A note on cusp forms as p-adic limits

# A note on cusp forms as p-adic limits

Scott Ahlgren Department of Mathematics, University of Illinois at Urbana-Champaign, Urbana, IL 61801, USA  and  Detchat Samart Department of Mathematics, University of Illinois at Urbana-Champaign, Urbana, IL 61801, USA
July 20, 2019
###### Abstract.

Several authors have recently proved results which express cusp forms as -adic limits of weakly holomorphic modular forms under repeated application of Atkin’s -operator. The proofs involve techniques from the theory of weak harmonic Maass forms, and in particular a result of Guerzhoy, Kent, and Ono on the -adic coupling of mock modular forms and their shadows. Here we obtain strengthened versions of these results using techniques from the theory of holomorphic modular forms.

###### Key words and phrases:
Modular forms, Cusp forms as -adic limits
###### 2010 Mathematics Subject Classification:
11F33, 11F11, 11F03
The first author was supported by a grant from the Simons Foundation (#208525 to Scott Ahlgren).

## 1. Introduction

In a recent paper [4], El-Guindy and Ono study a cusp form and a modular function related to the elliptic curve . Following their notation, define

 g(z) =η2(4z)η2(8z)=∑n≥1a(n)qn=q−2q5−3q9+6q13+⋯, (1.1) L(z) =η6(8z)η2(4z)η4(16z)=1q+2q3−q7−2q11+⋯, (1.2) F(z) =−g(z)L(2z)=∑n≥−1C(n)qn=−1q+2q3+q7−2q11+⋯. (1.3)

The main result of [4] states that if is a prime for which , then as a -adic limit, we have

 limm→∞F∣∣U(p2m+1)C(p2m+1)=g. (1.4)

The proof involves the theory of harmonic Maass forms, and in particular a result of Guerzhoy, Kent, and Ono [5] on the -adic coupling of mock modular forms and their shadows. Similar results were proved in [5] and [2].

Our goal is to prove strengthened versions of these results. We use a direct method; it does not involve harmonic Maass forms but rather an investigation of the action of the Hecke operators on a family of weakly holomorphic modular forms. A similar approach was recently employed in the study of the congruences of Honda and Kaneko [1]. For the modular forms described above, we prove the following, of which (1.4) is an immediate corollary. Note in addition that the case of (1.5) gives . Let denote the -adic valuation on .

###### Theorem 1.1.

Let be prime. Then for all integers we have

 vp(C(p2m+1)) =m, (1.5) vp(F|U(p2m+1)C(p2m+1)−g) ≥m+1. (1.6)

In Theorems 4.1 and 5.1 below we obtain similar improvements of results given in [5] and [2]. It is clear that the present approach would give similar results for a number of other spaces of modular forms.

## 2. Background

If is an integer, is a function of the upper half-plane, and , we define

 f(z)∣∣kγ:=(detγ)k/2(cz+d)−kf(az+bcz+d).

If , , and is a Dirichlet character modulo , let be the space consisting of functions which satisfy for all and which are holomorphic on the upper half plane and at the cusps. Let be the space of forms which are meromorphic at the cusps, and let denote the subspace of forms which are holomorphic at all cusps of other than . We drop the character from this notation when it is trivial. Each can be identified with its -expansion; with we have for some coefficients .

For each positive integer , the and -operators are defined on -expansions by

 ∑a(n)qn∣∣U(m) :=∑a(mn)qn, ∑a(n)qn∣∣V(m) :=∑a(n)qmn.

Let be the usual Hecke operator on . If is prime, then for and we have

 f|Tk,χ(pn)=n∑j=0χ(pj)p(k−1)jf|U(pn−j)|V(pj). (2.1)

Define

 Θ:=12πiddz=qddq.
###### Lemma 2.1.

If , then we have

 Tk,χ(m):M∞k(N,χ)→M∞k(N,χ). (2.2)

If then

 Θk−1:M∞2−k(N,χ)→M∞k(N,χ). (2.3)
###### Proof.

For the first statement, it suffices to show that for each prime we have

 Tk,χ(p):M∞k(N,χ)→M∞k(N,χ).

We have

 f|Tk,χ(p)=pk2−1(p−1∑j=0f∣∣k(1j0p)+χ(p)f∣∣k(p001)). (2.4)

Let be a cusp of inequivalent to and choose with . Given set . By a standard argument (see e.g. [6, §6.2]) we find that

 (1j0p)(abcd)=⎛⎝a+cjλ∗cpλ∗⎞⎠(λ∗0pλ)

where the first matrix on the right is in . It follows that each term from the sum on in (2.4) is holomorphic at cusps other than . To see that the last summand is also holomorphic at these cusps, let . Then

 (p001)(abcd)=(apλ′∗cλ′∗)(λ′∗0pλ′)

where the first matrix on the right is in .

Let be the Maass raising operator in weight , so that we have the basic relation

 Rk−2(f∣∣k−2γ)=(Rk−2f)∣∣kγ.

Bol’s identity (see for example [3, Lemma 2.1]) states that for we have

 Θk−1=1(−4π)k−1Rk−2∘Rk−4∘⋯∘R4−k∘R2−k.

It follows that

 Θk−1:M!2−k(N,χ)→M!k(N,χ)

and that

 (Θk−1f)∣∣kγ=Θk−1(f∣∣2−kγ).

The claim (2.3) follows from these two facts. ∎

If and , let denote the subset of consisting of forms whose coefficients are -integral rational numbers. If , define the filtration

 wp(f):=inf{k′:f≡g(modp) for some g∈M(p)k′(N)}.

We require two facts, which can be found for example in [7, §1]. First, if and , then . Also, we have

 wp(f|V(p))=pwp(f). (2.5)

## 3. Proof of Theorem 1.1

Recall the definitions (1.1)–(1.3), and note that and in the notation of the next proposition.

###### Proposition 3.1.

We have the following.

1. For every odd integer there exists a unique of the form

 Fm=−q−m+O(q3).
2. Suppose that is an odd prime and that . Then

 F|T2(pn)=pnFpn+C(pn)g.
###### Proof.

For each integer , let

 Er(z)=−g(z)Lr(2z)=−η2(4z)η6r(16z)η2r−2(8z)η4r(32z)=−q−2r+1+2q−2r+5+O(q−2r+9).

Using standard criteria (see, e.g. [8, Thm. 1.64, Thm. 1.65]) we find that . The forms can then be constructed as linear combinations of forms with . Uniqueness follows since the space is one-dimensional. This gives the first assertion.

From (2.1) we have

 F|T2(pn)=F|U(pn)+n−1∑j=1pjF|U(pn−j)|V(pj)+pnF|V(pn).

Observe that

 F|U(pn)=C(pn)q+O(q3)=C(pn)g+O(q3)

and that

 n−1∑j=1pjF|U(pn−j)|V(pj)+pnF|V(pn)=−pnq−pn+O(q3).

Assertion (2) follows from assertion (1) together with Lemma 2.1. ∎

Before proving Theorem 1.1 we require two lemmas.

###### Lemma 3.2.

For each prime and each integer we have

 C(p2m+1)≡(−1)mpmC(p)(modpm+1).
###### Proof.

Lemma 2.3 and Corollary 2.4 of [4] show that for each , there is a modular function of the form

 ϕp(z)=q−p+C(p)q+O(q3) (3.1)

(we have corrected a sign error in the proof of the corollary). From Lemma 2.1 we have

 Θ(ϕp)=−pq−p+C(p)q+O(q3)∈M∞2(32).

On the other hand, Proposition 3.1 gives

 F|T2(p)=−pq−p+C(p)q+O(q3).

Therefore

 F|T2(p)=Θ(ϕp), (3.2)

or equivalently

 F|U(p)=Θ(ϕp)−pF|V(p). (3.3)

Applying to both sides of (3.3) and arguing inductively, we obtain the following for each :

 F|U(p2m+1)=m∑k=0(−1)m−kpm−kΘ(ϕp)|U(p2k)+(−1)m+1pm+1F|V(p). (3.4)

For any we have . Therefore for each each we have

 F|U(p2m+1)≡(−1)mpmΘ(ϕp)(modpm+1). (3.5)

The lemma follows by comparing coefficients of in (3.5). ∎

The authors of [4] speculated that for every prime . We prove that this is the case.

###### Lemma 3.3.

For each prime we have .

###### Proof.

Assume to the contrary that From (3.2) and Proposition 3.1 it follows that

 Θ(ϕp)=F|T2(p)=pFp+C(p)g≡0(modp),

from which it follows that for some integral coefficients we have

 ϕp≡q−p+∞∑n=1Ap(np)qnp(modp).

Let

 f(z)=η8(32z)η4(16z)=q8+4q24+O(q40)∈M2(32).

Then has the form

 fp≡∞∑n=8Bp(np)qnp≡q8p+⋯(modp).

Since , we find that has the form

 hp≡∞∑n=7Dp(pn)qpn≡q7p+⋯(modp).

so that

 hp≡hp|U(p)|V(p)(modp). (3.6)

Using (2.5) we obtain

 wp(hp)=pwp(hp|U(p)).

Since and we must have , so that . Thus there exists such that

 h0≡hp|U(p)=q7+O(q8)(modp).

However, by examining a basis for the eight-dimensional space we find that there is no such form . This provides the desired contradiction. ∎

###### Proof of Theorem 1.1.

Assertion (1.5) follows from Lemmas 3.2 and 3.3. To prove (1.6), we use Proposition 3.1 and (2.1) to find that

 F|U(p2m+1)C(p2m+1)−g=1C(p2m+1)(p2m+1Fp2m+1−2m+1∑j=1pjF|U(p2m+1−j)|V(pj)). (3.7)

Using (2.1) we obtain

 F|T2(p2m)=2m+1∑j=1pj−1F|U(p2m+1−j)|V(pj−1).

Since for , we see from Proposition 3.1 that . It follows that

 2m+1∑j=1pjF|U(p2m+1−j)|V(pj)=p2m+1Fp2m|V(p)≡0(modp2m+1).

Assertion (1.6) now follows from (3.7) and (1.5). ∎

## 4. An example in weight 4 and level 9

In [5], the authors study the -adic coupling of mock modular forms and their shadows. As an application of their general result, they prove two -adic limit formulas involving the hypergeometric functions and evaluated at certain modular functions. We will use the following notation:

 g1(z) =η8(3z)=∑n≥1a(n)qn=q−8q4+20q7−70q13+⋯∈S4(9), L1(z) =η3(z)η3(9z)+3=1q+5q2−7q5+3q8+15q11+⋯, G(z) =g1(z)L21(z)=∑n≥−1C(n)qn=1q+2q2−49q5+48q8+771q11+⋯.

After rewriting using (3.3) and (3.4) of [5], we find that each of the two formulas in Theorem 1.3 of [5] is equivalent to the assertion that for every prime with we have

 limm→∞G|U(p2m+1)C(p2m+1)=g1(z). (4.1)

Here we prove a strengthened version of this result.

###### Theorem 4.1.

Let be a prime. Then for each integer we have

 vp(C(p2m+1)) ={3m+1,if p=2,3m,if p≠2. (4.2) vp(G|U(p2m+1)C(p2m+1)−g1) ≥{3m+2,if p=2,3m+3,if p≠2. (4.3)

The proof follows the argument in Section 3, so we give fewer details here.

###### Proposition 4.2.

We have the following.

1. For every integer with there exists a unique of the form

 Gm=q−m+O(q2).
2. Let be prime and let be a nonnegative integer. Then we have

 G|T4(pn)=p3nGpn+C(pn)g1.
###### Proof.

For each integer let

 Er(z)=g1(z)L1(z)r=q1−r+(5r−8)q4−r+O(q7−r).

Then . We construct each form by taking a linear combination of with Uniqueness follows since is spanned by the form .

We deduce assertion (2) as in the last section using (2.2), (2.1), and assertion (1).

###### Lemma 4.3.

If is prime, then

 C(p2m+1)≡(−1)mp3mC(p)(modp3m+3).
###### Proof.

Define

 ϕ2(z)=η2(3z)η6(9z)=∑n≥−2A2(n)qn=1q2−2q−q4+O(q5).

It is seen from the expression of as an infinite product that if . Similarly, if

 L1(z)=η3(z)η3(9z)+3=∑n≥−1b(n)qn,

then for all Therefore, for each positive integer there exist such that

 ϕl(z)=ϕ2(z)l−23∑j=0cjLl−2−3j1(z)=q−l+∑n≥1Al(n)qn∈M∞−2(9),

with and if (these coincide with the forms in [5, Prop. 3.1]). Since the constant term in the weight two modular form must be zero, we find as in the last section that . In particular, for any prime we have

 ϕp=q−p−C(p)q+O(q2).

By Lemma 2.1, we have

 Θ3(ϕp)=−p3q−p−C(p)q+O(q2)∈M∞4(9).

Hence it follows from Proposition 4.2 that

 Θ3(ϕp)=−p3Gp−C(p)g1=−G|T4(p)=−G|U(p)−p3G|V(p), (4.4)

so that

 G|U(p)=−Θ3(ϕp)−p3G|V(p). (4.5)

Applying iteratively leads to

 G|U(p2m+1)=m∑l=0(−1)m+1−lp3(m−l)Θ3(ϕp)|U(p2l)+(−1)m+1p3(m+1)G|V(p) (4.6)

for any non-negative integer . Since , we have from (4.6) that

 G|U(p2m+1)≡(−1)m+1p3mΘ3(ϕp)(modp3m+3). (4.7)

Comparing coefficients of in (4.7) gives the result. ∎

The authors of [5] verified that for every prime less than . Here we prove

###### Lemma 4.4.

For every odd prime , we have .

###### Proof.

Suppose by way of contradiction that is an odd prime with Then (4.4) gives

 Θ3(ϕp)≡0(modp),

which implies that for some coefficients we have

 ϕp≡q−p+∑n≥1Ap(np)qnp(modp).

Since has no zeros on the upper half plane (and does not vanish at any cusp), we have . Moreover,

 hp≡∑n≥pDp(pn)qpn≡qp+⋯(modp).

Therefore so that . Since and , we must have , but this is impossible since contains no non-constant elements. ∎

###### Proof of Theorem 4.1.

Assertion (4.2) follows from Lemma 4.3, Lemma 4.4, and the fact that . Next, we use Proposition 4.2 and (2.1) to write

 G|U(p2m+1)C(p2m+1)−g1=1C(p2m+1)(p6m+3Gp2m+1−2m+1∑j=1p3jG|U(p2m+1−j)|V(pj)). (4.8)

Since for any , Proposition 4.2 and (2.1) give

 2m+1∑j=1p3jG|U(p2m+1−j)|V(pj)=p3G|T4(p2m)|V(p)=p6m+3Gp2m|V(p)≡0(modp6m+3).

The result follows from (4.8) and (4.2). ∎

## 5. An example in weight 3 and level 16

In [2] the authors establish an analogous representation of a weight cusp form as a -adic limit. Let denote the non-trivial Dirichlet character modulo , and define

 g2(z) :=η6(4z)=∑n≥1a(n)qn=q−6q5+9q9+⋯∈S3(16,χ), L2(z) :=η6(8z)η2(4z)η4(16z)=1q+2q3−q7−2q11+⋯, H(z) :=g2(z)L22(z)=∑n≥−1C(n)qn=1q−2q3−13q7+26q11+⋯.

The two formulas stated in the main theorem of [2] involve the hypergeometric function after rewriting they are equivalent to the following statement: for every prime with we have

 limm→∞H|U(p2m+1)C(p2m+1)=g2(z).

Here we prove

###### Theorem 5.1.

For every prime and every integer we have

 vp(C(p2m+1)) =2m, (5.1) vp(H|U(p2m+1)C(p2m+1)−g2) ≥2m+2. (5.2)

We give only a sketch of the proof.

###### Proposition 5.2.

We have the following.

1. For every odd integer there exists a unique of the form

 Hm=q−m+O(q3).
2. Let be an odd prime and let be a nonnegative integer. Then we have

 H|T3,χ(pn)=χ(pn)p2nHpn+C(pn)g2.
###### Proof.

For each integer define

 Er(z):=g2(z)Lr2(z)=η6r(8z)η2r−6(4z)η4r(16z)∈M∞3(16,χ).

We construct the form with the desired properties by taking an appropriate linear combination of , and uniqueness follows since is one-dimensional. Assertion (2) is proved as before. ∎

###### Lemma 5.3.

If is prime and then

 C(p2m+1)≡p2mC(p)(modp2m+2).
###### Proof.

For each , let be the form given in [2, Lem. 3.3]. We have . For we have

 ϕl(z)=ϕ2(z)Pl(L2(z)),

where has Let be prime. As above we find that

 ϕp(z)=q−p−C(p)q+∑n≥5Ap(n)qn.

It follows from Proposition 2.1 that

 Θ2(ϕp)=p2q−p−C(p)q+O(q5)∈M∞3(16,χ), (5.3)

and we deduce using Proposition 5.2 that

 H|U(p)=H|T3,χ(p)+p2H|V(p)=−Θ2(ϕp)+p2H|V(p).

Iteratively applying results in

 H|U(p2m+1)=−m∑l=0p2(m−l)Θ2(ϕp)|U(p2l)+p2(m+1)H|V(p),

so we have

 H|U(p2m+1)≡−p2mΘ2(ϕp)(modp2m+2). (5.4)

Comparing coefficients gives the result. ∎

###### Lemma 5.4.

For every prime we have

###### Proof.

Suppose by way of contradiction that Then (5.3) and Lemma 5.2 show that whence

 ϕp≡q−p+∑n≥1Ap(np)qnp(modp).

Let Then has the form

 hp≡∑n≥5pDp(pn)qpn≡q5p+⋯(modp),

so that

 hp≡hp|U(p)|V(p)(modp).

Analyzing the filtration yields and . However, we find by examining a basis that there is no form with . This provides the desired contradiction. ∎

The proof of Theorem 5.1 follows as before.

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