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BY 4.0 license Open Access Published by De Gruyter Open Access October 28, 2022

Modular forms of half-integral weight on Γ0(4) with few nonvanishing coefficients modulo

  • Dohoon Choi and Youngmin Lee EMAIL logo
From the journal Open Mathematics

Abstract

Let k be a nonnegative integer. Let K be a number field and O K be the ring of integers of K . Let 5 be a prime and v be a prime ideal of O K over . Let f be a modular form of weight k + 1 2 on Γ 0 (4) such that its Fourier coefficients are in O K . In this article, we study sufficient conditions that if f has the form

f ( z ) n = 1 i = 1 t a f ( s i n 2 ) q s i n 2 ( mod v )

with square-free integers s i , then f is congruent to a linear combination of iterated derivatives of a single theta function modulo v .

MSC 2010: 11F33; 11F80

1 Introduction

The Fourier coefficients of modular forms of half-integral weight are related to various objects in number theory and combinatorics such as the algebraic parts of the central critical values of modular L-functions, orders of Tate-Shafarevich groups of elliptic curves, the number of partitions of a positive integer, and so on. With a lot of application to these objects, Bruinier [1], Bruinier and Ono [2], Ono and Skinner [3], Ahlgren and Boylan [4,5], and the others studied congruence properties modulo a power of a prime for Fourier coefficients of modular forms of half-integral weight. Many of them considered modular forms of half-integral weight whose the Fourier coefficients are supported on only finitely many square classes modulo a prime .

Let f be a modular form of half-integral weight on Γ 1 ( 4 N ) . Vignéras [6] proved that if the q -expansion of f has the form

f ( z ) = a f ( 0 ) + n = 1 i = 1 t a f ( s i n 2 ) q s i n 2 , q e 2 π i z

with a positive integer t and square-free integers s i , then f is a linear combination of single variable theta functions (a different proof of this result was given by Bruinier [1]). Many of the aforementioned results can be considered as positive characteristic extensions of Vignéras’ result on classification of modular forms of half-integral weight such that their nonvanishing Fourier coefficients lie in only finitely many square classes. Especially, Ahlgren et al. [7] obtained an explicit mod analog of the result of Vignéras for modular forms of half-integral weight on Γ 0 ( 4 ) satisfying the Kohnen-plus condition.

Let K be a number field and O K be the ring of integers of K . Let M k + 1 2 ( Γ 0 ( 4 ) ; O K ) (resp. S k + 1 2 ( Γ 0 ( 4 ) ; O K ) ) be the space of modular forms (resp. cusp forms) of weight k + 1 2 on Γ 0 ( 4 ) such that their Fourier coefficients are in O K and S k + 1 2 + ( Γ 0 ( 4 ) ; O K ) be the subspace of S k + 1 2 ( Γ 0 ( 4 ) ; O K ) consisting of f S k + 1 2 ( Γ 0 ( 4 ) ; O K ) satisfying the Kohnen-plus condition.

Let 5 be a prime and v be a prime ideal of O K over . For f S k + 1 2 + ( Γ 0 ( 4 ) ; O K ) , Ahlgren et al. [7] proved that if

(1.1) k + 1 2 < + 3 2

and

(1.2) f ( z ) n = 1 i = 1 t a f ( s i n 2 ) q s i n 2 ( mod v )

with square-free integers s i , then k is even and

f ( z ) a f ( 1 ) n = 1 n k q n 2 ( mod v ) .

In this article, we study sufficient conditions that if f has the form (1.2), then f is congruent to a linear combination of iterated derivatives of a single theta function modulo v .

For a positive number ε , let P ε be the set of primes such that for every f S k + 1 2 + ( Γ 0 ( 4 ) ; O K ) with k + 1 2 < 2 ( log ) 2 ε , if

f ( z ) n = 1 i = 1 t a f ( s i n 2 ) q s i n 2 ( mod v )

with square-free integers s i , then

f ( z ) a f ( 1 ) n = 1 n k q n 2 + a f ( ) n = 1 n k + 1 2 q n 2 ( mod v ) .

The following theorem proves that the portion of P ε in the set of primes is one.

Theorem 1.1

For a positive integer X, there is an absolute constant C such that

# { : P ε and X } C 0 X ( log X ) 1 + ε 2 1 + C log log X log X ,

where C 0 2 2 π 2 3 p > 2 p 2 p 2 1 .

For a nonnegative real number r , we define an operator Θ r on C [ [ q ] ] by

Θ r n = 0 a ( n ) q n n = 0 n r a ( n ) q n if r Z > 0 , 0 elsewhere.

For convenience, we let Θ Θ 1 . As in Theorem 1.1, the previous results on modular forms of half-integral weight having the form (1.2) such as [1,2,4,5,7] and so on imply that in many cases, if f has the form (1.2), then Θ ( f ) is congruent to a linear combination of iterated derivatives of a single theta function modulo v . These lead us to the following conjecture on modular forms f of half-integral weight having the form (1.2).

Conjecture 1.2

Let K be a number field and O K be the ring of integers of K . Let 5 be a prime and v be a prime ideal of O K over . Assume that f S k + 1 2 ( Γ 0 ( 4 ) ; O K ) has the form

Θ ( f ) ( z ) n = 1 s n 2 a f ( s n 2 ) q s n 2 ( mod v )

with a square-free integer s, then

Θ ( f ) ( z ) 1 2 a f ( 1 ) n Z n n k + 2 q n 2 ( mod v ) .

Assume that is a prime and m is a nonnegative integer. Let r ( m ) be the least positive integer such that

r ( m ) m ( mod 1 ) .

Let α ( , m ) be the smallest nonnegative integer i such that

m + 1 2 < 2 i r ( m ) + 1 2 + 1 2 ,

and β ( , m ) be the smallest nonnegative integer i such that

m + 1 2 < 2 i + 1 r m + 1 2 + 1 2 + 1 2 .

Let

T ( z ) 1 + 2 n = 1 q n 2 .

For convenience, let

n = a b a n n = a b a n if a b , 0 if a > b .

By using Conjecture 1.2, we have an explicit formula for modular forms of half-integral weight having the form (1.2).

Theorem 1.3

Let K , O K , , and v be as in Conjecture 1.2. Assume that f M k + 1 2 ( Γ 0 ( 4 ) ; O K ) . Conjecture 1.2implies that if f has the form

(1.3) f ( z ) a f ( 0 ) + n = 1 i = 1 t a f ( s i n 2 ) q s i n 2 ( mod v )

with square-free integers s i , then the following statements are true.

  1. If r ( k ) 1 and r ( k ) 1 2 , then

    f ( z ) 1 2 i = 0 α ( , k ) 1 a f ( 2 i ) Θ k 2 ( T ) ( 2 i z ) + 1 2 i = 0 β ( , k ) 1 a f ( 2 i + 1 ) Θ ( 2 k + 1 ) 4 ( T ) ( 2 i + 1 z ) ( mod v ) .

  2. If r ( k ) = 1 , then

    f ( z ) a f ( 0 ) T ( z ) + 1 2 i = 0 α ( , k ) 1 ( a f ( 2 i ) 2 a f ( 0 ) ) Θ k 2 ( T ) ( 2 i z ) + 1 2 i = 0 β ( , k ) 1 a f ( 2 i + 1 ) Θ ( 2 k + 1 ) 4 ( T ) ( 2 i + 1 z ) ( mod v ) .

  3. If r ( k ) = 1 2 , then

    f ( z ) a f ( 0 ) T ( z ) + 1 2 i = 0 α ( , k ) 1 a f ( 2 i ) Θ k 2 ( T ) ( 2 i z ) + 1 2 i = 0 β ( , k ) 1 ( a f ( 2 i + 1 ) 2 a f ( 0 ) ) Θ ( 2 k + 1 ) 4 ( T ) ( 2 i + 1 z ) ( mod v ) .

To give numerical evidence for Conjecture 1.2, we consider a basis of the space of modular forms of weight k + 1 2 on Γ 0 ( 4 ) . Let F 2 ( z ) = n = 0 σ ( 2 n + 1 ) q 2 n + 1 be the modular form of weight 2 on Γ 0 ( 4 ) , where σ ( n ) is the sum of positive divisors of n . Then

{ F 2 j T 2 k + 1 4 j } 0 j k 2

is a C -basis of the space of modular forms of weight k + 1 2 on Γ 0 ( 4 ) . Let A k , m be an m × k 2 + 1 matrix such that the ( i , j ) -entry of A k , m is the ( i 1 )th Fourier coefficient of F 2 j 1 T 2 k + 5 4 j modulo . Let B k , m be a submatrix of A k , m obtained by removing n 2 + 1 th rows for all nonnegative integers n with ( , n ) = 1 . Let Null ( B k , m ) be the null space of B k , m . With this notation, we give the following conjecture.

Conjecture 1.4

Let 5 be a prime. Let 1 + be the characteristic function of the set of positive real numbers. Then, for a positive even integer k, we have

lim m dim Null ( B k , m ) = 1 + ( α ( , k ) ) .

By comparing the intersection of the null spaces of B k , m and the space of mod v modular forms of weight k + 1 2 on Γ 0 ( 4 ) having the form

f ( z ) n a ( n 2 ) q n 2 ( mod v ) ,

we have the following theorem.

Theorem 1.5

Conjecture 1.2is equivalent to Conjecture 1.4.

Let us note that Null ( B k , m ) is stable for sufficiently large m . In the proof of Theorem 1.5, we prove that dim Null ( B k , m ) is larger than or equal to 1 + ( α ( , k ) ) for all positive integers m . Hence, if there is a positive integer m such that dim Null ( B k , m ) = 1 + ( α ( , k ) ) , then Conjecture 1.2 is true. To compute dim Null ( B k , m ) , we consider the row echelon form of B k , m . We use C++ in this process. Then we have the following theorem.

Theorem 1.6

Assume that k 1,000 , or that { 5 , 7 , 11 , 13 , 17 , 19 } and k 10,000 . Then, Conjecture 1.2is true.

The remainder of this article is organized as follows. In Section 2, we review some properties of f having the form (1.3) and the filtration for modular forms. In Section 3, we prove Theorems 1.1, 1.3, 1.5, and 1.6.

2 Preliminaries

In this section, we review some notions and properties of the filtration for modular forms, and then we introduce some properties about modular forms of half-integral weight on Γ 0 ( 4 ) such that their Fourier coefficients are supported on finitely many square classes modulo a prime . For further details, see [8].

Throughout the rest of this article, we fix the following notation. For a congruence subgroup Γ and w 1 2 Z , let M w ( Γ ) (resp. S w ( Γ ) ) be the space of modular forms (resp. cusp forms) of weight w on Γ . For a Dirichlet character χ modulo N , let M w ( Γ 0 ( N ) , χ ) (resp. S w ( Γ 0 ( N ) , χ ) ) be the space of modular forms (resp. cusp forms) of weight w on Γ 0 ( N ) with character χ .

Let k be a nonnegative integer and 5 be a prime. Let K be a number field and O K be the ring of integers of K . Let v be a prime ideal of O K over . Let M k + 1 2 ( Γ 0 ( 4 N ) ; O K ) (resp. S k + 1 2 ( Γ 0 ( 4 N ) ; O K ) ) be the space of modular forms (resp. cusp forms) of weight k + 1 2 on Γ 0 ( 4 N ) such that their Fourier coefficients are in O K and S k + 1 2 + ( Γ 0 ( 4 ) ; O K ) be the subspace of S k + 1 2 ( Γ 0 ( 4 ) ; O K ) consisting of f S k + 1 2 ( Γ 0 ( 4 ) ; O K ) satisfying the Kohnen-plus condition.

Now, we review the basic notions and properties about the Shimura correspondence. Assume that f is a cusp form of weight k + 1 2 on Γ 0 ( 4 ) . For a square-free integer t , we define A t ( n ) by

n = 1 A t ( n ) n s n = 1 ( 1 ) k t n 1 n s k + 1 n = 1 a t n 2 ( f ) n s .

Then, the Shimura lift Sh t ( f ) of f is defined by

Sh t ( f ) ( z ) n = 1 A t ( n ) q n .

Note that Sh t ( f ) S 2 k ( Γ 0 ( 2 ) ) . In particular, if f S k + 1 2 + ( Γ 0 ( 4 ) ) , then Sh t ( f ) S 2 k ( Γ 0 ( 1 ) ) . For each odd prime p with p t , we have

Sh t f T p 2 , k + 1 2 = Sh t ( f ) T p , 2 k ,

where T n , w denotes the n th Hecke operator on the space of modular forms of weight w . For each prime , operators U and V on formal power series are defined by

n = 0 a ( n ) q n U n = 0 a ( n ) q n

and

n = 0 a ( n ) q n V n = 0 a ( n ) q n .

2.1 Filtration for modular forms of half integral weight modulo a prime

The theory of filtration for modular forms of integral weight was developed by Serre [9], Swinnerton-Dyer [10], Katz [11], and Gross [12]. From this, the theory of filtration for modular forms of half-integral weight on Γ 0 ( 4 ) was studied. In this section, we review some properties of filtration for modular forms of half-integral weight on Γ 0 ( 4 ) . For the details, we refer to [13, Section 2].

We say that n = 0 a ( n ) q n is congruent to n = 0 b ( n ) q n modulo v , i.e.,

n = 0 a ( n ) q n n = 0 b ( n ) q n ( mod v ) ,

if a ( n ) b ( n ) ( mod v ) for all nonnegative integers n . For f M k + 1 2 ( Γ 0 ( 4 ) ; O K ) , we define a filtration ω ( f ) of f modulo v by

ω ( f ) inf k + 1 2 : there is f M k + 1 2 ( Γ 0 ( 4 ) ; O K ) such that f f ( mod v ) .

For convenience, if f 0 ( mod v ) , then let ω ( f ) = . We summarize the properties of ω ( f ) in the following lemma.

Lemma 2.1

Let f M k + 1 2 ( Γ 0 ( 4 ) ; O K ) . Then, the following statements are true.

  1. k ω ( f ) 1 2 ( mod 1 ) .

  2. ω ( f ) = ω ( f ) .

  3. There is a nonnegative integer k such that

    k k + 1 2 ( mod 1 ) ,

    and there is g M k + 1 2 ( Γ 0 ( 4 ) ; O K ) such that g f U ( mod v ) . Moreover, if f ( z ) n = 0 a f ( n ) q n ( mod v ) , then there is a nonnegative integer k such that

    k k + 1 2 ( mod 1 ) and k + 1 2 1 k + 1 2 ,

    and there is g M k + 1 2 ( Γ 0 ( 4 ) ; O K ) such that g f U ( mod v ) .

  4. There is h S k + + 3 2 ( Γ 0 ( 4 ) ) such that h Θ ( f ) ( mod v ) . In particular, if f S k + 1 2 + ( Γ 0 ( 4 ) ) , then h S k + + 3 2 + ( Γ 0 ( 4 ) ) .

Proof

The proofs of (1) and (2) are in [13, Proposition 2.2]. The proof of (3) is obtained by combining [7, Lemma 4.2] and [13, Proposition 2.2]. To prove (4), let

h k + 1 2 Θ ( E 1 ) f ( 1 ) E 1 Θ ( f ) ,

where E 1 denotes the Eisenstein series of weight 1 . Since E 1 1 ( mod v ) , we have h Θ ( f ) ( mod v ) . By [14, Corollary 7.2], we obtain h S k + + 3 2 ( Γ 0 ( 4 ) ) . When f satisfies the Kohnen-plus condition, the proof of ( 4 ) is in [7, Lemma 4.1].□

2.2 Modular forms of half-integral weight such that their Fourier coefficients are supported on finitely many square classes modulo

In this section, we introduce some properties of modular forms of half-integral weight on Γ 0 ( 4 ) such that their Fourier coefficients are supported on finitely many square classes modulo v .

Ahlgren and Boylan [4] obtained the necessary conditions for the weight of f M k + 1 2 ( Γ 0 ( 4 ) ) such that their Fourier coefficients are supported on finitely many square classes modulo v by using the theory of Galois representations. This was reproved in [15] by using only the theory of filtration for modular forms of integral weight. The Choi and Kilbourn [16] improved the necessary conditions for the weight by using only the theory of filtration for modular forms of integral weight. We review the results [4,16] in the following theorem.

Theorem 2.2

Let N be a positive integer and 5 be a prime with ( , N ) = 1 . Assume that f ( z ) M k + 1 2 ( Γ 1 ( 4 N ) ) O K [ [ q ] ] has the form

f ( z ) a f ( 0 ) + n = 1 i = 1 t a f ( s i n 2 ) q s i n 2 ( mod v )

with square-free integers s i . Let k ¯ and i k be nonnegative integers, which satisfy k = ( 1 ) i k + k ¯ and k ¯ < 1 . Then, the following statements are true.

  1. If n i for some i , then

    k ¯ 2 i k + 1 .

  2. If n i for all i and k ¯ 3 2 , then

    k ¯ i k + 1 2 .

  3. If n i for all i and k ¯ > 3 2 , then

    k ¯ i k + 1 2 .

Bruinier and Ono [2, Theorem 3.1] proved the following theorem by using an argument in [1].

Theorem 2.3

Let N be a positive integer and 5 be a prime with ( , N ) = 1 . Let χ be a real Dirichlet character modulo 4 N and f ( z ) S k + 1 2 ( Γ 0 ( 4 N ) , χ ) O K [ [ q ] ] . For each prime p with ( p , 4 N ) = 1 , if there exists ε p { ± 1 } such that

f ( z ) n p { 0 , ε p } a f ( n ) q n ( mod v ) ,

then

( p 1 ) f T p 2 , k + 1 2 ε p ( 1 ) k p χ ( p ) ( p k + p k 1 ) ( p 1 ) f ( mod v ) .

Ahlgren et al. [7] proved that if f S k + 1 2 + ( Γ 0 ( 4 ) ; O K ) and the Fourier coefficients of f are supported on finitely many square classes modulo v , then f has the form

f ( z ) n = 1 a f ( n 2 ) q n 2 + n = 1 a f ( n 2 ) q n 2 ( mod v ) .

By using the theory of Galois representations, we extend the result [7] to cusp forms of half-integral weight on Γ 0 ( 4 ) without the Kohnen-plus condition.

Proposition 2.4

Assume that f S k + 1 2 ( Γ 0 ( 4 ) ; O K ) has the form

(2.1) f ( z ) n = 1 i = 1 t a f ( s i n 2 ) q s i n 2 ( mod v )

with square-free integers s i . Then, the following statements are true.

  1. If 2 k and 1 ( mod 4 ) , then

    f ( z ) n = 1 a f ( n 2 ) q n 2 + n = 1 a f ( n 2 ) q n 2 ( mod v ) .

  2. If 2 k and 3 ( mod 4 ) , then

    f ( z ) n = 1 a f ( n 2 ) q n 2 ( mod v ) .

  3. If 2 k and 3 ( mod 4 ) , then

    f ( z ) n = 1 a f ( n 2 ) q n 2 ( mod v ) .

  4. If 2 k and 1 ( mod 4 ) , then

    f ( z ) 0 ( mod v ) .

Proof

Assume that for each i { 1 , , t } , there is a positive integer n i such that a f ( s i n i 2 ) 0 ( mod v ) . Following the proof of Lemma 4.1 in [4], there exist distinct odd primes p i , 1 , , p i , r i , each relatively to n i s i , and a modular form f i S k + 1 2 ( Γ 0 ( 4 j = 1 r i p i , j 2 ) ; O K ) such that

f i ( z ) n = 1 gcd ( n , j = 1 r i p i , j ) = 1 a f i ( s i n 2 ) q s i n 2 0 ( mod v ) .

By Theorem 2.3, for each prime p with p 2 s i j = 1 r i p i , j and p 1 ( mod ) , we have

f i T p 2 , k + 1 2 ( 1 ) k s i p ( p k + p k 1 ) f i ( mod v ) .

Since S 1 2 ( Γ 0 ( 4 ) ) = S 3 2 ( Γ 0 ( 4 ) ) = { 0 } , we may assume that k 2 . Let F i Sh s i ( f i ) S 2 k ( Γ 0 ( 2 j = 1 r i p i , j 2 ) ) be the Shimura lift of f i . Since the Shimura correspondence commutes with the Hecke operators, for each prime p with p 2 s i j = 1 r i p i , j and p 1 ( mod ) , we obtain

F i T p , 2 k ( 1 ) k s i p ( p k + p k 1 ) F i ( mod v ) .

Then, there is an integer N i such that N i 2 j = 1 r i p i , j 2 , and there is a newform G i S 2 k ( Γ 0 ( N i ) ) such that for each prime p with p 2 s i j = 1 r i p i , j and p 1 ( mod ) ,

λ i ( p ) ( 1 ) k s i p ( p k + p k 1 ) ( mod v ) .

Here, λ i ( p ) denotes the p th Hecke eigenvalue of G i . Let F v O K v . Note that there is a semi-simple Galois representation

ρ i : Gal ( Q ¯ Q ) GL 2 ( F v ) ,

such that for each prime p with p N i

tr ( ρ i ( Frob p ) ) λ i ( p ) ( mod v ) and det ( ρ i ( Frob p ) ) p 2 k 1 ( mod v ) ,

where Frob p denotes any Frobenius element at p . Let χ : Gal ( Q ¯ Q ) F be the mod- cyclotomic character. Following the argument of the proof of [5, Proposition 4.3], we have

(2.2) ρ i ( 1 ) k s i χ k 0 0 ( 1 ) k s i χ k 1 if s i , ( 1 ) k + 1 2 s i χ k + 1 2 0 0 ( 1 ) k + 1 2 s i χ k + 3 2 if s i ,

where s i = s i .

By the result of Carayol [17], the conductor of ρ i divides N i . By (2.2), we obtain that if s i , then s i 2 divides the conductor of ρ i , and if s i , then ( s i ) 2 divides the conductor of ρ i . Since N i 2 j = 1 r i p i , j 2 and gcd ( s i , j = 1 r i p i , j ) = 1 , we have s i { 1 , } . Moreover, the conductor of ρ i is not divided by 4. Therefore, we conclude that if k is odd, then s i 1 and if k + 1 2 is odd, then s i .□

We extend Proposition 2.4 to general modular forms of half-integral weight including noncusp forms in the following proposition.

Proposition 2.5

Assume that f M k + 1 2 ( Γ 0 ( 4 ) ; O K ) has the form

(2.3) f ( z ) a f ( 0 ) + n = 1 i = 1 t a f ( s i n 2 ) q s i n 2 ( mod v )

with square-free integers s i . Then,

f ( z ) a f ( 0 ) + n = 1 a f ( n 2 ) q n 2 + n = 1 a f ( n 2 ) q n 2 ( mod v ) .

Proof

Without loss of generality, we assume that there is a positive integer n 1 such that a f ( s 1 n 1 2 ) 0 ( mod v ) . Let a be the exponent of in s 1 n 1 2 . Then, there is a unique square-free integer s 1 such that s 1 n 1 2 = a s 1 m 1 2 for some positive integer m 1 . By Lemma 2.1 (3), there is an integer k and a modular form g M k + 1 2 ( Γ 0 ( 4 ) ) such that g f U a ( mod v ) . By Lemma 2.1 (4), there is h S k + + 3 2 ( Γ 0 ( 4 ) ) such that h Θ ( g ) ( mod v ) . Since a f ( s 1 n 1 2 ) 0 ( mod v ) , we have a h ( s 1 m 1 2 ) 0 ( mod v ) and then h has the form (2.1). Then, s 1 = 1 by Proposition 2.4. This implies that s 1 { 1 , } . Therefore, Proposition 2.5 is proved.□

Combining Theorem 2.2 and Proposition 2.5, we obtain an explicit formula of f M k + 1 2 ( Γ 0 ( 4 ) ) having the form (2.3) when k < 1 .

Lemma 2.6

Assume that f M k + 1 2 ( Γ 0 ( 4 ) ; O K ) has the form (2.3) and f 0 ( mod v ) . If k < 1 , then k { 0 , 1 2 } . Moreover,

f ( z ) a f ( 0 ) 1 + 2 n = 1 q n 2 ( mod v ) if k = 0

and

f ( z ) a f ( 0 ) 1 + 2 n = 1 q n 2 ( mod v ) if k = 1 2 .

Proof

We assume that k < 1 . By Theorem 2.2, we have k { 0 , 1 , 1 2 } . Note that M 1 2 ( Γ 0 ( 4 ) ) is generated by T . Thus, when k = 0 , we obtain that f is a constant multiple of T . If f has the form (2.3), then a f ( 2 ) 0 ( mod v ) by Proposition 2.5. Note that M 3 2 ( Γ 0 ( 4 ) ) is generated by T 3 and a T 3 ( 2 ) = 3 . Thus, when k = 1 , we have f 0 ( mod v ) . When k = 1 2 , we have by Theorem 2.2

f ( z ) n = 0 a f ( n ) q n ( mod v ) .

By Lemma 2.1 (3), there is g M 1 2 ( Γ 0 ( 4 ) ) such that g f U ( mod v ) . Since g is a constant multiple of T , f is congruent to a constant multiple of T V modulo v .□

3 Proof of Theorems

In this section, we prove Theorems 1.1, 1.3, 1.5, and 1.6. First, we prove Theorem 1.3.

Proof of Theorem 1.3

We fix a prime 5 . We prove Theorem 1.3 by induction on k . When k < 1 , Theorem 1.3 is true by Lemma 2.6. Thus, we assume that Theorem 1.3 is true when k < k 0 with a fixed positive integer k 0 , where k 0 is a positive integer larger than 1 .

To prove Theorem 1.3, it is enough to show that Theorem 1.3 is true when k = k 0 by induction on k . Assume that f M k 0 + 1 2 ( Γ 0 ( 4 ) ; O K ) has the form (1.3). Then by Lemma 2.5, f has the form

f ( z ) a f ( 0 ) + n = 1 a f ( n 2 ) q n 2 + n = 1 a f ( n 2 ) q n 2 ( mod v ) ,

and

Θ ( 1 ) 2 ( f ) ( z ) 1 2 n Z n a f ( n 2 ) q n 2 ( mod v ) .

By Lemma 2.1 (4), there is g 0 S k 0 + 2 2 ( Γ 0 ( 4 ) ) such that

g 0 Θ ( 1 ) 2 ( f ) ( mod v ) .

Let k 1 max k 0 + 1 2 , ω ( g 0 ) 1 2 . Then, there is g 1 M k 1 + 1 2 ( Γ 0 ( 4 ) ; O K ) such that

g 1 ( z ) ( f Θ ( 1 ) 2 ( f ) ) ( z ) a f ( 0 ) + n = 1 a f ( n 2 ) q n 2 + n = 1 a f ( 2 n 2 ) q 2 n 2 ( mod v ) .

Let k 2 be the largest integer satisfying

(3.1) k 2 + 1 2 1 k 1 + 1 2 and k 2 1 2 + k 1 1 2 + k 0 ( mod 1 ) .

By Lemma 2.1 (3), there is g 2 M k 2 + 1 2 ( Γ 0 ( 4 ) ; O K ) such that

g 2 ( z ) g 1 U ( z ) a f ( 0 ) + n = 1 a f ( n 2 ) q n 2 + n = 1 a f ( 2 n 2 ) q n 2 ( mod v ) .

Since k 0 > 2 , we have

k 2 + 1 2 1 k 1 + 1 2 1 k 0 + 2 2 < k 0 + 1 2 .

For a nonnegative integer k , we define a subset k of M k + 1 2 ( Γ 0 ( 4 ) ) by

k { Θ k 2 ( T ) V 2 i } 0 i < α ( , k ) { Θ ( 2 k + 1 ) 4 ( T ) V 2 i + 1 } 0 i < β ( , k ) { T } if r ( k ) = 1 , { Θ k 2 ( T ) V 2 i } 0 i < α ( , k ) { Θ ( 2 k + 1 ) 4 ( T ) V 2 i + 1 } 0 i < β ( , k ) { T V } if r ( k ) = 1 2 , { Θ k 2 ( T ) V 2 i } 0 i < α ( , k ) { Θ ( 2 k + 1 ) 4 ( T ) V 2 i + 1 } 0 i < β ( , k ) otherwise.

To prove Theorem 1.3, it is enough to show that if f M k 0 + 1 2 ( Γ 0 ( 4 ) ; O K ) has the form (1.3), then f is congruent to a linear combination of k 0 modulo v .

By Proposition 2.4, if k 0 is odd, then g 0 0 ( mod v ) . Combining the assumption that Conjecture 1.2 is true, we have

g 0 a f ( 1 ) 2 Θ k 0 2 ( T ) ( mod v ) .

Since k 2 k 0 + 1 2 ( mod 1 ) , it follows that Θ k 0 2 ( T ) Θ ( 2 k 2 + 1 ) 4 ( T ) ( mod v ) . By the induction hypothesis, g 2 is congruent to a linear combination of k 2 . Since

f f Θ 1 2 ( f ) + Θ 1 2 ( f ) g 2 V + g 0 ( mod v ) ,

we deduce that f is congruent to a linear combination of

{ Θ k 2 2 ( T ) V 2 i + 1 } 0 i < α ( , k 2 ) { Θ ( 2 k 2 + 1 ) 4 ( T ) V 2 i } 0 i < β ( , k 2 ) + 1 { T V } if r ( k 2 ) = 1 , { Θ k 2 2 ( T ) V 2 i + 1 } 0 i < α ( , k 2 ) { Θ ( 2 k 2 + 1 ) 4 ( T ) V 2 i } 0 i < β ( , k 2 ) + 1 { T V 2 } if r ( k 2 ) = 1 2 , { Θ k 2 2 ( T ) V 2 i + 1 } 0 i < α ( , k 2 ) { Θ ( 2 k 2 + 1 ) 4 ( T ) V 2 i } 0 i < β ( , k 2 ) + 1 otherwise.

If r ( k 2 ) = 1 2 , then

T V 2 T Θ ( 1 ) 2 ( T ) T Θ ( 2 k 2 + 1 ) 4 ( T ) ( mod v ) .

Thus, f is congruent to a linear combination of

{ Θ k 2 2 ( T ) V 2 i + 1 } 0 i < α ( , k 2 ) { Θ ( 2 k 2 + 1 ) 4 ( T ) V 2 i } 0 i < β ( , k 2 ) + 1 { T V } if r ( k 2 ) = 1 , { Θ k 2 2 ( T ) V 2 i + 1 } 0 i < α ( , k 2 ) { Θ ( 2 k 2 + 1 ) 4 ( T ) V 2 i } 0 i < β ( , k 2 ) + 1 { T } if r ( k 2 ) = 1 2 , { Θ k 2 2 ( T ) V 2 i + 1 } 0 i < α ( , k 2 ) { Θ ( 2 k 2 + 1 ) 4 ( T ) V 2 i } 0 i < β ( , k 2 ) + 1 otherwise.

To complete the proof, it is sufficient to show that

(3.2) α ( , k 2 ) β ( , k 0 ) and β ( , k 2 ) + 1 α ( , k 0 ) .

First, we assume that k 0 + 1 2 2 2 . Since Θ m ( T ) Θ ( 2 m + 1 ) 2 (