Independence between coefficients of two modular forms

TL;DR

This paper proves that two modular forms with finitely many ratios of Fourier coefficients are scalar multiples of each other, extending the result to weakly holomorphic forms and applying it to quadratic form representations.

Contribution

It establishes a rigidity result linking the finiteness of coefficient ratios to scalar multiples for modular forms, including weakly holomorphic forms.

Findings

Finite ratio set implies forms are scalar multiples

Extension to weakly holomorphic modular forms

Application to quadratic form representation counts

Abstract

Let $k$ be an even integer and $S_k$ be the space of cusp forms of weight $k$ on $\SL_2(\ZZ)$. Let $S = \oplus_{k\in 2\ZZ} S_k$. For $f, g\in S$, we let $R(f, g) = \{ (a_f(p), a_g(p)) \in \mathbb{P}^1(\CC)\ |\ \text{$p$ is a prime} \}$ be the set of ratios of the Fourier coefficients of $f$ and $g$, where $a_f(n)$ (resp. $a_g(n)$) is the $n$th Fourier coefficient of $f$ (resp. $g$). In this paper, we prove that if $f$ and $g$ are nonzero and $R(f,g)$ is finite, then $f = cg$ for some constant $c$. This result is extended to the space of weakly holomorphic modular forms on $\SL_2(\ZZ)$. We apply it to studying the number of representations of a positive integer by a quadratic form.

Equations

$$a_f(p{)}^2=a_g(p{)}^2,$$

$$f=g\otimes \chi .$$

$$M^{!}:={\oplus }_{k\in 2\mathbb{Z}}{M}_k^{!}.$$

$$R(f,g):=\left\{x\in {\mathbb{P}}^1(\mathbb{C})|x=[a_f(p):a_g(p)]\text{for some prime p}\right\}.$$

$$r_Q(n):=\left|\{(x_1,\dots ,x_d)\in {\mathbb{Z}}^d|Q(x_1,\dots ,x_d)=n\}\right|$$

$$1+\sum _{n=1}^{\infty }{r}_Q(n)q^n$$

$$R(Q_1,Q_2):=\left\{x\in {\mathbb{P}}^1(\mathbb{C})|x=[r_{Q_1}(p):r_{Q_2}(p)]\text{for some prime p}\right\}$$

$$R\left(f^{+},g^{+}\right):=\left\{x\in {\mathbb{P}}^1(\mathbb{C})|x=[a_{f^{+}}(p):a_{g^{+}}(p)]\text{for some prime p}\right\},$$

$${\mathrm{ord}}_{\infty }(f)\le o_k$$

$$f_{k,m}(z)=q^{-m}+O(q^{o_k+1}).$$

$$A_{m,c}(n):=\sum _{\genfrac{}{}{0ex}{}{-d=0}{(c,d)=1}}^{c-1}{e}^{\frac{2\pi i}{c}(nd-ma)},$$

$$I_n(z)$$

$$I_n(z)\sim \frac{e^z}{\sqrt{2\pi z}}$$

$$a_{k,m}(n)\sim C_k{A}_{m,1}(n){\left(\frac{n}{m}\right)}^{(k-1)/2}\frac{e^{4\pi \sqrt{mn}}}{(mn{)}^{1/4}}$$

$$a_{k,m}(n)\sim D_k{n}^{k-1}$$

$${\rho }_{f,\ell }:G_{\mathbb{Q}}\to {\mathrm{GL}}_2\left(E_f\otimes {\mathbb{Q}}_{\ell }\right)$$

$${\mathcal{A}}_{f,\ell }:=\left\{u\in {\mathrm{GL}}_2\left(E_f\otimes {\mathbb{Q}}_{\ell }\right)|\det (u)\in ({\mathbb{Z}}_{\ell }^{\times }{)}^{k-1}\right\},$$

$${\mathcal{G}}_{f,\ell }={\mathcal{A}}_{f,\ell }.$$

$$\mathcal{A}_{\rho_{f}\times\rho_{f^{\prime}}}=\left\{(u,u^{\prime})\in\mathcal{A}_{f,\ell}\times\mathcal{A}_{f^{\prime},\ell}\;\big{|}\;\mathrm{det}(u)=v^{k-1},\;\mathrm{det}(u^{\prime})=v^{k^{\prime}-1}\text{ for some }v\in\mathbb{Z}_{\ell}^{\times}\right\}.$$

$$\mathcal{U}={\mathcal{U}}_1\times \cdots \times {\mathcal{U}}_t,$$

$$\mathcal{U}:={\mathrm{SL}}_2\left(H_{f_1}\otimes {\mathbb{Z}}_{\ell }\right)\times \cdots \times {\mathrm{SL}}_2\left(H_{f_m}\otimes {\mathbb{Z}}_{\ell }\right)$$

$$\mathcal{H}:=\mathcal{U}\cap \mathcal{G}.$$

$${\mathrm{SL}}_2(H_{f_j}\otimes {\mathbb{Z}}_{\ell })\cong \prod _{v|\ell }{\mathrm{SL}}_2({\mathcal{O}}_v),$$

$${\mathrm{sl}}_2({\mathcal{O}}_v):=\left\{\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)|a,b,c,d\in {\mathcal{O}}_v\text{ and }a+c=0\right\}.$$

$$\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)=\left[\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\right],$$

$$\left(\begin{array}{cc}0& 2\\ 0& 0\end{array}\right)=\left[\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\right],$$

$$\left(\begin{array}{cc}0& 0\\ 2& 0\end{array}\right)=\left[\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\right].$$

$$\mathrm{Der}({\mathrm{sl}}_2({\mathcal{O}}_v))={\mathrm{sl}}_2({\mathcal{O}}_v).$$

$$U\not\subset \cup T_i.$$

$$U\not\subset \cup L_i.$$

Full text

Keywords: Fourier coefficient, Modular form, Galois representation

1. Introduction

The Fourier coefficients of a modular form play crucial roles in studying the theory of modular forms. In particular, the $q$-expansion principle (for example, see [1] or [4]) shows that a modular form is determined by its Fourier coefficients. A natural question is to find relations between two modular forms when a connection between their Fourier coefficients is given. It was proved by Ramakrishnan [3, Appendix] that if $f$ and $g$ are normalized Hecke eigenforms of the same weight such that for all primes $p$ outside a set $T$ of density $\delta(T)<\frac{1}{18}$

[TABLE]

then there exists a quadratic character $\chi$ such that

[TABLE]

Here, $a_{f}(n)$ (resp. $a_{g}(n)$) denotes the $n$th Fourier coefficient of $f$ (resp. $g$). This result was also known to Blasius and Serre.

Let $k$ be an even integer, and $M_{k}^{!}$ be the space of weakly holomorphic modular forms of weight $k$ on $\mathrm{SL}_{2}(\mathbb{Z})$. Let

[TABLE]

Suppose that $f$ and $g$ are weakly holomorphic modular forms in $M^{!}$. We define a subset $R(f,g)$ of $\mathbb{P}^{1}(\mathbb{C})$ by

[TABLE]

This is a set of ratios of the Fourier coefficients of $f$ and $g$. For example, if $a_{f}(p)^{2}=a_{g}(p)^{2}$ for every prime $p$, then $R(f,g)=\{[1:1],[1:-1]\}$. Therefore, if $f$ and $g$ are Hecke eigenforms and $R(f,g)=\{[1:1],[1:-1]\}$, then $f=cg$ for some constant $c$. In this vein, the objective of this paper is to classify $f$ and $g$ in $M^{!}$ such that $R(f,g)$ is a finite set. Our main result is as follows.

Theorem 1.1**.**

Suppose that $f$ and $g$ are nonzero weakly holomorphic modular forms in $M^{!}$. If $R(f,g)$ is a finite set, then $f=cg$ for some constant $c$.

This applies to study the number of representations of an integer by a quadratic form. Let $Q(x_{1},\ldots,x_{d})$ be a positive definite quadratic form over $\mathbb{Z}$ in $d$ variables with level one. For a positive integer $n$, let

[TABLE]

be the number of representations of the integer $n$ by a quadratic form $Q$. Note that

[TABLE]

is a modular form of weight $d/2$ on $\mathrm{SL}_{2}(\mathbb{Z})$. Here, $q$ denotes $e^{2\pi iz}$, where $z$ is a complex number whose imaginary part is positive. Therefore, Theorem 1.1 gives the following corollary.

Corollary 1.2**.**

Assume that $Q_{1}(x_{1},\ldots,x_{d})$ and $Q_{2}(x_{1},\ldots,x_{d})$ are positive definite quadratic forms over $\mathbb{Z}$ in $d$ variables with level one. If there is a positive integer $n$ such that the numbers of representations of $n$ by $Q_{1}$ and $Q_{2}$ are different, then the number of elements of the set

[TABLE]

is infinite.

The main ingredient of the proof of Theorem 1.1 is the result in [7] on the Galois representations attached to Hecke eigenforms. This is used to prove that if $f$ and $g$ are nonzero cusp forms on $\mathrm{SL}_{2}(\mathbb{Z})$ and $R(f,g)$ is finite, then $f=cg$ for some constant $c$. This is the main part of the proof of Theorem 1.1.

Remark 1.3**.**

In the same way, the result can be extended to harmonic weak Maass forms. In this case, if $f$ and $g$ are harmonic weak Maass forms whose shadows are cusp forms, then we need to look at the set

[TABLE]

where $f^{+}$ (resp. $g^{+}$) denotes the holomorphic part of $f$ (resp. $g$) and $a_{f^{+}}(n)$ (resp. $a_{g^{+}}(n)$) denotes the $n$th Fourier coefficient of $f^{+}$ (resp. $g^{+}$).

The remainder of this paper is organized as follows. In Section 2, we review some preliminaries concerning the Fourier coefficients of weakly holomorphic modular forms and the Galois representations attached to Hecke eigenforms. In Section 3, we prove the main theorem for the case of cusp forms. In Section 4, we prove the main theorem: Theorem 1.1.

2. Preliminaries

In this section, we review some basic material concerning the Fourier coefficients of weakly holomorphic modular forms and the Galois representations attached to Hecke eigenforms.

2.1. Fourier coefficients of weakly holomorphic modular forms

In this section, we review some results related to the asymptotic of the Fourier coefficients of weakly holomorphic modular forms based on [5] and [6].

Let $f\in M^{!}_{k}$. We write $k=12o_{k}+k^{\prime}$ with $o_{k}\in\mathbb{Z}$ and $k^{\prime}\in\{0,4,6,8,10,14\}$. Then, by the valence formula, we have

[TABLE]

if $f$ is nonzero. Moreover, for $m\geq-o_{k}$, there is a unique $f_{k,m}\in M^{!}_{k}$ such that

[TABLE]

Then, $\{f_{k,m}\ |\ m\geq-o_{k}\}$ forms a basis of $M^{!}_{k}$. In [2], Duke and Jenkins studied various properties of this basis.

For positive integers $m,n$, and $c$, let

[TABLE]

where $a$ is an integer such that $ad\equiv 1\ (\mathrm{mod}\ c)$. We introduce the Bessel function of the first kind (for example, see [9])

[TABLE]

Note that this Bessel function satisfies an asymptotic expansion

[TABLE]

as $z\to\infty$. Then, we have the following theorem.

Theorem 2.1**.**

[5, Theorem 1–3]**, [6, pp. 149–151] For $m\geq-o_{k}$, let $a_{k,m}(n)$ be the $n$th Fourier coefficient of $f_{k,m}$.

1. (1)

If $m>0$, then

[TABLE]

as $n\to\infty$, where $C_{k}$ is a constant dependent on $k$. 2. (2)

If $m=0$, then

[TABLE]

as $n\to\infty$, where $D_{k}$ is a constant dependent on $k$.

2.2. Galois representations attached to Hecke eigenforms

In this section, we introduce the result in [7] concerning the Galois representations attached to Hecke eigenforms. For a positive integer $n$, let $T_{n}$ be the $n$th Hecke operator. For a positive even integer $k$, let $S_{k}$ denote the space of cusp forms of weight $k$ on $\mathrm{SL}_{2}(\mathbb{Z})$. Let $f(z)=\sum_{n=1}^{\infty}a_{f}(n)q^{n}\in S_{k}$ be a normalized Hecke eigenform, i.e. $f|T_{p}=a_{f}(p)f$ for every prime $p$ and $a_{f}(1)=1$. Let $E_{f}$ be the field generated by all the Hecke eigenvalues $a_{f}(p)$ over $\mathbb{Q}$, and $H_{f}$ be the $\mathbb{Z}$-algebra generated by all the Hecke eigenvalues $a_{f}(p)$. Let $G_{\mathbb{Q}}$ denote $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$. For a prime $\ell$, let

[TABLE]

be the representation of $G_{\mathbb{Q}}$ attached to $f$. Note that if $p$ is a prime not equal to $\ell$ and $\mathrm{Frob}_{p}\in G_{\mathbb{Q}}$ is a Frobenius element at $p$, then the trace of $\rho_{f,\ell}(\mathrm{Frob}_{p})$ is $a_{f}(p)$ in $E_{f}\otimes\mathbb{Q}_{\ell}$ and the determinant of $\rho_{f,\ell}(\mathrm{Frob}_{p})$ is $p^{k-1}$ (for example, see page 261 in [7]).

Let $\mathcal{G}_{f,\ell}$ be the image of $\rho_{f,\ell}$ in $\mathrm{GL}_{2}\left(E_{f}\otimes\mathbb{Q}_{\ell}\right)$. If

[TABLE]

then $\mathcal{A}_{f,\ell}$ contains $\mathcal{G}_{f,\ell}$. Moreover, it was proved in [7] and [8] that for all but finitely many primes $\ell$, we have

[TABLE]

Let $f^{\prime}\in S_{k^{\prime}}$ be a normalized Hecke eigenform. Suppose that if $k=k^{\prime}$, then $f$ and $f^{\prime}$ are not conjugate under the action of $G_{\mathbb{Q}}$. Let $T_{f,f^{\prime}}$ be the $\mathbb{Z}$-subalgebra of $H_{f}\times H_{f^{\prime}}$ generated by the pairs $\left(a_{f}(p),a_{f^{\prime}}(p)\right)$. It should be noted that according to the assumption, $\left[H_{f}\times H_{f^{\prime}}:T_{f,f^{\prime}}\right]$ is finite (for example, see lines 8-10 on p.268 in [7]). Let $\mathcal{A}_{\rho_{f}\times\rho_{f^{\prime}}}$ be the image of $\rho_{f,\ell}\times\rho_{f^{\prime},\ell}$ in $\mathrm{GL}_{2}\left(E_{f}\otimes\mathbb{Q}_{\ell}\right)\times\mathrm{GL}_{2}\left(E_{f^{\prime}}\otimes\mathbb{Q}_{\ell}\right)$. Ribet [7] proved the following theorem.

Theorem 2.2** (Theorem 6.1 in [7]).**

If $\ell$ is a prime such that

• •

$\ell\geq k+k^{\prime}$,

• •

$\mathcal{G}_{f,\ell}=\mathcal{A}_{f,\ell}$,

• •

$\mathcal{G}_{f^{\prime},\ell}=\mathcal{A}_{f^{\prime},\ell}$,

• •

$\ell\nmid\left[H_{f}\times H_{f^{\prime}}:T_{f,f^{\prime}}\right]$,

then

[TABLE]

This theorem implies the following lemma.

Lemma 2.3**.**

For integers $j$, $1\leq j\leq m$, suppose that $f_{j}$ are normalized Hecke eigenforms in $S_{k_{j}}$ that are not conjugate to each other under the action of $G_{\mathbb{Q}}$. If $\ell$ is a sufficiently large prime, then $H_{f_{1}}\times\cdots\times H_{f_{m}}$ is dense in an open subset of $\left(H_{f_{1}}\otimes\mathbb{Q}_{\ell}\right)\times\cdots\times\left(H_{f_{m}}\otimes\mathbb{Q}_{\ell}\right)$.

To prove this lemma, we need the following result.

Lemma 2.4** (Lemma 3.4 in [7]).**

Let $\mathcal{U}_{1},\ldots,\mathcal{U}_{t}\;(t>1)$ be profinite groups. Assume that for each $i$ the following condition is satisfied: for each open subgroup $\mathcal{W}$ of $\mathcal{U}_{i}$, the closure of the commutator subgroup of $\mathcal{W}$ is open in $\mathcal{U}_{i}$. Let $\mathcal{H}$ be a closed subgroup of

[TABLE]

which maps to an open subgroup of each group $\mathcal{U}_{i}\times\mathcal{U}_{j}\;(i\neq j)$. Then, $\mathcal{H}$ is open in $\mathcal{U}$.

Now, we prove Lemma 2.3.

Proof.

Let $\mathcal{G}$ be the image of $\rho_{f_{1},\ell}\times\cdots\times\rho_{f_{m},\ell}$ in $\mathrm{GL}_{2}\left(H_{f_{1}}\otimes\mathbb{Z}_{\ell}\right)\times\cdots\times\mathrm{GL}_{2}\left(H_{f_{m}}\otimes\mathbb{Z}_{\ell}\right)$. We claim that $\mathcal{G}$ contains $\mathrm{SL}_{2}\left(H_{f_{1}}\otimes\mathbb{Z}_{\ell}\right)\times\cdots\times\mathrm{SL}_{2}\left(H_{f_{m}}\otimes\mathbb{Z}_{\ell}\right)$, which then provides the proof. Now, we prove the claim. Note that if $\ell$ is a sufficiently large prime, then, for all pairs $\left(j_{1},j_{2}\right)$, the prime $\ell$ satisfies the following conditions:

• •

$\mathcal{G}_{f_{j_{1}},\ell}=\mathcal{A}_{f_{j_{1}},\ell}$,

• •

$\mathcal{G}_{f_{j_{2}},\ell}=\mathcal{A}_{f_{j_{2}},\ell}$,

• •

$\ell\nmid\left[H_{f_{j_{1}}}\times H_{f_{j_{2}}}:T_{f_{j_{1}},f_{j_{2}}}\right]$.

Therefore, we assume that $\ell$ satisfies these conditions.

Let

[TABLE]

and

[TABLE]

Note that $\mathcal{H}$ is closed in $\mathrm{GL}_{2}\left(H_{f_{1}}\otimes\mathbb{Q}_{\ell}\right)\times\cdots\times\mathrm{GL}_{2}\left(H_{f_{m}}\otimes\mathbb{Q}_{\ell}\right)$ since $G_{\mathbb{Q}}$ is compact and $\rho_{f_{1},\ell}\times\cdots\times\rho_{f_{m},\ell}$ is continuous. Thus, we see that $\mathcal{H}$ is also closed in $\mathcal{U}$. For all pairs $(j_{1},j_{2})$, the projection of $\mathcal{H}$ to $\mathrm{SL}_{2}\left(H_{f_{j_{1}}}\otimes\mathbb{Z}_{\ell}\right)\times\mathrm{SL}_{2}\left(H_{f_{j_{2}}}\otimes\mathbb{Z}_{\ell}\right)$ is surjective by Theorem 2.2. For each $j$, we have

[TABLE]

where $v$ denote the prime ideals of $H_{f_{j}}$ above $\ell$, and $\mathcal{O}_{v}$ denotes the completion of $H_{f_{j}}$ at $v$. Note that $\mathrm{SL}_{2}(\mathcal{O}_{v})$ is a $\ell$-adic lie group, and the lie algebra of $\mathrm{SL}_{2}(\mathcal{O}_{v})$ is the same as its own derived algebra. This implies that for each open subgroup $\mathcal{W}$ of $\mathrm{SL}_{2}(H_{f_{j}}\otimes\mathbb{Z}_{\ell})$, the closure of the commutator subgroup of $\mathcal{W}$ is open in $\mathrm{SL}_{2}(H_{f_{j}}\otimes\mathbb{Z}_{\ell})$ (see Remark 3 on p. 253 in [7]). Therefore, by Lemma 2.4, we complete the proof of the claim. ∎

Remark 2.5**.**

For the convenience of readers, let us recall the lie algebra of $\mathrm{SL}_{2}(\mathcal{O}_{v})$ and its derived subalgebra. The lie algebra of $\mathrm{SL}_{2}(\mathcal{O}_{v})$ is isomorphic to

[TABLE]

The derived subalgebra $\mathrm{Der}(\mathrm{sl}_{2}(\mathcal{O}_{v}))$ of $\mathrm{sl}_{2}(\mathcal{O}_{v})$ is generated by all $[A,B]\;(A,B\in\mathrm{sl}_{2}(\mathcal{O}_{v}))$, where $[A,B]=AB-BA$. Note that

[TABLE]

[TABLE]

and

[TABLE]

Assume that $\ell\neq 2$. The matrices $\left(\begin{smallmatrix}1&0\\ 0&-1\\ \end{smallmatrix}\right)$, $\left(\begin{smallmatrix}0&2\\ 0&0\\ \end{smallmatrix}\right)$, and $\left(\begin{smallmatrix}0&0\\ 2&0\\ \end{smallmatrix}\right)$ consist a basis of $\mathrm{sl}_{2}(\mathcal{O}_{v})$. Therefore, we have

[TABLE]

2.3. Lemma for hyperplanes

For later use, we prove the following lemma.

Lemma 2.6**.**

Let $U$ be a subset of $V=\mathbb{Q}^{n}$. Suppose that $U$ is dense in an open subset of $V\otimes\mathbb{Q}_{\ell}$.

1. (1)

If $T_{1},\ldots,T_{m}$ are hyperplanes in $V$, then

[TABLE] 2. (2)

If $L_{1},\ldots,L_{m}$ are hyperplanes in $V\otimes\mathbb{C}$, then we have

[TABLE]

Proof.

(1) Suppose that

[TABLE]

This implies

[TABLE]

in $V\otimes\mathbb{Q}_{\ell}$. Then, $\cup\overline{T_{i}}$ contains an open set in $V\otimes\mathbb{Q}_{\ell}$. Note that $\overline{T_{i}}$ is a hyperplane in $V\otimes\mathbb{Q}_{\ell}$ for each $i$. This gives a contradiction.

(2) Due to (1), it is enough to prove that $L_{i}\cap V$ is contained in a hyperplane in $V$ for each $i$. Note that $L_{i}$ can be expressed as

[TABLE]

for $a_{i,j}\in\mathbb{C}$. Since $L_{i}$ is a hyperplane, we see that $(a_{i,1},\ldots,a_{i,n})\neq(0,\ldots,0)$. Without loss of generality, we may assume that $a_{i,n}\neq 0$. Then, $(y_{1},\ldots,y_{n})\in L_{i}$ is equivalent to

[TABLE]

for some $t_{1},\ldots,t_{n-1}\in\mathbb{C}$. Therefore, $(x_{1},\ldots,x_{n})\in L_{i}\cap V$ is equivalent to

[TABLE]

We can take $\mathbb{Q}$-linearly independent complex numbers $\alpha_{1}=1,\ldots,\alpha_{n-1}$ such that

[TABLE]

can be expressed as $\mathbb{Q}$-linear combinations of $\alpha_{1},\ldots,\alpha_{n-1}$. This implies that

[TABLE]

for some degree $1$ homogeneous polynomials $F_{i}$ with coefficients in $\mathbb{Q}$. Therefore, $x_{n}\in\mathbb{Q}$ is equivalent to $x_{n}=F_{1}(t_{1},\ldots,t_{n-1})$. From this, we see that $L_{i}\cap V$ is in the hyperplane

[TABLE]

∎

3. Coefficients of cusp forms

In this section, we prove that if $f$ and $g$ are cusp forms and $R(f,g)$ is finite, then $f=cg$ for some constant $c$. To prove this, we need the following lemmas.

Lemma 3.1**.**

Suppose that $f_{1},\ldots,f_{m}$ are normalized Hecke eigenforms on $\mathrm{SL}_{2}(\mathbb{Z})$ such that any two of them are not conjugate under the action of $G_{\mathbb{Q}}$. Then, there are no finite subsets $B$ of $(E_{f_{1}}\otimes\mathbb{C})\times\cdots\times(E_{f_{m}}\otimes\mathbb{C})$ such that $(0,\ldots,0)\not\in B$ and for any primes $p$, we have

[TABLE]

for some $(A_{1},\ldots,A_{m})\in B$.

Proof.

Suppose that $B$ is a finite subset of $(E_{f_{1}}\otimes\mathbb{C})\times\cdots\times(E_{f_{m}}\otimes\mathbb{C})$ such that $(0,\ldots,0)\not\in B$ and for any primes $p$, we have

[TABLE]

for some $(A_{1},\ldots,A_{m})\in B$. Then, the set

[TABLE]

is a finite union of hyperplanes $(E_{f_{1}}\otimes\mathbb{C})\times\cdots\times(E_{f_{m}}\otimes\mathbb{C})$.

By Lemma 2.3, we see that there is a prime $\ell$ such that $H_{f_{1}}\times\cdots\times H_{f_{m}}$ is dense in an open subset of $(H_{f_{1}}\otimes\mathbb{Q}_{\ell})\times\cdots\times(H_{f_{m}}\otimes\mathbb{Q}_{\ell})$. Since the trace of $\rho_{f_{i},\ell}(\mathrm{Frob}_{p})$ is $a_{f_{i}}(p)$ for primes $p\neq\ell$ and \{\mathrm{Frob}_{p}\ |\ \text{p is a prime}\} is dense in $G_{\mathbb{Q}}$ by Chebotarev’s density theorem, we see that the set

[TABLE]

is a dense subset of $H_{f_{1}}\times\cdots\times H_{f_{m}}$. Since we have

[TABLE]

and

[TABLE]

by Lemma 2.6, the set $T$ is not contained in any finite union of hyperplanes in $(E_{f_{1}}\otimes\mathbb{C})\times\cdots\times(E_{f_{m}}\otimes\mathbb{C})$. This is a contradiction since $T$ is contained in $C$ by the assumption. ∎

Let $f$ be a normalized Hecke eigenform on $\mathrm{SL}_{2}(\mathbb{Z})$. Let $m=[E_{f}:\mathbb{Q}]$ and $\sigma_{1},\ldots,\sigma_{m}$ be the embeddings from $E_{f}$ to $\bar{\mathbb{Q}}$. Note that $E_{f}\otimes\mathbb{Q}_{\ell}\cong(\mathbb{Q}_{\ell})^{m}$. Let $\{\tau_{1},\ldots,\tau_{m}\}$ be a basis of $E_{f}$ over $\mathbb{Q}$. Then, for each $n>0$, the coefficient $a_{f}(n)$ can be written as a linear combination of $\tau_{1},\ldots,\tau_{m}$, i.e.,

[TABLE]

for $a_{i}(n)\in\mathbb{Q}$. This means that

[TABLE]

for $\sigma\in G_{\mathbb{Q}}$. From this, we prove the following lemma.

Lemma 3.2**.**

There are no finite subsets $B$ of $\mathbb{C}^{m}$ such that $(0,\ldots,0)\not\in B$ and for any primes $p$, we have

[TABLE]

for some $(A_{1},\ldots,A_{m})\in B$.

Proof.

Suppose that $B$ is a finite subset of $\mathbb{C}^{m}$ such that $(0,\ldots,0)\not\in B$ and for any primes $p$, we have

[TABLE]

for some $(A_{1},\ldots,A_{m})\in B$. By (3.1), we see that the equation (3.2) is equivalent to

[TABLE]

Since $\{\tau_{1},\ldots,\tau_{m}\}$ is a basis of $E_{f}$ over $\mathbb{Q}$, the matrix

[TABLE]

is invertible. Therefore, we have

[TABLE]

since $(A_{1},\ldots,A_{m})\neq(0,\ldots,0)$. This means that the set

[TABLE]

is a subset of the set

[TABLE]

which is a finite union of hyperplanes in $E_{f}\otimes\mathbb{C}$.

By Lemma 2.3, we see that there is a prime $\ell$ such that $H_{f}$ is dense in an open subset of $H_{f}\otimes\mathbb{Q}_{\ell}$. Since the trace of $\rho_{f,\ell}(\mathrm{Frob}_{p})$ is $a_{f}(p)$ for primes $p\neq\ell$ and \{\mathrm{Frob}_{p}\ |\ \text{p is a prime}\} is dense in $G_{\mathbb{Q}}$ by Chebotarev’s density theorem, we see that the set

[TABLE]

is a dense subset of $H_{f}$. Then, by Lemma 2.6, the set $T$ is not contained in any finite union of hyperplanes in $E_{f}\otimes\mathbb{C}$. We consider the isomorphism from $E_{f}$ to $\mathbb{Q}^{m}$ defined by $a\mapsto(a_{1},\ldots,a_{m})$, where $a_{1},\ldots,a_{m}$ is determined by the decomposition

[TABLE]

By using this isomorphism, we see that the set

[TABLE]

is not contained in any finite union of hyperplanes in $\mathbb{C}^{m}$. This is a contradiction since $T^{\prime}$ is contained in $C$ by the assumption. ∎

Suppose that $f_{1},\ldots,f_{m}$ are normalized Hecke eigenforms such that any two of them are not conjugate under the action of $G_{\mathbb{Q}}$. Let $a_{i}(n)$ be the $n$th Fourier coefficient of $f_{i}$, i.e.,

[TABLE]

For each $i$, let $t_{i}=[E_{f_{i}}:\mathbb{Q}]$ and $\{\sigma_{i,1},\ldots,\sigma_{i,t_{i}}\}$ be the embeddings of $E_{f_{i}}$ to $\bar{\mathbb{Q}}$. Let $\{\tau_{i,1},\ldots,\tau_{i,t_{i}}\}$ be a basis of $E_{f_{i}}$ over $\mathbb{Q}$. Therefore, $a_{i}(n)$ can be written as a linear combination of $\tau_{i,1},\ldots,\tau_{i,t_{i}}$, i.e.,

[TABLE]

for $a_{i,j}(n)\in\mathbb{Q}$. From this, we see that

[TABLE]

for $\sigma\in G_{\mathbb{Q}}$. Then, we have the following lemma.

Lemma 3.3**.**

Let $t=t_{1}+\cdots+t_{m}$. There are no finite subsets $B$ of $\mathbb{C}^{t}$ such that $(0,\ldots,0)\not\in B$ and for any primes $p$, we have

[TABLE]

for some $(A_{1,1},\ldots,A_{1,t_{1}},\ldots,A_{m,1},\ldots,A_{m,t_{m}})\in B$.

Proof.

Suppose that $B$ is a finite subset of $\mathbb{C}^{t}$ such that $(0,\ldots,0)\not\in B$ and for any primes $p$, we have

[TABLE]

for some $(A_{1,1},\ldots,A_{1,t_{1}},\ldots,A_{m,1},\ldots,A_{m,t_{m}})\in B$.

By (3.4), we see that the equation (3.5) is equivalent to

[TABLE]

Note that the matrix

[TABLE]

is invertible since the matrix

[TABLE]

is invertible for each $i$. Therefore, if we let

[TABLE]

then $(A^{\prime}_{1,1},\ldots,A^{\prime}_{1,t_{1}},\ldots,A^{\prime}_{m,1},\ldots,A^{\prime}_{m,t_{m}})\neq(0,\ldots,0)$ since $(A_{1,1},\ldots,A_{1,t_{1}},\ldots,A_{m,1},\ldots,A_{m,t_{m}})\neq(0,\ldots,0)$. This means that the set

[TABLE]

is a subset of the set

[TABLE]

which is a finite union of hyperplanes in $\mathbb{C}^{t}$.

As in the proof of Lemma 3.1, there is a prime $\ell$ such that the set

[TABLE]

is not contained in any finite union of hyperplanes in $(E_{f_{1}}\otimes\mathbb{C})\times\cdots\times(E_{f_{m}}\otimes\mathbb{C})$. As in the proof of Lemma 3.2, we consider the isomorphism from $E_{f_{1}}\times\cdots\times E_{f_{m}}$ to $\mathbb{Q}^{t}$ defined by

[TABLE]

where $a_{1,1},\ldots,a_{1,t_{1}},\ldots,a_{m,1},\ldots,a_{m,t_{m}}$ are determined by the decomposition

[TABLE]

for each $i$. By this isomorphism, we see that

[TABLE]

is not contained in any finite union of hyperplanes in $\mathbb{C}^{t}$. This is a contradiction since $T^{\prime}$ is contained in $C$ by the assumption. ∎

From Lemma 3.3, we can prove the following theorem.

Theorem 3.4**.**

Suppose that $f$ and $g$ are nonzero cusp forms on $\mathrm{SL}_{2}(\mathbb{Z})$. If $f$ is not a constant multiple of $g$, then $R(f,g)$ defined in (1.1) is not finite.

Proof.

Since $f$ and $g$ are cusp forms, each function can be written as a linear combination of finitely many normalized Hecke eigenforms $\{f_{1},\ldots,f_{m}\}$, i.e.,

[TABLE]

for $a_{i},b_{i}\in\mathbb{C}$. For each $i$, we consider embeddings $\sigma_{i,1},\ldots,\sigma_{i,t_{i}}$ from $E_{f_{i}}$ to $\bar{\mathbb{Q}}$, where $t_{i}=[E_{f_{i}}:\mathbb{Q}]$. Consider the set

[TABLE]

We write

[TABLE]

Since $A$ contains $\{f_{1},\ldots,f_{m}\}$, both $f$ and $g$ can be written as linear combinations of elements of $A$, i.e.,

[TABLE]

for $\alpha_{i},\beta_{i}\in\mathbb{C}$.

Suppose that $R(f,g)$ is finite and that $f$ is not a constant multiple of $g$. Let $a_{i}(n)$ be the $n$th Fourier coefficient of $h_{i}$. Then, there is a finite subset $B$ of $\mathbb{C}^{t}$ such that $(0,\ldots,0)\not\in B$ and for each prime $p$, we have

[TABLE]

for some $(A_{1},\ldots,A_{t})\in B$. This is a contradiction due to Lemma 3.3. ∎

Remark 3.5**.**

Suppose that $f$ and $g$ are cusp forms. They may be zero. Then, Theorem 3.4 implies that if $R(f,g)$ is finite, then there are constants $\alpha$ and $\beta$ such that $\alpha f=\beta g$.

4. Proof of the main result

In this section, we prove Theorem 1.1. Suppose that $f$ is not a cusp form and $g$ is a cusp form. For a weakly holomorphic modular form $h\in M^{!}$, let $m_{h}=|\mathrm{ord}_{\infty}(h)|$ and $k_{h}$ be the weight of $h$. Since $g$ is a cusp form, the Fourier coefficients of $g$ should satisfy the Hecke bound

[TABLE]

as $n\to\infty$. Then, by Theorem 2.1, we have

[TABLE]

which means that the set $R(f,g)$ cannot be finite. This is a contradiction. Therefore, both $f$ and $g$ are cusp forms or none of them are cusp forms. If both $f$ and $g$ are cusp forms, then by Theorem 3.4, $f=cg$ for some constant $c$.

Suppose that neither $f$ nor $g$ is a cusp form. If $m_{f}\neq m_{g}$, then we may assume that $m_{f}>m_{g}$. By Theorem 2.1, we have

[TABLE]

which means that $R(f,g)$ is an infinite set. This is a contradiction. Therefore, $m_{f}=m_{g}$. In the same way, we see that $k_{f}=k_{g}$. By multiplying a nonzero constant to $g$, we may assume that

[TABLE]

Therefore, it is enough to show that $f=g$.

Suppose that $[\alpha:\beta]\in R(f,g)$ satisfying

[TABLE]

for infinitely many primes $p$. Such $[\alpha:\beta]$ exists since $R(f,g)$ is finite. By the asymptotic expansion given in Theorem 2.1, we see that both $f$ and $g$ have only finitely many zero coefficients. Therefore, both $\alpha$ and $\beta$ are nonzero.

Then, we have a strictly increasing sequence $\{p_{i}\}$ of primes satisfying (4.2). By Theorem 2.1 and (4.2), we have

[TABLE]

since $m_{f}=m_{g}$ and $k_{f}=k_{g}$. Therefore, we see that $[\alpha:\beta]=[1:1]\in R(f,g)$.

This implies that if $[\alpha:\beta]\in R(f,g)$ and $[\alpha:\beta]\neq[1:1]$, then the number of primes satisfying (4.2) is finite. Since $R(f,g)$ is finite, we see that

[TABLE]

for all but finitely many primes $p$.

Now, we prove that $f-g$ is a cusp form. Suppose that $f-g$ is not a cusp form. This means that the principal parts of $f$ and $g$ are not the same. Let

[TABLE]

Then, by (2.1), we see that $-m_{0}\leq o_{k}$. Let

[TABLE]

where $f_{k_{f},m}$ is a weakly holomorphic modular form as in (2.2). We denote by $a_{\hat{f}}(n)$ (resp. $a_{\tilde{f}}(n),a_{\hat{g}}(n)$, and $a_{\tilde{g}}(n)$) the $n$th Fourier coefficient of $\hat{f}$ (resp. $\tilde{f},\hat{g}$, and $\tilde{g}$). By the definition of $m_{0}$, we see that $\hat{f}=\hat{g}$. Then, for all but finitely many primes $p$, we have

[TABLE]

This implies that $R(\tilde{f},\tilde{g})$ is finite.

If $m_{0}>0$, then at least one of $\tilde{f}$ and $\tilde{g}$ is not a cusp form. Since $R(\tilde{f},\tilde{g})$ is finite, in the same way as above, we see that $m_{\tilde{f}}=m_{\tilde{g}}=m_{0}$. Note that we have a strictly increasing sequence $\{p_{i}\}$ of primes satisfying (4.4). Then, by the same argument as in (4.3), we have

[TABLE]

Therefore, we obtain $a_{\tilde{f}}(-m_{0})=a_{\tilde{g}}(-m_{0})$ which implies that $a_{f}(-m_{0})=a_{g}(-m_{0})$. This is a contradiction due to the definition of $m_{0}$.

If $m_{0}=0$, then both $\tilde{f}$ and $\tilde{g}$ are holomorphic modular forms and $a_{\tilde{f}}(0)\neq a_{\tilde{g}}(0)$. Then, we see that

[TABLE]

for some cusp forms $f_{c},g_{c}\in S_{k_{f}}$. By the Hecke bound, cusp forms $f_{c}(z)=\sum_{n>0}a_{f_{c}}(n)q^{n}$ and $g_{c}(z)=\sum_{n>0}a_{g_{c}}(n)q^{n}$ satisfy

[TABLE]

By (4.4), we have a strictly increasing sequence $\{p_{i}\}$ of primes satisfying

[TABLE]

where $a_{k_{f},0}(n)$ is the $n$th Fourier coefficient of $f_{k_{f},0}$. This is a contradiction since

[TABLE]

by Theorem 2.1 and (4.5). In conclusion, $f-g$ should be a cusp form.

If $S_{k_{f}}=\{0\}$, the proof is completed. Otherwise, we consider

[TABLE]

Both $F$ and $G$ are cusp forms and

[TABLE]

for all but finitely many primes $p$, where $a_{F}(n)$ (resp. $a_{G}(n)$) denotes the $n$th Fourier coefficient of $F$ (resp. $G$). Therefore, $R(F,G)$ is finite. By Theorem 3.4, there are constants $\alpha^{\prime},\beta^{\prime}$ such that $\beta^{\prime}F=\alpha^{\prime}G$. By (4.6), we see that $F=G$, and this implies that $f=g$ since $a_{f}(-m)=a_{g}(-m)$ for all $m\geq 0$. This completes the proof.

Acknowledgments

The authors are grateful to the referee for helpful comments and corrections. The authors also thank Jeremy Rouse for useful comments on the previous version of this paper.