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Congruence relations for rr-colored partitions

Robert Dicks Department of Mathematics
University of Illinois
Urbana, IL 61801
rdicks2@illinois.edu
Abstract.

Let 5\ell\geq 5 be prime. For the partition function p(n)p(n) and 5315\leq\ell\leq 31, Atkin found a number of examples of primes Q5Q\geq 5 such that there exist congruences of the form p(Q3n+β)0(mod).p(\ell Q^{3}n+\beta)\equiv 0\pmod{\ell}. Recently, Ahlgren, Allen, and Tang proved that there are infinitely many such congruences for every \ell. In this paper, for a wide range of c𝔽c\in\mathbb{F}_{\ell}, we prove congruences of the form p(Q3n+β0)cp(Qn+β1)(mod)p(\ell Q^{3}n+\beta_{0})\equiv c\cdot p(\ell Qn+\beta_{1})\pmod{\ell} for infinitely many primes QQ. For a positive integer rr, let pr(n)p_{r}(n) be the rr-colored partition function. Our methods yield similar congruences for pr(n)p_{r}(n). In particular, if rr is an odd positive integer for which >5r+19\ell>5r+19 and 2r+22±1(mod)2^{r+2}\not\equiv 2^{\pm 1}\pmod{\ell}, then we show that there are infinitely many congruences of the form pr(Q3n+β)0(mod)p_{r}(\ell Q^{3}n+\beta)\equiv 0\pmod{\ell}. Our methods involve the theory of modular Galois representations.

1. Introduction

Let n1n\geq 1 be an integer. A partition of nn is any non-increasing sequence of positive integers whose sum is nn. The partition function p(n)p(n) counts the number of partitions of nn. We agree that p(0)=1p(0)=1 and that p(n)=0p(n)=0 if n{0,1,2,}.n\not\in\{0,1,2,\dots\}. It is well-known that we have

n=0p(n)qn=n=1(1qn)1.\sum^{\infty}_{n=0}p(n)q^{n}=\prod^{\infty}_{n=1}(1-q^{n})^{-1}. (1.1)

The study of the arithmetic properties of p(n)p(n) has a long history. For example, Ramanujan [Ram21] proved the following congruences for p(n)p(n):

p(n+β)0(mod) for =5,7,11,p(\ell n+\beta_{\ell})\equiv 0\pmod{\ell}\ \ \ \ \ \text{ for $\ell=5,7,11,$} (1.2)

where β:=124(mod)\beta_{\ell}:=\frac{1}{24}\pmod{\ell}. Since the work of Ramanujan, many authors found more examples of congruences for p(n)p(n). These take the form

p(Qmn+β)0(mod),p(\ell Q^{m}n+\beta)\equiv 0\pmod{\ell}, (1.3)

where 5\ell\geq 5 is prime, QQ is a prime distinct from \ell and m1m\geq 1 is an integer. In his breakthrough work, Ono [Ono00] showed that for every 5\ell\geq 5, there are infinitely many primes QQ for which (1.3) holds with m=4m=4. Such arithmetic phenomena also occur for the coefficients of a wide class of weakly holomorphic modular forms (see e.g. [Tre06] and [Tre08]). In contrast to this, Ahlgren and Boylan [AB03] showed that if (1.3) holds for m=0m=0, then =5,7, or 11\ell=5,7,\text{ or }11. For m=1,2m=1,2, Ahlgren, Beckwith, and Raum [ABR22] have shown that congruences of the form (1.3) are scarce in a precise sense.

In the 1960s, Atkin discovered examples of congruences (1.3) with m=3m=3. In recent work, Ahlgren, Allen, and Tang [AAT21] study the existence of such congruences. These come from two families which we describe below.

For primes 133113\leq\ell\leq 31, Atkin [Atk68a, eq. 5252] gave examples of primes QQ for which

p(Q2n+124)0(mod) if (nQ)=ϵQ,p\left(\frac{\ell Q^{2}n+1}{24}\right)\equiv 0\pmod{\ell}\ \ \ \text{ if }\left(\frac{n}{Q}\right)=\epsilon_{Q}, (1.4)

where ϵQ{±1}\epsilon_{Q}\in\{\pm 1\}. By fixing nn in one of the allowable residue classes modulo 24Q24Q, we obtain a congruence (1.3) with m=3m=3. We refer to these as type I congruences. For each of =5,7, and 13\ell=5,7,\text{ and }13, Atkin [Atk68a, Thm. 11, 22] showed that for Q2(mod)Q\equiv-2\pmod{\ell}, we have

p(Q2n+124)0(mod) if (n)=1 and (nQ)=1.p\left(\frac{Q^{2}n+1}{24}\right)\equiv 0\pmod{\ell}\ \ \text{ if }\left(\frac{-n}{\ell}\right)=-1\text{ and }\left(\frac{-n}{Q}\right)=-1. (1.5)

We refer to these as type II congruences. By using modular Galois representations, Ahlgren, Allen and Tang proved that there are infinitely many type I congruences for every prime 5\ell\geq 5 and that there are infinitely many type II congruences for 17/24\geq 17/24 of the primes \ell.

For a positive integer rr, define the rr-colored partition function by

n=0pr(n)qn:=n=1(1qn)r=1+rq+.\sum^{\infty}_{n=0}p_{r}(n)q^{n}:=\prod^{\infty}_{n=1}(1-q^{n})^{-r}=1+rq+\cdots.

We agree that pr(n)=0p_{r}(n)=0 if n{0,1,2,}n\not\in\{0,1,2,...\}. Let 5\ell\geq 5 be prime. Many congruences have been proven for pr(n)p_{r}(n) (see e.g. [Atk68b, Thm. 11] and [Gor83, Thm. 22]). In this paper, for a wide range of c𝔽c\in\mathbb{F}_{\ell}, we prove congruences of the form pr(Q3n+β0)cpr(Qn+β1)(mod)p_{r}(\ell Q^{3}n+\beta_{0})\equiv c\cdot p_{r}(\ell Qn+\beta_{1})\pmod{\ell} for infinitely many primes QQ. In particular, when c=0c=0, we get congruences analogous to (1.4). Our first result relies on the assumption that rr and \ell are compatible (we choose to delay the precise definition to Section 22 for ease of exposition). Moreover, we will prove in Section 55 that rr and \ell are compatible if we have 2r+22±1(mod)2^{r+2}\not\equiv 2^{\pm 1}\pmod{\ell} and >5r+19\ell>5r+19. We will also prove in Section 55 that if 1r391\leq r\leq 39 and >r+4\ell>r+4, then rr and \ell are compatible whenever 2r+22±1(mod)2^{r+2}\not\equiv 2^{\pm 1}\pmod{\ell}. Stating the result requires some notation. Define

χ(r):={(4)if 3r,(12)if 3r.\chi^{(r)}:=\begin{cases}\left(\frac{-4}{\bullet}\right)&\text{if }3\mid r,\\ \left(\frac{12}{\bullet}\right)&\text{if }3\nmid r.\end{cases}

Our first result states that we have type I congruences modulo \ell for the functions pr(n)p_{r}(n) when rr and \ell are compatible. We state these theorems under the assumption that r<4r<\ell-4. When r=4r=\ell-4, the situation is much simpler and will be discussed in the last section.

Theorem 1.1.

Suppose that 5\ell\geq 5 is prime, that r<4r<\ell-4 is an odd positive integer and that rr and \ell are compatible. Then there exists a positive density set SS of primes such that if QSQ\in S, then we have Q1(mod)Q\equiv 1\pmod{\ell} and

pr(Q2n+r24)0(mod) if (nQ)=χ(r)(Q)(12Q)(1Q)r22.p_{r}\left(\frac{\ell Q^{2}n+r}{24}\right)\equiv 0\pmod{\ell}\ \ \ \text{ if }\ \ \ \left(\frac{n}{Q}\right)=\chi^{(r)}(Q)\left(\frac{12}{Q}\right)\left(\frac{-1}{Q}\right)^{\frac{\ell-r-2}{2}}.

Thus, by choosing nn in any of the allowable residue classes modulo 24Q24Q, we obtain congruences of the form

pr(Q3n+β)0(mod).p_{r}(\ell Q^{3}n+\beta)\equiv 0\pmod{\ell}.
Remark.

In the above theorem and the theorems which follow, our definition of density is that of natural density.

Remark.

When rr and \ell are close, the modular form fr,f_{r,\ell} which captures the relevant values of pr(n)p_{r}(n) modulo \ell has small weight. We illustrate this when r=6r=\ell-6. In this case, the form f6,f_{\ell-6,\ell} is in S52(1,νη61)S_{\frac{5}{2}}(1,\nu^{6\ell-1}_{\eta}) (see Section 22 for a definition of this space), which has dimension 0 if 1(mod4)\ell\not\equiv 1\pmod{4}. For such \ell, we have f6,=0f_{\ell-6,\ell}=0, which implies that

p6(n+624)0(mod).p_{\ell-6}\left(\frac{\ell n+\ell-6}{24}\right)\equiv 0\pmod{\ell}.

We prove a more general result from which Theorem 1.1 follows by setting α=1\alpha=1.

Theorem 1.2.

Suppose that 5\ell\geq 5 is prime, that r<4r<\ell-4 is an odd positive integer and that rr and \ell are compatible. Suppose that α2(mod)\alpha\not\equiv-2\pmod{\ell} is an integer. Then there exists a positive density set SS of primes such that if QSQ\in S, then Q1(mod)Q\equiv 1\pmod{\ell} and

pr(Q2n+r24)(α1)χ(r)(Q)pr(n+r24)(mod) if (nQ)=χ(r)(Q)(12Q)(1Q)r22.p_{r}\left(\frac{\ell Q^{2}n+r}{24}\right)\equiv(\alpha-1)\chi^{(r)}(Q)p_{r}\left(\frac{\ell n+r}{24}\right)\pmod{\ell}\ \text{ if }\ \ \left(\frac{n}{Q}\right)=\chi^{(r)}(Q)\left(\frac{12}{Q}\right)\left(\frac{-1}{Q}\right)^{\frac{\ell-r-2}{2}}.

Thus, by choosing nn in any of the allowable residue classes modulo 24Q24Q we obtain congruences of the form

pr(Q3n+β0)(α1)χ(r)(Q)pr(Qn+β1)(mod).p_{r}(\ell Q^{3}n+\beta_{0})\equiv(\alpha-1)\chi^{(r)}(Q)p_{r}(\ell Qn+\beta_{1})\pmod{\ell}.
Remark.

If we make the further assumption that there is no congruence modulo any prime above \ell between distinct newforms in S(r,r12)S\left(-r\ell,\frac{\ell-r-1}{2}\right) (see Section 66 for the definition), Theorem 1.2 also holds when α2(mod)\alpha\equiv-2\pmod{\ell}.

The next result gives congruences analogous to (1.4)(1.4) and (1.5)(1.5) involving primes QQ in a different residue class modulo \ell. Our result does not rely on the assumption of compatibility.

Theorem 1.3.

Suppose that rr is an odd positive integer. Suppose that 5\ell\geq 5 is prime, that r<4r<\ell-4 and that

 there exists an integer a with 2a1(mod).\text{ there exists an integer $a$ with $2^{a}\equiv-1\pmod{\ell}$}. (1.6)
  1. (1)

    There exists a positive density set SS of primes with the property that if QSQ\in S, then Q2(mod)Q\equiv-2\pmod{\ell} and there exists ϵQ{±1}\epsilon_{Q}\in\{\pm 1\} such that

    pr(Q2n+r24)0(mod) if 
    (nQ)
    =ϵQ
    .
    p_{r}\left(\frac{\ell Q^{2}n+r}{24}\right)\equiv 0\pmod{\ell}\ \ \ \text{ if }\\ \ \left(\frac{n}{Q}\right)=\epsilon_{Q}.
  2. (2)

    Suppose that all of the integers rw<0-r\leq w<0 with wr(mod24)w\equiv-r\pmod{24} satisfy (rw)=1\left(\frac{-rw}{\ell}\right)=1. Then there exists a positive density set SS of primes with the property that if QSQ\in S, then Q2(mod)Q\equiv-2\pmod{\ell} and there exists ϵQ{±1}\epsilon_{Q}\in\{\pm 1\} such that

    pr(Q2n+r24)0(mod) if (rn)=1 and (nQ)=ϵQ.p_{r}\left(\frac{Q^{2}n+r}{24}\right)\equiv 0\pmod{\ell}\ \ \text{ if }\ \ \left(\frac{-rn}{\ell}\right)=-1\ \ \ \text{ and }\ \left(\frac{n}{Q}\right)=\epsilon_{Q}.
Remark.

The value of ϵQ\epsilon_{Q} in (1)(1) and (2)(2) can be made explicit for any rr and \ell using the arguments at the end of Section 66.

Remark.

The condition on the integers rw<0-r\leq w<0 with wr(mod24)w\equiv-r\pmod{24} in (2)(2) is necessary to ensure that the form gr,g_{r,\ell} which we construct in Section 33 is a cusp form. We can use the same method to prove an analogous result for primes 5\ell\geq 5 with the property that 3a2(mod)3^{a}\equiv-2\pmod{\ell} for some integer aa. However, we would need to add the condition that there are no congruences between distinct newforms in the spaces under consideration.

Remark.

If r23r\leq 23, then the condition on the integers rw<0-r\leq w<0 with wr(mod24)w\equiv-r\pmod{24} is always satisfied, since the only choice for ww is r-r. It also holds for r=25r=25 (since it is a square) and r=27r=27 (since (3)=(27)\left(\frac{3}{\ell}\right)=\left(\frac{27}{\ell}\right)). For a fixed rr, we can use the quantitative version of Dirichlet’s theorem on primes in arithmetic progressions, quadratic reciprocity, and the Chinese remainder theorem to prove that the condition holds for a positive proportion of the primes \ell.

The organization of this paper is as follows. In Section 22, we give some background on modular forms and Galois representations. In Section 33, we construct for each rr and \ell important modular forms whose coefficients capture the relevant values of pr(n)p_{r}(n) modulo \ell. In the remaining sections (save for Section 77), our main tool is the theory of modular Galois representations. In Section 44, we prove that a wide range of rr and \ell are compatible. In Section 55, we prove the main technical result which we will need in order to prove Theorem 1.2. In Section 66, we prove Theorems 1.21.31.2-1.3. Finally, in Section 77, we discuss congruences for pr(n)p_{r}(n) when r=4r=\ell-4; in particular, we give a short proof of [Boy04, Thm. 2.12.1, (2)(2)].

2. Background

We follow the exposition in [ABR22] and [AAT21]. Throughout, let 5\ell\geq 5 be prime and set q:=e2πizq:=e^{2\pi iz}. Suppose that k12k\in\frac{1}{2}\mathbb{Z} and that NN is a positive integer. For a function f(z)f(z) on the upper half plane and

γ=(abcd)GL2+(),\gamma=\left(\begin{matrix}a&b\\ c&d\end{matrix}\right)\in\operatorname{GL}_{2}^{+}(\mathbb{Q}),

we have the weight kk slash operator

f(z)|kγ:=det(γ)k2(cz+d)kf(az+bcz+d).f(z)\big{|}_{k}\gamma:=\det(\gamma)^{\frac{k}{2}}(cz+d)^{-k}f\left(\frac{az+b}{cz+d}\right).

Let AA\subseteq\mathbb{C} be a subring. If ν\nu is a multiplier system on Γ0(N)\Gamma_{0}(N), we denote by Mk(N,ν,A)M_{k}(N,\nu,A) (resp. Sk(N,ν,A)S_{k}(N,\nu,A)) the space of modular forms (resp. cusp forms) of weight kk and multiplier ν\nu on Γ0(N)\Gamma_{0}(N) whose Fourier coefficients are in AA. When ν=1\nu=1 or AA is the subring of algebraic numbers that are integral at all of the primes above \ell, we omit them from the notation. Forms in these spaces satisfy the transformation law

f|kγ=ν(γ)f for γ=(abcd)Γ0(N)f\big{|}_{k}\gamma=\nu(\gamma)f\ \ \ \text{ for }\ \ \ \gamma=\left(\begin{matrix}a&b\\ c&d\end{matrix}\right)\in\Gamma_{0}(N)

and the appropriate conditions at the cusps of Γ0(N)\Gamma_{0}(N).

If kk is even, we denote by Sknew(N,)S^{\text{new}}_{k}(N,\mathbb{C}) the new subspace. Let Sknew(N,A)S^{\text{new}}_{k}(N,A) be defined by Sk(N,A)Sknew(N,)S_{k}(N,A)\cap S^{\text{new}}_{k}(N,\mathbb{C}). When NN is square-free, there is an Atkin-Lehner involution WpW_{p} on Sk(N,)S_{k}(N,\mathbb{C}) for every prime divisor pp of NN. Given a tuple ϵ=(ϵp)pN\epsilon=(\epsilon_{p})_{p\mid N} (where each ϵp{±1}\epsilon_{p}\in\{\pm 1\}), let Sknew(N,,ϵ)S^{\operatorname{new}}_{k}(N,\mathbb{C},\epsilon) be the subspace of forms ff with f|kWp=ϵpff\big{|}_{k}W_{p}=\epsilon_{p}f for pNp\mid N. If \ell and NN are coprime, then we let Sknew(N,ϵ)S^{\operatorname{new}}_{k}(N,\epsilon) be the subspace of Sknew(N)S^{\operatorname{new}}_{k}(N) attached to the tuple ϵ\epsilon (for such \ell, the operator WpW_{p} acts on Sknew(N)S^{\operatorname{new}}_{k}(N)). We define the eta function by

η(z):=q124n=1(1qn)\eta(z):=q^{\frac{1}{24}}\prod_{n=1}^{\infty}(1-q^{n})

and the theta function by

θ(z):=n=qn2.\theta(z):=\sum_{n=-\infty}^{\infty}q^{n^{2}}.

The eta function has a multiplier νη\nu_{\eta} satisfying

η(γz)=νη(γ)(cz+d)12η(z),γ=(abcd)SL2();\eta(\gamma z)=\nu_{\eta}(\gamma)(cz+d)^{\frac{1}{2}}\eta(z),\ \ \ \ \ \gamma=\left(\begin{matrix}a&b\\ c&d\end{matrix}\right)\in\operatorname{SL}_{2}(\mathbb{Z});

throughout, we choose the principal branch of the square root. For c>0c>0, we have the formula [Kno70,  §\mathsection4.14.1]

νη(γ)={(dc)e(124((a+d)cbd(c21)3c))if c is odd,(cd)e(124((a+d)cbd(c21)+3d33cd))if c is even.\nu_{\eta}(\gamma)=\begin{cases}\left(\frac{d}{c}\right)e\left(\frac{1}{24}((a+d)c-bd(c^{2}-1)-3c)\right)&\text{if }c\text{ is odd,}\\ \left(\frac{c}{d}\right)e\left(\frac{1}{24}((a+d)c-bd(c^{2}-1)+3d-3-3cd)\right)&\text{if }c\text{ is even}.\end{cases} (2.1)

For the multiplier of the theta function we have

νθ(γ):=(cz+d)12θ(γz)θ(z)=(cd)ϵd1,γ=(abcd)Γ0(4),\nu_{\theta}(\gamma):=(cz+d)^{-\frac{1}{2}}\frac{\theta(\gamma z)}{\theta(z)}=\left(\frac{c}{d}\right)\epsilon_{d}^{-1},\ \ \ \ \ \gamma=\left(\begin{matrix}a&b\\ c&d\end{matrix}\right)\in\Gamma_{0}(4),

where

ϵd={1if d1(mod4),iif d3(mod4).\epsilon_{d}=\begin{cases}1&\text{if }d\equiv 1\pmod{4},\\ i&\text{if }d\equiv 3\pmod{4}.\end{cases}

For odd values of dd, we have the formula

e(1d8)=(2d)ϵd.e\left(\frac{1-d}{8}\right)=\left(\frac{2}{d}\right)\epsilon_{d}. (2.2)

For rr\in\mathbb{Z}, we have

Mk(1,νηr)={0}if 2kr0(mod4).M_{k}(1,\nu^{r}_{\eta})=\{0\}\ \ \ \text{if}\ \ \ 2k-r\not\equiv 0\pmod{4}. (2.3)

If fMk(1,νηr)f\in M_{k}(1,\nu^{r}_{\eta}), then ηrf\eta^{-r}f is a weakly holomorphic form on SL2()\operatorname{SL}_{2}(\mathbb{Z}). This implies that each fMk(1,νηr)f\in M_{k}(1,\nu^{r}_{\eta}) has a Fourier expansion of the form

f=nr(mod24)a(n)qn24.f=\sum_{n\equiv r\pmod{24}}a(n)q^{\frac{n}{24}}. (2.4)

We have ηrfMkr2(1)\eta^{-r}f\in M_{k-\frac{r}{2}}(1) when 0<r<240<r<24.

We next recall the UU and VV operators. For a positive integer mm, we define them on Fourier expansions by

(n=1a(n)qn24)|Um:=n=1a(mn)qn24,\left(\sum_{n=1}^{\infty}a(n)q^{\frac{n}{24}}\right)\big{|}U_{m}:=\sum_{n=1}^{\infty}a(mn)q^{\frac{n}{24}},
(n=1a(n)qn24)|Vm:=n=1a(n)qmn24.\left(\sum_{n=1}^{\infty}a(n)q^{\frac{n}{24}}\right)\big{|}V_{m}:=\sum_{n=1}^{\infty}a(n)q^{\frac{mn}{24}}.

If k12\k\in\frac{1}{2}\mathbb{Z}\backslash\mathbb{Z}, then a computation using (2.1), (2.2) and (7.1) implies that

fSk(1,νηr)f|V24Sk(576,(12)νθr) if 3rf\in S_{k}(1,\nu^{r}_{\eta})\implies f\big{|}V_{24}\in S_{k}\left(576,\left(\frac{12}{\bullet}\right)\nu^{r}_{\theta}\right)\ \ \ \text{ if $3\nmid r$} (2.5)

and

fSk(1,νηr)f|V8Sk(64,νθr) if 3r.f\in S_{k}(1,\nu^{r}_{\eta})\implies f\big{|}V_{8}\in S_{k}\left(64,\nu^{r}_{\theta}\right)\ \ \ \text{ if $3\mid r$.} (2.6)

If k12\k\in\frac{1}{2}\mathbb{Z}\backslash\mathbb{Z}, then for each prime Q3Q\geq 3 we have the Hecke operator

TQ2:Sk(1,νηr)Sk(1,νηr).T_{Q^{2}}:S_{k}(1,\nu^{r}_{\eta})\rightarrow S_{k}(1,\nu^{r}_{\eta}).

For f=a(n)qn24Sk(1,νηr)f=\displaystyle\sum a(n)q^{\frac{n}{24}}\in S_{k}(1,\nu^{r}_{\eta}) and Q5Q\geq 5, we have

f|TQ2=(a(Q2n)+Qk32(1Q)k12(12nQ)a(n)+Q2k2a(nQ2))qn24.f\big{|}T_{Q^{2}}=\sum\left(a(Q^{2}n)+Q^{k-\frac{3}{2}}\left(\frac{-1}{Q}\right)^{k-\frac{1}{2}}\left(\frac{12n}{Q}\right)a(n)+Q^{2k-2}a\left(\frac{n}{Q^{2}}\right)\right)q^{\frac{n}{24}}. (2.7)

(see e.g. [Yan14, Proposition 1111]). Yang only states the result for rr such that 3r3\nmid r. The result when 3r3\mid r follows from (2.6) and [Shi73, Thm. 1.71.7]. When 3r3\mid r, we also have (2.7) for Q=3Q=3.

For k12\k\in\frac{1}{2}\mathbb{Z}\backslash\mathbb{Z} such that k52k\geq\frac{5}{2} and each odd squarefree tt, if 3r3\nmid r (resp. 3r3\mid r), then we have a Shimura lift on Sk(1,νηr)S_{k}(1,\nu^{r}_{\eta}) defined via (2.5) (resp. (2.6)) and the usual Shimura lift [Shi73] on Sk(576,(12)νθr)S_{k}\left(576,\left(\frac{12}{\bullet}\right)\nu^{r}_{\theta}\right) (resp. Sk(64,νθr)S_{k}\left(64,\nu^{r}_{\theta}\right)). Define

ψ(r):={(4)if 3r,(12)if 3r.\psi^{(r)}:=\begin{cases}\left(\frac{4}{\bullet}\right)&\text{if }3\mid r,\\ \left(\frac{12}{\bullet}\right)&\text{if }3\nmid r.\end{cases}

and

wr:={3if 3r,1if 3r.w_{r}:=\begin{cases}3&\text{if }3\mid r,\\ 1&\text{if }3\nmid r.\end{cases}

The action on Fourier expansions is given by

Sht(a(n)qn24)=At(n)qn,\text{Sh}_{t}\left(\sum a(n)q^{\frac{n}{24}}\right)=\sum A_{t}(n)q^{n},

where

At(n)=dn(1d)k12ψ(r)(d)(td)dk32a(wrtn2d2).A_{t}(n)=\sum_{d\mid n}\left(\frac{-1}{d}\right)^{k-\frac{1}{2}}\psi^{(r)}(d)\left(\frac{t}{d}\right)d^{k-\frac{3}{2}}a\left(\frac{w_{r}tn^{2}}{d^{2}}\right).

We have (see e.g. [ABR22, (2.13)(2.13)])

f0(mod)Sht(f)0(mod) for all squarefree t.f\equiv 0\pmod{\ell}\iff\text{Sh}_{t}(f)\equiv 0\pmod{\ell}\ \ \ \text{ for all squarefree $t$}. (2.8)

They only state their result for when 3r3\nmid r, but the same argument applies when 3r3\mid r. From [Yan14, Thm. 11, 22], it follows that

Sht:Sk(1,νηr)S2k1new(6,(8r),(12r))(12)if 3r\text{Sh}_{t}:S_{k}(1,\nu^{r}_{\eta})\rightarrow S^{\text{new}}_{2k-1}\left(6,-\left(\frac{8}{r}\right),-\left(\frac{12}{r}\right)\right)\otimes\left(\frac{12}{\bullet}\right)\ \ \text{if $3\nmid r$}

and

Sht:Sk(1,νηr)S2k1new(2,(8r))(4)if 3r\text{Sh}_{t}:S_{k}(1,\nu^{r}_{\eta})\rightarrow S_{2k-1}^{\text{new}}\left(2,\left(\frac{8}{r}\right)\right)\otimes\left(\frac{-4}{\bullet}\right)\ \ \text{if $3\mid r$}

(note that Yang uses 3r3r in place of rr). Moreover, we have

Sht(f|TQ2)=(Shtf)|TQ,\operatorname{Sh}_{t}\left(f\big{|}T_{Q^{2}}\right)=\left(\operatorname{Sh}_{t}f\right)\big{|}T_{Q},

where TQT_{Q} is the Hecke operator of index QQ on the integral weight space.

We now summarize some facts about modular Galois representations. See [Hid00] and [Edi92] for more details. We begin with some notation. Recall that 5\ell\geq 5 is prime. Let kk be an even integer and N+N\in\mathbb{Z}^{+} with N\ell\nmid N. Let ¯\overline{\mathbb{Q}}\subseteq\mathbb{C} be the algebraic closure of \mathbb{Q} in \mathbb{C}. If pp is prime, let ¯p\overline{\mathbb{Q}}_{p} be a fixed algebraic closure of p\mathbb{Q}_{p} and fix an embedding ιp:¯¯p\iota_{p}:\overline{\mathbb{Q}}\hookrightarrow\overline{\mathbb{Q}}_{p}. The embedding ι\iota_{\ell} allows us to view the coefficients of forms in Sk(N)S_{k}(N) as elements of ¯\overline{\mathbb{Q}}_{\ell}, and for each prime pp, the embedding ιp\iota_{p} allows us to view Gp:=Gal(¯p/p)G_{p}:=\operatorname{Gal}(\overline{\mathbb{Q}}_{p}/\mathbb{Q}_{p}) as a subgroup of G:=Gal(¯/)G_{\mathbb{Q}}:=\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}). If IpGpI_{p}\subseteq G_{p} is the inertia subgroup, we denote the coset of absolute Frobenius elements above pp in Gp/IpG_{p}/I_{p} by Frobp\text{Frob}_{p}. For any finite extension K/K/\mathbb{Q}, denote by Frobp|K\text{Frob}_{p}|_{K} the restrictions to KK of elements in Frobp\text{Frob}_{p}. For representations ρ:GGL2(¯)\rho:G_{\mathbb{Q}}\rightarrow\operatorname{GL}_{2}(\overline{\mathbb{Q}}_{\ell}) which are unramified at pp (which is to say that IpkerρI_{p}\subseteq\ker{\rho}), the image of any σGp\sigma\in G_{p} under ρ\rho only depends on σIpGp/Ip\sigma I_{p}\in G_{p}/I_{p} (the same conclusion holds for representations of Gal(K/)\operatorname{Gal}(K/\mathbb{Q}) and σ|K\sigma|_{K}); for any such representation, we denote by ρ(Frobp)\rho(\text{Frob}_{p}) the image of any element in Frobp\text{Frob}_{p} under ρ\rho and make a similar definition for representations of Gal(K/)\operatorname{Gal}(K/\mathbb{Q}) and Frobp|K\operatorname{Frob}_{p}|_{K}.

We denote by χ:G\chi:G_{\mathbb{Q}}\rightarrow\mathbb{Z}^{*}_{\ell} and ω:G𝔽\omega:G_{\mathbb{Q}}\rightarrow\mathbb{F}^{*}_{\ell} the \ell-adic and mod \ell cyclotomic characters, respectively. We let ω2,ω2:I𝔽2\omega_{2},\omega^{\prime}_{2}:I_{\ell}\rightarrow\mathbb{F}^{*}_{\ell^{2}} denote Serre’s fundamental characters of level 22 (see [Ser87,  §\mathsection2.12.1]). Both characters have order 21\ell^{2}-1, and we have ω2+1=ω2+1=ω\omega^{\ell+1}_{2}=\omega^{\prime\ell+1}_{2}=\omega.

The following theorem is due to Deligne, Fontaine, Langlands, Ribet, and Shimura (see [AAT21, Thm. 2.1]).

Theorem 2.1.

Let f=q+n2a(n)qnSk(N)f=q+\sum_{n\geq 2}a(n)q^{n}\in S_{k}(N) be a normalized Hecke eigenform. There is a continuous irreducible representation ρf:GGL2(¯)\rho_{f}:G_{\mathbb{Q}}\rightarrow\operatorname{GL}_{2}(\overline{\mathbb{Q}}_{\ell}) with semisimple mod \ell reduction ρ¯f:GGL2(𝔽¯)\overline{\rho}_{f}:G_{\mathbb{Q}}\rightarrow\operatorname{GL}_{2}(\overline{\mathbb{F}}_{\ell}) satisfying the following properties.

  1. (1)

    If pNp\nmid\ell N, then ρf\rho_{f} is unramified at pp and the characteristic polynomial of ρf(Frobp)\rho_{f}(\operatorname{Frob}_{p}) is X2ι(a(p))X+pk1X^{2}-\iota_{\ell}(a(p))X+p^{k-1}.

  2. (2)

    If qNq\mid N and q2Nq^{2}\nmid N, then ρf|Iq\rho_{f}|_{I_{q}} is unipotent. In particular, the prime-to-\ell Artin conductor N(ρ¯f)N(\overline{\rho}_{f}) of ρ¯f\overline{\rho}_{f} is not divisible by q2q^{2}. If further ff is qq-new, then we have

    ρf|Gq(χψ0ψ),\rho_{f}|_{G_{q}}\cong\left(\begin{matrix}\chi\psi&*\\ 0&\psi\end{matrix}\right),

    where ψ:Gq¯\psi:G_{q}\rightarrow\overline{\mathbb{Q}}^{*}_{\ell} is the unramified character with ψ(Frobq)=ι(aq)\psi({\operatorname{Frob}}_{q})=\iota_{\ell}(a_{q}).

  3. (3)

    Assume that 2k+12\leq k\leq\ell+1. Then

    • If ι(a)\iota_{\ell}(a_{\ell}) is an \ell-adic unit, then ρf|G\rho_{f}|_{G_{\ell}} is reducible and we have

      ρf|I(χk101).\rho_{f}|_{I_{\ell}}\cong\left(\begin{matrix}\chi^{k-1}&*\\ 0&1\end{matrix}\right).
    • If ι(a)\iota_{\ell}(a_{\ell}) is not an \ell-adic unit, then ρ¯f|G\overline{\rho}_{f}\big{|}_{G_{\ell}} is irreducible and ρ¯f|Iω2k1ω2(k1)\overline{\rho}_{f}|_{I_{\ell}}\cong\omega^{k-1}_{2}\oplus\omega^{\prime(k-1)}_{2}.

Remark.

The Galois representations depend on the choice of embedding ι:¯¯\iota_{\ell}:\overline{\mathbb{Q}}\hookrightarrow\overline{\mathbb{Q}}_{\ell}, but we have suppressed this from the notation.

We now define what it means for rr and \ell to be compatible. Define

Nr:={2if 3r,6if 3rN_{r}:=\begin{cases}2&\text{if }3\mid r,\\ 6&\text{if }3\nmid r\end{cases}

and

ϵr:={(8r)if 3r,((8r),(12r))if 3r.\epsilon_{r}:=\begin{cases}\left(\frac{8}{r\ell}\right)&\text{if }3\mid r,\\ (-\left(\frac{8}{r\ell}\right),-\left(\frac{12}{r\ell}\right))&\text{if }3\nmid r.\end{cases}
Definition 2.2.

Suppose that 5\ell\geq 5 is prime and that rr is an odd positive integer such that r<4r<\ell-4. We say that rr and \ell are compatible if for every newform fSr2new(Nr,ϵr)f\in S^{\operatorname{new}}_{\ell-r-2}(N_{r},\epsilon_{r}) the image of ρ¯f\overline{\rho}_{f} contains a conjugate of SL2(𝔽)\operatorname{SL}_{2}(\mathbb{F}_{\ell}).

We now discuss filtrations. Recall that 5\ell\geq 5 is prime. If fMk(1,)f\in M_{k}(1,\mathbb{Z}) is given by f=n=0a(n)qnf=\displaystyle\sum^{\infty}_{n=0}a(n)q^{n}, then we define

f¯:=n=0a(n)¯qn𝔽[[q]]\overline{f}:=\sum^{\infty}_{n=0}\overline{a(n)}q^{n}\in\mathbb{F}_{\ell}[[q]]

and

w(f¯):=inf{k:there exists gMk(1,) with f¯=g¯}.w_{\ell}(\overline{f}):=\inf\{k^{\prime}:\ \ \ \text{there exists $g\in M_{k^{\prime}}(1,\mathbb{Z})$ with $\overline{f}=\overline{g}$}\}.

We also define

Θ:=12πiddz=qddq.\Theta:=\frac{1}{2\pi i}\frac{d}{dz}=q\frac{d}{dq}. (2.9)

We require the following facts (see e.g. [Ser73, §\mathsection2.22.2] and [Joc82, §\mathsection11]).

Lemma 2.3.

Let kk and kk^{\prime} be even integers. Let fMk(1,)f\in M_{k}(1,\mathbb{Z}) and 5\ell\geq 5 be prime.

  1. (1)

    We have

    w(f|U)+w(f)1.w_{\ell}(f\big{|}U_{\ell})\leq\ell+\frac{w_{\ell}(f)-1}{\ell}.
  2. (2)

    If gMk(1,)g\in M_{k^{\prime}}(1,\mathbb{Z}) satisfies f¯=g¯\overline{f}=\overline{g}, then we have kk(mod1)k\equiv k^{\prime}\pmod{\ell-1}.

  3. (3)

    There exists a form gSk++1(1,)g\in S_{k+\ell+1}(1,\mathbb{Z}) with gΘf(mod)g\equiv\Theta f\pmod{\ell}.

  4. (4)

    For i+i\in\mathbb{Z}^{+}, we have w(fi)=iw(f)w_{\ell}(f^{i})=iw_{\ell}(f).

Finally, for an even integer kk, denote the weight kk Eisenstein series on SL2()\operatorname{SL}_{2}(\mathbb{Z}) by EkE_{k}.

3. The modular forms fr,f_{r,\ell} and gr,g_{r,\ell}

Let

Δ:=qn=1(1qn)24\Delta:=q\prod^{\infty}_{n=1}(1-q^{n})^{24}

be the unique normalized cusp form of weight 1212 on SL2()\operatorname{SL}_{2}(\mathbb{Z}). Set

δ:=2124.\delta_{\ell}:=\frac{\ell^{2}-1}{24}.

By studying the filtration w(Δrδ|U)w_{\ell}(\Delta^{r\delta_{\ell}}\big{|}U_{\ell}), we prove the following result.

Proposition 3.1.

Suppose that 5\ell\geq 5 is prime. Assume that rr is an odd positive integer with r4r\leq\ell-4. Then there exists a modular form fr,Sr12(1,νηr,)f_{r,\ell}\in S_{\frac{\ell-r-1}{2}}(1,\nu^{-r\ell}_{\eta},\mathbb{Z}) with

fr,pr(nrδ)qnr24pr(n+r24)qn24(mod).f_{r,\ell}\equiv\sum p_{r}\left(\ell n-r\delta_{\ell}\right)q^{n-\frac{r\ell}{24}}\equiv\sum p_{r}\left(\frac{\ell n+r}{24}\right)q^{\frac{n}{24}}\pmod{\ell}. (3.1)
Remark.

Boylan [Boy04, Theorem 1.31.3] gave the complete list of pairs (r,)(r,\ell) with r47r\leq 47 for which fr,0(mod)f_{r,\ell}\equiv 0\pmod{\ell}.

Proof.

Note that

Δrδ=qrδn=1(1qn)r2(1qn)rn=1(1qn)rn=0pr(nrδ)qn(mod).\Delta^{r\delta_{\ell}}=q^{r\delta_{\ell}}\prod^{\infty}_{n=1}\frac{(1-q^{n})^{r\ell^{2}}}{(1-q^{n})^{r}}\equiv\prod^{\infty}_{n=1}(1-q^{\ell n})^{r\ell}\cdot\sum^{\infty}_{n=0}p_{r}(n-r\delta_{\ell})q^{n}\pmod{\ell}. (3.2)

Thus, we have

Δrδ|Un=1(1qn)rn=0pr(nrδ)qn(mod).\Delta^{r\delta_{\ell}}\big{|}U_{\ell}\equiv\prod^{\infty}_{n=1}(1-q^{n})^{r\ell}\cdot\sum^{\infty}_{n=0}p_{r}(\ell n-r\delta_{\ell})q^{n}\pmod{\ell}. (3.3)

Define

k:=w(Δrδ|U).k:=w_{\ell}\left(\Delta^{r\delta_{\ell}}\big{|}U_{\ell}\right).

If k=k=-\infty, then the right hand side of (3.3) is congruent to 0(mod)0\pmod{\ell}, so there is nothing to prove. Therefore, we assume that kk\neq-\infty. By definition, there exists a form Gr,Mk(1,)G_{r,\ell}\in M_{k}(1,\mathbb{Z}) with

Gr,Δrδ|U(mod).G_{r,\ell}\equiv\Delta^{r\delta_{\ell}}\big{|}U_{\ell}\pmod{\ell}.

By (4)(4) of Lemma 2.3, we see that w(Δrδ)=12rδw_{\ell}(\Delta^{r\delta_{\ell}})=12r\delta_{\ell}. By (1)(1) of Lemma 2.3, we have

kr+22r+22.k\leq\frac{r+2}{2}\ell-\frac{r+2}{2\ell}.

Since Δrδ|UΔrδ|T(mod)\Delta^{r\delta_{\ell}}\big{|}U_{\ell}\equiv\Delta^{r\delta_{\ell}}\big{|}T_{\ell}\pmod{\ell} and Δrδ|T\Delta^{r\delta_{\ell}}\big{|}T_{\ell} has weight r(212)0(mod1)r(\frac{\ell^{2}-1}{2})\equiv 0\pmod{\ell-1}, we see by (2)(2) of Lemma 2.3 that

k0(mod1).k\equiv 0\pmod{\ell-1}. (3.4)

If r4r\leq\ell-4, we have

r+22r+22<r+32(1)\frac{r+2}{2}\ell-\frac{r+2}{2\ell}<\frac{r+3}{2}(\ell-1)

which implies by (3.4) that

kr+12(1).k\leq\frac{r+1}{2}(\ell-1). (3.5)

Let d=dim(Mk(1,)).d=\dim(M_{k}(1,\mathbb{Z})). The space Mk(1,)M_{k}(1,\mathbb{Z}) has a basis {f0,,fd}\{f_{0},...,f_{d}\} of forms such that

fi(z)=qi+O(qi+1), 0id1.f_{i}(z)=q^{i}+O(q^{i+1}),\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\leq i\leq d-1. (3.6)

After subtracting an appropriate integral linear combination of these basis elements from Gr,G_{r,\ell}, we may assume that ord(Gr,)r(2124)\operatorname{ord}_{\infty}(G_{r,\ell})\geq\lceil r(\frac{\ell^{2}-1}{24\ell})\rceil. If nn\in\mathbb{Z} satisfies r(21)24n<r24\frac{r(\ell^{2}-1)}{24\ell}\leq n<\frac{r\ell}{24}, then r2r24n<r2r\ell^{2}-r\leq 24\ell n<r\ell^{2}; this is a contradiction since r4r\leq\ell-4. This implies that ord(Gr,)r(2124)r+124\operatorname{ord}_{\infty}(G_{r,\ell})\geq\lceil r(\frac{\ell^{2}-1}{24\ell})\rceil\geq\frac{r\ell+1}{24}, so

fr,:=Gr,ηrSkr2(1,νηr).f_{r,\ell}:=\frac{G_{r,\ell}}{\eta^{r\ell}}\in S_{k-\frac{r\ell}{2}}(1,\nu^{-r\ell}_{\eta}).

This form satisfies (3.1). Since ord(Gr,)r+124\operatorname{ord}_{\infty}(G_{r,\ell})\geq\frac{r\ell+1}{24} and ord(Gr,)k12\operatorname{ord}_{\infty}(G_{r,\ell})\leq\frac{k}{12}, it follows that kr+12>r12(1)k\geq\frac{r\ell+1}{2}>\frac{r-1}{2}(\ell-1). By (3.5), we get k=r+12(1)k=\frac{r+1}{2}(\ell-1); the result follows. ∎

We now construct the half-integral weight forms which we will need in order to prove the second assertion of Theorem 1.3.

Proposition 3.2.

Let rr be an odd positive integer. Let 5\ell\geq 5 be a prime such that r4r\leq\ell-4. Suppose that all of the integers rw<0-r\leq w<0 satisfying wr(mod24)w\equiv-r\pmod{24} have the property that (rw)=1\left(\frac{-rw}{\ell}\right)=1. Then there exists a form gr,g_{r,\ell} in S2r12(1,νηr,)S_{\frac{\ell^{2}-r-1}{2}}(1,\nu^{-r}_{\eta},\mathbb{Z}) satisfying

gr,(rn)=1pr(n+r24)qn24.g_{r,\ell}\equiv\sum_{\left(\frac{-rn}{\ell}\right)=-1}p_{r}\left(\frac{n+r}{24}\right)q^{\frac{n}{24}}. (3.7)
Proof.

By (3)(3) of Lemma 2.3, there exists a form Hr,Sr(21)2+(+1)12(1,)H_{r,\ell}\in S_{\frac{r(\ell^{2}-1)}{2}+(\ell+1)\frac{\ell-1}{2}}(1,\mathbb{Z}) with

Hr,(r)Θ12Δrδ(mod).H_{r,\ell}\equiv\left(\frac{-r}{\ell}\right)\Theta^{\frac{\ell-1}{2}}\Delta^{r\delta_{\ell}}\pmod{\ell}.

Recall that E11(mod)E_{\ell-1}\equiv 1\pmod{\ell}. Define H^r,Sr(21)2+(+1)12(1)\widehat{H}_{r,\ell}\in S_{\frac{r(\ell^{2}-1)}{2}+(\ell+1)\frac{\ell-1}{2}}(1) by

H^r,:=ΔrδE1+12Hr,.\widehat{H}_{r,\ell}:=\Delta^{r\delta_{\ell}}E^{\frac{\ell+1}{2}}_{\ell-1}-H_{r,\ell}.

By (3.2), we have

H^r,(1(rn))pr(nrδ)qnn=1(1qn)r(mod).\widehat{H}_{r,\ell}\equiv\sum\left(1-\left(\frac{-rn}{\ell}\right)\right)p_{r}(n-r\delta_{\ell})q^{n}\cdot\prod^{\infty}_{n=1}\left(1-q^{\ell n}\right)^{r\ell}\pmod{\ell}.

This implies that

H^r,ηr2((rn)=0pr(n+r24)qn24+2(rn)=1pr(n+r24)qn24)(mod).\widehat{H}_{r,\ell}\equiv\eta^{r\ell^{2}}\left(\sum_{\left(\frac{-rn}{\ell}\right)=0}p_{r}\left(\frac{n+r}{24}\right)q^{\frac{n}{24}}+2\sum_{\left(\frac{-rn}{\ell}\right)=-1}p_{r}\left(\frac{n+r}{24}\right)q^{\frac{n}{24}}\right)\pmod{\ell}.

Since (rw)=1\left(\frac{-rw}{\ell}\right)=1 for all of the integers rw<0-r\leq w<0 which satisfy wr(mod24)w\equiv~{}-r~{}\pmod{24}, we conclude that H^r,\widehat{H}_{r,\ell} vanishes to order >r224>\frac{r\ell^{2}}{24} modulo \ell. By arguing as in the proof of Proposition 3.1 using (3.6), we may assume that H^r,\widehat{H}_{r,\ell} vanishes to order >r224>\frac{r\ell^{2}}{24}, so

H^r,ηr2S2r12(1,νηr,).\frac{\widehat{H}_{r,\ell}}{\eta^{r\ell^{2}}}\in S_{\frac{\ell^{2}-r-1}{2}}(1,\nu^{-r}_{\eta},\mathbb{Z}).

We have a form fr,Sr12(1,νηr)f_{r,\ell}\in S_{\frac{\ell-r-1}{2}}(1,\nu^{-r\ell}_{\eta}) which satisfies (3.1), so the form fr,S2r2(1,νηr)f^{\ell}_{r,\ell}\in S_{\frac{\ell^{2}-r\ell-\ell}{2}}(1,\nu^{-r}_{\eta}) has the property that

fr,(rn)=0pr(n+r24)qn24(mod).f^{\ell}_{r,\ell}\equiv\sum_{\left(\frac{-rn}{\ell}\right)=0}p_{r}\left(\frac{n+r}{24}\right)q^{\frac{n}{24}}\pmod{\ell}.

Let ss\in\mathbb{Z} satisfy s21(mod)s\equiv 2^{-1}\pmod{\ell}, and define gr,S2r12(1,νηr,)g_{r,\ell}\in S_{\frac{\ell^{2}-r-1}{2}}(1,\nu^{-r}_{\eta},\mathbb{Z}) by

gr,:=s(H^r,ηr2fr,E1r+12)(rn)=1p(n+r24)qn24(mod).g_{r,\ell}:=s\left(\frac{\widehat{H}_{r,\ell}}{\eta^{r\ell^{2}}}-f^{\ell}_{r,\ell}E^{\frac{r+1}{2}}_{\ell-1}\right)\equiv\sum_{\left(\frac{-rn}{\ell}\right)=-1}p\left(\frac{n+r}{24}\right)q^{\frac{n}{24}}\pmod{\ell}.

This concludes the proof. ∎

4. Compatibility

Suppose that 5\ell\geq 5 is prime and that rr is an odd positive integer such that r<4r<\ell-4. Recall that

Nr={2if 3r,6if 3rN_{r}=\begin{cases}2&\text{if }3\mid r,\\ 6&\text{if }3\nmid r\end{cases}

and that

ϵr={(8r)if 3r,((8r),(12r))if 3r.\epsilon_{r}=\begin{cases}\left(\frac{8}{r\ell}\right)&\text{if }3\mid r,\\ (-\left(\frac{8}{r\ell}\right),-\left(\frac{12}{r\ell}\right))&\text{if }3\nmid r.\end{cases}

Recall also that rr and \ell are compatible if for every newform fSr2new(Nr,ϵr)f\in S^{\operatorname{new}}_{\ell-r-2}(N_{r},\epsilon_{r}), the image of ρ¯f\overline{\rho}_{f} contains a conjugate of SL2(𝔽)\operatorname{SL}_{2}(\mathbb{F}_{\ell}). Finally, recall for each newform f=a(n)qnf=\sum a(n)q^{n} that we have a mod \ell reduction f¯=a(n)¯qn𝔽¯[[q]]\overline{f}=\sum\overline{a(n)}q^{n}\in\overline{\mathbb{F}}_{\ell}[[q]]. We now show that we have compatibility for a wide range of rr and \ell.

Proposition 4.1.

If 2r+22±1(mod)2^{r+2}\not\equiv 2^{\pm 1}\pmod{\ell} and >5r+19\ell>5r+19, then rr and \ell are compatible.

Proof.

Let f=q+n2a(n)qnSr2new(Nr,ϵr)f=q+\sum_{n\geq 2}a(n)q^{n}\in S^{\text{new}}_{\ell-r-2}(N_{r},\epsilon_{r}) be a newform. By [DDT97, Theorem 2.47(b)2.47(b)], there are four possibilities for the image of ρ¯f\overline{\rho}_{f}:

  1. (1)

    ρ¯f\overline{\rho}_{f} is reducible.

  2. (2)

    ρ¯f\overline{\rho}_{f} is dihedral, i.e. ρ¯f\overline{\rho}_{f} is irreducible but ρ¯f|GK\overline{\rho}_{f}|_{G_{K}} is reducible for some quadratic K/K/\mathbb{Q}.

  3. (3)

    ρ¯f\overline{\rho}_{f} is exceptional, i.e. the projective image of ρ¯f\overline{\rho}_{f} is conjugate to one of A4A_{4}, S4S_{4}, A5A_{5}.

  4. (4)

    The image of ρ¯f\overline{\rho}_{f} contains a conjugate of SL2(𝔽)\operatorname{SL}_{2}(\mathbb{F}_{\ell}).

Since 2r+22±1(mod)2^{r+2}\not\equiv 2^{\pm 1}\pmod{\ell}, we have 2r32±1(mod)2^{\ell-r-3}\not\equiv 2^{\pm 1}\pmod{\ell}. By [AAT21, Lemma 3.2], we conclude that ρ¯f\overline{\rho}_{f} is irreducible. By the condition >5r+19\ell>5r+19, we have r2+12,+32\ell-r-2\neq\frac{\ell+1}{2},\frac{\ell+3}{2}; by the same lemma, we see that ρ¯f\overline{\rho}_{f} is not dihedral.

Thus, in order to show that rr and \ell are compatible, it suffices to show that the image of ρ¯f\overline{\rho}_{f} cannot be exceptional. To this end, we show that the projective image contains an element of order >5>5. Suppose that ι(a)\iota_{\ell}(a_{\ell}) is an \ell-adic unit. Recall that χ\chi is the \ell-adic cyclotomic character. By part 33 of Theorem 2.1, we know that

ρf|I(χr301).\rho_{f}|_{I_{\ell}}\cong\left(\begin{matrix}\chi^{\ell-r-3}&*\\ 0&1\end{matrix}\right).

Recall that ω\omega is the mod \ell cyclotomic character. Since ω\omega has order 1\ell-1, we see that the projective image of ρ¯f\overline{\rho}_{f} contains an element of order 1gcd(1,r3)\geq\frac{\ell-1}{\text{gcd}(\ell-1,\ell-r-3)}. Since >5r+19\ell>5r+19, we see that 1gcd(1,r3)>5\frac{\ell-1}{\text{gcd}(\ell-1,\ell-r-3)}>5.

Recall that ω2\omega_{2} and ω2\omega_{2}^{\prime} are Serre’s fundamental characters of level 22. If ι(a)\iota_{\ell}(a_{\ell}) is not an \ell-adic unit, then part 33 of Theorem 2.1 implies that

ρ¯f|I(ω2r300ω2r3).\overline{\rho}_{f}|_{I_{\ell}}\cong\left(\begin{matrix}\omega_{2}^{\ell-r-3}&0\\ 0&\omega_{2}^{\prime\ell-r-3}\end{matrix}\right).

Since ω2/ω2\omega_{2}/\omega^{{}^{\prime}}_{2} has order +1\ell+1, we know that the projective image of ρ¯f\overline{\rho}_{f} contains an element of order +1gcd(+1,r3)\frac{\ell+1}{\text{gcd}(\ell+1,\ell-r-3)}. By the fact that >5r+19\ell>5r+19, we see that +1gcd(+1,r3)>5\frac{\ell+1}{\text{gcd}(\ell+1,\ell-r-3)}>5. ∎

By computing in Magma, for small rr we can establish compatibility for a wide range of \ell.

Proposition 4.2.

Let rr be an odd positive integer satisfying 1r391\leq r\leq 39.

  1. (1)

    If >r+4\ell>r+4 is a prime such that 2r+22±1(mod)2^{r+2}\not\equiv 2^{\pm 1}\pmod{\ell}, then rr and \ell are compatible.

  2. (2)

    In particular, if r=1r=1 or 33 and >r+4\ell>r+4 is prime, then rr and \ell are always compatible.

Proof.

Fix an odd integer rr such that 1r391\leq r\leq 39. Our strategy to prove (1)(1) is to compute in Magma to rule out each of the first three possibilities in Proposition 4.1 as we vary over the primes >r+4\ell>r+4 such that 2r+22±1(mod)2^{r+2}\not\equiv 2^{\pm 1}\pmod{\ell}. By the same proposition, we may assume that r+4<5r+19r+4<\ell\leq 5r+19. Fix such a prime \ell. Arguing as in the proof of Proposition 4.1, we conclude that ρ¯f\overline{\rho}_{f} is irreducible for each newform fSr2new(Nr,ϵr)f\in S^{\text{new}}_{\ell-r-2}(N_{r},\epsilon_{r}).

For each newform f=q+n2af(n)qnSr2new(Nr,ϵr)f=q+\sum_{n\geq 2}a_{f}(n)q^{n}\in S^{\text{new}}_{\ell-r-2}(N_{r},\epsilon_{r}), by [AAT21, Lemma 3.2], we see that ρ¯f\overline{\rho}_{f} is dihedral only if ρ¯fρ¯fω12\overline{\rho}_{f}\cong\overline{\rho}_{f}\otimes\omega^{\frac{\ell-1}{2}} and =2r+5\ell=2r+5 or =2r+7\ell=2r+7. If pp\neq\ell is prime, then χ(Frobp)=p\chi(\operatorname{Frob}_{p})=p. if =2r+5\ell=2r+5 or 2r+72r+7, then we compute in Magma to conclude that there exists p{5,7}p\in\{5,7\} with the properties that pp\neq\ell, ω12(Frobp)=1\omega^{\frac{\ell-1}{2}}(\operatorname{Frob}_{p})=-1 (which would mean that tr ρ¯f(Frobp)=0\text{tr }\overline{\rho}_{f}(\operatorname{Frob}_{p})=0) and af(p)¯0\overline{a_{f}(p)}\neq 0. However, this contradicts the fact that tr ρ¯f(Frobp)=af(p)¯\text{tr }\overline{\rho}_{f}(\operatorname{Frob}_{p})=\overline{a_{f}(p)}. Therefore, ρ¯f\overline{\rho}_{f} is not dihedral for each newform fSr2new(Nr,ϵr)f\in S^{\text{new}}_{\ell-r-2}(N_{r},\epsilon_{r}).

Thus, to prove (1)(1), it suffices to show for each newform fSr2new(Nr,ϵr)f\in S^{\text{new}}_{\ell-r-2}(N_{r},\epsilon_{r}) that ρ¯f\overline{\rho}_{f} does not have exceptional image. From the proof of Proposition 4.1, we know that this condition holds if 1gcd(1,r3),+1gcd(+1,r3)>5.\frac{\ell-1}{\text{gcd}(\ell-1,\ell-r-3)},\frac{\ell+1}{\text{gcd}(\ell+1,\ell-r-3)}>5. If either of these inequalities fails to hold, then let p6p\nmid 6\ell be a prime. Define uf(p):=af(p)2¯/pr3u_{f}(p):=\overline{a_{f}(p)^{2}}/p^{\ell-r-3}. If the projective image of ρ¯f\overline{\rho}_{f} is A4,S4, or A5A_{4},S_{4},\text{ or }A_{5}, then we have

uf(p)=0,1,2,4 or uf(p)23uf(p)+1=0,u_{f}(p)=0,1,2,4\ \ \ \text{ or }u_{f}(p)^{2}-3u_{f}(p)+1=0, (4.1)

depending on the order of the image of ρ¯f(Frobp)\overline{\rho}_{f}(\operatorname{Frob}_{p}) in PGL2(𝔽¯)\operatorname{PGL}_{2}(\overline{\mathbb{F}}_{\ell}) (see e.g. [Rib75, p. 264] and [Rib85, p. 189]). To prove our result, it suffices to check for each newform fSr2new(Nr,ϵr)f\in S^{\text{new}}_{\ell-r-2}(N_{r},\epsilon_{r}) that there exists a prime p6p\nmid 6\ell such that (4.1) does not hold. For such ff, we compute with Magma to show that there exists p{5,7}p\in\{5,7\} such that (4.1) does not hold for uf(p)u_{f}(p).

To prove (2)(2), note that if r=1r=1 or 33, then 2r+22±1(mod)2^{r+2}\not\equiv 2^{\pm 1}\pmod{\ell} only fails when =5\ell=5 or 77, respectively; this contradicts the assumption that >r+4\ell>r+4. ∎

5. The main technical result

Choose a number field EE containing all of the coefficients of all of the normalized Hecke eigenforms in Sr2new(Nr,ϵr)S^{\text{new}}_{\ell-r-2}(N_{r},\epsilon_{r}). Recall that for each prime 5\ell\geq 5 we have fixed an embedding ι:¯¯\iota_{\ell}:\overline{\mathbb{Q}}\rightarrow\overline{\mathbb{Q}}_{\ell}. If λ\lambda is the prime of EE induced by the embedding ι:¯¯\iota_{\ell}:\overline{\mathbb{Q}}\rightarrow\overline{\mathbb{Q}}_{\ell}, then let EλE_{\lambda} be the completion of EE at λ\lambda with ring of integers 𝒪λ\mathcal{O}_{\lambda}. For a fixed α\alpha\in\mathbb{Z}, we can assume without loss of generality that 𝒪λ\mathcal{O}_{\lambda} has the property that

 the polynomial x2αx+1 factors in 𝒪λ with roots α1 and α2.\text{ the polynomial $x^{2}-\alpha x+1$ factors in $\mathcal{O}_{\lambda}$ with roots $\alpha_{1}$ and $\alpha_{2}$}. (5.1)

We now prove the main technical result which we will need in order to prove Theorem 1.2.

Theorem 5.1.

Let 5\ell\geq 5 be prime. Suppose that λ\lambda, α\alpha, α1\alpha_{1}, and α2\alpha_{2} are defined as above and that α±2(mod)\alpha\not\equiv\pm 2\pmod{\ell}. Let m1m\geq 1 be an integer. Suppose that rr and \ell are compatible. Then there exists a positive density set SS of primes such that if QSQ\in S, then Q1(modm)Q\equiv 1\pmod{\ell^{m}} and for all of the newforms fSr2new(Nr,ϵr)f\in S^{\operatorname{new}}_{\ell-r-2}(N_{r},\epsilon_{r}), we have

f|TQ(α1m1+α2m1)f(modλm).f\big{|}T_{Q}\equiv(\alpha_{1}^{\ell^{m-1}}+\alpha_{2}^{\ell^{m-1}})f\pmod{\lambda^{m}}.
Proof.

By the assumption that α±2(mod)\alpha\not\equiv\pm 2\pmod{\ell}, we know that α1α2(modλ)\alpha_{1}\not\equiv\alpha_{2}\pmod{\lambda}. For each newform fSr2new(Nr,ϵr)f\in S^{\operatorname{new}}_{\ell-r-2}(N_{r},\epsilon_{r}), the Galois representations ρf\rho_{f} and ρ¯f\overline{\rho}_{f} can be defined over 𝒪λ\mathcal{O}_{\lambda} and 𝒪λ/λ\mathcal{O}_{\lambda}/\lambda, respectively. Define α¯:=α(mod)\overline{\alpha}:=\alpha\pmod{\ell}. Since rr and \ell are compatible, we know that the image of each ρ¯f\overline{\rho}_{f} contains a conjugate of SL2(𝔽)\operatorname{SL}_{2}(\mathbb{F}_{\ell}), so we can use [AAT21,  Proposition 3.83.8] to find an element σGal(¯/(ζ))\sigma\in\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}(\zeta_{\ell})) such that each ρ¯f(σ)\overline{\rho}_{f}(\sigma) is conjugate to (α¯110)\left(\begin{matrix}\overline{\alpha}&1\\ -1&0\end{matrix}\right). It follows that the characteristic polynomial of ρf(σ)\rho_{f}(\sigma) is congruent to x2αx+1x^{2}-\alpha x+1 modulo λ\lambda. Since α1α2(modλ)\alpha_{1}\not\equiv\alpha_{2}\pmod{\lambda}, we can use Hensel’s lemma to factor the characteristic polynomial of ρf(σ)\rho_{f}(\sigma) over 𝒪λ\mathcal{O}_{\lambda}. Thus, ρf(σ)\rho_{f}(\sigma) is conjugate to a diagonal matrix with entries β1\beta_{1} and β2\beta_{2} in 𝒪λ\mathcal{O}_{\lambda} with the properties that β1α1(modλ)\beta_{1}\equiv\alpha_{1}\pmod{\lambda} and β2α2(modλ)\beta_{2}\equiv\alpha_{2}\pmod{\lambda}. By induction, we see that β1m1α1m1(modλm)\beta^{\ell^{m-1}}_{1}\equiv\alpha^{\ell^{m-1}}_{1}\pmod{\lambda^{m}} and β2m1α2m1(modλm)\beta^{\ell^{m-1}}_{2}\equiv\alpha^{\ell^{m-1}}_{2}\pmod{\lambda^{m}}. This implies that the characteristic polynomial of ρf(σm1)\rho_{f}(\sigma^{\ell^{m-1}}) is congruent to x2(α1m1+α2m1)x+1(modλm).x^{2}-(\alpha^{\ell^{m-1}}_{1}+\alpha^{\ell^{m-1}}_{2})x+1\pmod{\lambda^{m}}.

Since Gal((ζm)/(ζ))\operatorname{Gal}(\mathbb{Q}(\zeta_{\ell^{m}})/\mathbb{Q}(\zeta_{\ell})) has order m1\ell^{m-1}, we have σm1Gal(¯/(ζm))\sigma^{\ell^{m-1}}\in\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}(\zeta_{\ell^{m}})). By composing ρf|Gal(¯/(ζm))\rho_{f}|_{\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}(\zeta_{\ell^{m}}))} with the projection τm:GL2(𝒪λ)GL2(𝒪λ/λm)\tau_{m}:\operatorname{GL}_{2}(\mathcal{O}_{\lambda})\rightarrow\operatorname{GL}_{2}(\mathcal{O}_{\lambda}/\lambda^{m}), we have a representation ρ¯f,m:Gal(¯/(ζm))GL2(𝒪λ/λm)\overline{\rho}_{f,m}:\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}(\zeta_{\ell^{m}}))\rightarrow\operatorname{GL}_{2}(\mathcal{O}_{\lambda}/\lambda^{m}). This representation has finite image, so there exists a finite extension L/(ζm)L/\mathbb{Q}(\zeta_{\ell^{m}}) such that if τ:Gal(¯/(ζm))Gal(L/(ζm))\tau:\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}(\zeta_{\ell^{m}}))\rightarrow\operatorname{Gal}(L/\mathbb{Q}(\zeta_{\ell^{m}})) is the restriction map, then there exists a representation ρ~f,m:Gal(L/(ζm))GL2(𝒪λ/λm)\widetilde{\rho}_{f,m}:\operatorname{Gal}(L/\mathbb{Q}(\zeta_{\ell^{m}}))\rightarrow\operatorname{GL}_{2}(\mathcal{O}_{\lambda}/\lambda^{m}) such that ρ~f,mτ=ρ¯f,m\widetilde{\rho}_{f,m}\circ\tau=\overline{\rho}_{f,m}. The Chebotarev density theorem implies that there is a positive density set SS of primes such that if QSQ\in S, then ρ~f,m(FrobQ|L)\widetilde{\rho}_{f,m}(\text{Frob}_{Q}|_{L}) is conjugate to ρ~f,m(σm1|L)\widetilde{\rho}_{f,m}(\sigma^{\ell^{m-1}}\big{|}_{L}); the fact that FrobQ|LGal(L/(ζm))\text{Frob}_{Q}|_{L}\subset\operatorname{Gal}(L/\mathbb{Q}(\zeta_{\ell^{m}})) for such QQ implies that Q1(modm)Q\equiv 1\pmod{\ell^{m}}. For such QQ and any newform fSr2new(Nr,ϵr)f\in S^{\text{new}}_{\ell-r-2}(N_{r},\epsilon_{r}), we have

f|TQ(tr ρ~f,m(FrobQ|L))f(α1m1+α2m1)f(modλm).f\big{|}T_{Q}\equiv(\text{tr }\widetilde{\rho}_{f,m}(\text{Frob}_{Q}|_{L}))f\equiv(\alpha^{\ell^{m-1}}_{1}+\alpha_{2}^{\ell^{m-1}})f\pmod{\lambda^{m}}.

Remark.

We assume that α±2(mod)\alpha\not\equiv\pm 2\pmod{\ell} so that we can apply Hensel’s lemma. However, we do not need Hensel’s lemma when m=1m=1. Thus, if m=1m=1, then we have the result when α±2(mod)\alpha\equiv\pm 2\pmod{\ell}.

6. Proofs of Theorems 1.21.2 and 1.31.3

We now prove Theorems 1.21.31.2-1.3. First, we introduce some notation. Let 5\ell\geq 5 be prime. Suppose that fSk(1,νηr,)f\in S_{k}(1,\nu^{r}_{\eta},\mathbb{Z}) with f0(mod)f\not\equiv 0\pmod{\ell}, where rr is an odd positive integer and k12\k\in\frac{1}{2}\mathbb{Z}\backslash\mathbb{Z} satisfies k52k\geq\frac{5}{2}. We define

S(r,k):={S2k1new(2,(8r))if 3r,S2k1new(6,(8r),(12r))if 3rS(r,k):=\begin{cases}S^{\operatorname{new}}_{2k-1}\left(2,\left(\frac{8}{r}\right)\right)&\text{if }3\mid r,\\ S^{\operatorname{new}}_{2k-1}\left(6,-\left(\frac{8}{r}\right),-\left(\frac{12}{r}\right)\right)&\text{if }3\nmid r\end{cases}

and (as in the introduction)

χ(r):={(4)if 3r,(12)if 3r.\chi^{(r)}:=\begin{cases}\left(\frac{-4}{\bullet}\right)&\text{if }3\mid r,\\ \left(\frac{12}{\bullet}\right)&\text{if }3\nmid r.\end{cases}

For each squarefree tt, let FtS(r,k)F_{t}\in S(r,k) be the form with Shtf=Ftχ(r)\text{Sh}_{t}f=F_{t}\otimes\chi^{(r)}. As tt ranges over all of the squarefree integers, there are only finitely many non-zero possibilities for Ft(mod)F_{t}\pmod{\ell}. Let S:={Ft1,,Ftk}S:=\{F_{t_{1}},...,F_{t_{k}}\} be a collection which represents all of these possibilities. By (2.8), we see that SS is not empty.

The space S(r,k)S(r,k) is spanned by newforms g1,,gdg_{1},...,g_{d}. For j{1,,k}j\in\{1,...,k\}, write

Ftj=i=1dci,jgi,F_{t_{j}}=\sum^{d}_{i=1}c_{i,j}g_{i},

and let EE be a number field which contains the coefficients of each gig_{i} as well as all of the coefficients ci,jc_{i,j}. Fix a prime λ\lambda of EE over \ell and define

m(f):=max(1,1min(ordλ(ci,j)))m(f):=\text{max}(1,1-\text{min}(\operatorname{ord}_{\lambda}(c_{i,j})))

(the definition depends on the choice of λ\lambda, but this is not important to us). Before we prove Theorem 1.2, we require the following lemma.

Lemma 6.1.

Suppose that 5\ell\geq 5 is prime and that fSk(1,νηr,)f\in S_{k}(1,\nu^{r}_{\eta},\mathbb{Z}), where rr is an odd positive integer. Suppose that the newforms gig_{i} and the integer m(f)m(f) are defined as above. Suppose that Q5Q\geq 5 is a prime and that λQ\lambda_{Q} is an algebraic integer in EE with the property that gi|TQλQgi(modλm(f))g_{i}\big{|}T_{Q}\equiv\lambda_{Q}g_{i}\pmod{\lambda^{m(f)}} for all ii. Then

f|TQ2χ(r)(Q)λQf(modλ).f\big{|}T_{Q^{2}}\equiv\chi^{(r)}(Q)\lambda_{Q}f\pmod{\lambda}.
Proof.

For each tjt_{j} we have

Ftjχ(r)=i=1dci,jgiχ(r),F_{t_{j}}\otimes\chi^{(r)}=\sum^{d}_{i=1}c_{i,j}g_{i}\otimes\chi^{(r)},

and for each ii we have

(giχ(r))|TQ=χ(r)(Q)(gi|TQ)χ(r)χ(r)(Q)λQgiχ(r)(modλm(f)).\left(g_{i}\otimes\chi^{(r)}\right)\big{|}T_{Q}=\chi^{(r)}(Q)\left(g_{i}\big{|}T_{Q}\right)\otimes\chi^{(r)}\equiv\chi^{(r)}(Q)\lambda_{Q}g_{i}\otimes\chi^{(r)}\pmod{\lambda^{m(f)}}.

By the definition of m(f)m(f), it follows for each tjt_{j} that

(Ftjχ(r))|TQχ(r)(Q)λQFtjχ(r)(modλ).\left(F_{t_{j}}\otimes\chi^{(r)}\right)\big{|}T_{Q}\equiv\chi^{(r)}(Q)\lambda_{Q}F_{t_{j}}\otimes\chi^{(r)}\pmod{\lambda}.

Thus, for each squarefree tt, we have

Sht(f|TQ2)=(Shtf)|TQχ(r)(Q)λQShtf(modλ).\operatorname{Sh}_{t}\left(f\big{|}T_{Q^{2}}\right)=\left(\operatorname{Sh}_{t}f\right)\big{|}T_{Q}\equiv\chi^{(r)}(Q)\lambda_{Q}\operatorname{Sh}_{t}f\pmod{\lambda}.

The result follows from (2.8). ∎

The next result explains how to produce congruences from Lemma 6.1.

Lemma 6.2.

Suppose that 5\ell\geq 5 is prime, that rr is an odd positive integer, and that f=a(n)qn24Sk(1,νηr,)f=\sum a(n)q^{\frac{n}{24}}\in S_{k}(1,\nu^{r}_{\eta},\mathbb{Z}). Suppose that Q5Q\geq 5 is prime. Suppose that there exists αQ{±1}\alpha_{Q}\in\{\pm 1\} and α\alpha\in\mathbb{Z} with

f|TQ2αQαQk32f(mod).f\big{|}T_{Q^{2}}\equiv\alpha_{Q}\alpha Q^{k-\frac{3}{2}}f\pmod{\ell}.

Then we have

a(Q2n)(α1)αQQk32a(n)(mod) if (nQ)=αQ(12Q)(1Q)k12.a(Q^{2}n)\equiv(\alpha-1)\alpha_{Q}Q^{k-\frac{3}{2}}a(n)\pmod{\ell}\ \ \ \text{ if }\ \ \left(\frac{n}{Q}\right)=\alpha_{Q}\left(\frac{12}{Q}\right)\left(\frac{-1}{Q}\right)^{k-\frac{1}{2}}.
Proof.

This follows from the definition of the Hecke operator in (2.7). If f|TQ2=b(n)qn24f\big{|}T_{Q^{2}}=\sum b(n)q^{\frac{n}{24}}, then the third term of b(n)b(n) does not contribute and if (nQ)=αQ(12Q)(1Q)k12\left(\frac{n}{Q}\right)=\alpha_{Q}\left(\frac{12}{Q}\right)\left(\frac{-1}{Q}\right)^{k-\frac{1}{2}}, then the middle term of b(n)b(n) becomes αQQk32a(n)\alpha_{Q}Q^{k-\frac{3}{2}}a(n). ∎

We now prove Theorem 1.2.

Proof of Theorem 1.2.

Suppose that 5\ell\geq 5 is prime, that r<4r<\ell-4 is an odd positive integer and that rr and \ell are compatible. By Proposition 3.1, there is a modular form fr,Sr12(1,νηr)f_{r,\ell}\in S_{\frac{\ell-r-1}{2}}(1,\nu^{-r\ell}_{\eta}) such that

fr,pr(n+r24)qn24(mod).f_{r,\ell}\equiv\sum p_{r}\left(\frac{\ell n+r}{24}\right)q^{\frac{n}{24}}\pmod{\ell}.

Since χ(r)=χ(r)\chi^{(-r\ell)}=\chi^{(r)}, each of its Shimura lifts lands in the space S(r,r12)χ(r)S\left(-r\ell,\frac{\ell-r-1}{2}\right)\otimes\chi^{(r)}.

For each squarefree tt, let F~tS(r,r12)\tilde{F}_{t}\in S(-r\ell,\frac{\ell-r-1}{2}) be the form with Shtfr,=F~tχ(r)\text{Sh}_{t}f_{r,\ell}=\tilde{F}_{t}\otimes\chi^{(r)}. By [Ser76, Exercise 6.46.4], we see that there exists a positive density set SS of primes such that if QSQ\in S, then Q1(mod)Q\equiv 1\pmod{\ell} and

F~t|TQ2F~t(mod).\tilde{F}_{t}\big{|}T_{Q}\equiv 2\tilde{F}_{t}\pmod{\ell}.

for all squarefree tt. Thus, for all squarefree tt, we have

(F~tχ(r))|TQ2χ(r)(Q)F~tχ(r)(mod),\left(\tilde{F}_{t}\otimes\chi^{(r)}\right)\big{|}T_{Q}\equiv 2\chi^{(r)}(Q)\tilde{F}_{t}\otimes\chi^{(r)}\pmod{\ell},

which means that

Sht(f|TQ2)=(Shtf)|TQ2χ(r)(Q)Shtf(mod).\operatorname{Sh}_{t}\left(f\big{|}T_{Q^{2}}\right)=\left(\operatorname{Sh}_{t}f\right)\big{|}T_{Q}\equiv 2\chi^{(r)}(Q)\operatorname{Sh}_{t}f\pmod{\ell}.

It follows from (2.8) that

fr,|TQ22χ(r)(Q)fr,(mod).f_{r,\ell}\big{|}T_{Q^{2}}\equiv 2\chi^{(r)}(Q)f_{r,\ell}\pmod{\ell}.

Thus, for α2(mod)\alpha\equiv 2\pmod{\ell}, the result follows from Lemma 6.2.

Recall the definitions of α1\alpha_{1} and α2\alpha_{2} in (5.1). If α±2(mod)\alpha\not\equiv\pm 2\pmod{\ell}, then Theorem 5.1 implies that there is a positive density set SS of primes such that if QSQ\in S, then Q1(mod)Q\equiv 1\pmod{\ell} and for all of the newforms gig_{i} in S(r,r12)S(-r\ell,\frac{\ell-r-1}{2}), we have

gi|TQ(α1m(fr,)1+α2m(fr,)1)gi(modλm(fr,)).g_{i}\big{|}T_{Q}\equiv(\alpha^{\ell^{m(f_{r,\ell})-1}}_{1}+\alpha_{2}^{\ell^{m(f_{r,\ell})-1}})g_{i}\pmod{\lambda^{m(f_{r,\ell})}}. (6.1)

Since

(α1m(fr,)1+α2m(fr,)1)(α1+α2)m(fr,)1αm(fr,)1α(modλ),(\alpha^{\ell^{m(f_{r,\ell})-1}}_{1}+\alpha_{2}^{\ell^{m(f_{r,\ell})-1}})\equiv(\alpha_{1}+\alpha_{2})^{\ell^{m(f_{r,\ell})-1}}\equiv\alpha^{\ell^{m(f_{r,\ell})-1}}\equiv\alpha\pmod{\lambda},

it follows from Lemma 6.1 that fr,|TQ2αχ(r)(Q)fr,(modλ)f_{r,\ell}\big{|}T_{Q^{2}}\equiv\alpha\chi^{(r)}(Q)f_{r,\ell}\pmod{\lambda}. Therefore,

fr,|TQ2αχ(r)(Q)fr,(mod),f_{r,\ell}\big{|}T_{Q^{2}}\equiv\alpha\chi^{(r)}(Q)f_{r,\ell}\pmod{\ell},

and the result follows from Lemma 6.2. ∎

Remark.

If there are no congruences between distinct newforms in S(r,r12)S(-r\ell,\frac{\ell-r-1}{2}), then m(fr,)=1m(f_{r,\ell})=1 by [AAT21, Lemma 5.35.3]. By the remark following the proof of Theorem 5.1, this implies that (6.1) holds when α2(mod)\alpha\equiv-2\pmod{\ell}, which gives Theorem 1.2 for this case.

To prove Theorem 1.3, we make use of the following consequence of [AAT21, Thm. 4.24.2]. They only state the result for fSknew(6)f\in S^{\operatorname{new}}_{k}(6), but the same argument applies with Sknew(6)S^{\operatorname{new}}_{k}(6) replaced by Sknew(2)S^{\operatorname{new}}_{k}(2).

Theorem 6.3.

Suppose that 5\ell\geq 5 is prime, that rr is an odd positive integer, and that there exists an integer aa for which 2a1(mod)2^{a}\equiv-1\pmod{\ell}. Let k12\k\in\frac{1}{2}\mathbb{Z}\backslash\mathbb{Z}. Then there exists a positive density set SS of primes such that if QSQ\in S, then Q2(mod)Q\equiv-2\pmod{\ell} and for every fS(r,k)f\in S(r,k) we have f|TQ(ϵ2)aQk32f(mod)f\big{|}T_{Q}\equiv-(-\epsilon_{2})^{a}Q^{k-\frac{3}{2}}f\pmod{\ell}, where ϵ2\epsilon_{2} is the W2W_{2} eigenvalue of ff.

Proof of Theorem 1.3.

Suppose that rr is an odd positive integer. Suppose that 5\ell\geq 5 is prime, that r<4r<\ell-4 and that there exists an integer aa with 2a1(mod)2^{a}\equiv-1\pmod{\ell}. For each squarefree tt, let F~tS(r,r12)\tilde{F}_{t}\in S(-r\ell,\frac{\ell-r-1}{2}) be the form with Shtfr,=F~tχ(r)\text{Sh}_{t}f_{r,\ell}=\tilde{F}_{t}\otimes\chi^{(r)}. By Theorem 6.3, we see that there exists γ{±1}\gamma\in\{\pm 1\} and a positive density set SS of primes such that if QSQ\in S, then Q2(mod)Q\equiv-2\pmod{\ell} and

F~t|TQγQr42F~t(mod).\tilde{F}_{t}\big{|}T_{Q}\equiv\gamma Q^{\frac{\ell-r-4}{2}}\tilde{F}_{t}\pmod{\ell}.

for all tt. Thus, for all tt, we have

(F~tχ(r))|TQγχ(r)(Q)Qr42F~tχ(r)(mod),\left(\tilde{F}_{t}\otimes\chi^{(r)}\right)\big{|}T_{Q}\equiv\gamma\chi^{(r)}(Q)Q^{\frac{\ell-r-4}{2}}\tilde{F}_{t}\otimes\chi^{(r)}\pmod{\ell},

which means that

Sht(fr,|TQ2)=(Shtfr,)|TQγχ(r)(Q)Qr42Shtfr,(mod).\operatorname{Sh}_{t}\left(f_{r,\ell}\big{|}T_{Q^{2}}\right)=\left(\operatorname{Sh}_{t}f_{r,\ell}\right)\big{|}T_{Q}\equiv\gamma\chi^{(r)}(Q)Q^{\frac{\ell-r-4}{2}}\operatorname{Sh}_{t}f_{r,\ell}\pmod{\ell}.

It follows from (2.8) that

fr,|TQ2γχ(r)(Q)Qr42fr,(mod).f_{r,\ell}\big{|}T_{Q^{2}}\equiv\gamma\chi^{(r)}(Q)Q^{\frac{\ell-r-4}{2}}f_{r,\ell}\pmod{\ell}.

It then follows (after setting αQ=γχ(r)(Q)\alpha_{Q}=\gamma\chi^{(r)}(Q) and β=1\beta=1) from Lemma 6.2 that there exists ϵQ{±1}\epsilon_{Q}\in\{\pm 1\} with the property that

pr(Q2n+r24)0(mod) if (nQ)=ϵQ.p_{r}\left(\frac{\ell Q^{2}n+r}{24}\right)\equiv 0\pmod{\ell}\ \ \ \text{ if }\ \ \left(\frac{n}{Q}\right)=\epsilon_{Q}.

This proves part (1)(1) of Theorem 1.3.

We now prove part (2)(2). Under the hypotheses on the integers rw<0-r\leq w<0 which satisfy wr(mod24)w\equiv-r\pmod{24}, we have the form gr,=a(n)qn24S2r12(1,νηr,)g_{r,\ell}=\sum a(n)q^{\frac{n}{24}}\in S_{\frac{\ell^{2}-r-1}{2}}(1,\nu^{-r}_{\eta},\mathbb{Z}) constructed in Proposition 3.2. It satisfies

a(n)pr(n+r24)(mod) when (rn)=1.a(n)\equiv p_{r}\left(\frac{n+r}{24}\right)\pmod{\ell}\ \ \ \ \text{ when }\left(\frac{-rn}{\ell}\right)=-1.

We now argue as in the proof of part (1)(1) to conclude that there exists ϵQ{±1}\epsilon_{Q}\in\{\pm 1\} for which

pr(Q2n+r24)0(mod)if(rn)=1 and (nQ)=ϵQ.p_{r}\left(\frac{Q^{2}n+r}{24}\right)\equiv 0\pmod{\ell}\ \ \ \ \text{if}\ \ \left(\frac{-rn}{\ell}\right)=-1\ \ \text{ and }\ \ \ \left(\frac{n}{Q}\right)=\epsilon_{Q}.

The result follows. ∎

7. The case when r=4r=\ell-4

When r=4r=\ell-4, we obtain results stronger than Theorems 1.11.31.1-1.3. The following result gives a short proof of [Boy04, Thm. 2.12.1, (2)(2)].

Lemma 7.1.

If 1(mod6)\ell\not\equiv 1\pmod{6} is prime, then f4,=0f_{\ell-4,\ell}=0. Thus, we have

p4(n+424)0(mod)for all n.p_{\ell-4}\left(\frac{\ell n+\ell-4}{24}\right)\equiv 0\pmod{\ell}\ \ \ \text{for all $n$}.
Proof.

Note that

f4,S32(1,νη41).f_{\ell-4,\ell}\in S_{\frac{3}{2}}(1,\nu^{4\ell-1}_{\eta}). (7.1)

Let ss\in\mathbb{Z} be such that s41(mod24)s\equiv 4\ell-1\pmod{24} and 0<s<240<s<24. By (2.4), we see that f4,f_{\ell-4,\ell} vanishes to order s\geq s. Since 1(mod6)\ell\not\equiv 1\pmod{6}, we have s>3s>3, so η3f4,S0(1,νη44)={0}\eta^{-3}f_{\ell-4,\ell}\in S_{0}(1,\nu_{\eta}^{4\ell-4})=\{0\}. The result follows. ∎

When 1(mod6)\ell\equiv 1\pmod{6}, we can use the fact that η3\eta^{3} is an eigenform for the Hecke operators TQ2T_{Q^{2}} for all primes Q3Q\geq 3 to produce the following congruences.

Lemma 7.2.

If 1(mod6)\ell\equiv 1\pmod{6} is prime, then f4=cη3f_{\ell-4}=c\eta^{3} for some cc\in\mathbb{Z}. For all primes Q3Q\geq 3, if Q2nQ^{2}\nmid n, then

pr(Q2n+r24)(1Q)(Q+1(12nQ))pr(n+r24)(mod).p_{r}\left(\frac{\ell Q^{2}n+r}{24}\right)\equiv\left(\frac{-1}{Q}\right)\left(Q+1-\left(\frac{12n}{Q}\right)\right)p_{r}\left(\frac{\ell n+r}{24}\right)\pmod{\ell}.
Proof.

Since 1(mod6)\ell\equiv 1\pmod{6}, we know that 3413\mid 4\ell-1. We also see by (7.1) that

η3f4,M0(1,νη44)=M0(1).\eta^{-3}f_{\ell-4,\ell}\in M_{0}(1,\nu^{4\ell-4}_{\eta})=M_{0}(1).

Since f4,f_{\ell-4,\ell} has integral coefficients, we conclude that f4,=cη3f_{\ell-4,\ell}=c\eta^{3} for some cc\in\mathbb{Z}. It follows from (2.7) for each prime Q3Q\geq 3 that

η3|TQ2=(1Q)(Q+1)η3.\eta^{3}\big{|}T_{Q^{2}}=\left(\frac{-1}{Q}\right)(Q+1)\eta^{3}.

Thus, when Q2nQ^{2}\nmid n, we have

pr(Q2n+r24)(1Q)(Q+1(12nQ))pr(n+r24)(mod).p_{r}\left(\frac{\ell Q^{2}n+r}{24}\right)\equiv\left(\frac{-1}{Q}\right)\left(Q+1-\left(\frac{12n}{Q}\right)\right)p_{r}\left(\frac{\ell n+r}{24}\right)\pmod{\ell}.

8. Acknowledgements

The author would like to thank Scott Ahlgren for suggesting this project and making many helpful comments. The author would also like to thank the Graduate College Fellowship program at the University of Illinois at Urbana-Champaign and the Alfred P. Sloan Foundation for their generous research support.

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