo-Hae Im , Hojin Kim and Wonwoong Lee

Published/Copyright: October 3, 2024

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$Open Mathematics$

From the journal Open Mathematics Volume 22 Issue 1

Abstract

In this article, we study common zeros of the iterated derivatives of the Eisenstein series for Γ 0 + ( N ) of level N = 1 , 2 , and 3 , which are quasi-modular forms. More precisely, we investigate the common zeros of quasi-modular forms and prove that all the zeros of the iterated derivatives of the Eisenstein series θ m E k ( N ) of weight k = 2 , 4 , 6 for Γ 0 + ( N ) of level N = 2 , 3 are simple by generalizing the results of Meher and Gun-Oesterlé for SL 2 ( Z ) .

Keywords: modular forms; quasi-modular forms; Fricke group

MSC 2010: 11F11; 11F99

1 Introduction and the main results

The zeros of certain modular forms for SL 2 ( Z ) have been studied actively. Their study dates back to F. K. C. Rankin and Swinnerton-Dyer’s celebrated proof [1] of Wohlfahrt’s conjecture [2]. They proved that all the zeros of the Eisenstein series E k for k ≥ 4 lie on the unit circle { τ ∈ H : ∣ τ ∣ = 1 } in the standard fundamental domain τ ∈ H : − 1 2 ≤ ℜ τ ≤ 1 2 , ∣ τ ∣ ≥ 1 . Since then, extensive and various studies on the zeros of modular and quasi-modular forms have been conducted in this regard. Getz [3] has generalized F. K. C. Rankin and Swinnerton-Dyer’s result [1] to the modular form of arbitrary weight k ≥ 4 for SL 2 ( Z ) . They proved that if a holomorphic modular form f ( τ ) has a Fourier expansion of the form f ( τ ) = 1 + c q dim S k + 1 + … for some c ∈ C , where q ≔ e 2 π i τ and S k is the space of cusp forms of weight k for SL 2 ( Z ) , then the zeros of f ( τ ) have the same property, namely, all the zeros of f ( τ ) in the standard fundamental domain lie on the unit circle { τ ∈ H : ∣ τ ∣ = 1 } . On the other hand, R. A. Rankin [4] has extended F. K. C. Rankin and Swinnerton-Dyer’s results [1] to the Poincaré series whose order is a rational function with real coefficients. Gun [5] has generalized the argument of R. A. Rankin [4] to prove a similar result of [3] for certain cusp forms. Kohnen [6] gave the explicit formula of zeros of E 2 lying on the unit circle in the right half plane.

Also, for the aspect of our interest, we note that Saber and Sebbar [7] have studied the zeros of θ f of a modular form f for SL 2 ( Z ) , where

θ ≔ 1 2 π i d d τ = q d d q

is a normalized derivation. For an even integer k ≥ 4 , Balasubramanian and Gun [8] have studied the transcendence of the zeros of θ E k ( τ ) , and Miezaki et al. [9] have studied the location of the zeros of the Eisenstein series E k ( N ) for Γ 0 + ( N ) , for N = 2 , 3 .

In [2], together with the famous conjecture resolved in [1], Wohlfahrt has proved that all the zeros of the Eisenstein series E k for weight 4 ≤ k ≤ 26 except for those equivalent to e π i ⁄ 3 are simple. El Basraoui and Sebbar [10] have paid attention to the Eisenstein series E 2 , which is not a modular form. They showed that there are infinitely many SL 2 ( Z ) -inequivalent zeros of E 2 , and they also proved that these zeros are all simple.

On the other hand, Choi and the first author have traced the location of the Eisenstein series of weight 2 for the Fricke group Γ 0 + ( N ) in [11,12]. In particular, they have proved that the Eisenstein series E 2 ( N ) has infinitely many Γ 0 + ( N ) -inequivalent zeros in the strip { τ ∈ H : − 1 2 < ℜ τ ≤ 1 2 } and for N = 2 there is a fundamental domain of Γ 0 + ( N ) in which E 2 ( N ) does not have zeros. They have also proved that, up to Γ 0 + ( N ) -equivalence, there is exactly one zero lying on each of the mixed Ford circles for N = 2 , 3 . Here, a mixed Ford circle for Γ 0 + ( 2 ) or Γ 0 + ( 3 ) is defined as either a circle centered at a c , 1 2 s c 2 with radius 1 2 s c 2 or a circle centered at a c , N 2 s c 2 with radius N 2 s c 2 , where a b c d ∈ Γ 0 + ( N ) and s ≥ 1 ⁄ 2 is a real number (see [12] for details).

Meher [13] has focused on the multiplicity of zeros. More specifically, Meher has studied several properties on the common zeros of quasi-modular forms for SL 2 ( Z ) and then has proved that all the zeros of E 2 , θ E 2 , θ 2 E 2 , and θ 3 E 2 are simple, which is a generalization of the result in [10]. Then, Gun and Oesterlé [14] have generalized Meher’s result. More precisely, they have provided certain conditions which imply that two quasi-modular forms for SL 2 ( Z ) have no common zero and proved that for k ≥ 2 , the iterated derivative θ j E k of the Eisenstein series for an arbitrary order j ≥ 0 has only simple zeros.

In this article, we generalize the result by Gun and Osterlé for the Fricke groups Γ 0 + ( 2 ) and Γ 0 + ( 3 ) and prove the following result.

Theorem 1.1

For N = 2 , 3 , all zeros of θ j E 2 ( N ) , θ j E 4 ( N ) , θ j E 6 ( N ) are simple for all integers j ≥ 0 .

We give the proof of Theorem 1.1 in Section 4 and Appendix A. The idea of the proof relies on the algebraic independence of the values of the Eisenstein series, which is initially proved for the full modular group by Chudnovsky and Nesterenko in [15] and [16], respectively, and for arithmetic Hecke triangle groups by the first and third authors in [17]. Since Γ 0 + ( 2 ) and Γ 0 + ( 3 ) are arithmetic Hecke triangle groups, the problem on the common zeros of quasi-modular forms may be reduced to the problem on certain polynomials and their common divisors. Gun and Oesterlé proved the simplicity of zeros of the iterative derivatives of the Eisenstein series for N = 1 in [14]. However, contrary to the case N = 1 , the Ramanujan identities for N = 2 , 3 (see (2) and (3) in Section 3) cannot be expressed in terms of polynomials, but rather as rational functions in the Eisenstein series. This is a serious obstruction to resolving the problem. In Section 4, we show how to overcome this obstruction by a delicate analysis of the arithmetic nature of coefficients that appear in certain polynomials associated with the iterated derivatives of the Eisenstein series for Γ 0 + ( N ) for N = 2 , 3 . We note that some calculations for Γ 0 + ( 3 ) are postponed to Appendix A, since the main idea is the same as the case for Γ 0 + ( 2 ) , except that the required calculations are more complicated for Γ 0 + ( 3 ) . We also treat the case when N = 1 for the reader’s convenience.

In Section 3, we investigate the common zeros of various quasi-modular forms for Γ 0 + ( N ) of level N = 1 , 2 , 3 . Particularly, we prove that there are no common zeros of E 2 ( N ) and θ m E 2 ( N ) for any positive integer m . This can be shown by two different methods (Propositions 3.5 and 3.10). We also prove that if j N ( α ) is an algebraic number, then α can never be a zero of θ n E 2 ( N ) for n ≥ 0 (see Proposition 3.6). These are the generalizations of all of Meher’s results in [13], and the proofs are similar. We also give various pairs of quasi-modular forms for SL 2 ( Z ) which have no common zero. We show that there is no common zero of any holomorphic modular form and a quasi-modular form of maximal depth, both with algebraic Fourier coefficients (Example 3.15). Also, we deduce that if k = 4 or 6, then there is no common zero of θ m E k and an arbitrary non-zero holomorphic modular form with algebraic Fourier coefficients for SL 2 ( Z ) (Example 3.16).

2 Preliminaries

Let Γ denote a Fuchsian group of the first kind. For a positive integer N , we consider the group Γ 0 + ( N ) generated by the Hecke congruence group Γ 0 ( N ) and the Fricke involution w N ≔ 0 − 1 N N 0 . Also, let H be the complex upper half-plane.

First, we recall the definition of a quasi-modular form for Γ as introduced by Kaneko and Zagier [18].

Definition 2.1

For a positive integer k and a non-negative integer ℓ , a quasi-modular form of weight k and depth ℓ for Γ is a holomorphic function f on H satisfying the following conditions:

There exist holomorphic functions Q i ( f ) for i = 0 , 1 , … , ℓ that satisfy
f [ γ ] k = ∑ i = 0 ℓ Q i ( f ) X ( γ ) i , with Q ℓ ( f ) ≢ 0 , for all γ = a b c d ∈ Γ ,
where the operator [ γ ] k is defined by
f [ γ ] k ( τ ) ≔ ( c τ + d ) − k f ( γ τ ) ,
and the function X ( γ ) is defined by
X ( γ ) ( τ ) ≔ τ c τ + d .
A function f is polynomially bounded, i.e., there exists a constant α > 0 such that f ( τ ) = O ( ( 1 + ∣ τ ∣ 2 ) ⁄ v ) α as v → ∞ and v → 0 , where v = ℑ ( τ ) .

One may define the quasi-modular form using the notion of almost holomorphic modular forms, that is, the constant term F 0 ( τ ) in the variable v of an almost holomorphic modular form F ( τ ) , given by F ( τ ) ≔ ∑ i = 0 l F i ( τ ) ( − 4 π v ) − i . In fact, the notions of quasi-modular forms and almost holomorphic modular forms are equivalent since if f = F 0 is a quasi-modular form inherited from an almost holomorphic modular form F ( τ ) ≔ ∑ i = 0 l F i ( τ ) ( − 4 π v ) − i , then Q i ( f ) = F i (see [19, Section 5.3] for more details).

Note that if f and g are quasi-modular forms of weight k and k ′ and depth ℓ and ℓ ′ , respectively, then f g is also a quasi-modular form of weight k + k ′ and depth ℓ + ℓ ′ . The following proposition shows the additional structure of the space of quasi-modular forms.

Proposition 2.2

[19, Proposition 20] Let ℓ be a non-negative integer and Γ be a non-cocompact Fuchsian group of the first kind. For a non-negative integer k, denote by M k ( Γ ) the space of modular forms of weight k for Γ and by M ˜ k ( ≤ ℓ ) ( Γ ) the space of quasi-modular forms of weight k and depth ≤ ℓ for Γ . Let θ ≔ 1 2 π i d d τ and ϕ be a quasi-modular form of weight 2 for Γ which is not modular. Then, we have the following:

θ ( M ˜ k ( ≤ ℓ ) ( Γ ) ) ⊆ M ˜ k + 2 ( ≤ ℓ + 1 ) ( Γ ) .
M ˜ k ( ≤ ℓ ) ( Γ ) = ⊕ r = 0 ℓ M k − 2 r ( Γ ) ⋅ ϕ r .

Remark 2.3

A quasi-modular form ϕ of weight 2 which is not a modular form must be of depth 1. To see why, we first note that if f ∈ M ˜ k ( ≤ ℓ ) ( Γ ) , then Q 0 ( θ f ) = Q 0 ( θ f ) , Q i ( θ f ) = θ ( Q n ( f ) ) + ( k − i + 1 ) 2 π i Q i − 1 ( f ) for 1 ≤ i ≤ ℓ , and Q ℓ + 1 ( θ f ) = k − ℓ 2 π i Q ℓ ( f ) . Assuming that ϕ has depth ℓ ≥ 3 , applying θ to ϕ yields a non-zero quasi-modular form of weight 4 and depth ℓ + 1 , since Q ℓ + 1 ( θ f ) ≠ 0 . However, every quasi-modular form for Γ is of depth which is divisible by ℓ due to Proposition 2.2 (ii), leading to a contradiction. Therefore, ϕ has depth at most 2. Now, suppose that ϕ is of depth 2. As the weight and depth of ϕ are the same, applying θ to ϕ yields a quasi-modular form of depth at most 2 as well. By Proposition 2.2 (ii), this means that the depth of θ ϕ is either 0 or 2. However, since the weight of θ ϕ is larger than its depth, a quasi-modular form θ 2 ϕ must be of depth 1 or 3, contradicting the fact that every quasi-modular form has even depth in this case.

Let j 2 and j 3 be the Hauptmoduln for Γ 0 + ( 2 ) and Γ 0 + ( 3 ) , respectively, which can be defined by (cf. [20, Table 3])

j 2 ( τ ) = η ( τ ) η ( 2 τ ) 24 + 24 + 2 12 η ( 2 τ ) η ( τ ) 24 , j 3 ( τ ) = η ( τ ) η ( 3 τ ) 12 + 12 + 3 6 η ( 3 τ ) η ( τ ) 12 ,

where η ( τ ) is the Dedekind eta function,

η ( τ ) = q 1 ⁄ 24 ∏ n = 1 ∞ ( 1 − q n ) .

For each positive even integer k , let

E k ( N ) ( τ ) = 1 1 + N k ⁄ 2 ( E k ( τ ) + N k ⁄ 2 E k ( N τ ) ) for N = 1 , 2 , 3 .

We note that E k ( N ) is a modular form of weight k for Γ 0 + ( N ) if k ≥ 4 , and in particular for N = 1 ,

E k ( 1 ) ( τ ) ≔ E k ( τ ) = 1 − 2 k B k ∑ n = 1 ∞ σ k − 1 ( n ) τ n

is the standard Eisenstein series of weight k for SL 2 ( Z ) , where σ k − 1 ( n ) ≔ ∑ d ∣ n d k − 1 and B k is the k th Bernoulli number. One can express j N as a rational function in E 4 ( N ) and E 6 ( N ) (see [20]).

For N = 2 , 3 and a non-negative integer ℓ ∈ Z ≥ 0 , Proposition 2.2 implies that every quasi-modular form f of weight k and depth ≤ ℓ for Γ 0 + ( N ) can be written as

f ( τ ) = f 0 ( τ ) + f 1 ( τ ) E 2 ( N ) ( τ ) + … + f ℓ ( τ ) ( E 2 ( N ) ( τ ) ) ℓ ,

where f i is a modular form of weight k − 2 i for Γ 0 + ( N ) for 0 ≤ i ≤ ℓ .

3 Common zeros of quasi-modular forms for N = 1 , 2 , 3

3.1 Common zeros for N = 2 , 3

In 1916, Ramanujan proved the following Ramanujan identities for SL 2 ( Z ) :

(1) θ E 2 = 1 12 ( E 2 2 − E 4 ) , θ E 4 = 1 3 ( E 2 E 4 − E 6 ) , θ E 6 = 1 2 ( E 2 E 6 − E 4 2 ) .

Zudilin [21] showed the analogues of the Ramanujan identities for Γ 0 + ( 2 ) and Γ 0 + ( 3 ) , that is,

(2) θ E 2 ( 2 ) = 1 8 ( ( E 2 ( 2 ) ) 2 − E 4 ( 2 ) ) , θ E 4 ( 2 ) = 1 2 ( E 2 ( 2 ) E 4 ( 2 ) − E 6 ( 2 ) ) , θ E 6 ( 2 ) = 1 4 3 E 2 ( 2 ) E 6 ( 2 ) − 2 ( E 4 ( 2 ) ) 2 − ( E 6 ( 2 ) ) 2 E 4 ( 2 )

and

(3) θ E 2 ( 3 ) = 1 6 ( ( E 2 ( 3 ) ) 2 − E 4 ( 3 ) ) , θ E 4 ( 3 ) = 2 3 ( E 2 ( 3 ) E 4 ( 3 ) − E 6 ( 3 ) ) , θ E 6 ( 3 ) = 1 2 2 E 2 ( 3 ) E 6 ( 3 ) − ( E 4 ( 3 ) ) 2 − ( E 6 ( 3 ) ) 2 E 4 ( 3 ) .

In this section, for N = 2 , 3 , using the Ramanujan identities (2) and (3) for Γ 0 + ( N ) , we investigate common zeros of quasi-modular forms for Γ 0 + ( N ) . We first establish Proposition 3.3, which let us reduce the question on the common zeros of quasi-modular forms whose Fourier coefficients are algebraic numbers to the question about the greatest common divisor of certain polynomials in Q ¯ [ x , y , z ] , based on the idea of Gun and Oesterlé [14]. Using Proposition 3.3, we investigate common zeros of various quasi-modular forms, as an extension of Meher’s result in [13].

Lemma 3.1

Let N = 1 , 2 , 3 . For any α ∈ H , the ideal

P α ≔ { F ∈ Q ¯ [ x , y , z ] : F ( E 2 ( N ) ( α ) , E 4 ( N ) ( α ) , E 6 ( N ) ( α ) ) = 0 }

is a principal ideal of Q ¯ [ x , y , z ] .

In order to prove Lemma 3.1, we need the following recent result on the algebraic independence for the Eisenstein series, which is a generalization of Nesterenko’s theorem [16] for the Fricke groups.

Theorem 3.2

[17, Theorem 4.6] Let N = 1 , 2 , 3 . Then, for any α ∈ H , at least three of the numbers e 2 π i α , E 2 ( N ) ( α ) , E 4 ( N ) ( α ) , E 6 ( N ) ( α ) are algebraically independent over Q , hence over Q ¯ .

Proof of Lemma 3.1

The idea of the proof follows the proof of [14, Proposition 2], but we give its self-contained proof for convenience. Note that a map

Q ¯ [ x , y , z ] ⟶ Q ¯ [ E 2 ( N ) ( α ) , E 4 ( N ) ( α ) , E 6 ( N ) ( α ) ] ,

which sends x , y , z to E 2 ( N ) ( α ) , E 4 ( N ) ( α ) , E 6 ( N ) ( α ) , respectively, induces an isomorphism

Q ¯ [ x , y , z ] ⁄ P α ≅ Q ¯ [ E 2 ( N ) ( α ) , E 4 ( N ) ( α ) , E 6 ( N ) ( α ) ] .

For a finitely generated Q ¯ -algebra R which is an integral domain, we denote by trdeg Q ¯ R the transcendence degree of a field of fractions Frac ( R ) of R over Q , i.e., the number of elements of a maximal algebraically independent subset of Frac ( R ) over Q ¯ . By Theorem 3.2, we know that

trdeg Q ¯ Q ¯ [ E 2 ( N ) ( α ) , E 4 ( N ) ( α ) , E 6 ( N ) ( α ) ] ≥ 2 ,

so the height ht ( P α ) of a prime ideal P α ([22, Definition in p. 6]) is bounded by

ht ( P α ) ≤ dim Q ¯ [ x , y , z ] − dim Q ¯ [ x , y , z ] ⁄ P α = 3 − trdeg Q ¯ Q ¯ [ E 2 ( N ) ( α ) , E 4 ( N ) ( α ) , E 6 ( N ) ( α ) ] ≤ 1 .

If ht ( P α ) = 0 , then P α = { 0 } , so it is principal. If ht ( P α ) = 1 , then since Q ¯ [ x , y , z ] is a Noetherian unique factorization domain, P α is a principal ideal of Q ¯ [ x , y , z ] .□

We note that if N = 1 , 2 , 3 , every quasi-modular form for Γ 0 + ( N ) can be expressed in terms of rational functions in E 2 ( N ) , E 4 ( N ) , and E 6 ( N ) . Indeed, the algebras of modular forms for Γ 0 + ( 2 ) and Γ 0 + ( 3 ) are given by C [ E 4 ( 2 ) , E 6 ( 2 ) , E 8 ( 2 ) ] and C [ E 4 ( 3 ) , E 6 ( 3 ) , E 8 ( 3 ) , E 10 ( 3 ) , E 12 ( 3 ) ] , respectively ([20, Table 2]; see also [23, Theorem 4 (iv)]). Note that Γ 0 + ( N ) for N = 1 , 2 , 3 are SL 2 ( Z ) -conjugate to the Hecke triangle groups of type ( 2 , 3 , ∞ ) , ( 2 , 4 , ∞ ) and ( 2 , 6 , ∞ ) , respectively ([23, Table 1]). Thus, the algebras of modular forms can also be given by C [ f 4 ( 2 ) , f 6 ( 2 ) , f 8 ( 2 ) ] and C [ f 4 ( 3 ) , f 6 ( 3 ) , f 8 ( 3 ) , f 10 ( 3 ) , f 12 ( 3 ) ] , respectively, where f 2 ℓ ( N ) are certain modular forms (the definitions are detailed in [23, (17)] or [24, (4.5)]). Moreover, it is known that f 4 ( N ) = E 4 ( N ) , f 6 ( N ) = E 6 ( N ) (cf. [23, Theorem 4 (iii)]), and f 2 ℓ ( N ) satisfy the differential equations listed in [24, Table 2]:

ϑ ˜ f 6 ( 2 ) = 1 4 ( − 3 ( f 4 ( 2 ) ) 2 + f 8 ( 2 ) ) , ϑ ˜ f 6 ( 3 ) = 1 2 ( − 2 ( f 4 ( 3 ) ) 2 + f 8 ( 3 ) ) , ϑ ˜ f 8 ( 3 ) = − 1 3 f 10 ( 3 ) , ϑ ˜ f 10 ( 3 ) = 1 6 ( − 4 ( f 4 ( 3 ) ) 3 + 4 ( f 6 ( 3 ) ) 2 + f 12 ( 3 ) ) ,

where ϑ ˜ is defined as in (6). In particular, by comparing these equations to (2) and (3), we observe that every modular form for Γ 0 + ( N ) can be expressed as a rational function in E 4 ( N ) and E 6 ( N ) . Consequently, Proposition 2.2 (ii) implies that quasi-modular forms for Γ 0 + ( N ) can be expressed as rational functions in E 2 ( N ) , E 4 ( N ) , and E 6 ( N ) .

Let

(4) ρ : Q ¯ ( E 2 ( N ) , E 4 ( N ) , E 6 ( N ) ) → Q ¯ ( x , y , z )

be a natural isomorphism which sends E 2 ( N ) , E 4 ( N ) , E 6 ( N ) to x , y , z , respectively. Such an isomorphism exists since the functions E 2 ( N ) , E 4 ( N ) , E 6 ( N ) are algebraically independent over C ( e 2 π i τ ) by [21, Proposition 6]. This isomorphism will be used several times throughout this article as it is useful for studying the common zero, based on the following proposition.

Proposition 3.3

For N = 1 , 2 , 3 , let f and g be quasi-modular forms in Q ¯ ( E 2 ( N ) , E 4 ( N ) , E 6 ( N ) ) which have a common zero in H . Write ρ ( f ) = f 1 f 0 and ρ ( g ) = g 1 g 0 in reduced forms, where f 0 , f 1 , g 0 , g 1 ∈ Q ¯ [ x , y , z ] . Then, f 1 and g 1 have a non-unit common divisor.

Proof

Denote the common zero of f and g by α . It is obvious that f 1 , g 1 ∈ P α , so they have a non-unit common divisor, namely, a generator of P α .□

Corollary 3.4

For N = 1 , 2 , 3 , let f and g be quasi-modular forms with algebraic Fourier coefficients for Γ 0 + ( N ) . Let h ≔ ρ − 1 ( ngcd ( ρ ( f ) , ρ ( g ) ) ) , where ngcd ( F , G ) is the gcd of the numerators of reduced forms of F and G in Q ¯ ( x , y , z ) . Then

Z ( f , g ) = Z ( h ) ,

where Z ( f 1 , f 2 , … , f r ) denotes the common zero set of f 1 , f 2 , … , f r in H .

Since x and c 0 ( y , z ) + c 1 ( y , z ) x + … + c n ( y , z ) x n for some c i ( y , z ) ∈ Q ¯ [ y , z ] have no non-unit common divisor if c 0 ( y , z ) ≠ 0 , we obtain the following proposition.

Proposition 3.5

Let N = 2 or 3. Let f = f 0 + f 1 E 2 ( N ) + … + f n ( E 2 ( N ) ) n be a quasi-modular form of depth n for Γ 0 + ( N ) such that f 0 is not identically zero and f i ∈ Q ¯ ( E 4 ( N ) , E 6 ( N ) ) for 1 ≤ i ≤ n . Then, there is no common zero of f and the Eisenstein series E 2 ( N ) . In particular, there is no common zero of θ m E 2 ( N ) and E 2 ( N ) for any positive integer m.

Proof

Referring to (4) for ρ , since ρ ( E 2 ( N ) ) = x and the numerator of a reduced form of ρ ( f ) = ρ ( f 0 ) + ρ ( f 1 ) x + … + ρ ( f n ) x n has no non-unit common divisor, Proposition 3.3 implies that E 2 ( N ) and f have no common divisor.□

Proposition 3.6

Let N = 2 or 3 and α ∈ H be such that j N ( α ) is an algebraic number. Then, θ n E 2 ( N ) ( α ) is transcendental and hence non-zero for any positive integer n.

Proposition 3.6 follows by using a similar argument as in the proof of [13, Proposition 2.9] along with the fact that j N belongs to Q ¯ ( E 4 ( N ) , E 6 ( N ) ) . Indeed, [20, Section 5.1, 5.2] provides explicit formulas of j 2 and j 3 in terms of E 4 ( 2 ) , E 6 ( 2 ) , E 8 ( 2 ) and E 4 ( 3 ) , E 6 ( 3 ) , E 12 ( 3 ) , respectively. Note that E 8 ( 2 ) and E 12 ( 3 ) can be expressed in terms of rational functions in E 4 ( 2 ) , E 6 ( 2 ) and E 4 ( 3 ) , E 6 ( 3 ) , respectively, because if N = 2 , 3 , then the algebra of modular forms for Γ 0 + ( N ) with algebraic Fourier coefficients is contained in Q ¯ ( E 4 ( N ) , E 6 ( N ) ) .

Let

(5) m = 3 if N = 1 , 4 if N = 2 , 6 if N = 3 , and d = 12 if N = 1 , 8 if N = 2 , 6 if N = 3 .

For N = 1 , 2 , 3 , let f be a modular form of weight k for Γ 0 + ( N ) . Define

(6) ϑ ˜ f ≔ θ f − k d E 2 ( N ) f .

This operator ϑ ˜ sends f to a modular form of weight k + 2 for Γ 0 + ( N ) . This derivation is a kind of the Serre derivative. In general, the Serre derivative of a non-cocompact Fuchsian group Γ is defined as follows: Let ϕ be a quasi-modular form of weight 2 and depth 1 for Γ which is not modular. Recall Q 1 ( ϕ ) given in Definition 2.1. Since ϕ is of depth 1, a function Q 1 ( ϕ ) is a constant, so by multiplying by an appropriate constant, we may assume that Q 1 ( ϕ ) = 1 . Then, the Serre derivative of weight k for Γ is

ϑ ˜ Γ ≔ θ − k 2 π i ϕ .

See [19, p. 48 and p. 62] for more details.

Let Δ N = ( E 4 ( N ) ) 3 − ( E 6 ( N ) ) 2 . For a positive integer r , we denote by C Δ 1 r the one-dimensional C -vector space generated Δ 1 r .

Lemma 3.7

For N = 1 , 2, or 3, let f be a modular form of weight k for Γ 0 + ( N ) , which is expressed as a polynomial in E 4 ( N ) and E 6 ( N ) over C .

If N = 1 , then ϑ ˜ f is identically zero if and only if f ∈ C Δ 1 r for some r ≥ 1 .
If N = 2 or 3, then ϑ ˜ f is not identically zero.

Note that Lemma 3.7 is stated under the assumption that f can be expressed as a polynomial in E 4 ( N ) and E 6 ( N ) . This assumption is necessary because Lemma 3.7 relies on a certain D N ′ -eigenvalue property of a weighted homogeneous polynomial in a polynomial ring as described in the following proposition.

Proposition 3.8

Let N = 2 , 3 . Suppose F ∈ C [ q , x , y , z ] is a nonzero weighted homogeneous polynomial, i.e., the associated one-variable polynomial in t

F ( q , t x , t 2 y , t 3 z )

is a homogeneous polynomial. If

D N ′ F = B F ,

for some B ∈ C [ q , x , y , z ] , then there exist a 2 ′ , a 3 ′ , c ∈ C , and α ∈ Z ≥ 0 such that B and F are given by

B ( q , x , y , z ) = a 2 ′ x y + a 3 ′ z + α y , F ( q , x , y , z ) = c q α Δ a 2 ′ + a 3 ′ y − 2 m − 3 m − 2 a 2 ′ − 3 a 3 ′ .

Proof

The proof is given in [17, p. 875] as part of the proof of [17, Theorem 3.2].□

Proof of Lemma 3.7

(a) Suppose ϑ ˜ f ≡ 0 , i.e., θ f = k 12 E 2 ( 1 ) f , and define a derivation

D ′ = q ∂ ∂ q + 1 12 ( x 2 − y ) ∂ ∂ x + 1 3 ( x y − z ) ∂ ∂ y + 1 2 ( x z − y 2 ) ∂ ∂ z .

By the Ramanujan identity (1), we have

D ′ ρ ( f ) = k 12 x ρ ( f ) .

By the proof of [16, Lemma 4.1], ρ ( f ) is of the form c Δ r y s for r , s ∈ Z ≥ 0 and a constant c ∈ C , where Δ = ρ ( Δ 1 ) = y 3 − z 2 . Moreover, since ρ ( f ) ∈ C [ y , z ] , ρ ( f ) is of the form c Δ r . Conversely, let f = c Δ 1 r for r ≥ 1 . Then, it is easy to see that

θ f = θ ( c Δ 1 r ) = r c E 2 ( 1 ) Δ 1 r = r E 2 ( 1 ) f .

Since Δ 1 has weight 12 for SL 2 ( Z ) , f has weight k = 12 r , so r = k ⁄ 12 . This proves ϑ ˜ f ≡ 0 .

(b) Let m and d be integers as in (5). Suppose ϑ ˜ f ≡ 0 , i.e., θ f = k d E 2 ( N ) f , and let D N ′ be a derivation as defined in Proposition 3.8. By the Ramanujan identities (2) and (3), we have ρ ( E 4 ( N ) θ f ) = D N ′ ρ ( f ) , thus

D N ′ ρ ( f ) = k d x y ρ ( f ) .

By Proposition 3.8, f must be of the form

f = c Δ N k d ( E 4 ( N ) ) − 2 k ( m − 3 ) d ( m − 2 ) for some c ∈ C .

Since ρ ( f ) ∈ C [ y , z ] , we have that k d ∈ Z . Thus, for N = 2 , we have

f = c Δ 2 E 4 ( 2 ) k 8 ,

which is not a holomorphic form. Similarly for N = 3 ,

f = c Δ 3 ( E 4 ( 3 ) ) 3 2 k 6 ,

which is not holomorphic. This contradicts the assumption that f is a modular form of weight k . Hence, ϑ ˜ f is not identically zero.□

Proposition 3.9

For N = 1 , 2, or 3, let f be a modular form of weight k for Γ 0 + ( N ) , which is expressed as a polynomial in E 4 ( N ) and E 6 ( N ) over C . Then,

For N = 1 , then f ′ and E 2 ( N ) = E 2 have infinitely many common SL 2 ( Z ) -inequivalent zeros if and only if f ∈ C Δ 1 r for some r ≥ 1 , and
For N = 2 or 3, then f ′ and E 2 ( N ) have finitely many common Γ 0 + ( N ) -inequivalent zeros.

Proof

(a) Suppose f ′ and E 2 ( N ) have infinitely many common SL 2 ( Z ) -inequivalent zeros. Then, ϑ ˜ f = θ f − k 12 E 2 ( N ) f also has infinitely many SL 2 ( Z ) -inequivalent zeros. Since any non-zero modular form has only finitely many SL 2 ( Z ) -inequivalent zeros, ϑ ˜ f must be 0. By Lemma 3.7, f ∈ C Δ 1 r . Conversely, if f = c Δ 1 r for some c ∈ C , then θ f = r c E 2 ( 1 ) Δ 1 r , so every inequivalent zero of E 2 ( 1 ) is a zero of f ′ , and there are infinitely many of them.

(b) Suppose that f ′ and E 2 ( N ) have infinitely many common Γ 0 + ( N ) -inequivalent zeros. By the same argument as in the proof of (a), θ f must be 0, which is a contradiction to Lemma 3.7 (b).□

Proposition 3.10

For N = 1 , 2 , or 3, let f be a modular form of weight k for Γ 0 + ( N ) , which is expressed as a polynomial in E 4 ( N ) and E 6 ( N ) over Q ¯ .

For N = 1 , then f ′ and E 2 ( N ) = E 2 have no common zero if and only if f ∉ C Δ 1 r for any r ≥ 1 .
For N = 2 or 3, then f ′ and E 2 ( N ) have no common zero.

Proof

(a) Suppose f ∈ C Δ 1 r . Then, by Proposition 3.9 (a), f ′ and E 2 ( 1 ) have infinitely many common zeros. Suppose f ∉ C Δ 1 r . Then, ϑ ˜ f is a non-zero modular form, by Lemma 3.7 (a). Since f ∈ Q ¯ [ E 4 ( 1 ) , E 6 ( 1 ) ] , ϑ ˜ f must be in Q ¯ [ E 4 ( 1 ) , E 6 ( 1 ) ] . Assume that f ′ and E 2 ( 1 ) have a common zero, say α . Then, E 2 ( 1 ) ( α ) = 0 , so E 4 ( 1 ) ( α ) and E 6 ( 1 ) ( α ) must be algebraically independent by Theorem 3.2. However, ϑ ˜ f ( α ) = 0 implies that E 4 ( 1 ) ( α ) and E 6 ( 1 ) ( α ) are algebraically dependent. This is a contradiction; hence, f ′ and E 2 ( 1 ) have no common zero.

(b) Suppose f ′ and E 2 ( N ) have a common zero, say β . Then, β is a zero of non-zero modular form ϑ ˜ f ∈ Q ¯ [ E 4 ( N ) , E 6 ( N ) ] . This yields a contradiction by the same method as in the proof of (a).□

We mention that in [13, Theorem 3.5 (iii)], the condition f ∉ C Δ is supposed to be f ∉ C Δ r for any r ≥ 1 as in Proposition 3.10 (a).

3.2 Common zeros for N = 1

In this subsection, we provide further investigations on quasi-modular forms for SL 2 ( Z ) with algebraic Fourier coefficients which have no common zeros.

For a non-cocompact Fuchsian group Γ , we denote the space of quasi-modular forms of weight k and depth ≤ ℓ for Γ by M ˜ k ( ≤ ℓ ) ( Γ ) . Also, we denote the space of modular forms of weight k for Γ by M k ( Γ ) . In the case when Γ = SL 2 ( Z ) , we omit Γ and write M k ≔ M k ( Γ ) and M ˜ k ( ≤ ℓ ) ≔ M ˜ k ( ≤ ℓ ) ( Γ ) .

Let M Q ¯ ≔ ( ⊕ k ≥ 0 M k ) ∩ ρ − 1 ( Q ¯ [ y , z ] ) be the ring of modular forms for SL 2 ( Z ) with algebraic Fourier coefficients, and let M ˜ Q ¯ ≔ ( ⋃ ℓ ≥ 0 ⊕ k ≥ 0 M ˜ k ( ≤ ℓ ) ) ∩ ρ − 1 ( Q ¯ [ x , y , z ] ) be the ring of quasi-modular forms for SL 2 ( Z ) with algebraic Fourier coefficients.

Proposition 3.11

Let f be a quasi-modular form with algebraic Fourier coefficients. Suppose f = Δ e f 1 e 1 … f r e r where f 1 , … , f r are quasi-modular forms that are irreducible in M ˜ Q ¯ and e , e 1 , … , e r are non-negative integers. Then, the zero set of f is given by

Z ( f ) = Z ( f 1 ) ⊔ … ⊔ Z ( f r )

and if τ ∈ Z ( f i ) , then ord τ ( f ) = e i .

Proof

First, it is obvious that

Z ( f ) = Z ( f 1 ) ⊔ … ⊔ Z ( f r ) ,

so it is enough to prove the remaining part. Since Δ has no zero in H , we may replace f by f ⁄ Δ e so that Δ ∤ f . Let τ be a zero of f i . Since f i ∣ θ j f for j = 0 , 1 , … , e i − 1 , we have ord τ ( f ) ≥ e i . Assume ord τ ( f ) > e i so that θ e i f ( τ ) = 0 . Since f i ( τ ) = 0 , we have

θ e i f ( τ ) = e i ! f 1 ( τ ) e 1 … f i − 1 ( τ ) e i − 1 ( θ f i ) e i f i + 1 ( τ ) e i + 1 … f r ( τ ) e r = 0 .

Note that if f j ( τ ) = 0 for some j ≠ i , then τ is a common zero of f i and f j which implies that ρ ( f i ) and ρ ( f j ) have the common factor in Q ¯ [ x , y , z ] by Proposition 3.3. Thus, we obtain θ f i ( τ ) = 0 , and hence ρ ( f i ) and ρ ( θ f i ) = D ′ ρ ( f ) have the common factor in Q ¯ [ x , y , z ] . As ρ ( f i ) is irreducible, this implies that ρ ( f i ) ∣ D ′ ρ ( f i ) . The irreducibility of f i implies that ρ ( f i ) ∣ D ′ ρ ( f i ) , in particular ( ρ ( f i ) ) is a principal prime ideal in Q ¯ [ q , x , y , z ] which is invariant under the derivation D ′ . By [16, Lemma 4.1], ( ρ ( f i ) ) is either generated by q or y 2 − z , but we have f i ∈ ( Δ ) as ρ ( f i ) ∈ Q ¯ [ x , y , z ] , which contradicts the assumption Δ ∤ f .□

Proposition 3.12

Let f and g be quasi-modular forms for SL 2 ( Z ) , both with algebraic Fourier coefficients. Let F be a quasi-modular form of depth ≥ 1 and write g as

g = g 0 + g 1 F + … + g ℓ F ℓ ∈ Q ¯ [ F , E 4 , E 6 ] ⊆ Q ¯ [ E 2 , E 4 , E 6 ]

for g i ∈ M Q ¯ . If there exists a modular form h ∈ f M ˜ Q ¯ + g M ˜ Q ¯ such that h is relatively prime to g i in the ring of modular forms M Q ¯ for some i ∈ { 0 , 1 , … , ℓ } , then f and g have no common zero.

Proof

As the algebra of quasi-modular forms for SL 2 ( Z ) with algebraic Fourier coefficients is Q ¯ [ E 2 , E 4 , E 6 ] , we use gcd instead of ngcd (which are the same). Let gcd ( ρ ( f ) , ρ ( g ) ) = d so that d ∣ ρ ( h ) . Since h ∈ f M ˜ Q ¯ + g M ˜ Q ¯ and is a modular form, ρ ( h ) belongs to Q ¯ [ y , z ] , so d must as well. Note that ρ ( F ) is transcendental over a polynomial ring Q ¯ [ y , z ] as it is non-constant as a polynomial in x . Thus, from

d ∣ ρ ( g 0 ) + ρ ( g 1 ) ρ ( F ) + … + ρ ( g ℓ ) ρ ( F ) ℓ ∈ R [ ρ ( F ) ]

with R ≔ Q ¯ [ y , z ] , we have d ∣ ρ ( g j ) for any j ∈ { 0 , 1 , … , ℓ } , so d divides both of ρ ( h ) and ρ ( g i ) . Therefore, d ∈ Q ¯ and f and g have no common zero by Proposition 3.3.□

Corollary 3.13

Let f and g be quasi-modular forms for SL 2 ( Z ) , both with algebraic Fourier coefficients. Suppose that f is of depth ≥ 1 and g has an f-expansion, i.e.

g = g 0 + g 1 f + … + g ℓ f ℓ

for some ℓ ≥ 1 and g 0 , g 1 , … , g ℓ ∈ M Q ¯ . If g 0 is relatively prime to g i in M Q ¯ for some i ∈ { 1 , 2 , … , ℓ } , then f and g have no common zero.

Proof

It follows from Proposition 3.12 with h = g 0 = g − ( g 1 + g 2 f + … + g ℓ f ℓ − 1 ) f .□

If we let f = E 2 , then Corollary 3.13 recovers Proposition 3.5 for N = 1 ([13, Proposition 3.2]).

Corollary 3.14

Let f be a modular form of arbitrary weight for SL 2 ( Z ) and g be a quasi-modular form for SL 2 ( Z ) , both with algebraic Fourier coefficients. If there is a quasi-modular form F ∈ M ˜ Q ¯ of depth ≥ 1 such that g has the F-expansion g = g 0 + g 1 F + … g ℓ F ℓ with ℓ ≥ 1 for which g i ’s are coprime in M Q ¯ , then f and g have no common zero.

Proof

By Proposition 3.11, if f and g have a common zero, then there is an irreducible factor f i of f such that f i and g have a common zero. Thus, we may assume f is irreducible in M ˜ Q ¯ .

Note that since ℓ ≥ 1 and g 0 , g 1 , … , g ℓ are coprime in M Q ¯ , f is relatively prime to g i for some 0 ≤ i ≤ ℓ . By applying Proposition 3.12 with h = f , we conclude that f and g have no common zero.□

Example 3.15

There is no common zero of any holomorphic modular form (equivalently, a quasi-modular form of the minimal depth r = 0 ) and a quasi-modular form of the maximal depth (i.e., a quasi-modular form of weight k and depth r = k 2 ) for SL 2 ( Z ) , both with algebraic Fourier coefficients.

Example 3.16

Let k = 2 , 4 , 6 . For each integer n ≥ 1 , there is no common zero of θ n E k and any non-zero holomorphic modular form of arbitrary weight for SL 2 ( Z ) with algebraic Fourier coefficients.

Remark 3.17

Let I ⊆ C [ x , y , z ] be the vanishing ideal of a singleton { ( 1 , 1 , 1 ) } ⊂ A 3 , and let J be a weighted homogeneous ideal contained in I , where a weighted degree in C [ x , y , z ] is given by

wtdeg ≔ deg x + 2 deg y + 3 deg z .

Then, the first elimination ideal J 1 ≔ J ∩ C [ y , z ] of J is contained in ( y 3 − z 2 ) C [ y , z ] . Indeed, if we pick an arbitrary generator F of the ideal J , then f ≔ ρ − 1 ( F ) is a quasi-modular form of weight 2 wtdeg ( F ) and is cuspidal. Since J is weighted homogeneous, so is J 1 , which means that any weighted homogeneous element of J 1 is also cuspidal, i.e., it corresponds to a holomorphic cusp form. Note that any cusp form is written as Δ 1 ⋅ g for some modular form g and the weight 12 cusp form Δ 1 ≔ E 4 3 − E 6 2 . Hence, any generator of an ideal J 1 is divided by the polynomial y 3 − z 2 .

4 Simplicity of zeros of quasi-modular forms: Proof of Theorem 1.1

In this section, we show the simplicity of zeros of quasi-modular forms (Theorem 1.1). Proposition 3.3 provides a useful method to prove the non-existence of common zeros of certain quasi-modular forms. For example, one can easily deduce the following proposition from Proposition 3.3.

Proposition 4.1

For N = 2 , 3 , all the zeros of each of

E 2 ( N ) , θ E 2 ( N ) , θ 2 E 2 ( N ) , θ 3 E 2 ( N )

are simple. Moreover, there are no common zeros of any two of

E 2 ( N ) , θ E 2 ( N ) , θ 2 E 2 ( N ) , θ 3 E 2 ( N ) , θ 4 E 2 ( N ) .

Proof

By the Ramanujan identity (2) for Γ 0 + ( 2 ) , we have

θ E 2 ( 2 ) = 1 8 ( ( E 2 ( 2 ) ) 2 − E 4 ( 2 ) ) , θ 2 E 2 ( 2 ) = 1 32 ( ( E 2 ( 2 ) ) 3 − 3 E 2 ( 2 ) E 4 ( 2 ) + 2 E 6 ( 2 ) ) , θ 3 E 2 ( 2 ) = 1 256 3 ( E 2 ( 2 ) ) 4 − 18 ( E 2 ( 2 ) ) 2 E 4 ( 2 ) + 24 E 2 ( 2 ) E 6 ( 2 ) − 5 ( E 4 ( 2 ) ) 2 − 4 ( E 6 ( 2 ) ) 2 E 4 ( 2 ) , θ 4 E 2 ( 2 ) = 1 512 3 ( E 2 ( 2 ) ) 5 − 30 ( E 2 ( 2 ) ) 3 E 4 ( 2 ) + 60 ( E 2 ( 2 ) ) 2 E 6 ( 2 ) − 25 E 2 ( 2 ) ( E 4 ( 2 ) ) 2 + 12 E 4 ( 2 ) E 6 ( 2 ) − 20 E 2 ( 2 ) ( E 6 ( 2 ) ) 2 E 4 ( 2 ) ,

so the numerators of reduced forms of ρ ( θ i E 2 ( 2 ) ) for i = 0 , 1 , 2 , 3 , 4 are

x , x 2 − y , x 3 − 3 x y + 2 z , 3 x 4 y − 18 x 2 y 2 + 24 x y z − 5 y 3 − 4 z 2 , 3 x 5 y − 30 x 3 y 2 + 60 x 2 y z − 25 x y 3 + 12 y 2 z − 20 x z 2 ,

respectively. The proof for the case when N = 2 follows from the fact that any two numerators listed above

are pairwise coprime. Similarly, for N = 3 , by the Ramanujan identity (3) for Γ 0 + ( 3 ) , we have

θ E 2 ( 3 ) = 1 6 ( ( E 2 ( 3 ) ) 2 − E 4 ( 3 ) ) , θ 2 E 2 ( 3 ) = 1 18 ( ( E 2 ( 3 ) ) 3 − 3 E 2 ( 3 ) E 4 ( 3 ) + 2 E 6 ( 3 ) ) , θ 3 E 2 ( 3 ) = 1 36 ( E 2 ( 3 ) ) 4 − 6 ( E 2 ( 3 ) ) 2 E 4 ( 3 ) − ( E 4 ( 3 ) ) 2 + 8 E 2 ( 3 ) E 6 ( 3 ) − 2 ( E 6 ( 3 ) ) 2 E 4 ( 3 ) , θ 4 E 2 ( 3 ) = 1 54 ( E 2 ( 3 ) ) 5 − 10 ( E 2 ( 3 ) ) 3 E 4 ( 3 ) − 5 E 2 ( 3 ) ( E 4 ( 3 ) ) 2 + 20 ( E 2 ( 3 ) ) 2 E 6 ( 3 ) + 3 E 4 ( 3 ) E 6 ( 3 ) − 10 E 2 ( 3 ) ( E 6 ( 3 ) ) 2 E 4 ( 3 ) + ( E 6 ( 3 ) ) 3 ( E 4 ( 3 ) ) 2 ,

so the same argument completes the proof for N = 3 .□

This method can be adapted to show the simplicity of zeros of all derivatives of quasi-modular forms E 2 ( N ) , E 4 ( N ) , E 6 ( N ) by analyzing the rational functions ρ ( E 2 ( N ) ) , ρ ( E 4 ( N ) ) , ρ ( E 6 ( N ) ) and the derivations D ( N ) in Q ( x , y , z ) corresponding to θ for N = 1 , 2 , 3 .

In [14, Theorem 7], the authors consider the case N = 1 and proved that all the zeros of θ r E k are simple for all integers r ≥ 1 and even integers k ≥ 2 . Their method, which relies on the properties of certain polynomial rings, has some obstruction to be generalized to the cases when N ≥ 2 , since the Ramanujan identities (2) and (3) for N = 2 , 3 are not of polynomial form in terms of E k ( N ) ’s anymore. We conducted a thorough analysis of this issue to extend the simplicity result to cases where N = 2 , 3 . For the readers’ convenience, we convey its complete proof including the case N = 1 .

With the natural isomporphism ρ : Q ( E 2 ( N ) , E 4 ( N ) , E 6 ( N ) ) → Q ( x , y , z ) such that ρ ( E 2 ( N ) ) = x , ρ ( E 4 ( N ) ) = y , ρ ( E 6 ( N ) ) = z given in (4), the derivation θ on the space of quasi-modular forms can be represented as the derivation D ( N ) on Q ( x , y , z ) as follows:

(7) D ( N ) f ≔ x 2 − y 12 ∂ ∂ x f + x y − z 3 ∂ ∂ y f + z x − y 2 2 ∂ ∂ z f if N = 1 , x 2 − y 8 f x + x y − z 2 f y + 3 x z − 2 y 2 − z 2 ⁄ y 4 f z if N = 2 , x 2 − y 6 f x + 2 ( x y − z ) 3 f y + 2 x z − y 2 − z 2 ⁄ y 2 f z if N = 3 , = p ( N ) f x + q ( N ) f y + r ( N ) f z ,

where

p ( 1 ) ≔ x 2 − y 12 , q ( 1 ) ≔ x y − z 3 , r ( 1 ) ≔ z x − y 2 2 , p ( 2 ) ≔ x 2 − y 8 , q ( 2 ) ≔ x y − z 2 , r ( 2 ) ≔ 3 x z − 2 y 2 − z 2 ⁄ y 4 , p ( 3 ) ≔ x 2 − y 6 , q ( 3 ) ≔ 2 ( x y − z ) 3 , r ( 3 ) ≔ 2 x z − y 2 − z 2 ⁄ y 2 ,

referring to (1), (2), and (3). Recalling m and d in (5) depending on N , we note that d = 4 m m − 2 for each N = 1 , 2 , 3 . Then, the above equations can be summarized as

D ( N ) f = 1 d ( x 2 − y ) f x + 4 ( x y − z ) f y + 6 x z − 2 m m − 2 y 2 − 4 ( m − 3 ) m − 2 z 2 ⁄ y f z .

Note that for N = 2 , 3 , y D ( N ) are operators in Q [ x , y , z ] , but D ( N ) are not. Before further discussion on common zeros, we need to proceed with some careful analyses on the degrees of the denominators of ( D ( 2 ) ) n and ( D ( 3 ) ) n at x , y , and z , for n ≥ 1 .

4.1 Degree analysis on ( D ( N ) ) n

This section is dedicated to some necessary analyses of the degrees of ( D ( N ) ) n x , ( D ( N ) ) n y , and ( D ( N ) ) n z for N = 2 , 3 , and especially the degrees of their denominators.

We consider the case when N = 2 in detail here. The results for the case when N = 3 will be stated in this section, and their proofs are given in Appendix A.

We clear denominators to obtain an operator with integral coefficients, namely

D ( 2 ) ≔ d D ( 2 ) = 8 D ( 2 ) = ( x 2 − y ) ∂ ∂ x + ( 4 x y − 4 z ) ∂ ∂ y + ( 6 x z − 4 y 2 − 2 y − 1 z 2 ) ∂ ∂ z .

Let deg x , deg y , deg z be the degrees of monomials in Q [ x , y , z , 1 ⁄ y ] with respect to x , y , and z , respectively. Also, let

wtdeg = deg x + 2 deg y + 3 deg z

be a weighted degree. Note that all monomials in ( D ( N ) ) n x , ( D ( N ) ) n y , ( D ( N ) ) n z have the same wtdeg , so wtdeg can be extended to them, for n ≥ 0 . Also, we note that wtdeg ( D ( 2 ) f ) = wtdeg ( f ) + 1 for a weighted homogeneous f ∈ Q [ x , y , z , 1 ⁄ y ] .

Since

D ( 2 ) ( x a y b z c ) = ( a + 4 b + 6 c ) x a + 1 y b z c − ( 4 b + 2 c ) x a y b − 1 z c + 1 − 4 c x a y b + 2 z c − 1 − a x a − 1 y b + 1 z c ,

when we represent (some coefficient) ⋅ x a y b z c as a point ( a , b , c ) T ∈ Z 3 (here, T stands for the transpose) D ( 2 ) transfers ( a , b , c ) T to (at most) four points as

a b c ⟼ D ( 2 ) a b c + 0 1 ⁄ 2 0 + ± 0 − 3 ⁄ 2 1 , ± 1 − 1 ⁄ 2 0 .

We introduce the matrix

(8) T = 1 14 13 − 2 − 3 3 6 − 5 1 2 3 with its inverse T − 1 = 1 0 1 − 1 2 3 2 2 0 − 1 3 ,

which is determined to have the following relation and transforms the above four candidate points into the unit vectors on a plane:

T a b c ⟼ T D ( 2 ) T − 1 T a b c + T 0 1 ⁄ 2 0 + ± 0 1 0 , ± 1 0 0 ,

and, by this way, we can represent the monomial x a y b z c in ( D ( 2 ) ) r ( x ) as ( λ , ν ) ∈ Z 2 with relation

(9) ( a , b , c ) T = ( 1 , r ⁄ 2 , 0 ) T + T − 1 ( λ , ν , 0 ) T ,

i.e., ( a , b , c ) = ( λ + 1 , ( − λ + 3 ν + r ) ⁄ 2 , − ν ) . After applying ( D ( 2 ) ) r , the monomials of ( D ( 2 ) ) r x are represented as elements in

{ ( 1 , r ⁄ 2 , 0 ) T + T − 1 ( λ , ν , 0 ) T : ∣ λ ∣ + ∣ ν ∣ ≤ r , λ + ν ≡ r ( mod 2 ) } .

This set is not necessarily the whole set of all monomials in ( D ( 2 ) ) r x ; for example, there are some obvious restrictions like the non-negativity of deg x and deg z , which means λ ≥ − 1 and ν ≤ 0 . We can prove another condition and find the formula for the minimal deg y of monomials of ( D ( 2 ) ) r x .

Let x a y b z c be a monomial appearing in ( D ( 2 ) ) r x . If D ( 2 ) ( x a y b z c ) contains a monomial with deg y less than b , then the monomial with decreased deg y is − ( 4 b + 2 c ) x a y b − 1 z c + 1 . This term is non-zero only when c ≠ − 2 b . When c = − 2 b , since a + 2 b + 3 c = wtdeg ( ( D ( 2 ) ) r x ) = r + 1 , we have a = r + 1 − 2 b − 3 c = r + 1 − 4 c . Referring to (9), the corresponding condition for ( λ , ν ) is that ν = 1 2 λ − r 2 , i.e., only the monomials (represented as a lattice point ( λ , ν ) ) off this line can provide the monomials whose deg y drops by 1 after applying D ( 2 ) .

Figure 1 shows the situation for ( D ( 2 ) ) 10 x . Each point ( λ , ν ) on the figure represents each monomial x a y b z c appearing in ( D ( 2 ) ) 10 x by the relation given in (9). The gray-filled point at ( λ , ν ) = ( − 1 , − 5 ) (which represents y − 2 z 5 -term) is off the dashed line ν = 1 2 λ − r 2 . So, D ( 2 ) ( y − 2 z 5 ) provides a function that contains a monomial whose deg y is − 3 ; one can check that D ( 2 ) ( y − 2 z 5 ) = 22 x z 5 y − 2 − 2 z 6 y − 3 − 20 z 4 .

$Figure 1 Monomials in ( D ( 2 ) ) 10 ( x ) {({{\mathcal{D}}}^{\left(2)})}^{10}\left(x) , represented as lattice points ( λ , ν ) \left(\lambda ,\nu ) . Note that two bold axes intersect at the origin ( 0 , 0 ) \left(0,0) , and the gray-filled point at ( ‒ 1 , ‒ 5 ) \left(‒1,‒5) represents y ‒ 2 z 5 {y}^{‒2}{z}^{5} .$

Figure 1

Monomials in ( D ( 2 ) ) 10 ( x ) , represented as lattice points ( λ , ν ) . Note that two bold axes intersect at the origin ( 0 , 0 ) , and the gray-filled point at ( ‒ 1 , ‒ 5 ) represents y ‒ 2 z 5 .

From the discussion so far, we conclude the following lemma.

Lemma 4.2

Let x a y b z c be some monomial in ( D ( 2 ) ) r x (up to nonzero coefficients over Q ).

Its corresponding lattice point ( λ , ν ) via (9) satisfies that ν ≥ 1 2 ( λ − r ) .
2 b + c ≥ 0 and b ≥ − r + 1 4 .

Proof

(a) This holds for the initial case r = 0 . As r increases, all points continue to lie above or on the line ν = 1 2 ( λ − r ) , since the line itself is shifted by − 1 ⁄ 2 in the y-axis direction on each step, and the point on the line does not provide a monomial with decreased deg y , i.e., decreased ν coordinate.

(b) Referring to (9), since a + 2 b + 3 c = r + 1 , the inequality ν ≥ λ 2 − r 2 is equivalent to 2 b + c ≥ 0 , which implies the second inequality since a ≥ 0 and a + 2 b + 3 c = r + 1 .□

In the remaining section, we prove that the second inequality of Lemma 4.2 (b) is sharp, in the sense that when r ≡ 3 ( mod 4 ) , there exists a monomial with deg y = − r + 1 4 in ( D ( 2 ) ) r ( x ) .

To prove this, we need to show that the coefficient of the y − k z 2 k + 1 -term in ( D ( 2 ) ) 4 k + 2 is non-zero for all integers k ≥ 0 . It can be shown by proving that their coefficients are all 1 modulo 5.

Let a i , j , k ( r ) denote the coefficient of the x i y j z k -term in ( D ( 2 ) ) r x . By keeping track of the recurrences for the coefficients, we reduce the degrees as follows:

a 0 , − k , 2 k + 1 ( 4 k + 2 ) = ( − 4 ( − k + 1 ) − 2 ( 2 k ) ) a 0 , − k + 1 , 2 k ( 4 k + 1 ) = − 4 a 0 , − k + 1 , 2 k ( 4 k + 1 ) ≡ a 0 , − k , 2 k ( 4 k + 1 ) ( mod 5 ) = ( − 4 ( − k + 2 ) − 2 ( 2 k − 1 ) ) a 0 , − k + 2 , 2 k − 1 ( 4 k ) − a 1 , − k , 2 k ( 4 k ) = − 6 a 0 , − k + 2 , 2 k − 1 ( 4 k ) − a 1 , − k , 2 k ( 4 k ) ≡ − ( a 0 , − k + 2 , 2 k − 1 ( 4 k ) + a 1 , − k , 2 k ( 4 k ) ) ( mod 5 ) .

Since

a 0 , − k + 2 , 2 k − 1 ( 4 k ) = − 8 a 0 , − k + 3 , 2 k − 2 ( 4 k − 1 ) − a 1 , − k + 1 , 2 k − 1 ( 4 k − 1 ) − 8 r a 0 , − k , 2 k ( 4 k − 1 ) , a 1 , − k , 2 k ( 4 k ) = − 2 a 1 , − k + 1 , 2 k − 1 ( 4 k − 1 ) + 8 r a 0 , − k , 2 k ( 4 k − 1 ) ,

we have

a 0 , − k , 2 k + 1 ( 4 k + 2 ) ≡ 3 ( a 1 , − k + 1 , 2 k − 1 ( 4 k − 1 ) + a 0 , − k + 3 , 2 k − 2 ( 4 k − 1 ) ) ( mod 5 ) .

Similarly, since

a 0 , − k + 3 , 2 k − 2 ( 4 k − 1 ) = ( − 8 k + 4 ) a 0 , − k + 1 , 2 k − 1 ( 4 k − 2 ) − 10 a 0 , − k + 4 , 2 k − 3 ( 4 k − 2 ) − a 1 , − k + 2 , 2 k − 2 ( 4 k − 2 ) , a 1 , − k + 1 , 2 k − 1 ( 4 k − 1 ) = − 2 a 1 , − k + 2 , 2 k − 2 ( 4 k − 2 ) + ( 8 k − 2 ) a 0 , − k + 1 , 2 k − 1 ( 4 k − 2 ) ,

we have

a 0 , − k , 2 k + 1 ( 4 k + 2 ) ≡ a 0 , − k + 1 , 2 k − 1 ( 4 k − 2 ) ( mod 5 ) .

Since a 0 , 0 , 1 ( 2 ) ≢ 0 ( mod 5 ) , the coefficient a 0 , − k , 2 k + 1 ( 4 k + 2 ) never vanishes.

We conclude that y ℓ r ⋅ ( D ( 2 ) ) r x ∈ Q [ x , y , z ] \ y Q [ x , y , z ] , with ℓ r = ⌊ ( r + 1 ) ⁄ 4 ⌋ .

The analyses for ( D ( 2 ) ) r y and ( D ( 2 ) ) r z are exactly the same; in conclusion, we have the following lemma.

Lemma 4.3

For each integer r ≥ 0 , if we let ℓ r = ⌊ ( r + 1 ) ⁄ 4 ⌋ , then

y ℓ r ⋅ ( D ( 2 ) ) r x , y ℓ r + 1 ⋅ ( D ( 2 ) ) r y , y ℓ r + 2 ⋅ ( D ( 2 ) ) r z ∈ Q [ x , y , z ] \ y Q [ x , y , z ] .

We obtain similar results for the case when N = 3 as follows and we postpone their proofs to Appendix A so that the readers can see the proof of Theorem 1.1 right away.

Lemma 4.4

Let x a y b z c be some monomial in ( D ( 3 ) ) r x (up to some nonzero Q coefficient).

Its corresponding lattice point ( λ , ν ) via (9) satisfies that ν ≥ 2 3 ( λ − r ) .
4 b + 3 c ≥ 0 and b ≥ − r + 1 2 .

Lemma 4.5

For r ≥ 0 , let ℓ r = ⌊ ( r + 1 ) ⁄ 2 ⌋ − 1 if ℓ ≡ 1 , 2 , 3 ( mod 6 ) , ⌊ ( r + 1 ) ⁄ 2 ⌋ otherwise, then,

y ℓ r ⋅ ( D ( 3 ) ) r x , y ℓ r + 1 ⋅ ( D ( 3 ) ) r y , y ℓ r + 2 ⋅ ( D ( 3 ) ) r z ∈ Q [ x , y , z ] \ y Q [ x , y , z ] .

4.2 Proof of Theorem 1.1

With the results in Section 4.1, we prove Theorem 1.1.

Recall (7) for the definition of D ( N ) . We define the auxiliary differential operators D t ( N ) for t = x , y , z , as

D t ( N ) ≔ p t ( N ) f x + q t ( N ) f y + r t ( N ) f z ,

for each N = 1 , 2 , 3 , where p t , q t and r t are the partial derivatives with respect to the variable t = x , y , z . Particularly, D x ( N ) plays a special role among others as we see later in this section; thus, we let D ( N ) ˜ ≔ D x ( N ) .

We note that for N = 1 , 2 , 3 ,

D ( N ) ˜ f = ( 2 x f x + 4 y f y + 6 z f z ) ⁄ d ,

and particularly,

(10) D ( N ) ˜ x = ( 2 ⁄ d ) x , D ( N ) ˜ y = ( 4 ⁄ d ) y , D ( N ) ˜ z = ( 6 ⁄ d ) z .

We generalize [14, Lemmas 16 and 17] for the case N = 1 and obtain the following two lemmas for N = 2 , 3 .

Lemma 4.6

We have the following relations of D ( N ) , D ( N ) ˜ and ∂ ∂ x for each N = 1 , 2 , 3 :

∂ ∂ x D ( N ) f = D ( N ) ˜ f + D ( N ) ( f x ) for f ∈ Q [ x , y , z , 1 ⁄ y ] .
D ( N ) ˜ ( D ( N ) ) n = ( D ( N ) ) n D ( N ) ˜ + 2 n d ( D ( N ) ) n for all n ≥ 1 .

Proof

(a) follows from a direct calculation; for t = x , y , z ,

∂ ∂ t D ( N ) f = p t ( N ) f x + q t ( N ) f y + r t ( N ) f z + p ( N ) f x t + q ( N ) f y t + r ( N ) f z t = D t ( N ) f + D ( N ) ( f t ) .

In order to prove (b), we can verify the following:

A ≔ p x ( N ) p y ( N ) p z ( N ) q x ( N ) q y ( N ) q z ( N ) r x ( N ) r y ( N ) r z ( N ) = 1 d 2 x − 1 0 4 y 4 x − 4 6 z 4 ( m − 3 ) z 2 ( m − 2 ) y 2 − d y 6 x − 8 ( m − 3 ) z ( m − 2 ) y , A p x ( N ) q x ( N ) r x ( N ) = D ˜ p ( N ) D ˜ q ( N ) D ˜ r ( N ) = 2 d 2 p 3 q 4 r ,

and

B ≔ p x x p x y p x z q x x q x y q x z r x x r x y r x z = 2 d 1 0 0 0 2 0 0 0 3 , B p q r = D p x D q x D r x = 2 d p 2 q 3 r .

With these equations, the direct calculation shows that D ( N ) ˜ D ( N ) f − D ( N ) D ( N ) ˜ f = 2 d D ( N ) f . With this and (a), we can show (b) inductively.□

Lemma 4.7

Let N = 1 , 2 , 3 . For each integer n ≥ 1 , we have

(11) ∂ ∂ x ( ( D ( N ) ) n x ) = n ( n + 1 ) d ( D ( N ) ) n − 1 x ,

(12) ∂ ∂ x ( ( D ( N ) ) n y ) = n ( n + 3 ) d ( D ( N ) ) n − 1 y ,

(13) ∂ ∂ x ( ( D ( N ) ) n z ) = n ( n + 5 ) d ( D ( N ) ) n − 1 z ,

i.e., ∂ ∂ x ( D ( N ) ) n x is a scalar multiple of ( D ( N ) ) n − 1 x , and so are those at y and z.

Proof

We prove the following by induction on n :

∂ ∂ x ( ( D ( N ) ) n f ) = n ( D ( N ) ) n − 1 ( D ( N ) ) ˜ f + ( D ( N ) ) n f x + n ( n − 1 ) d ( D ( N ) ) n − 1 f .

For n = 1 , we have that ∂ ∂ x D ( N ) f = D ( N ) ˜ f + D ( N ) ( f x ) by Lemma 4.6. To proceed by induction, suppose it holds for n − 1 . Then, by Lemma 4.6, we have that

∂ ∂ x ( ( D ( N ) ) n f ) = ∂ ∂ x D ( N ) ( ( D ( N ) ) n − 1 f ) = D ( N ) ˜ ( ( D ( N ) ) n − 1 f ) + D ( N ) ∂ ∂ x ( D ( N ) ) n − 1 f = ( D ( N ) ) n − 1 D ( N ) ˜ f + 2 ( n − 1 ) d ( D ( N ) ) n − 1 f + ( n − 1 ) ( D ( N ) ) n − 1 D ( N ) ˜ f + ( D ( N ) ) n f x + ( n − 1 ) ( n − 2 ) d ( D ( N ) ) n − 1 f = n ( D ( N ) ) n − 1 D ( N ) ˜ f + ( D ( N ) ) n f x + n ( n − 1 ) d ( D ( N ) ) n − 1 f .

Substituting x , y , z for f in turn yields the desired result, referring to (10).□

Lemma 4.8

Let N = 1 , 2 , 3 . Let f ∈ Q [ x , y , z ] be a prime factor of the numerator of a reduced form of one of ( D ( N ) ) n x , ( D ( N ) ) n y , and ( D ( N ) ) n z , for some integer n ≥ 1 . Then, f divides neither y nor the numerator of a reduced form of D ( N ) f , and f x ≠ 0 .

Proof

Note that the only principal ideals I with the property y D ( N ) ( I ) ⊆ I are I = ( y ) or I = ( y 3 − z 2 ) (see [17, Theorem 3.2]).

Let n ≥ 1 be a given integer. Let F ⁄ y ℓ be a reduced form of ( D ( N ) ) n x in Q ( x , y , z ) for some integer ℓ ≥ 0 . We claim that F ∉ y Q [ x , y , z ] . When N = 1 , since the coefficient of x n + 1 -term in D ( n ) x is not zero, F ∉ y Q [ x , y , z ] . For N = 2 , 3 , we have shown that ℓ ≥ 1 , i.e., y ∤ F , in Lemmas 4.3 and 4.5 when n ≥ 3 . Since ( D ( N ) ) x and ( D ( N ) ) 2 x do not have y as a factor of each, F ∉ y Q [ x , y , z ] for n ≥ 1 .

Let f be a prime factor of F . If f divides the numerator of a reduced form of D ( N ) f , then f ∣ y D ( N ) f (in Q [ x , y , z ] ), i.e., f ∈ ( y ) or y ∈ ( y 3 − z 2 ) . Since f ∉ ( y ) , f = k ( y 3 − z 2 ) for some k ∈ Q . Thus, we have ( y 3 − z 2 ) ∣ F .

Note that there is a non-vanishing x n + 1 -term in ( D ( N ) ) n x . Since y 3 − z 2 divides F , ( D ( N ) ) n x has a non-zero x n + 1 y − 3 z 2 -term with the same (non-zero) coefficient. This is clearly impossible when N = 1 , and also impossible when N = 2 , 3 , referring to Lemmas 4.2 (b) and 4.4 (b).

Also, since f ∣ F = y ℓ ( D ( N ) ) n x has a x n + 1 y ℓ -term, we can show that f x ≠ 0 by a similar argument.

The proof for the cases of ( D ( N ) ) n y or ( D ( N ) ) n z follows in a similar manner.□

Now, we are ready to prove the following proposition.

Proposition 4.9

Let N = 1 , 2 , 3 . For each integer n ≥ 1 , we have

ngcd ( ( D ( N ) ) n x , ( D ( N ) ) n + 1 x ) = ngcd ( ( D ( N ) ) n y , ( D ( N ) ) n + 1 y ) = ngcd ( ( D ( N ) ) n z , ( D ( N ) ) n + 1 z ) = 1 .

Here, ngcd ( f , g ) is the gcd of the numerators of reduced forms of f and g, where f , g ∈ Q [ x , y , z , 1 ⁄ y ] .

Proof

The key idea for the proof is to use the interplaying properties such as Lemmas 4.6 and 4.7, and to consider ∂ ∂ x ( ( D ( N ) ) n + 1 x ) and ( D ( N ) ) ( ( D ( N ) ) n x ) .

Assume that a prime polynomial f ∈ Q [ x , y , z ] is a common factor of the numerators of reduced forms of ( D ( N ) ) n x and ( D ( N ) ) n + 1 x . Then, we write

( D ( N ) ) n x = f k g ⁄ y ℓ , ( D ( N ) ) n + 1 x = f k h ⁄ y ℓ ′

for some integers k ≥ 1 and ℓ , ℓ ′ ≥ 0 , and for some g , h ∈ Q [ x , y , z ] such that f ∤ gcd ( g , h ) .

Note that ℓ and ℓ ′ ∈ { ℓ , ℓ + 1 } are integers ≥ 0 . Referring to Lemma 4.8, we see that y ∤ f .

Referring to Lemma 4.7 (11), we have

∂ ∂ x ( D ( N ) ) n + 1 x = ( f k h x + k f k − 1 f x h ) ⁄ y ℓ ′ = ( n + 1 ) ( n + 2 ) d f k g ⁄ y ℓ ,

( n + 1 ) ( n + 2 ) d g y ℓ ′ − ℓ − h x f = k f x h

in Q [ x , y , z ] , thus f ∣ h , since f is a prime and f x ≠ 0 by Lemma 4.8.

Now, we consider

( D ( N ) ) n + 1 x = f k h ⁄ y ℓ ′ = D ( N ) ( f k ⋅ g y − ℓ ) = k f k − 1 g y − ℓ D ( N ) ( f ) + f k D ( N ) ( g y − ℓ ) ,

i.e.,

k g y ℓ ′ − ℓ D ( N ) ( f ) = f h − f y ℓ ′ D ( N ) ( g y − ℓ ) .

Note that y ℓ ′ − ℓ D ( N ) ( f ) or D ( N ) ( g y − ℓ ) might have some powers of y in the denominators of their reduced forms, so we let y ℓ ″ be the largest power dividing their denominators. Then, we have

k g y ( ℓ ″ + ℓ ′ − ℓ ) D ( N ) ( f ) = ( h y ℓ ″ − y ℓ ″ + ℓ ′ D ( N ) ( g y − ℓ ) ) ⋅ f

in Q [ x , y , z ] . Thus,

f ∣ g y ℓ ″ + ℓ ′ − ℓ D ( f ) .

Note that f does not divide the denominator of a reduced form of D ( f ) by Lemma 4.8 and also f ∤ y , therefore f ∣ g . This contradicts that f ∤ gcd ( g , h ) . This proves ngcd ( ( D ( N ) ) n x , ( D ( N ) ) n + 1 x ) = 1 .

We apply the same argument with respect to y and z to complete the proof.□

We now have all ingredients to prove Theorem 1.1.

Proof of Theorem 1.1

Assume that there is a non-simple zero of θ j E k ( N ) ( τ ) for some integer j ≥ 0 , N = 1 , 2 , 3 , and k = 2 , 4 , 6 . By Proposition 3.3, the numerators of θ j E k ( N ) ( τ ) and θ j + 1 E k ( N ) ( τ ) in the reduced forms have a non-unit common divisor, which is a contradiction by Proposition 4.9.□

Acknowledgements

The authors are grateful for the reviewer’s valuable comments that improved the manuscript. We also appreciate Professor SoYoung Choi for suggesting this problem.

Funding information: Bo-Hae Im was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIT) (NRF-2023R1A2C1002385).
Author contributions: All authors have accepted responsibility for the entire content of this manuscript and consented to its submission to the journal, reviewed all the results and approved the final version of the manuscript.
Conflict of interest: Authors state no conflict of interest.

Appendix A Degree analysis on ( D ( 3 ) ) n : Proofs of Lemmas 4.4 and 4.5

In this appendix, we give an analysis on the degrees of denominators of ( D ( 3 ) ) n at x , y , z , and we prove Lemmas 4.4 and 4.5, which have been postponed from Section 4.1 since they can be obtained in a similar manner as done for ( D ( 2 ) ) n in Section 4.1. One may note from the computations presented in this appendix that the case for D ( 3 ) is more complicated than the case for D ( 2 ) .

In order to prove Lemmas 4.4 and 4.5, we will give some conditions for the monomials appearing in each of ( D ( 3 ) ) n x , ( D ( 3 ) ) n y , ( D ( 3 ) ) n z , and show that each of the minimal degrees with respect to y among the monomials of ( D ( 3 ) ) n ( x ) , ( D ( 3 ) ) n ( y ) , and ( D ( 3 ) ) n ( z ) decreases by 3 as n increases by 6, for n ≥ 0 .

We clear denominators to obtain an operator with integral coefficients, namely

D ( 3 ) ≔ d D ( 3 ) = 6 D ( 3 ) = ( x 2 − y ) ∂ ∂ x + ( 4 x y − 4 z ) ∂ ∂ y + ( 6 x z − 3 y 2 − 3 y − 1 z 2 ) ∂ ∂ z

and

D ( 3 ) ( x a y b z c ) = ( a + 4 b + 6 c ) x a + 1 y b z c − 3 c x a y b + 2 z c − 1 − ( 4 b + 3 c ) x a y b − 1 z c + 1 − a x a − 1 y b + 1 z c .

We choose the same T as defined in (8), and let an integral lattice point ( λ , ν ) represent the monomial x a y b z c via relation (9) as it is done for N = 2 case, i.e.,

( a , b , c ) T = ( 1 , r ⁄ 2 , 0 ) T + T − 1 ( λ , ν , 0 ) T .

Proof of Lemma 4.4

Let x a y b z c be a monomial in ( D ( 3 ) ) r x . If D ( 3 ) ( x a y b z c ) contains a monomial whose deg y is less than b , then such a monomial is − ( 4 b + 3 c ) x a y b − 1 z c + 1 , which is non-zero only when 4 b + 3 c ≠ 0 . Since the condition 4 b + 3 c = 0 is equivalent to ν = 2 3 ( λ − r ) via (9), the points on this line do not produce a monomial with decreased deg y through the differential operator.

(a) holds for the initial case r = 0 . As r increases, all points keep lying above or on the line ν = 2 3 ( λ − r ) ; so we have ν ≥ 2 3 ( λ − r ) for all r . This proves (a).

The inequality from (a) is equivalent to 4 b + 3 c ≥ 0 by (9), and we conclude (b) since 4 b + 3 c ≥ 0 , a ≥ 0 , and a + 2 b + 3 c = r + 1 .□

Proof of Lemma 4.5

Let a i , j , k ( r ) denote the coefficient of x i y j z k in ( D ( 3 ) ) r x .

We prove the lemma by showing that the minimal value among deg y of the monomials in ( D ( 3 ) ) r ( x ) decreases exactly when applying D ( 3 ) to each of ( D ( 3 ) ) ( 6 k + 2 ) ( x ) , ( D ( 3 ) ) ( 6 k + 3 ) ( x ) , and ( D ( 3 ) ) ( 6 k + 4 ) ( x ) . It is enough to show that the coefficients a 0 , − 3 k − 1 , 4 k + 2 ( 6 k + 3 ) , a 0 , − 3 k , 4 k + 1 ( 6 k + 2 ) , a 0 , − 3 k − 2 , 4 k + 3 ( 6 k + 4 ) never vanish for k ≥ 0 . We show by the non-triviality of those coefficients modulo 7.

By some tedious calculations, we obtain

2 A ≔ a 0 , − 3 k − 1 , 4 k + 2 ( 6 k + 3 ) = − 3 a 0 , − 3 k , 4 k + 1 ( 6 k + 2 ) , and a 0 , − 3 k , 4 k + 1 ( 6 k + 2 ) = − 4 a 0 , − 3 k + 1 , 4 k ( 6 k + 1 ) ,

therefore, A = 12 a 0 , − 3 k + 1 , 4 k ( 6 k + 1 ) . Similarly, since a 0 , − 3 k + 1 , 4 k ( 6 k + 1 ) = − 5 a 0 , − 3 k + 2 , 4 k − 1 ( 6 k ) − a 1 , − 3 k , 4 k ( 6 k ) , we have

A = − 12 ( 5 a 0 , − 3 k + 2 , 4 k − 1 ( 6 k ) + a 1 , − 3 k , 4 k ( 6 k ) ) .

Since

a 0 , − 3 k + 2 , 4 k − 1 ( 6 k ) = − 6 a 0 , − 3 k + 3 , 4 k − 2 ( 6 k − 1 ) − 12 k a 0 , − 3 k , 4 k ( 6 k − 1 ) − a 1 , − 3 k + 1 , 4 k − 1 ( 6 k − 1 ) , a 1 , − 3 k , 4 k ( 6 k ) = 12 k a 0 , − 3 k , 4 k ( 6 k − 1 ) − a 1 , − 3 k + 1 , 4 k − 1 ( 6 k − 1 ) ,

we have

A = 12 × 6 ( 5 a 0 , − 3 k + 3 , 4 k − 2 ( 6 k − 1 ) + 8 k a 0 , − 3 k , 4 k ( 6 k − 1 ) + a 1 , − 3 k + 1 , 4 k − 1 ( 6 k − 1 ) ) .

Since

a 0 , − 3 k + 3 , 4 k − 2 ( 6 k − 1 ) = − 7 a 0 , − 3 k + 4 , 4 k − 3 ( 6 k − 2 ) + ( − 12 k + 3 ) a 0 , − 3 k + 1 , 4 k − 1 ( 6 k − 2 ) − a 1 , − 3 k + 2 , 4 k − 2 ( 6 k − 2 ) , a 0 , − 3 k , 4 k ( 6 k − 1 ) = − a 0 , − 3 k + 1 , 4 k − 1 ( 6 k − 2 ) , a 1 , − 3 k + 1 , 4 k − 1 ( 6 k − 1 ) = ( 12 k − 2 ) a 0 , − 3 k + 1 , 4 k − 1 ( 6 k − 2 ) − 2 a 1 , − 3 k + 2 , 4 k − 2 ( 6 k − 2 ) ,

we have

A = 12 × 6 ( − 35 a 0 , − 3 k + 4 , 4 k − 3 ( 6 k − 2 ) + ( − 56 k + 13 ) a 0 , − 3 k + 1 , 4 k − 1 ( 6 k − 2 ) − 7 a 1 , − 3 k + 2 , 4 k − 2 ( 6 k − 2 ) ) ≡ − 2 a 0 , − 3 k + 1 , 4 k − 1 ( 6 k − 2 ) ( mod 7 ) .

Finally, since a 0 , − 3 k + 1 , 4 k − 1 ( 6 k − 2 ) = − 2 a 0 , − 3 k + 2 , 4 k − 2 ( 6 k − 3 ) , we also have

(A1) A = a 0 , − 3 k − 1 , 4 k + 2 ( 6 k + 3 ) ≡ 4 a 0 , − 3 ( k − 1 ) − 1 , 4 ( k − 1 ) + 2 ( 6 ( k − 1 ) + 3 ) ( mod 7 ) .

Recalling that D ( 3 ) ( x ) = 6 x 4 − 36 x 2 y + 48 x z − 6 y 2 − 12 z 2 y , since a 0 , − 1 , 2 ( 3 ) = − 12 ≢ 0 ( mod 7 ) , the coefficient a 0 , − 3 k − 1 , 4 k + 2 ( 6 k + 3 ) never vanishes. Many parts of the calculations for the recurrences of the coefficients are recyclable for other two cases. In short, we obtain

a 0 , − 3 k , 4 k + 1 ( 6 k + 2 ) = − 4 a 0 , − 3 k + 1 , 4 k ( 6 k + 1 ) = 4 ( 5 a 0 , − 3 k + 2 , 4 k − 1 ( 6 k ) + a 1 , − 3 k , 4 k ( 6 k ) ) = − 24 ( 5 a 0 , − 3 k + 3 , 4 k − 2 ( 6 k − 1 ) + 8 k a 0 , − 3 k , 4 k ( 6 k − 1 ) + a 1 , − 3 k + 1 , 4 k − 1 ( 6 k − 1 ) ) = − 24 ( − 35 a 0 , − 3 k + 4 , 4 k − 3 ( 6 k − 2 ) + ( − 56 k + 13 ) a 0 , − 3 k + 1 , 4 k − 1 ( 6 k − 2 ) − 7 a 1 , − 3 k + 2 , 4 k − 2 ( 6 k − 2 ) ) ≡ 3 a 0 , − 3 k + 1 , 4 k − 1 ( 6 k − 2 ) ≡ a 0 , − 3 k + 2 , 4 k − 2 ( 6 k − 3 ) ≡ 4 a 0 , − 3 k + 3 , 4 k − 3 ( 6 k − 4 ) ( mod 7 ) , and a 0 , − 3 k − 2 , 4 k + 3 ( 6 k + 4 ) = − 2 a 0 , − 3 k − 1 , 4 k + 2 ( 6 k + 3 ) = 6 a 0 , − 3 k , 4 k + 1 ( 6 k + 2 ) = − 24 a 0 , − 3 k + 1 , 4 k ( 6 k + 1 ) = 24 ( 5 a 0 , − 3 k + 2 , 4 k − 1 ( 6 k ) + a 1 , − 3 k , 4 k ( 6 k ) ) = − 24 × 6 ( 5 a 0 , − 3 k + 3 , 4 k − 2 ( 6 k − 1 ) + 8 k a 0 , − 3 k , 4 k ( 6 k − 1 ) + a 1 , − 3 k + 1 , 4 k − 1 ( 6 k − 1 ) ) ≡ 4 a 0 , − 3 k + 1 , 4 k − 1 ( 6 k − 2 ) ( mod 7 ) ,

thus

(A2) a 0 , − 3 k , 4 k + 1 ( 6 k + 2 ) ≡ 4 a 0 , − 3 ( k − 1 ) , 4 ( k − 1 ) + 1 ( 6 ( k − 1 ) + 2 ) ( mod 7 ) ,

(A3) a 0 , − 3 k − 2 , 4 k + 3 ( 6 k + 4 ) ≡ 4 a 0 , − 3 ( k − 1 ) − 2 , 4 ( k − 1 ) + 3 ( 6 ( k − 1 ) + 4 ) ( mod 7 ) .

Since a 0 , 0 , 1 ( 2 ) = 4 ≢ 0 ( mod 7 ) and a 0 , − 2 , 3 ( 4 ) = 24 ≢ 0 ( mod 7 ) , we conclude that the coefficients a 0 , − 3 k − 1 , 4 k + 2 ( 6 k + 3 ) , a 0 , − 3 k , 4 k + 1 ( 6 k + 2 ) , and a 0 , − 3 k − 2 , 4 k + 3 ( 6 k + 4 ) never vanish for k ≥ 0 , i.e., they provide non-zero monomials with decreased deg y after applying D ( 3 ) . With Lemma 4.4 (b), the minimal deg y among monomials appearing in ( D ( 3 ) ) r + 1 ( x ) drops by 1 than those for ( D ( 3 ) ) r ( x ) , only when r ≡ 2 , 3 , 4 ( mod 6 ) as shown in (A1), (A2), and (A3). Therefore, we conclude that y ℓ r ⋅ ( D ( 3 ) ) r x ∈ Q [ x , y , z ] \ y Q [ x , y , z ] , with ℓ 1 , … , ℓ 6 = 0 , 0 , 1 , 2 , 3 , 3 and ℓ r + 6 = ℓ r + 3 for all r ≥ 1 .

The analyses for ( D ( 3 ) ) r y and ( D ( 3 ) ) r z can be done in the same way, and this completes the proof.□

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Received: 2024-03-26

Revised: 2024-07-27

Accepted: 2024-08-30

Published Online: 2024-10-03

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

Articles in the same Issue

https://doi.org/10.1515/math-2024-0065

Keywords for this article

modular forms; quasi-modular forms; Fricke group

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BY 4.0