Home Mathematics Explicit construction of mock modular forms from weakly holomorphic Hecke eigenforms
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Explicit construction of mock modular forms from weakly holomorphic Hecke eigenforms

  • SoYoung Choi and Chang Heon Kim EMAIL logo
Published/Copyright: April 19, 2022

Abstract

Extending our previous work we construct weakly holomorphic Hecke eigenforms whose period polynomials correspond to elements in a basis consisting of odd and even Hecke eigenpolynomials induced by only cusp forms. As an application of our results, we give an explicit construction of the holomorphic parts of harmonic weak Maass forms that are good for Hecke eigenforms. Moreover, we give an explicit construction of the Hecke-equivariant map between the space of weakly holomorphic cusp forms and two copies of the spaces of cusp forms, and show that the map is compatible with the corresponding map on the spaces of period polynomials.

1 Introduction and statement of results

Let p be one or a prime and Γ 0 + ( p ) be the group generated by the congruence subgroup Γ 0 ( p ) and the Fricke involution W p = 0 1 / p p 0 . For any even integer k , let M k ! ( p ) be the space of weakly holomorphic (i.e., meromorphic with poles only at the cusps) modular forms f of weight k for Γ 0 ( p ) . For ε { ± 1 } , let M k ! , ε ( p ) be the subspace of M k ! ( p ) with f k W p = ε f . Each f M k ! , ε ( p ) has a Fourier development of the form

f ( z ) = n n 0 a f ( n ) q n ,

where the parameter q stands for exp ( 2 π i z ) , as usual. We set ord f = n 0 if a f ( n 0 ) 0 . Let S be the set consisting of values of p for which the genus of Γ 0 + ( p ) is zero, that is,

S = { 1 , 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 41 , 47 , 59 , 71 } .

When p S , the space M k ! , ε ( p ) has a canonical basis: see [1,2, 3,4]. We define

m k ε max { ord f f 0 M k ! , ε ( p ) } .

When k > 2 , we note that m k ε is given by dim S k ε ( p ) . Indeed, for every integer m with m m k ε , there exists a unique weakly holomorphic modular form f k , m ε M k ! , ε ( p ) with Fourier expansion of the form

(1) f k , m ε ( τ ) = q m + n > m k ε a k ε ( m , n ) q n

and together they form a basis for M k ! , ε ( p ) . Indeed the basis element f k , m ε can be given explicitly in the form f k , m ε = f k , m k ε ε F k , m + m k ε ( j p + ) , where j p + is the Hauptmodul for Γ 0 + ( p ) and F k , D ( x ) is a monic polynomial in x of degree D . Moreover, the Fourier coefficients a k ε ( m , n ) of q n in f k , m ε turn out to be rational integers. Throughout this paper, we simply write f k , m f k , m + .

As usual, we denote by S k + ( p ) the space of holomorphic cusp forms of weight k for Γ 0 + ( p ) . Let S k ! , + ( p ) be the subspace of M k ! , + ( p ) consisting of weakly holomorphic modular forms for Γ 0 + ( p ) with zero constant term in the Fourier expansion. We now consider Hecke operators on S k + ( p ) . For each positive integer n relatively prime to p , the usual Hecke operator T n on S k + ( p ) extends to S k ! , + ( p ) . In particular, for prime indices l ( p ) , the Hecke operators T on S k ! , + ( p ) are given as follows: for f S k ! , + ( p ) ,

(2) T f = l k / 2 1 a d = l b ( mod d ) f k a b 0 d = n ( a f ( l n ) + l k 1 a f ( n / l ) ) q n .

Common eigenforms of all Hecke operators T n on S k + ( p ) with n coprime to p are called Hecke eigenforms. For later use, we let t = dim S k + ( p ) and

f n = m > 0 λ ( n , m ) q m n = 1 , 2 , , t

be a basis of S k + ( p ) consisting of normalized Hecke eigenforms. Following [5,6] we call f S k ! , + ( p ) a weakly holomorphic Hecke eigenform with respect to S k ! , + ( p ) / D k 1 ( M 2 k ! , + ( p ) ) if for every Hecke operator T n with ( n , p ) = 1 there is a complex number λ n for which

T n f λ n f D k 1 ( M 2 k ! , + ( p ) ) ,

where D stands for the differential operator 1 2 π i d d z .

In the work of Bringmann et al. [5, Theorem 1.5] weakly holomorphic Hecke eigenforms are constructed in the level 1 case by making use of harmonic weak Maass forms which are preimages of Hecke eigenforms under the differential operator ξ k 2 i y k z ¯ ¯ originated from [7]. Thus, their construction is not explicit. In [6, Theorem 1.2], we extended it to higher level cases to the primes for which Γ 0 + ( p ) has genus zero (primes up to 71 excluding 37, 43, 61, and 67). The construction was given explicitly in terms of weakly holomorphic modular forms without relying on the theory of harmonic weak Maass forms, and we gave an explicit description of the “polar” eigenform h n in terms of a linear combination of cuspidal eigenforms and the dual form f n , as in (EF1)–(EF3). Here the duality is with respect to a certain pairing of functions introduced by Guerzhoy as follows.

Following [1,5,6,8,9], for f , g M k ! , + ( p ) , we define a pairing { f , g } originated from Bruinier and Funke [7] by

(3) { f , g } n Z , n 0 a f ( n ) a g ( n ) n k 1 .

It is antisymmetric (since k is even), bilinear, and Hecke equivariant. Specifically, for any prime ( p )

{ T f , g } = { f , T g } .

Moreover, D k 1 ( M 2 k ! , + ( p ) ) is perpendicular to all of S k ! , + ( p ) , they are the only weakly holomorphic modular forms with that property (by the Serre duality theorem), and two elements of S k + ( p ) are perpendicular to one another. Let

(4) f n = m t μ ( m , n ) q m

with μ ( m , n ) C be a linear combination of f k , 1 , , f k , t , which is dual to f n with respect to the pairing, i.e., { f m , f n } = δ m n , where δ m n is the Kronecker delta function. Then such functions f n are unique. Moreover, it follows from [1,9] that T f n and λ ( n , ) f n represent the same coset in S k ! , + ( p ) / ( D k 1 ( M 2 k ! , + ( p ) ) S k + ( p ) ) . Thus, we can write

(5) T f n = λ ( n , ) f n + D k 1 g n , + j = 1 t a j n ( ) f j

for some g n , M 2 k ! , + ( p ) and a j n ( ) C .

Let p S , t = dim S k + ( p ) , and be a prime different from p . We then obtain from [6, Theorem 1.2] the following assertions.

(EF1) Let i , n { 1 , , t } with i n . Let r be a prime ( p ) such that λ ( i , r ) λ ( n , r ) and put

x i ( n ) a n i ( r ) λ ( i , r ) λ ( n , r ) .

Then x i ( n ) is independent of the choice of r and the quantity a n i ( r ) is given by

a n i ( r ) = { f n , T r f i } .

(EF2) For each n with 1 n t , let

h n i = 1 i n t x i ( n ) f i + f n .

Then h n is a Hecke eigenform with respect to S k ! , + ( p ) / D k 1 ( M 2 k ! , + ( p ) ) having the same eigenvalues as those of f n . More explicitly one has

T ( h n ) = λ ( n , ) h n + D k 1 ( g n , ) ,

where g n , is the modular form defined in (5) and computed as

g n , = 1 s t s > t μ ( s , n ) s k 1 f 2 k , s l ,

where μ ( , n ) is the Fourier coefficient of f n , defined in (4).

(EF3) The set

{ [ f 1 ] , , [ f t ] , [ h 1 ] , , [ h t ] }

forms a basis for S k ! , + ( p ) / D k 1 ( M 2 k ! , + ( p ) ) , where [ f ] stands for the class of f .

For Γ { Γ 0 ( p ) , Γ 0 + ( p ) } , let H 2 k ( Γ ) be the space of harmonic weak Maass forms of weight 2 k for Γ . Following [10,11], we say that F ( z ) H 2 k ( Γ 0 ( p ) ) is “good” for the Hecke eigenform f c ( z ) f ( z ¯ ) ¯ S k ( p ) if it satisfies the following:

  1. The principal part of F at the cusp belongs to K f [ q 1 ] . Here K f denotes the number field obtained by adjoining to Q the Fourier coefficients of f .

  2. The principal part of F at the cusp 0 is constant.

  3. We have ξ 2 k F = f c ( f c , f c ) .

Remark 1.1

  1. The existence of F , which is good for a Hecke eigenform f c is guaranteed by [10, Proposition 5.1].

  2. Let F 0 be good for a Hecke eigenform f c and denote t = dim S k + ( p ) and t = dim S k ( p ) . Let M 2 k ( p ) be the space of weakly holomorphic modular forms of weight 2 k for Γ 0 ( p ) with poles allowed only at the cusp . For p { 1 , 2 , 3 , 5 , 7 , 13 } , it follows from [12, Theorem 1.1, Theorem 1.7, and line 6 in p. 123] that

    max { ord f f 0 M 2 k ( p ) } = 1 t t ,

    and for each integer m with m 1 t t , there exists f 2 k , m = q m + O ( q t t ) M 2 k ( p ) with integral Fourier coefficients. By subtracting a suitable linear combination of f 2 k , m ’s from F 0 we can take a unique F , which is good for f c and F + = O ( q t t ) .

  3. In [11] a direct method for relating the coefficients of f c and F is provided by means of p -adic coupling and an algebraic regularized mock modular form F α . More precisely, if we let α be the coefficient of q 1 in F + , then F α is given by

    F α = D k 1 F + α f = n c α ( n ) q n .

    Moreover, F α has coefficients in K f and

    (6) lim w + n c α ( w n ) q n c α ( w ) = f ( z ) β 1 k 1 f ( z ) ,

    where is a prime number and β , β are the roots of the equation X 2 a f ( ) X + χ ( ) k 1 = ( X β ) ( X β ) ordered so that ord ( β ) ord ( β ) , χ is a trivial character modulo p , and we assume that β 0 in the case = p . We take the limit in (6) in -adic topology.

  4. In [13], the structure of half-integral weight weakly holomorphic Hecke eigenforms was developed, and in [14] half-integral weight p -adic coupling was investigated.

Let f , g M k ! ( p ) . We define a regularized inner product as follows. For T > 0 , we denote by T the truncated fundamental domain for SL 2 ( Z )

T = { z H x 1 / 2 , z 1 , and y T } .

Moreover, we define the truncated fundamental domain for Γ 0 ( p ) by

T ( Γ 0 ( p ) ) = γ V γ T ,

where V is a fixed set of representatives of Γ 0 ( p ) \ SL 2 ( Z ) . Now we define the regularized inner product ( f , g ) reg as the constant term in the Laurent expansion at s = 0 of the function

1 [ SL 2 ( Z ) : Γ 0 ( p ) ] lim T T ( Γ 0 ( p ) ) f ( z ) g ( z ) ¯ y k s d x d y y 2 .

As explained in [10] and [15], ( f , g ) reg exists if f or g is a holomorphic modular form. If both f and g are holomorphic modular forms such that f g is a cusp form, then ( f , g ) reg reduces to the Petersson inner product ( f , g ) .

In the next theorems, we give an explicit construction of D k 1 F in terms of polar eigenforms and a canonical basis. Theorem 1.2 applies for p = 1 only, while Theorems 1.3 and 1.5 cover higher levels.

Theorem 1.2

Let 2 < k 2 Z , and t = dim S k + ( 1 ) . Then for each n { 1 , , t } ,

h n + ( f n , f n ) reg ( f n , f n ) f n

is equal to D k 1 F n for the unique F n H 2 k ( Γ 0 ( 1 ) ) , which is good for f n c and F n + = O ( q t ) .

Theorem 1.3

Let p be a prime for which Γ 0 ( p ) is of genus zero, i.e., p { 2 , 3 , 5 , 7 , 13 } , 2 < k 2 Z , t = dim S k + ( p ) , t = dim S k ( p ) , and n { 1 , , t } . Then the following assertions are true.

  1. Let A be a t × t matrix whose i j -entry is given by CT ( f i f 2 k , t + j ) , where CT ( f ) denotes the constant term of the Fourier expansion of f . Then the matrix A is invertible.

  2. Let β i j be the i j -entry of the matrix A 1 . Take the unique weakly holomorphic modular form w n M 2 k ! , + ( p ) such that

    w n j = 1 t β j n f 2 k , t + j = O ( q t ) .

    Then

    h n + ( f n , f n ) reg ( f n , f n ) f n + D k 1 w n + j = 1 t β j n f 2 k , t + j

    is equal to D k 1 F n for the unique F n H 2 k ( Γ 0 ( p ) ) , which is good for f n c and F n + = O ( q t t ) .

Remark 1.4

In Theorem 1.3, we observe from duality that m 2 k ε = dim S k ε ( p ) 1 . This observation makes the existence and uniqueness of w n much clearer.

Theorem 1.5

Let p { 1 , 2 , 3 , 5 , 7 , 13 } , 2 < k 2 Z , t = dim S k + ( p ) , and n { 1 , , t } . Then the coefficients of f n are in K f n .

Let P k 2 denote the space of all polynomials of degree at most k 2 . For any meromorphic function f on the complex upper half plane H , we define the action of γ = a b c d G L 2 + ( R ) by

( f k γ ) ( z ) = ( det γ ) k / 2 ( c z + d ) k f ( γ z ) .

For p { 1 , 2 , 3 } , a subspace W k 2 + ( p ) of P k 2 is defined by

W k 2 + ( p ) = { g P k 2 g + g 2 k W p = 0 = g + g 2 k U + g 2 k U 2 + + g 2 k U n p 1 }

with T = 1 1 0 1 , U = T W p , and n p = 3 if p = 1 2 p if p = 2 , 3 . We call the elements of the space W k 2 + ( p ) period polynomials. Period polynomials have been investigated in relation to Eichler integrals, cusp forms via Eichler-Shimura isomorphisms, and to special values of modular L -functions. Indeed, Eichler [16] discovered relations between periods of cusp forms, and Shimura [17] extended them. Later, Manin [18] made more explicit the connection of these relations with the Fourier coefficients, by using the Hecke operators and continued fractions. For more discussion on the classical theory of period polynomials, one is referred to [18,19,20, 21,22] and [23, Chapter 12]. The period polynomials are also related to weakly holomorphic modular forms as follows. For each f = n a f ( n ) q n M k ! , + ( p ) we define the period polynomial for f by

(7) r + ( f ) = r + ( f , z ) c k ( f f 2 k W p ) ( z ) ,

where c k = Γ ( k 1 ) ( 2 π i ) k 1 and f denotes the Eichler integral

(8) f ( z ) n n 0 a f ( n ) n 1 k q n .

Let S k ! , + ( p ) be the subspace of M k ! , + ( p ) consisting of those f M k ! , + ( p ) with a f ( 0 ) = 0 . As described in [24, Section 4], r + gives a map from S k ! , + ( p ) to W k 2 + ( p ) . For p { 1 , 2 , 3 } and even k > 2 , in virtue of [5, Theorem 1.6], [8, Theorem 1.1], [24, Theorem 1.2], and [25, Theorem 2], we have the following exact sequence:

(9) 0 D k 1 ( M 2 k ! , + ( p ) ) S k ! , + ( p ) r + W k 2 + ( p ) / ( p z ) k 2 1 0 .

Let p { 1 , 2 , 3 } , n be a positive integer relatively prime to p , and k be an even integer greater than 2. Following Knopp [19,20] we define a Hecke operator T ^ n on the space W k 2 + ( p ) . Suppose that q ( z ) W k 2 + ( p ) and F ( z ) is a meromorphic function on H satisfying

(10) F 2 k T = F and F 2 k W p = F + q ( z ) .

We also assume that F ( z ) is meromorphic in the local uniformizing variable at the cusp of a fundamental region for Γ 0 + ( p ) . In such a case, F ( z ) is called a modular integral for q ( z ) of weight 2 k . We define T ^ n by

(11) T ^ n ( q ( z ) ) = F n 2 k ( W p 1 ) ,

where

F n = n k / 2 a d = n b ( d ) F 2 k a b 0 d .

We then obtain that T ^ n ( q ( z ) ) W k 2 + ( p ) and it follows from [6, Theorem 1.3] the following diagram

is commutative, i.e., r + is a Hecke equivariant homomorphism. We recall the Eisenstein series

E k ( z ) = 1 2 k B k n = 1 σ k 1 ( n ) q n , G k ( z ) = B k 2 k E k ( z )

and

G k + ( z ) = G k , p + ( z ) = G k ( z ) , if p = 1 G k ( z ) + p k / 2 G k ( p z ) , if p = 2 or 3 ,

where B k is the kth Bernoulli number and σ k 1 denotes the usual divisor sum.

With the same notation as above, let p { 1 , 2 , 3 } and l be a prime different from p . We set p 0 ( z ) ( p z ) k 2 1 . We then obtain from [6, Theorem 1.4] the following assertions.

(EP1) T ^ ( p 0 ) = ( 1 + 1 k ) p 0 and T ^ ( r + ( f n ) ) = 1 k λ ( n , ) r + ( f n ) for each n { 1 , , t } .

(EP2) For each n with 1 n t we define

c 0 ( n ) = c k t ( 0 ) m = 1 t μ ( m , n ) m k 1 t ( m ) ,

where c k is the constant defined in (7) and t ( m ) denotes the coefficient of q m in G k + ( z ) . Then we have

T ^ ( c 0 ( n ) p 0 + r + ( h n ) ) = 1 k λ ( n , ) ( c 0 ( n ) p 0 + r + ( h n ) ) .

(EP3) The set

{ p 0 ( z ) , r + ( f 1 ) , , r + ( f t ) , c 0 ( 1 ) p 0 + r + ( h 1 ) , , c 0 ( t ) p 0 + r + ( h t ) }

forms a basis for W k 2 + ( p ) consisting of Hecke eigenpolynomials. As we see, the basis in (EP3) is constructed by using weakly holomorphic modular forms that are not holomorphic ones. Every period polynomial decomposes into an even and an odd part. By the Eichler-Shimura isomorphism, the space of even period polynomials consists of p 0 and those coming from cusp forms. Also the space of odd period polynomials can be constructed by using cusp forms. So it is natural to find a basis for the space W k 2 + ( p ) consisting of odd and even period polynomials induced by only cusp forms that are Hecke eigenpolynomials. Although this result is well-known from the literature [18,21,22], we state it in the next theorem and reprove it by using the theory of harmonic weak forms because its proof will be used to prove our main results. For f S k + ( p ) , r + ( f ) denotes r ( f , z ) r ( f , z ) 2 and r + + ( f ) stands for r ( f , z ) + r ( f , z ) 2 i .

Theorem 1.6

With the same notations as above, let p { 1 , 2 , 3 } , 2 < k 2 Z , t = dim S k + ( p ) , and l be a prime different from p . Then for each n { 1 , , t } , the following assertions are true.

  1. The polynomial r + ( f n , z ) belongs to the space W k 2 + ( p ) and is a Hecke eigenpolynomial with the same eigenvalues as those of r + ( f n , z ) , i.e.,

    T ^ ( r + ( f n , z ) ) = l 1 k λ ( n , l ) r + ( f n , z ) .

  2. Both polynomials r + + ( f n , z ) and r + ( f n , z ) belong to the space W k 2 + ( p ) and are Hecke eigenpolynomials with the same eigenvalues as those of r + ( f n , ± z ) .

  3. The set

    { p 0 ( z ) , r + + ( f 1 ) , , r + + ( f t ) , r + ( f 1 ) , , r + ( f t ) }

    forms a basis for W k 2 + ( p ) consisting of Hecke eigenpolynomials.

Since r + ( f n , z ) W k 2 + ( p ) , it follows from (9) that there exists g S k ! , + ( p ) such that

r + ( g ) r + ( f n , z ) ( mod p 0 ) .

When p = 1 , by utilizing harmonic weak Maass forms, Bringmann et al. [5] constructed such g . So their construction is not explicit. In Theorem 1.8, we will provide an explicit construction of such g for all p S without using harmonic weak Maass forms.

Remark 1.7

Theorem 1.6 is related to the results in [18,21]. In [22], more general level is considered by using a cohomological approach.

Theorem 1.8

Let p S , 2 < k 2 Z , and t = dim S k + ( p ) . For each n { 1 , , t } , choose G n ( z ) H 2 k ( Γ + 0 ( p ) ) such that ξ 2 k G n ( z ) = f n ( z ) . Let

g n ( z ) D k 1 G n c and g n ( f n , f n ) b k h n + ( f n , f n ) reg b k f n with b k = Γ ( k 1 ) ( 4 π ) k 1 .

Then the following assertions are true.

  1. The class [ g n ] is independent of the choice of G n and uniquely determined by f n . Moreover, we have [ g n ] = [ b k g n ] .

  2. Let p { 1 , 2 , 3 } . Then we have

    r + ( g n , z ) = r + ( f n , z ) + ( f n , f n ) b k c 0 ( n ) p 0 .

Remark 1.9

When p = 1 , the formula in Theorem 1.8(ii) is related to the results in [23, Theorem 12.10] and the following remark.

We define a map r : S k + ( p ) × S k + ( p ) W k 2 + ( p ) by r ( f , g ) = r + ( f ) + i r + + ( g ) and W 0 = im r . Then it is well known that W k 2 + ( p ) = W 0 p 0 and therefore we can consider a map P : W k 2 + ( p ) W 0 , which is the projection to the first component. When p = 1 , Bringmann et al. [5] suggested the following diagram of exact sequences

where a map Φ : S k ! , + ( p ) S k + ( p ) × S k + ( p ) is not given explicitly. So they did not show that the diagram is commutative in the part.

In the next two theorems, we will give an explicit construction of such a Hecke-equivariant map Φ for p { 1 , 2 , 3 } , which makes the diagram commutative. To this end, we are in need of the following proposition.

Proposition 1.10

Let p S , 2 < k 2 Z , and t = dim S k + ( p ) . With the same notations as above, we have the following assertions.

  1. The set { [ g 1 ] , [ g 2 ] , , [ g t ] , [ f 1 ] , [ f 2 ] , , [ f t ] } forms a basis for the space S k ! , + ( p ) / D k 1 ( M 2 k ! , + ( p ) ) .

  2. The set { [ g 1 + f 1 ] , [ g 2 + f 2 ] , , [ g t + f t ] , [ f 1 ] , [ f 2 ] , , [ f t ] } forms a basis for the space S k ! , + ( p ) / D k 1 ( M 2 k ! , + ( p ) ) .

We note from Proposition 1.10 that for any F S k ! , + ( p ) , there exists unique constants α n , β n C such that

(12) F = n α n ( g n + f n ) + n β n f n + D k 1 g F

for some g F M 2 k ! , + ( p ) . Now we define a map Φ : S k ! , + ( p ) S k + ( p ) × S k + ( p ) by

(13) Φ ( F ) = 2 n α n f n + n β n f n , n β n f n ,

where α n and β n are the coefficients appearing in (12). We then easily check that the map Φ is a linear map.

Theorem 1.11

Let p { 1 , 2 , 3 } and 2 < k 2 Z . With the same notations as above, we have the following assertions.

  1. The maps Φ and r are Hecke equivariant homomorphisms, i.e., for F S k ! , + ( p ) and f , g S k + ( p ) ,

    Φ ( T ( F ) ) = ( T × T ) ( Φ ( F ) )

    and

    1 k r ( T f , T g ) = T ^ r ( f , g ) .

  2. ker Φ = D k 1 ( M 2 k ! , + ( p ) ) .

  3. P r + = r Φ .

This paper is organized as follows. In Section 2, we give examples which illustrate Theorems 1.2, 1.3, and 1.5. In Sections 3 and 4, we prove Theorems 1.6 and 1.8, respectively. Next, proofs of Theorems 1.2, 1.3, and 1.5 are given in Section 5. Finally, in Section 6 we prove Proposition 1.10 and Theorem 1.11.

2 Examples

Example 2.1

(Cf. [6, Example 2.1] and [11, Examples in p. 6170]) Let p = 1 and k = 12 . In this case, we have t = dim S 12 + ( 1 ) = 1 . Let η ( z ) be the Dedekind eta function defined by η ( z ) = q 1 24 n = 1 ( 1 q n ) . Using Δ ( z ) = η ( z ) 24 S 12 + ( 1 ) and the Hauptmodul j 1 ( z ) = E 4 ( z ) 3 / Δ ( z ) 744 for Γ 0 + ( 1 ) one can express f 12 , m ( 1 m 1 ) as follows:

f 12 , 1 ( z ) = Δ ( z ) = q 24 q 2 + 252 q 3 1472 q 4 + 4830 q 5 6048 q 6 16744 q 7 + 84480 q 8 113643 q 9 + O ( q 10 ) f 12 , 0 ( z ) = Δ ( j 1 + 24 ) = 1 + 196560 q 2 + 16773120 q 3 + 398034000 q 4 + 4629381120 q 5 + 34417656000 q 6 + O ( q 7 ) f 12 , 1 ( z ) = Δ ( j 1 2 + 24 j 1 393444 ) = 1 q + 47709536 q 2 + 39862705122 q 3 + 7552626810624 q 4 + 609136463852480 q 5 + O ( q 6 ) .

It is immediate from the definitions of f 1 , f 1 , h 1 that

(14) f 1 = f 12 , 1 and h 1 = f 1 = f 12 , 1 .

Then one has from Remark 1.1(iii) and Theorem 1.2,

F α = h 1 = f 12 , 1 = q 1 n = 2 a 12 ( 1 , n ) q n ,

and if we take l = 3 and w = 1 in (6), by using the Sturm bound one verifies that

n = 1 a 12 ( 1 , 3 n ) q n a 12 ( 1 , 3 ) = q + 27947672851540608 39862705122 q 2 + 340389905850815087232 39862705122 q 3 + 652352555863500246844416 39862705122 q 4 + Δ ( mod 3 10 ) .

Example 2.2

Let p = 5 and k = 10 . In this case, one has t = dim S 10 + ( 5 ) = 1 and t = dim S 10 ( 5 ) = 2 . Let Δ 5 + ( z ) = ( η ( z ) η ( 5 z ) ) 4 be the unique cusp form in S 4 + ( 5 ) and E 6 ( z ) be the normalized Eisenstein series of weight six. We also let j 5 + ( z ) be the Hauptmodul for Γ 0 + ( 5 ) , given by

j 5 + ( z ) = η ( z ) η ( 5 z ) 6 + 6 + 5 3 η ( 5 z ) η ( z ) 6 .

One then computes that

f 10 , 1 = Δ 5 + E 6 + = q 8 q 2 114 q 3 448 q 4 625 q 5 + 912 q 6 + 4242 q 7 + 7680 q 8 6687 q 9 + , f 10 , 1 = Δ 5 + E 6 + ( j 5 + 2 + 8 j 5 + 90 ) = 1 q 192 q 2 14511 q 3 370176 q 4 5152500 q 5 , f 8 , 2 = ( Δ 5 + ) 2 = 1 q 2 + 8 q + 44 + 192 q + 726 q 2 + 2472 q 3 + 7768 q 4 + 22880 q 5 + , f 8 , 3 = ( Δ 5 + ) 2 ( j 5 + 8 ) = 1 q 3 + 114 q + 1672 + 14511 q + 94848 q 2 + 515774 q 3 + 2454144 q 4 + 10533315 q 5 + , f 8 , 3 = ( Δ 5 + ) 3 E 4 = 1 q 3 + 2 q 2 120 q 1740 14855 q 96200 q 2 520532 q 3 2469320 q 4 10578425 q 5 + ,

where E 6 + = 1 1 + 5 3 ( E 6 + E 6 6 W 5 ) and E 4 = 1 1 5 2 ( E 4 E 4 4 W 5 ) . We then obtain that

f 1 = f 10 , 1 , h 1 = f 1 = f 10 , 1 , and w 1 = β 11 f 8 , 3 + 2 β 11 f 8 , 2 ,

where β 11 = 1 CT ( f 1 f 8 , 3 ) = 1 250 . Then one has from Remark 1.1(iii) and Theorem 1.3(ii),

F α = h 1 + D 9 ( w 1 + β 11 f 8 , 3 ) + 4 25 f 1 = n = 3 c α ( n ) q n ,

and if we take l = 3 and w = 1 in (6), by using the Sturm bound one verifies that

n = 1 c α ( 3 n ) q n c α ( 3 ) = 6561 6308 q 18528264 1577 q 2 + 808269273 1577 q 3 + 68622811200 1577 q 4 533626633125 1577 q 5 + 2832551189208 1577 q 6 f 1 ( mod 3 8 ) .

Example 2.3

(Cf. [1, Example 2.10]) Let p = 5 and k = 12 . In this case, t = dim S 12 + ( 5 ) = 3 and the space S 12 + ( 5 ) is spanned by

f 12 , 3 ( z ) = Δ 5 + ( z ) 3 = q 3 12 q 4 + 54 q 5 88 q 6 + , f 12 , 2 ( z ) = Δ 5 + ( z ) 3 ( j 5 + ( z ) + 12 ) = q 2 + 44 q 4 288 q 5 + 306 q 6 + , f 12 , 1 ( z ) = Δ 5 + ( z ) 3 ( j 5 + ( z ) 2 + 12 j 5 + ( z ) 178 ) = q + 2608 q 4 65 q 5 + 23472 q 6 + .

It then follows from [1, Example 2.10] that the Hecke eigenforms are given by

f 1 = f 12 , 1 24 f 12 , 2 + 252 f 12 , 3 , f 2 = f 12 , 1 + ( 10 + 6 151 ) f 12 , 2 + ( 110 + 32 151 ) f 12 , 3 , f 3 = f 12 , 1 + ( 10 6 151 ) f 12 , 2 + ( 110 32 151 ) f 12 , 3 ,

so that K f 1 = Q and K f 2 = K f 3 = Q ( 151 ) . Now utilizing [6, Theorem 1.2(i)] one finds that

f 1 = 17 131 f 12 , 1 16384 655 f 12 , 2 + 531441 1310 f 12 , 3 , f 2 = 3 ( 2869 + 43 151 ) 19781 f 12 , 1 + 512 ( 2416 + 181 151 ) 98905 f 12 , 2 + 177147 ( 453 + 7 151 ) 395620 f 12 , 3 , f 3 = 3 ( 2869 43 151 ) 19781 f 12 , 1 + 512 ( 2416 181 151 ) 98905 f 12 , 2 + 177147 ( 453 7 151 ) 395620 f 12 , 3 ,

so that K f i = K f i for each i { 1 , 2 , 3 } , as expected from Theorem 1.5.

3 Proof of Theorem 1.6

We recall that every H 2 k ( Γ 0 ( p ) ) has a canonical decomposition [23, Section 4.2]

( z ) = ( z ) + + ( z ) ,

where (respectively, + ) is nonholomorphic (respectively, holomorphic) on the complex upper-half plane H . The holomorphic part + has a Fourier expansion

+ ( z ) = n a + ( n ) q n ( q n e 2 π i z ) .

Then we call + a mock modular form if 0 . We define

H 2 k ( Γ 0 + ( p ) ) { H 2 k ( Γ 0 ( p ) ) 2 k W p = } .

For each H 2 k ( Γ 0 + ( p ) ) we define the W p -mock modular period function for + by

P ( + , W p ; z ) b k 1 ( + + 2 k W p ) ( z ) ,

where b k = Γ ( k 1 ) ( 4 π ) k 1 . Let L k be the Maass lowering operator L k defined by

L k 2 i y 2 τ ¯ = i y 2 x + i y .

Then for k 2 the differential operator ξ 2 k y k L 2 k ¯ defines antilinear maps

ξ 2 k : H 2 k ( Γ 0 + ( p ) ) S k ( Γ 0 + ( p ) ) .

Lemma 3.1

Choose G n H 2 k ( Γ + 0 ( p ) ) such that ξ 2 k ( G n ) = f n . We then have

  1. ξ 2 k ( G n c ) = f n c = f n and

  2. G n c H 2 k ( Γ 0 + ( p ) ) .

Proof

  1. For simplicity, we let G = G n . If we write the Fourier expansion of G as

    G = G + + n 0 c G ( n ) Γ ( k 1 , 4 π n y ) q n ,

    then

    G c = ( G + ) c + n 0 c G ( n ) ¯ Γ ( k 1 , 4 π n y ) q n ,

    so that

    ξ 2 k ( G c ) = n > 0 ( 4 π n ) k 1 c G ( n ) q n .

    Since ξ 2 k ( G ( z ) ) = f n ( z ) = n > 0 ( 4 π n ) k 1 c G ( n ) ¯ q n , we must have ξ 2 k ( G c ) = f n c . Moreover since f n R [ [ q ] ] (see [26, p.263]), we readily have f n c = f n .

  2. To show that G c H 2 k ( Γ 0 + ( p ) ) , we first easily check that

    G c 2 k γ = G c for all γ Γ 0 + ( p ) .

    Second, we obtain that

    Δ 2 k ( G c ) = ξ k ξ 2 k ( G c ) = ξ k ( f n c ) = 0 .

    Third, the growth condition is immediate from that of G . Thus, the assertion is proved.□

Lemma 3.2

With the same notations as above, we have

r + ( f n , z ) = c k P ( G n + ( z ¯ ) ¯ , W p ; z ) ,

where c k = Γ ( k 1 ) ( 2 π i ) k 1 .

Proof

By Theorem 1.5 in [8] we have P ( G n + , W p ; z ) ¯ = c k 1 r + ( f n , z ¯ ) . Thus, we obtain that

(15) r + ( f n , z ) = c k P ( G n + , W p ; z ¯ ) ¯ .

Since P ( G n + , W p ; z ) is, by definition, b k 1 ( G n + G n + 2 k W p ) ( z ) , we compute that

(16) P ( G n + , W p ; z ¯ ) ¯ = P ( G n + ( z ¯ ) ¯ , W p ; z ) .

Combining (15) and (16), we obtain the assertion.□

Lemma 3.3

With the same notations as above, we write the Fourier expansion of G n + as

G n + ( z ) = m c G n + ( m ) q m .

We then have

(17) r + ( f n , z ) = r + ( D k 1 ( G n c ) , z ) c G n + ( 0 ) c k p 0 b k W k 2 + ( p )

and

(18) T ^ ( r + ( f n , z ) ) = l 1 k r + ( T ( D k 1 ( G n c ) ) ) ( 1 + l 1 k ) c G n + ( 0 ) c k p 0 b k .

Proof

For simplicity, we let G = G n and a = c G + ( 0 ) . Note that D k 1 ( G c ) S k ! , + ( p ) has Fourier expansion

D k 1 ( G c ) = n 0 n k 1 c G + ( n ) ¯ q n

and

D k 1 ( G c ) = n 0 c G + ( n ) ¯ q n = G + ( z ¯ ) ¯ a .

Thus, we come up with

b k 1 P ( G + ( z ¯ ) ¯ , W p ; z ) = D k 1 ( G c ) D k 1 ( G c ) 2 k W p + a a 2 k W p = c k 1 r + ( D k 1 ( G c ) , z ) a p 0 .

Then (17) is immediate from the above equation and Lemma 3.2. Applying the Hecke operator T ^ to (17) and then employing the fact that T ^ r + = r + 1 k T with (EP1) we obtain (18).□

In what follows, we will simply denote G = G n and λ = λ ( n , ) . It then follows from [27, Theorem 7.10] that

ξ 2 k ( T ( G c ) ) = 1 k λ ξ 2 k ( G c ) = 1 k λ f n .

Thus, we have

(19) T ( G c ) = 1 k λ G c + h

for some h M 2 k ! , + ( p ) . Comparing the constant terms of holomorphic parts in both sides of (19) we obtain that

(20) c G + ( 0 ) + 1 k c G + ( 0 ) = 1 k λ l c G + ( 0 ) + c h ( 0 ) ,

where c h ( 0 ) denotes the constant term of h . Now, applying D k 1 to both sides of (19) we obtain that

(21) D k 1 ( T ( G c ) ) = 1 k λ D k 1 ( G c ) + D k 1 ( h ) .

Meanwhile, we compute that

(22) T ( D k 1 ( G c ) ) = T n n k 1 c G + ( n ) ¯ q n = n ( n ) k 1 c G + ( n ) ¯ + k 1 n k 1 c G + n ¯ q n = k 1 D k 1 n c G + ( n ) ¯ + 1 k c G + n ¯ q n = k 1 D k 1 ( T ( G c ) ) .

Therefore, we find that

(23) 1 k λ D k 1 ( G c ) + D k 1 ( h ) = 1 k T ( D k 1 ( G c ) ) ,

which gives rise to

r + ( T ( D k 1 ( G c ) ) ) = k 1 r + ( D k 1 ( h ) ) + λ r + ( D k 1 ( G c ) ) = k 1 r + ( D k 1 ( h ) ) + λ ( b k r + ( f n , z ) + a c k p 0 ) by ( 17 ) ,

where a = c G + ( 0 ) . Thus, we obtain that

b k T ^ ( r + ( f n , z ) ) = 1 k r + ( T ( D k 1 ( G c ) ) ) ( 1 + 1 k ) a c k p 0 by ( 18 ) = 1 k ( k 1 r + ( D k 1 ( h ) ) + λ ( b k r + ( f n , z ) + a c k p 0 ) ) ( 1 + 1 k ) a c k p 0 = r + ( D k 1 ( h ) ) + 1 k λ b k r + ( f n , z ) + 1 k λ a c k p 0 ( 1 + 1 k ) a c k p 0 = c k c h ( 0 ) p 0 + 1 k λ l b k r + ( f n , z ) + 1 k λ a c k p 0 ( 1 + 1 k ) a c k p 0 by [8, p. 3373 line 9] = 1 k λ b k r + ( f n , z ) by (20).

This proves Theorem 1.6(i). Theorem 1.6(ii) is an immediate consequence of Theorem 1.6(i). Now it remains to prove Theorem 1.6(iii). Since we know that dim W k 2 + ( p ) = 2 t + 1 , it suffices to show that the set

{ p 0 ( z ) , r + + ( f 1 ) , , r + + ( f t ) , r + ( f 1 ) , , r + ( f t ) }

is linearly independent. Suppose that

(24) n = 1 t a n r + + ( f n ) + n = 1 t b n r + ( f n ) + c p 0 = 0

for some a n , b n , c C . Thus, we have

n = 1 t b n r + ( f n ) = n = 1 t a n r + + ( f n ) c p 0 .

Since the left side of the above equation is an odd polynomial while the right side is even, we must have

n = 1 t b n r + ( f n ) = 0 = n = 1 t a n r + + ( f n ) c p 0 .

It follows from [28, Theorem 1.1 and Remark 1.3] that the maps r + + and r + are injective and the image of r + + does not contain p 0 . Thus, we have c = 0 = n a n f n = n b n f n and therefore c = 0 = a n = b n = 0 for all n { 1 , , t } .

4 Proof of Theorem 1.8

  1. Suppose that H n H 2 k ( Γ 0 + ( p ) ) such that

ξ 2 k H n = f n = ξ 2 k G n .

Thus, we see that H n G n M 2 k ! , + and therefore D k 1 ( H n ) D k 1 ( G n ) D k 1 ( M 2 k ! , + ) . We then have [ D k 1 ( H n ) ] = [ g n ] , which implies that the class [ g n ] is independent of the choice of G n and uniquely determined by f n . Since g n belongs to the space S k ! , + ( p ) we can write g n as

g n = m = 1 t a m ( n ) f m + m = 1 t b m ( n ) f m + D k 1 v n

for some a m ( n ) , b m ( n ) C , and v n M 2 k ! , + ( p ) . We observe from [8, Theorem 1.1] that

(25) { D k 1 v n , f } = 0 for every f S k ! , + ( p ) .

For each m { 1 , , t } , pairing with f m yields

{ g n , f m } = j = 1 t a j ( n ) { f j , f m } + j = 1 t b j ( n ) { f j , f m } + { D k 1 v n , f m } since our pairing is bilinear = a m ( n ) + { D k 1 v n , f m } since { f j , f m } = δ j m and { f j , f m } = 0 = a m ( n ) by (25).

Similarly, pairing with f m we obtain that

{ g n , f m } = b m ( n ) .

Moreover, one has

a m ( n ) = { g n , f m } = { f m , g n } = { f m , D k 1 G n c } = ( f m , ξ 2 k G n c ) reg by [5, (1.16)] or [8, Lemma 2.2] or [29, Theorem 3.1] = ( f m , f n ) reg by Lemma 3.1(i) = ( f m , f n ) = 0 , if m n , ( f n , f n ) , if m = n .

Similarly, one finds that

b m ( n ) = { f m , g n } = ( f m , f n ) reg .

Thus, g n can be rewritten as

(26) g n = a n ( n ) f n + m = 1 t b m ( n ) f m + D k 1 v n = a n ( n ) ( h n i = 1 i n t x i ( n ) f i ) + m = 1 t b m ( n ) f m + D k 1 v n = a n ( n ) h n + b n ( n ) f n + i = 1 i n t ( b i ( n ) a n ( n ) x i ( n ) ) f i + D k 1 v n .

Now let

(27) C n i = 1 i n t ( b i ( n ) a n ( n ) x i ( n ) ) f i .

Then for each prime l with gcd ( l , p ) = 1 , we observe that

T ( C n ) = T ( g n a n ( n ) h n b n ( n ) f n D k 1 v n ) = λ ( n , ) ( g n a n ( n ) h n b n ( n ) f n ) + D k 1 E for some E M 2 k ! , + ( p ) , by (23) and (EF2) = λ ( n , ) C n + D k 1 E ˜ for some E ˜ M 2 k ! , + ( p ) .

This implies that

D k 1 E ˜ = T C n λ ( n , l ) C n S k + ( p ) D k 1 M 2 k ! , + ( p ) = { 0 } ,

and therefore

T C n = λ ( n , l ) C n for each prime l with gcd ( l , p ) = 1 .

Thus, by multiplicity one theorem has C n f n , which together with (27) yields that

C n f n f 1 , , f n 1 , f n + 1 , , f t = { 0 } .

It then follows from (26) that

g n = a n ( n ) h n + b n ( n ) f n + D k 1 v n .

Thus, we have

(28) g n = ( f n , f n ) h n + ( f n , f n ) reg f n + D k 1 v n = b k g n + D k 1 v n ,

which means that [ g n ] = [ b k g n ] .

(ii) We obtain from (17) that

(29) b k r + ( f n , z ) + c G n + ( 0 ) c k p 0 = r + ( g n ) = a n ( n ) r + ( h n ) + b n ( n ) r + ( f n ) + c k c v n ( 0 ) p 0 ,

where c v n ( 0 ) stands for the constant term in the Fourier expansion of v n . Thus, we have an equality

A n b k r + ( f n , z ) a n ( n ) ( r + ( h n ) + c 0 ( n ) p 0 ) b n ( n ) r + ( f n ) = ( c k c v n ( 0 ) a n ( n ) c 0 ( n ) c G n + ( 0 ) c k ) p 0 B n ,

where c 0 ( n ) is the constant appeared in (EP2). Applying T ^ and utilizing (EP1) and (EP2) we obtain that

(30) 1 k λ ( n , ) A n = ( 1 + 1 k ) B n .

Now choose a prime l different from p such that 1 k λ ( n , ) 1 + 1 k . Indeed, if there is no such , then we have λ ( n , ) = k 1 + 1 for all , which contradicts a well-known estimate λ ( n , ) 2 k 1 2 (see [26, (15)]). For such , since A n = B n , we obtain from (30) that A n = B n = 0 , which renders

b k r + ( f n , z ) a n ( n ) ( r + ( h n ) + c 0 ( n ) p 0 ) b n ( n ) r + ( f n ) = 0

and

(31) ( c k c v n ( 0 ) a n ( n ) c 0 ( n ) c G n + ( 0 ) c k ) p 0 = 0 .

Combining (29) and (31) we obtain the assertion.

5 Proofs of Theorems 1.2, 1.3, and 1.5

Proof of Theorem 1.2

We recall that ξ 2 k G n c ( f n , f n ) = f n c ( f n , f n ) . It follows from (28) that

D k 1 ( G n c ) = ( f n , f n ) h n + ( f n , f n ) reg f n + D k 1 v n

for some v n M 2 k ! , + ( 1 ) . Thus, we obtain that

D k 1 G n c ( f n , f n ) v n ( f n , f n ) = h n + ( f n , f n ) reg ( f n , f n ) f n ,

whose principal part is equal to prin ( f n ) K f n [ q 1 ] . Here prin ( ) means the principal part at the cusp except for the constant term. Let F n G n c ( f n , f n ) v n ( f n , f n ) . As mentioned in Remark 1.1(ii), let Q H 2 k ( Γ 0 ( 1 ) ) , which is good for f n c with Q + = O ( q t ) . Since ξ 2 k ( Q F n ) = 0 , we obtain Q F n M 2 k ! ( 1 ) . Moreover, we observe Q F n = Q + F n + O ( q t ) since the order at of the holomorphic part of a harmonic weak Maass form of weight 2 k is the same as the order of the pole of its image under D k 1 . But it follows from [2, Remark 3.8] that

max { ord f f 0 M 2 k ! ( 1 ) } = 1 t ,

which forces Q F n to be zero. Thus, CT ( F n + ) K f n and therefore F n is good for f n c and D k 1 ( F n ) = h n + ( f n , f n ) reg ( f n , f n ) f n , as desired.□

Proof of Theorem 1.3

(i) Let Q = Q n be good for f n c . Write Q = Q + Q W p 2 + Q Q W p 2 . Since ξ 2 k ( Q ) = ξ 2 k Q + Q W p 2 = f n c ( f n , f n ) , we obtain Q Q W p 2 M 2 k ! , ( p ) . It then follows from Theorem 1.8 that

D k 1 Q + Q W p 2 = h n + ( f n , f n ) reg ( f n , f n ) f n ,

so that

D k 1 Q + Q W p 2 = h n + ( f n , f n ) reg ( f n , f n ) f n + D k 1 ( v n )

for some v n M 2 k ! , + ( p ) . We then obtain that

prin D k 1 Q Q W p 2 = prin D k 1 Q + Q W p 2 = prin ( f n ) + prin ( D k 1 ( v n ) ) ,

which renders

(32) prin D k 1 Q Q W p 2 v n = prin ( f n ) .

Now we find

CT f m Q Q W p 2 = CT f m Q Q W p 2 v n = { f m , f n } (by (32)) = δ m n ,

where the first equality follows from the residue theorem because the residue of the meromorphic 1-form d ( f m v n ) is given by CT ( f m v n ) . Utilizing Remark 1.1(ii) we can take Q such that Q + = O ( q t t ) . Thus, we can write

(33) Q Q W p 2 = j = 1 t b j n f 2 k , t + j for some b j n C .

Then we have

(34) j = 1 t CT ( f m f 2 k , t + j ) b j n = δ m n ,

which proves the assertion.

(ii) We write

D k 1 w n j = 1 t β j n f 2 k , t + j = i = 1 t α i n q i + O ( 1 ) for some α i n C

so that

CT f m w n j = 1 t β j n f 2 k , t + j = i = 1 t λ ( m , i ) α i n ( i ) k 1 .

Meanwhile, by the residue theorem the left-hand side of the above identity reduces to

CT f m j = 1 t β j n f 2 k , t + j = δ m n .

Comparing this with the identities in [6, (20)–(22)] we obtain

(35) prin D k 1 w n j = 1 t β j n f 2 k , t + j = prin ( f n ) .

We recall from (28) that

D k 1 ( G n c ) = ( f n , f n ) h n + ( f n , f n ) reg f n + D k 1 v n

for some v n M 2 k ! , + ( p ) . Now we let

F n G n c ( f n , f n ) v n ( f n , f n ) + w n + j = 1 t β j n f 2 k , t + j .

We then have ξ 2 k F n = f n c ( f n , f n ) and

(36) D k 1 F n = h n + ( f n , f n ) reg ( f n , f n ) f n + D k 1 w n + j = 1 t β j n f 2 k , t + j

whose principal part is equal to

prin ( f n ) + prin D k 1 w n + j = 1 t β j n f 2 k , t + j = O ( q t t ) .

Now we investigate the principal part of F n at the cusp 0. To this end, we consider

D k 1 F n 2 k W p 1 / p 0 0 p = D k 1 ( F n ) k W p 1 / p 0 0 p by Bol’s identity = h n + ( f n , f n ) reg ( f n , f n ) f n k W p 1 / p 0 0 p + D k 1 ( w n ) k W p 1 / p 0 0 p + D k 1 j = 1 t β j n f 2 k , t + j k W p 1 / p 0 0 p = p k h n z p + ( f n , f n ) reg ( f n , f n ) f n z p + p k D k 1 ( w n ) z p D k 1 j = 1 t β j n f 2 k , t + j z p

whose principal part is equal to that of p k f n z p + p k f n z p = 0 by (35). This means that the principal part of F n at the cusp 0 is constant. As mentioned in Remark 1.1(ii) and in the proof of the assertion (i), let Q H 2 k ( p ) , which is good for f n c with Q + = O ( q t t ) . Since ξ 2 k ( Q F n ) = 0 and the principal part of Q F n at the cusp 0 is constant, we obtain Q F n M 2 k ( p ) . Moreover, we observe Q F n = Q + F n + O ( q t t ) . But it follows from [12] that

max { ord f f 0 M 2 k ( p ) } = 1 t t ,

which forces Q F n to be zero. Thus, CT ( F n + ) K f n and hence F n is good for f n c . Now the assertion follows from (36).□

Proof of Theorem 1.5

First we prove that prin ( h n ) K f n [ q 1 ] . In the case of p = 1 , this immediately follows from Theorem 1.2. Now let p { 2 , 3 , 5 , 7 , 13 } . We adopt the same notation as in Theorem 1.3. Since Q is good, prin ( Q ) K f n [ q 1 ] and Q W p has constant principal part. Meanwhile (33) implies that

prin Q 2 = prin Q Q W p 2 = j = 1 t b j n q t + j + O ( q t ) K f n [ q 1 ] since m 2 k = 1 t by [3] .

We note from (34) that β j n = b j n K f n . Thus, by (36) we obtain that

(37) prin ( h n ) + prin ( D k 1 w n ) K f n [ q 1 ] .

Since β j n K f n and f 2 k , m has rational Fourier coefficients, the definition of w n implies that

(38) prin ( w n ) K f n [ q 1 ] .

By (37) and (38) one has prin ( h n ) K f n [ q 1 ] . Observing that prin ( h n ) = prin ( f n ) , we obtain that

(39) prin ( f n ) K f n [ q 1 ] .

Since f n = m = 1 t μ ( m , n ) f k , m , it then follows from (39) that μ ( m , n ) K f n for each n { 1 , t } and hence every coefficient of f n is contained in K f n .□

6 Proofs of Proposition 1.10 and Theorem 1.11

Proof of Proposition 1.10

The assertion follows immediately from the statement (EF3) and the relation between G n and h n .□

Proof of Theorem 1.11

(i) It follows from (12) that

(40) T ( F ) = n α n ( T ( g n ) + λ f n ) + n β n λ f n + T ( D k 1 g F ) ,

for some g F M 2 k ! , + ( p ) . Meanwhile, we know from (21) and (22) that

l 1 k T ( D k 1 G n c ) = D k 1 ( T ( G n c ) ) = 1 k λ D k 1 ( G n c ) + D k 1 ( h )

for some h M 2 k ! , + ( p ) . Thus, we find that

(41) T ( g n ) = λ g n + D k 1 ( k 1 h ) .

Combining (40) and (41) we obtain that

T ( F ) = n α n λ ( g n + f n ) + n β n λ f n + D k 1 H

for some H M 2 k ! , + ( p ) . It then follows from the definition of the map Φ that

Φ ( T ( F ) ) = 2 n α n λ f n + n β n λ f n , n β n λ f n = ( T × T ) ( Φ ( F ) ) .

Next for f , g S k + ( p ) we write f = n a n f n and g = n b n f n for some a n , b n C . We then have

T ^ r ( f , g ) = T ^ r n a n f n , n b n f n = T ^ r + n a n f n + i r + + n b n f n = n a n T ^ ( r + ( f n ) ) + i n b n T ^ ( r + + ( f n ) ) = n a n 1 k λ ( n , ) r + ( f n ) + i n b n 1 k λ ( n , ) r + + ( f n ) by Theorem 1.6(ii) = n a n 1 k r + ( T f n ) + i n b n 1 k r + + ( T f n ) = 1 k r + ( T n a n f n ) + i r + + T n b n f n = 1 k ( r + ( T f ) + i r + + ( T g ) ) = 1 k r ( T f , T g ) , as desired.

(ii) Given F S k ! , + ( p ) , by (12) we can write

(42) F = n α n ( g n + f n ) + n β n f n + D k 1 g F , for some g F D k 1 M 2 k ! , + ( p ) .

We then observe from (13) that

Φ ( F ) = ( 0 , 0 ) 2 n α n f n + n β n f n = 0 and n β n f n = 0 β n = α n = 0 for all n { 1 , 2 , , t } F = D k 1 g F ,

which proves the assertion.

(iii) First we note that

r + ( f n , z ) + r + ( f n , z ) = 2 i r + + ( f n , z ) r + ( f n , z ) + r + ( f n , z ) = 2 r + ( f n , z ) r + ( f n , z ) = r + ( f n , z ) + i r + + ( f n , z ) .

Now we compute that

P r + ( F ) = P r + n α n ( g n + f n ) + n β n f n + D k 1 g F = P n α n ( r + ( f n , z ) + r + ( f n , z ) ) + n β n r + ( f n , z ) + a p 0 for some a C by Theorem 1.8(ii) = P n α n 2 r + ( f n , z ) + n β n ( r + ( f n , z ) + i r + + ( f n , z ) ) + a p 0 = n α n 2 r + ( f n , z ) + n β n ( r + ( f n , z ) + i r + + ( f n , z ) ) .

Meanwhile,

r Φ ( F ) = r 2 n α n f n + n β n f n , n β n f n = 2 n α n r + ( f n , z ) + n β n r + ( f n , z ) + i n β n r + + ( f n , z ) .

Thus, we have P r + ( F ) = r Φ ( F ) .□

Acknowledgements

We would like to thank the referees for valuable comments that help to improve our manuscript.

  1. Funding information: Choi was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIT) (No. 2020R1A2C1A01007112). Kim was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIP) (NRF-2021R1A2C1003998 and 2016R1A5A1008055).

  2. Conflict of interest: Authors state no conflict of interest.

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Received: 2021-09-18
Revised: 2022-01-18
Accepted: 2022-01-18
Published Online: 2022-04-19

© 2022 SoYoung Choi and Chang Heon Kim, published by De Gruyter

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

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