Eta-quotients

# On spaces of modular forms spanned by eta-quotients

Jeremy Rouse Department of Mathematics, Wake Forest University, Winston-Salem, NC 27109  and  John J. Webb Department of Mathematics, Wake Forest University, Winston-Salem, NC 27109
Department of Mathematics & Statistics, James Madison University, Harrisonburg, VA 22802
###### Abstract.

An eta-quotient of level is a modular form of the shape . We study the problem of determining levels for which the graded ring of holomorphic modular forms for is generated by (holomorphic, respectively weakly holomorphic) eta-quotients of level . In addition, we prove that if is a holomorphic modular form that is non-vanishing on the upper half plane and has integer Fourier coefficients at infinity, then is an integer multiple of an eta-quotient. Finally, we use our results to determine the structure of the cuspidal subgroup of .

###### 2010 Mathematics Subject Classification:
Primary 11F20, 11F30; Secondary 11G18

## 1. Introduction

In 1877, Richard Dedekind [3] defined the function

 (1) η(z)=q1/24∞∏n=1(1−qn),q=e2πiz,

for , which is now known as the Dedekind eta-function. The reciprocal of the Dedekind eta-function is

 1η(z)=q−1/24∞∑n=0p(n)qn,

where is the number of partitions of . The modular transformations of the Dedekind eta-function, which take the form

 η(az+bcz+d)=ϵ(a,b,c,d)(cz+d)1/2η(z) % for [abcd]∈SL2(Z) and ϵ(a,b,c,d)24=1,

play an important role in Hardy and Ramanujan’s exact formula for , derived using the circle method.

An eta-quotient of level is a function of the shape

 f(z)=∏δ|Nη(δz)rδ,

where . As a consequence of the product definition for , any eta-quotient is non-vanishing on . The following are examples of well-known eta-quotients:

 Δ(z)=η(z)24=∞∑n=1τ(n)qn,η(2z)5η(z)2η(4z)2=∑n∈Zqn2,η(4z)8η(2z)4=∞∑n=0σ1(2n+1)q2n+1,

where for all positive integers , , we have .

Let

A weight weakly holomorphic modular form is a function on that obeys the weight modular transformation law for , is holomorphic on , but may possess poles at the cusps. In the 1950s, Morris Newman (see [12] and [13]) used the Dedekind eta-function to systematically build weakly holomorphic modular forms for . A key ingredient in his work is a classification of when is actually a modular form for . If

 (2) ∑δ|Nδrδ≡0(mod24),∑δ|NNδrδ≡0(mod24), and ∏δ|Nδrδ is the square of a % rational number,

then transforms like a weight modular form for . We denote the set of weight modular forms on by .

Let be the algebraic curve , and let be its compactification. This algebraic curve has a model defined over given by the classical modular equation (see for example [16], pg. 109-110). In [8], Ligozat determines the order of vanishing of an eta quotient at the cusps of the level modular curve .

###### Theorem.

Let , and be positive integers with and . If is an eta-quotient, then the order of vanishing of at the cusp is

 N24∑δ|Ngcd(d,δ)2rδgcd(d,N/d)dδ.
###### Remark.

The formula for the order of vanishing only depends on , and not on .

In [14], Ken Ono observes that every holomorphic modular form for can be expressed as a linear combination of eta-quotients of level , and poses the following problem: “Classify the spaces of modular forms which are generated by eta-quotients.” The goals of the present paper are to address this question, to give an intrinsic characterization of level eta-quotients, and to apply this information to the study of the cuspidal subgroup of the Jacobian of .

## 2. Generating spaces with holomorphic and weakly holomorphic eta-quotients

Combining the hypotheses in the theorems of Newman and Ligozat show that the sequences of integers , which are the exponents of an eta-quotient in , correspond to integer points inside a -dimensional convex polytope, where is the number of divisors of .

Example. Using the lattice point enumeration program LattE [1], we are able to determine precisely the number of eta-quotients in for various values of . These eta-quotients correspond to lattice points in an -dimensional polytope whose volume increases quickly with . There are 4,988 eta-quotients in and there are 703,060,312 eta-quotients in .

###### Theorem 1.

There are precisely positive integers so that the graded ring of modular forms for is generated by eta-quotients.

A necessary condition for eta-quotients to generate the graded ring of modular forms on if is composite is for to be spanned by eta-quotients. Numerical evidence suggests that it is only possible to find enough eta-quotients to span if is sufficiently composite. We make this precise in our next result.

###### Theorem 2.

Suppose that . Then we have

 ∑δ|N|rδ|≤2k∏p|N(p+1p−1)min{2,ordp(N)}.
###### Remark.

The bound is sharp. For each , there is a weight and an eta-quotient for which the bound is achieved.

If and the number of divisors of are fixed, the number of sequences of integers that satisfy the inequality in Theorem 2 is bounded. Since tends to infinity with , it follows that for a fixed and a given number of prime divisors of , there are only finitely many spaces that are spanned by eta-quotients. Accordingly, a space will be spanned by eta-quotients only if is “sufficiently composite” in relation to its size.

###### Remark.

Theorem 2 and the consequences regarding being spanned by eta-quotients were first obtained by Soumya Bhattacharya in 2011 who is (as of this writing) a Ph.D. student of Professor Don Zagier at the University of Bonn. Bhattacharya’s work is not yet available in manuscript form.

Our next result gives examples of spaces that are not spanned by eta-quotients.

###### Corollary 3.

If is prime, then is spanned by eta-quotients if and only if .

Instead of insisting that every element of be expressible as a linear combination of eta-quotients which are holomorphic on , we could instead ask if every element of is expressible as a linear combination of weakly holomorphic eta-quotients of weight and level with a pole only at infinity.

For example, has dimension , but only contains eta-quotients, namely

 η(2z)4η(22z)4η(z)2η(11z)2,η(z)2η(11z)2,η(2z)2η(22z)2, and η(z)4η(11z)4η(2z)2η(22z)2.

The basis element is not expressible as a linear combination of these four. However, if , then and every holomorphic modular form in is a linear combination of (holomorphic) eta-quotients. Since is non-vanishing at all cusps except infinity, this implies that is a linear combination of weakly holomorphic eta-quotients with poles only at infinity.

Next, we will study the levels for which every form in is a linear combination of weakly holomorphic eta-quotients with a pole only at infinity. To state our result, let be the vector space of all weakly holomorphic modular forms that have a pole only at the cusp at infinity. Let be the subspace of consisting of forms that are linear combinations of eta-quotients with a pole only at infinity. Note that and are rings, and and naturally have the structure of modules over and , respectively.

Our next result is the following. Let and denote the number of orbits of elliptic points of order and for , respectively.

###### Theorem 4.
1. Suppose that is composite or . Suppose also that either or , and either or . Then has finite codimension in .

2. If is prime and or , then has infinite codimension in for all non-negative even integers .

Our next result gives a classification of the levels for which every element of has an expression in terms of weakly holomorphic eta-quotients.

###### Theorem 5.

Let be a positive integer. Then the following are equivalent.

1. For all positive even integers , . In other words, every element of is expressible as a linear combination of weakly holomorphic weight eta-quotients of level with poles only at infinity.

2. for all non-negative even integers .

3. .

Theorem 4 and the proof of Theorem 5 show that in order for to equal , it is necessary for to be composite and for to have no elliptic points. Our next result shows that for most small , this is sufficient.

###### Theorem 6.

If is composite, has no elliptic points, and , then .

###### Remark.

It appears likely that for and , the equivalent conditions of Theorem 5 are false.

There are two significant consequences of being able to span spaces of modular forms using eta-quotients. First, this provides a means of computing the Fourier expansions of a basis for . The theory of modular symbols can also be used to do this, but this process is somewhat inefficient in that it is first necessary to compute the Fourier expansions of Hecke eigenforms, and then to build a basis for using those. There are a number of situations where we are able to quickly construct bases for spaces with dimension greater than where the modular symbols algorithm is much less efficient. Second, if , it is sometimes desirable to compute the Fourier expansion of at other cusps of . There are no general algorithms to accomplish this task. However, if is an eta-quotient (or a linear combination of eta-quotients), it is straightforward to do this, since the transformation formula for is known for every matrix in .

## 3. Modular forms non-zero on H

Looking at (1), it is not difficult to see that any eta-quotient must have integral Fourier coefficients and be non-zero on the upper-half plane. We will examine to what degree the converse is true. Kohnen [7] studied a related problem, which we now describe. Let

 f(z)=cqh∏n≥1(1−qn)c(n)∈M!k(Γ0(N))

where denotes the space of weight weakly holomorphic modular forms on , is a non-zero constant, and each . Then he proves the following.

###### Theorem (Theorem 2 in [7]).

Suppose that is square-free. Then has no zeros or poles on if and only if depends only on the greatest common divisor .

A direct consequence of his proof is that for some that has no zeros or poles on and where -square free, there exists some positive integer such that is a constant times an eta-quotient. If we replace the condition that is square-free with , we prove the following result.

###### Theorem 7.

Suppose has the property that is non-zero on . Then where and is an eta-quotient.

Let be an even integer and be non-zero on . Since is zero at every cusp of , for sufficiently large , is a holomorphic modular form on with integral Fourier coefficients and non-zero on the upper-half plane. By Theorem 7, this implies is a constant times an eta-quotient which then implies the following.

###### Corollary 8.

Suppose has the property that is non-zero on . Then where and is an eta-quotient.

On the other hand, the condition in Theorem 7 is necessary. To see this, suppose that is an odd prime with . Then by the a theorem of Manin and Drinfeld [5], [10], the cuspidal subgroup of the Jacobian of is finite which implies that there exists a whose zeros are only supported at the cusp and whose poles are only supported at the cusp . It follows that for a sufficiently large , we have . Such a form is non-zero on . Since , the remark following the theorem of Ligozat implies that . We also have and . It follows that . Using the remark after the theorem of Ligozat again, we conclude that is not an eta-quotient. By Corollary 8, such a modular function must have non-integral Fourier coefficients. Another way of seeing this it to note that the cusps and on are not defined over and are in the same orbit under . Therefore, a modular form with a zero at one and no zero at the other cannot be fixed by and this implies it cannot have rational Fourier coefficients.

The multiplicative group of is often called the modular units on . If is a modular unit, then by definition as well, which implies that must be non-zero on . Consider . This is the subgroup of the modular units on with integral coefficients and a leading coefficient of . A special case of Corollary 8 is the following.

###### Corollary 9.

Suppose . Then where is an eta-quotient.

Our next result concerns the structure of the rational cuspidal subgroup when is a power of 2. Let denote the degree 0 divisors of the modular curve fixed by every element of . Then is isomorphic to . Ling [9] computed the structure of where is prime and . Using Corollary 9, we compute . Let and define the set as follows. For , define

 Ik=⎧⎨⎩∅ if 1≤k≤4,{2} if k=5,{1,2,2} if k=6.

For , if is odd, define

 it,k={⌊t−12⌋+k−12−2 if 1≤t≤k−1,k−3 if t=k−2,

and if is even, define

 it,k={⌊t2⌋+k2−3 if 1≤t≤k−4,k−4 if t=k−3 or k−2.

Then let

 Ik={it,k}k−2t=1.
###### Theorem 10.

For ,

 J0(2k)(Q)cusp≅⨁i∈IkZ/2iZ.

We now give a formula for the number of weight eta-quotients of level assuming that has genus zero (and hence is trivial).

###### Theorem 11.

Assume that is a positive integer so that has genus zero, and there is a holomorphic eta-quotient in . Then, the number of eta-quotients in is equal to the number of tuples of non-negative integers with

 ∑d|Ncdϕ(gcd(d,Nd))=k12⋅[SL2(Z):Γ0(N)].

Here is the usual Euler totient function.

###### Remark.

As a consequence, there are eta-quotients in , eta-quotients in , and eta-quotients in .

Acknowledgments We used Magma [2] version 2.19-8 and LattE [1] version 1.5 for computations. Details about the computations that were done are available at
http://users.wfu.edu/rouseja/eta/. The authors wish to thank the anonymous referees for helpful comments which have improved the exposition.

## 4. Background

Given a prime number and a non-zero integer , we define to be the highest power of that evenly divides .

For a positive even integer , let

 Ek(z)=1−2kBk∞∑n=1σk−1(n)qn

be the usual weight Eisenstein series, where is the sum of the st powers of the divisors of . Here is the th Bernoulli number. If , then . Let

 j(z)=E4(z)3Δ(z)=q−1+744q+196884q+⋯

be the usual modular -function. If is a positive integer, define the operator by

 f(z)=(∞∑n=0a(n)qn)|V(d)=f(dz)=∞∑n=0a(n)qdn.

It is well-known that if , then . (See for example Proposition 2.22 of [14].)

A level modular function is a meromorphic modular form of weight zero for . It is known that a level modular function is a rational function in and . (See Proposition 7.5.1 of Diamond and Shurman’s book [4], page 279.)

An elliptic point of is a number so that there is an with and . In this case has order or in , and we call an elliptic point of order or . Let denote the number of orbits of elliptic points of order , and the number of orbits of elliptic points of order . Diamond and Shurman ([4], Corollary 3.7.2, page 96) give the formulas

 ϵ2(Γ0(N)) ={∏p|N(1+(−1p)) if 4∤N0 if 4|N ϵ3(Γ0(N)) ={∏p|N(1+(−3p)) if p∤N0 if 9|N.

The action of on the upper half-plane extends to an action on . The cusps of are the -orbits of . Given a number (with and ), there is a matrix with . If , the order of vanishing of at the cusp is the smallest power of with a nonzero coefficient in the expansion of

 (cz+d)−kf(az+bcz+d)=∞∑n=0a(n)qn/N.

This will be denoted by . It does not depend on the matrix chosen, and if and are -equivalent, then .

The graded ring of modular forms for is . We require some results about the weights in which this graded ring is generated. Let be the sheaf of -forms on . Since is non-singular, is invertible and is isomorphic to . Multiplication of modular forms corresponds to maps

 H0(X0(N),L⊗a)⊗H0(X0(N),L⊗b)→H0(X0(N),L⊗a+b).

In [11], page 55, Mumford shows that if is an invertible sheaf on a curve of genus and with degree , then is ample with normal generation. This implies that the tensor product map above is surjective, for all and hence the graded ring of modular forms is generated in weight . This result seems to have been rediscovered by Rustom [15] and Khuri-Makdisi (see Proposition 2.1 of [6]).

If has no elliptic points, the degree of the invertible sheaf is where is the number of cusps of . If is composite, then . Hence, if is composite we have that the multiplication map is surjective for all and with . This shows that the graded ring of modular forms is generated in weight as long as is composite and has no elliptic points.

## 5. Proofs

First, we will analyze when there are eta-quotients of weight for . This is determined by the presence of elliptic points for . Recall that is the space of weakly-holomorphic eta quotients of weight and level .

###### Lemma 12.

Fix a positive integer . Then for even positive integers , is non-empty precisely when

1. if and ,

2. if and ,

3. if and , and

4. if .

Moreover, if is non-empty, then there is a holomorphic eta-quotient in .

###### Proof.

First, suppose that is a weakly holomorphic modular form of weight for . If is an elliptic point for , then for . It follows from this that . The transformation law gives that

 f(z)=(cz+d)kf(z).

Thus, either or . Corollary 2.3.4 of [4] (page 55) implies that any elliptic point of order has the form or and where . It follows from this fact that if is an elliptic point of order , then or , and if is an elliptic point of order then . It follows that if and is an elliptic point of order , then . Also, if and is an elliptic point of order , then .

Assuming that is a weakly holomorphic eta-quotient, then we must have that is non-vanishing on and this shows that in order for an eta-quotient to exist, the conditions in the lemma must be satisfied.

It suffices to construct holomorphic eta-quotients in the remaining cases. Since for all , this shows that there are always eta-quotients of weight .

Assume that . Then either , in which case , or there is a prime with . In this case and so there are eta-quotients of weight .

Assume that . Then either , in which case , or there is a prime with . If then and . If then . Thus, there is an eta-quotient of weight when .

Assume now that . In the cases that or we have already constructed weight eta-quotients. Otherwise, there is a prime with and a prime with . If , then and . If and then , and if and then . Thus, there is an eta-quotient of weight when . ∎

Now, we will prove Theorem 1.

###### Proof of Theorem 1.

If the graded ring of modular forms for is generated by eta-quotients, then must be spanned by eta-quotients. Since has positive dimension if , and there can be no weight eta-quotients if has elliptic points by Lemma 12, it follows that has no elliptic points. (The only level eta-quotients are powers of , so the same conclusion follows when .)

In each case, we attempt to find enough eta-quotients to span . If we succeed and is composite, then the fact that the graded ring of level modular forms is generated in weight proves the desired result. In cases where we fail to find enough eta-quotients, we enumerate all vectors in the corresponding -dimensional lattice that correspond to weight eta-quotients, and check that the dimension of the space they span is less than to prove that is not generated by eta-quotients.

There are no prime levels for which is generated by eta-quotients. (In fact, there are no prime levels whatsoever, by Theorem 4.) Details about the computations are available at http://users.wfu.edu/rouseja/eta/. ∎

Before we prove Theorem 2, we need two lemmas first.

###### Lemma 13.

Suppose that is a level eta quotient. Suppose that is a positive integer and . Let and view as an eta quotient of level . Let be a divisor of and write , where and . Then we have

 ord1d(f(z)|V(r))=egcd(d22,r2)rgcd(d22,e)ord1d1(f(z)).
###### Proof.

We have , where if and . Otherwise . The order of vanishing at the cusp is then

 Ne24∑δ|Ner|δgcd(d,δ)2sδgcd(d,Ned)dδ=Ne24∑δ|Ngcd(d,rδ)2rδgcd(d,Ned)drδ =Ne24r∑δ|Ngcd(d1,δ)2gcd(d2,r)2rδgcd(d1,Nd1)gcd(d2,ed2)d1d2δ=egcd(d22,r2)rgcd(d22,e)⋅⎛⎜ ⎜⎝N24∑δ|Ngcd(d1,δ)2rδgcd(d1,Nd1)d1δ⎞⎟ ⎟⎠ =egcd(d22,r2)rgcd(d22,e)ord1d1(f(z)).

The eta-quotients constructed in the next lemma will play a role in the proofs of many of the theorems.

###### Lemma 14.

If is a positive integer, then for each divisor of , there is a holomorphic eta-quotient that vanishes only at the cusps . Moreover, if , we have

 12kd∑δ|N|rδ|≤∏p|N(p+1p−1)min{2,ordp(N)}.

We follow the convention that an empty product on the right hand side takes the value .

###### Proof.

We will prove our result by induction on the number of distinct prime factors of . The base case is . In this case we have which vanishes at the only cusp of and we have equality in the claimed inequality.

Assume now that is a positive integer with distinct prime factors, one of which is and the highest power of dividing is . By induction, for each divisor of , there is a form

 F(z)=Ed,N/pm(z)

that has all of its zeros at cusps with denominator on .

Let and let and be divisors of and respectively. We now apply Lemma 13 with , or . Then we get that

 ord1/d1d2(Ed,N(z))=((pm+1gcd(d22,pm))−(pmgcd(d22,p2)pgcd(d22,pm)))ord1/d1(Ed,N(z)).

It is clear that this quantity is always non-negative. Moreover, the expression in parentheses is equal to zero if . If , then . This implies that all of the zeros of