ETA-QUOTIENTS AND THETA FUNCTIONS

ETA-QUOTIENTS AND THETA FUNCTIONS ROBERT J. LEMKE OLIVER Abstract. The Jacobi Triple Product Identity gives a closed form for many infinite product generating functions that arise naturally in combinatorics and number theory. Of particular interest is its application to Dedekind s eta-function ηz, defined via an infinite product, giving it as a certain kind of infinite sum known as a theta function. Using the theory of modular forms, we classify all eta-quotients that are theta functions. 1. Introduction and Statement of Results Jacobi s Triple Product Identity states that 1.1 1 x n 1 + x n 1 z 1 + x n 1 z = z m x m, which is surprising because it gives a striking closed form expression for an infinite product. Using 1.1, one can derive many elegant q-series identities. For example, one has Euler s identity 1. q 1 q 4n = 1 k q 6k+1 and Jacobi s identity 1.3 k= 1 q n 5 1 q n 1 q 4n = k= Both 1. and 1.3 can be viewed as identities involving Dedekind s eta-function ηz, which is defined by 1.4 ηz := q 1/4 1 q n, where q := e πiz. It is well known that ηz is essentially a half-integral weight modular form, a fact which Dummit, Kisilevsky, and McKay [5] exploited to classify all the eta-products functions of the form s i=1 ηn iz t i, where each n i and each t i is a positive integer whose q-series have multiplicative coefficients. Martin [7] later obtained the complete list of integer weight eta-quotients permitting the t i to be negative with multiplicative coefficients. The right hand sides of both 1. and 1.3 also have an interpretation in terms of halfintegral weight modular forms: they are examples of theta functions. Given a Dirichlet character ψ, the theta function θ ψ z of ψ is given by 1.5 θ ψ z := n ψnn δ q n, q k. Key words and phrases. Modular forms; q-series; eta-quotients; theta functions. 1

ROBERT J. LEMKE OLIVER where δ = 0 or 1 according to whether ψ is even or odd. The summation over n in 1.5 is over the positive integers, unless ψ is the trivial character, in which case the summation is over all integers. With this language, 1. becomes η4z = θ χ1 z, where χ 1 n = 1 n and is the Jacobi symbol. This fact is subsumed into the theorem of Dummit, Kisilevsky, and McKay, as η4z is an eta-product and any theta function necessarily has multiplicative coefficients. However, we note that 1.3 is equivalent to ηz 5 ηz η4z = θ 1z, which is covered neither by the theorem of Dummit, Kisilevsky, and McKay as the left-hand side is a quotient of eta-functions, not merely a product, nor is it covered by the theorem of Martin as the modular forms involved are of half-integral weight. It is therefore natural to ask which eta-quotients are theta functions. Theorem 1.1. 1. The following eta-quotients are the only ones which are theta functions for an even character: ηz 5 ηz η4z = η8zη3z η16z η16z η8z = = q n, q n, n n q n, η6z η9zη36z η3zη1zη18z = n 3 η4z = η48z 3 η4zη96z = η48zη7z η4zη144z = η4zη96zη144z 5 η48z η7z η88z = 1 n 4 n q n, q n, q n, q n, n 6 18 q n. n. The following eta-quotients are the only ones which are theta functions for an odd character: 4 η8z 3 = nq n, n

ETA-QUOTIENTS AND THETA FUNCTIONS 3 η16z 9 η8z 3 η3z = nq n, 3 n η3z η1z n = nq n, 3 η6z η48z 13 η4z 5 η96z = 6 nq n, 5 n η4z 5 η48z = n nq 1 n. In fact, we establish a broader classification theorem. Given a positive integer m, let Θ 0 m denote the linear span of the set of all theta functions associated to an even character ψ whose modulus is m together with its twists by χ,0, χ 3,0, and χ 6,0, where χ r,0 denotes the principal character modulo r, and let Θ 1 m denote the analogous space associated to odd characters of modulus m. Here, the twist of a theta function associated to ψ by another character χ is the theta function associated to ψχ. For convenience, let Θ m denote the union of Θ 0 m and Θ 1 m. We call an element of Θ m monic if its q-expansion has the form 1 + Oq or q + Oq 4. Theorem 1.. 1. The only eta-quotients which are monic elements of Θ 0 m for some m are those in Theorem 1.1 together with n 1 q n = 1 n q n, ηz ηz = ηzη4zη6z ηzη3zη1z = ηz η3z ηzη6z = η8z 5 η4z η16z = η9z η18z = η18z 5 η9z η36z = η4zη16zη4z η8zη1zη48z = η7z 5 η36z η144z = 1 3 n q n, 3 n 3 1 n 1 q n, n 3 + 3 n 6 q n, n n n 1 + q n, 3 6 n 1 q n, 3 n 3 1 3 n + q n, 6 n 3 n n n 1 + q n, 3 6

4 ROBERT J. LEMKE OLIVER η3zη18z η6zη9z = η8z η48z η16zη4z = n n q n, 6 3 n n 3 q n. 6. The only eta-quotients which are monic elements of Θ 1 m for some m are those in Theorem 1.1 together with η6z 5 η3z = n n nq 1 n. 3 Remark. A theorem of Mersmann [8] classifying holomorphic eta-quotients implies that there are essentially only finitely many eta-quotients which could be in any Θ m, even allowing non-monic elements. Unfortunately, while Mersmann s result can be made effective, the computations necessary to prove either case of Theorem 1. in this way would be prohibitively large. Consequently, our proof proceeds along fundamentally different lines. We note that Mersmann s result is slightly misquoted in [3] - the theorem credited to Mersmann on Page 30 of [3] is stronger than what he proves in his thesis. Our proof proceeds as follows. Instead of using the method employed by Mersmann [8] - essentially a careful study of the order of vanishing of eta-quotients - we make use of the combinatorial properties of eta-quotients and the constraints on the q-series of theta functions. Combined with the theory of modular forms, in particular the Fricke involution W k,m, asymptotic formulae, Eisenstein series, and Shimura s correspondence, the classification in Theorems 1.1 and 1. reduces to a case by case analysis. In this analysis, we make great use of the simple observation that if a > b, then 1 + q a 1 + q b = 1 + q b + Oq a. In this regard, we also need the solution to a classical Diophantine problem. Acknowledgements The author would like to thank Ken Ono, who suggested this problem to him, Marie Jameson, Jeremy Rouse, and Karl Mahlburg, for useful discussions, and the anonymous referees, who suggested considerable improvements.. Preliminary Facts We begin by recalling some basic facts about modular forms. A holomorphic function f : H C is a weakly holomorphic modular form of weight k 1 Z for the subgroup a b Γ SL Z if fγz = ɛ γ cz + d k fz for all γ = Γ, acting in the usual way c d by fractional linear transformation, where ɛ γ is a suitable fourth root of unity. Moreover, we require for each γ SL Z that f k γz := cz + d k fγz is represented by a Fourier series of the form f k γz = a γ nqn, n n n 0 where q N := e πi/n. In fact, there are only finitely many such series required, one for each cusp of Γ \ H, that is, an element ρ Γ \ Q. In this case, we let q ρ := q N. The space of all weakly holomorphic modular forms of weight k on Γ is denoted by M k! Γ; its subspace

ETA-QUOTIENTS AND THETA FUNCTIONS 5 consisting of all forms which are holomorphic resp. vanishing at the cusps is denoted by M k Γ resp. S k Γ these are the spaces of modular forms and cusp forms, respectively. For the subgroups we are concerned with, namely the congruence subgroups of level N, Γ 1 N := { a b c d a b SL Z : c d 1 0 1 } mod N, weakly holomorphic modular forms are fixed under the substitution z z + 1, and so they have a Fourier series at infinity with respect to the variable q := e πiz. Although we will briefly need the Fourier expansions at other cusps, it is this Fourier series also called a q-expansion that is of the most interest to us. If fz M! k Γ 1N, then fz is said to be modular of level N. If fz is holomorphic, then necessarily k 0 and the space M k Γ 1 N decomposes naturally into two pieces: the previously mentioned cusp space S k Γ 1 N and the so-called Eisenstein space E k Γ 1 N. If k Z, exploiting this decomposition, the size of the Fourier coefficients a f n of fz is well-understood. In particular, letting a f n = a cusp n + a Eis n, then we have that both a Eis n f,ɛ n k 1+ɛ, due to the explicit nature of the coefficients see.5 below, and a cusp n f,ɛ n k 1/+ɛ, which is the celebrated bound of Deligne. If k 1/ + Z, then the coefficients of both the Eisenstein series and the cuspidal part are not understood nearly as well, as both frequently encode values of L-functions. Nevertheless, polynomial bounds are known for each. In particular, we have the trivial bound that valid for all k 1/ and a cusp n f n k/, a Eis n f,ɛ n k 1+ɛ for k 3/, and a Eis n n ɛ if k = 1/. Although stronger bounds are known most recently, due to Blomer and Harcos [1], it is only the fact that each is polynomially bounded that will be relevant to us. This is because if fz is weakly holomorphic, but not holomorphic, then the coefficients of fz are of a fundamentally different size. Namely, for n in certain arithmetic progressions depending on the level, they satsify log a f n n 1/. This is due, in various settings, to Rademacher and Zuckerman [11], [1], [14], [15], and, more recently, to Bringmann and Ono []. We will find this vast difference in size useful later on. We recall that, given a Dirichlet character ψ of modulus r, the function θ ψ z is defined by θ ψ z := n ψnnδ q n, where δ = 0 or 1 according to whether ψ is even or odd. In both cases, the summation over n is assumed to be over the positive integers unless r = 1, in which case the sum is over all integers. It is classical that θ ψ z is a modular form of weight 1/ if ψ is even and of weight 3/ if ψ is odd. Each θ ψ z is of level 4r and, moreover, if r 1, then it is a cusp form. Regardless of the parity of ψ, we refer to θ ψ z as a theta function of modulus r. The twist of a theta function θ ψ z by a character χ is the theta function associated to ψχ. Given a positive integer m, we let Θ 0 m denote the linear span of the set of all weight 1/

6 ROBERT J. LEMKE OLIVER theta functions whose moduli are m and their twists by χ,0, χ 3,0, and χ 6,0, and we let Θ 1 m denote the analogous space for weight 3/ theta functions. Let Θ m be the union of Θ 0 m and Θ 1 m. Dedekind s eta-function ηz is defined by ηz := q 1/4 n 1 q n. It is almost a modular form of weight 1/ on SL Z, in the sense that 1.1 η = iz 1/ ηz, z but it fails to transform suitably under z z +1. However, since ηz is non-vanishing away from the cusps, a function of the form. fz = d N ηdz r d will be a weakly holomorphic modular form on Γ 1 4N if d N dr d 0 mod 4. This level may not be sharp, in the sense that fz may be a weakly holomorphic modular form on Γ 1 M for some proper divisor M of 4N, but what is important for our purposes is that the only primes dividing the level of fz are those dividing N together with and 3. We call a function of the form. satisfying this condition an eta-quotient. The order of vanishing of fz at the cusp ρ := α is given by [10, Theorem 1.65] δ.3 ord z=ρ fz = N 4 d N d, δ r d δ, N δ dδ. Lemma.1. Suppose that fz = d N ηdzr d is an element of Θm. Set a := 1 + max1, ν m and b := max1, ν 3 m, where ν p is the standard p-adic valuation, and let m 0 be the maximal divisor of m coprime to 6. Then r d = 0 for d a 3 b m 0. Proof. Given a theta function θz of modulus r and weight k, it is well known that θz W k,4r := rz k θ 1 4r z is again a modular form of weight k whose Fourier series has integral exponents, where W k,4r is the usual Fricke involution see [10, Proposition 3.8]. This property also holds for θz W k,4r t for any t, and so we see that the operator W k, a 3 b m 0 sends Θ m to the union of two spaces of modular forms whose Fourier series have integral exponents. If fz is in Θ m and has weight k, therefore, we must have that fz W k, a 3 b m 0 is a modular form of weight k with only integral exponents. On the other hand, we compute using.1 that fz W k, a 3 b m 0 = a 3 b m 0 z k d η a 3 b m d N 0z = C a 3 b m rd η 0z =: C d fz d N for some constant C. But the only way for fz to have a Fourier series with integral exponents is if for each d a 3 b m 0 we have that r d = 0. rd

ETA-QUOTIENTS AND THETA FUNCTIONS 7 The above lemma limits the eta-quotients which are in Θ m for a fixed m, but we still need a way to control the possible values of m. The following proposition permits us to do that. First, though, we fix notation. Given a weakly holomorphic modular form fz = n anqn of level N, the U p operator for a prime p is defined by fz U p := apnq n. n It is well known that fz U p is again a weakly holomorphic modular form of level N if p N and level pn if p N. Proposition.. If f M! k Γ 1N and p N is prime, then fz U p = 0 if and only if fz = 0. Before proving Proposition., we deduce its application to the problem at hand. Corollary.3. If fz = d N ηdzr d m are and 3. is in Θm for some m, then the only primes dividing Proof. We first show that any eta-quotient fz is not annihilated by the U p operator for any p 5. Fix such a prime and write fz = f 1 zf pz, with f 1 z := ηdz r d, and d N,p d f z := d N,p d η rd dz, p where an empty product has value 1. We now have that f4z U p = f 1 4z U p f 4z. Since f 1 4z is a weakly holomorphic modular form of level indivisible by p, we see by Proposition. that f 1 4z U p is non-zero, and since f 4z 0, we also have that f4z U p is non-zero. Since p 4 by assumption, it follows that that fz U p 0. On the other hand, functions in Θ m have the property that their coefficients are supported on exponents which are coprime to m. Hence, for any prime divisor p of m, we must have that U p annihilates Θ m. Thus, if fz is in Θ m, the only way it can have this property is if the only primes dividing m are and 3. The proof of Proposition. relies upon a lemma on sums of almost-everywhere multiplicative functions, which we define to be functions satisfying fmn = fmfn for any coprime m and n, neither of which is divisible by any of a finite set of primes called the bad primes. As an example, if fn is a multiplicative function such that ft 0 for some t, then ftn/ft is not generically multiplicative. It is, however, almost-everywhere multiplicative away from the primes dividing t. Lemma.4. Suppose that f 1,, f s are almost-everywhere multiplicative functions which are each non-zero for an infinite set of primes. Moreover, assume that, for each i j, f i p f j p for an infinite number of primes p. If c 1 f 1 n +... + c s f s n = 0 for all n indivisible by every bad prime, then each c i = 0. Proof. We proceed by induction, noting that the result is obviously true if s = 1.

8 ROBERT J. LEMKE OLIVER If s, we may assume by way of contradiction that each c i 0, so we have that s 1 c i.4 f s n = f i n c s for every n not divisible by any bad prime. Let m and n be coprime integers not divisible by any bad prime. We then have that both and f s mn = i=1 i=1 s 1 c i f s mn = f i mf i n c s = s 1 s 1 i=1 i=1 c i c s c i c s f i m s 1 j=1 s 1 c j c s f j m i=1 f i n. c i c s f i n Equating these two expressions for f s mn, we obtain that s 1 c i f i m + c s 1 i c j f j m f i n = 0, c s c s c s i=1 which, by our induction hypothesis, can only happen if for each i and m, we have that s 1 c j f i m + f j m = 0. c s c i c s j=1 j=1 Since c i 0 for each i, we then get a linear combination of f j m equaling 0, and again using the induction hypothesis, we find that c i = c s and all other c j = 0. Since we assumed that each c j 0, this can only happen if s =, and in that case,.4 yields that f 1 n = f n for all n away from the set of bad primes. But since these functions were assumed to be distinct, this cannot happen. Proof of Proposition.. If the Fourier expansion of fz at the cusp ρ is given by fz = a ρ nqρ n+κρ, then the principal part of fz at ρ is f ρ z = n n+κ ρ<0 a ρ nq n+κρ ρ. Following either the classical work of Rademacher and Zuckerman [11], [1], [14], [15] or the recent work of Bringmann and Ono [], we can write fz = f z+f hol z, where f z is a linear combination of so-called Maass-Poincaré series which matches the principal part of fz at each cusp and f hol z is a holomorphic modular form. Since the coefficients of the Maass-Poincaré series grow superpolynomially along certain arithmetic progressions modulo N and the coefficients of f hol z are polynomially bounded see the above discussion, in order for fz U p to be 0, we must have that f z = 0 and fz = f hol z. In the case that k < 0, we are now done, as there are no holomorphic modular forms of negative weight. If

ETA-QUOTIENTS AND THETA FUNCTIONS 9 k = 0, the only holomorphic modular forms are constant and are preserved under U p. Hence, in this case too, we must have that fz = 0. If k = 1/, a deep theorem of Serre and Stark [13] states that fz must be a linear combination of weight 1/ theta functions θ χ z of level dividing N and their dilates θ χ tz, with t condχ N. Thus, if n, N = 1, we have that 0 = a f p n = χ c χ χpχn, where the sum runs over the characters of conductor dividing N, and each c χ is a constant. But by the linear independence of characters, we must have that each c χ χp = 0, whence c χ = 0 since p, condχ = 1. Considering iteratively a f tp n in the same way, the result follows if k = 1/. We may now suppose that k 1. In the case that k is an integer, following [4], a basis for the Eisenstein space of M k Γ 1 N is given by where we define E ε,ψ,t k z using the series.5 E ε,ψ k z = c k,ε,ψ + {E ε,ψ,t k z : ε, ψ, t A k,n }, εn/dψdd k 1 q n d n by E ε,ψ,t k z = E ε,ψ k tz for k or ε, ψ 1, 1, and E 1,1,t z = E 1,1 z te 1,1 tz for t 1. In the above, c k,ε,ψ is a constant and A k,n is the set of triples ε, ψ, t where ε and ψ are primitive Dirichlet characters of conductor u and v, respectively, with εψ 1 = 1 k, and t is a positive integer such that tuv N. In the case k = we exclude the triple 1, 1, 1, and in the case k = 1, we require the first two elements of a triple to be unordered. We note that the Fourier coefficients of the Eisenstein series E ε,ψ k z are multiplicative; we denote these coefficients by σ ε,ψ k 1 n. For the cusp space S k Γ 1 N, we may choose a basis of Hecke eigenforms, so that any holomorphic modular form is a linear combination of forms g 1 z,, g s z, each with Fourier coefficients a 1 n,, a s n that are essentially multiplicative: in the case that a i n arises from a newform that is, a form not coming from some level M N, it is legitimately multiplicative, but if a i n arises from a non-newform, then it is of the form a i tn = a j n for some t N and the coefficients a j n of some newform the role of t will be handled easily in the proof below. In particular, if fz = anq n, we may, for n, N = 1, write an = c 1 f 1 n + c r f r n for some constants c 1,, c r and multiplicative f i n, coming either from an Eisenstein series or an eigenform. Since fz U p = 0, we must have that apn = 0 for all n. In particular, for n, pn = 1, we must have that 0 = c 1 f 1 pn + + c r f r pn = c 1 f 1 pf 1 n + + c r f s pf r n, and we may omit any f i n arising from an Eisenstein series E ε,ψ k tz with t > 1 or from a non-newform since n, N = 1 and t N, the omitted coefficients are 0 and will have no effect on apn. By Lemma.4, we must have that each c i f i p = 0. Now, f i p may be zero for some i, but in that case we necessarily have that f i p 0 this follows from the Euler product expansion in the case of a cusp form and by a direct computation in

10 ROBERT J. LEMKE OLIVER the case of Eisenstein series, which can only have this property if k = 1. Thus, by also considering ap n, we see that each c i not arising from an E ε,ψ k tz or a non-newform must be 0. Iteratively letting t be the smallest divisor of N not yet considered and repeating the argument above for atpn and atp n, we see that all c i = 0, whence fz = 0 identically. In the case that the weight is half-integral, we may still choose as a basis of the cusp space a sequence of Hecke eigenforms, but we no longer know that there is a basis of the Eisenstein space with multiplicative Fourier coefficients. We proceed, therefore, to show that fz is a cusp form. The argument in the integer weight case then applies, showing that fz = 0. Consider the image of fz under the Shimura map S λ,τ : M λ+ 1 Γ 1 N M λ Γ 1 N, where λ := k 1/ and τ is any squarefree positive integer often, this map is only defined for the cusp-space; see, for example, work of Jagathesan and Manickam [6] on extensions of this. If fz = anq n, the non-constant terms of S λ,τ fz = bnq n =: F z are given by the Dirichlet series formula [10, Theorem 3.14] bnn s = Ls + 1 λ, χ τ aτn n s, where χ τ is a Dirichlet character. Hence, we have that bn = τn d λ 1 χ τ da. d d n In particular if p, n = 1, then bp m n = d p m n = d n τp d λ 1 m n χ τ da d τn p m d λ 1 χ τ p m da = p mλ 1 χ τ p m bn, since apr = 0 for any r. Let F 0 z denote the projection of F z into the Eisenstein space of M λ Γ 1 N. We can express F 0 z as.6 F 0 z = a ε,ψ,t E ε,ψ λ tz, ε,ψ,t where we are summing over all triples ε, ψ, t such that ε and ψ are primitive characters of conductor dividing N and t is any divisor of N thus, we include triples ε, ψ, t which do not arise in A λ,n. We require that a ε,ψ,t = 0 when ε, ψ, t A λ,n, unless λ = 1, where we let a 1,1,1 absorb the coefficients of E 1,1 z arising from the terms E 1,1,t z in the basis expansion of the Eisenstein space of M Γ 1 N. We then have that, if n, pn = 1, bp m n = ε,ψ d a ε,ψ,1 σ ε,ψ λ 1 pm n + O p m n λ 1/+ɛ by Deligne s bound. Similarly, we also have for such n that bn = ε,ψ a ε,ψ,1 σ ε,ψ λ 1 n + O n λ 1/+ɛ.

ETA-QUOTIENTS AND THETA FUNCTIONS 11 Since bp m n = p mλ 1 χ τ p m bn, we must have that.7 ã ε,ψ σ ε,ψ λ 1 n = O p m n λ 1/+ɛ, where ε,ψ ã ε,ψ = a ε,ψ,1 σ ε,ψ λ 1 pm p mλ 1 χ τ p m. Let n = l 1 l, where l 1 and l are large primes such that l l 1/ 1. Then.5 implies that σ ε,ψ λ 1 l 1l = ψl 1 l λ 1 1 + εl 1 ψl l λ 1 + εl = ψl 1 ψl l 1 l λ 1 + ψl 1 εl l λ 1 1 + O Using this in.7 and dividing by l 1 l λ 1, we see that ã ε,ψ ψl 1 ψl = O l λ+1/ 1 + p mλ 1/+ɛ l λ+1/+ɛ 1, ε,ψ l λ 1/ 1 and letting l 1 and l tend to infinity along fixed arithmetic progressions modulo N, we see that, in fact, ã ε,ψ ψl 1 ψl = 0. ε,ψ Hence,.7 and the expansion of σ λ 1 l 1 l now yield that ã ε,ψ ψl 1 εl = O l λ+1/ 1 + p mλ 1+ɛ l 1 λ 1/+ɛ 1, ε,ψ and again letting l 1 and l tend to infinity along fixed arithmetic progressions, we see that.8 ã ε,ψ ψl 1 εl = 0. ε,ψ Since l 1 and l were chosen to be in arbitrary arithmetic progressions modulo N, this can be viewed as an equation in terms of the matrix K N K N, where K N is the φn φn matrix whose components are χa as a runs over elements of Z/NZ and χ runs over its characters. Since K N is invertible it is the tensor product of Vandermonde matrices arising from the cyclic factors of Z/NZ, K N K N is as well. Hence, the only way for.8 to hold is if each ã ε,ψ = 0. Recall that ã ε,ψ = a ε,ψ,1 σ ε,ψ λ 1 pm p mλ 1 χ τ p m, and since σ ε,ψ λ 1 pm p mλ 1, by considering large enough m, we conclude that each a ε,ψ,1 = 0. By iteratively letting t be the smallest divisor of N not yet considered and looking at btp m n, the above argument shows that each a ε,ψ,t = 0. Consequently, we must have that F 0 z = 0 and F z is a cusp form. But since this is true independent of the choice of τ in the map S λ,τ, we also have that fz is itself a cusp form. We now proceed as in the integer-weight case. Strictly speaking, the coefficients of half-integer weight eigenforms are only almost-everywhere multiplicative in square classes that is, for t squarefree, atn /at is multiplicative in n, provided that n, N = 1, but by considering each square class separately, the result follows. Before we can prove Theorem 1., we need one further lemma..

1 ROBERT J. LEMKE OLIVER Lemma.5. The only a = i 3 j which are one less than a square are a = 3, 8, 4, 48, and 88. Proof. Suppose a = n 1 = n + 1n 1. If a is odd, then we must have that both n 1 and n + 1 are powers of 3, and so a = 3. If a is not divisible by 3, then both n 1 and n + 1 must be powers of, and so a = 8. Lastly, suppose that a is divisible by both and 3. Since both n 1 and n + 1 are then required to be even, the pair n 1, n+1 must be either i, 3 j or 3 j, i. It is a classical result due to Levi ben Gerson that the only powers of and 3 which differ by one are, 3, 3, 4, and 8, 9. These lead to a = 4, 48, and 88, respectively. Remark. Of course, by Mihăilescu s resolution of Catalan s conjecture [9] it is known that 8 and 9 are the only consecutive perfect powers. 3. Proof of Theorem 1. We begin by considering a few concrete cases of Theorem 1.1. Although we could prove Theorem 1. without doing so, this allows us to illustrate the constructive approach we shall take. Lemma.1 and Corollary.3 together tell us that if fz = d N ηdzr d is an eta-quotient which is also a theta function θ ψ z of modulus r, then r must be divisible only by the primes and 3, and we may take N = 4r. Since the coefficients of any eta-quotient are real, we must also have that ψ is a quadratic character. The key observation which leads to Theorem 1.1 is that if we know the modulus of a quadratic theta function, then we know the first few terms in its Fourier series, at least up to a sign. First, we consider whether θ 1 z = 1+ qn is an eta-quotient. Let η 0 z := 1 q n, so that fz = q a f d N η 0dz r d, where af = dr d d N. In order for fz to be equal to 4 θ 1 z, we must have that a f = 0, as η 0 dz = 1 + Oq d. In fact, we know more: 3.1 η 0 dz r d = 1 r d q d + Oq d. Consequently, in order for the Fourier expansion of fz to match that of θ 1 z = 1 + q + q 4 + q 9 + Oq 16, we must have that r 1 = or, equivalently, that is the exact power of ηz dividing fz we say that ηz divides fz. The Fourier series of η 0 z is given by η 0 z = 1 + q + 5q + Oq 3, and since there is no q term in the Fourier expansion of θ 1 z, we see that ηz 5 must also divide fz in order to cancel the 5q term in the Fourier series for η 0 z. This leads us to consider η 0 z 5 η 0 z = 1 + q 4q5 + Oq 6. Since we now need to add q 4 to this Fourier expansion, we see that η4z must also divide fz. We compute that η 0 z 5 η 0 z η 0 4z = 1 + q + q4 + q 9 + Oq 16,

ETA-QUOTIENTS AND THETA FUNCTIONS 13 which matches perfectly the Fourier expansion of θ 1 z! Since for fz = ηz5, we have ηz η4z that a f = 0, fz is a candidate for an eta-quotient representation of θ 1 z. One easily verifies via.3 that fz is holomorphic, and then the Sturm bound [10, Theorem.58] implies that fz = θ 1 z. We now suppose that ψ is a character whose modulus is a positive power of and that the weight of θ ψ z is 1/. The Fourier series of θ ψ z must start as θ ψ z = q ± q 9 ± q 5 ± q 49 + Oq 81. Consequently, with the notation from before, we must have that a f = 1, and from 3.1, that the term ηdz r d dividing fz with smallest d must be either η8z or η8z 1. We consider only the first case before turning to the general situation of Theorem 1.. The Fourier expansion of η 0 8z is given by η 0 8z = 1 q 8 q 16 + Oq 40. As before, we now see that η16z 1 must divide fz, and η 0 8z η 0 16z = 1 q8 q 4 + q 3 + Oq 40. Since the modulus of ψ is a power of, the proof of Lemma.1 implies that we cannot change the coefficient of q 4, as it would require the level to be divisible by 3. Of course, this is acceptable, as the coefficient of q 5 in the expansion of θ ψ z is permitted to be 1. Continuing, we see that η3z must divide fz, which leads us to η 0 8zη 0 3z η 0 16z = 1 q 8 q 4 + q 48 + q 80 q 10 q 168 + Oq 4, a very promising Fourier series. We compute that for fz = η8zη3z, we have that a η16z f = 1, fz is holomorphic, and fz = θ χ8 z where χ 8 = is a primitive character of conductor 8. A priori it is conceivable that there are other theta functions expressible as eta-quotients divisible by η8z. However, since we have now reached a Fourier series whose exponents are supported on the squares, if fz were additionally divisible by some ηdz r d with d > 3 chosen minimally, then we must have both that d is a power of, based on the level, and that d is one less than a square, based on its effect on the Fourier series. But Lemma.5 implies that there are no such d, so we have produced the only eta-quotient divisible by η8z which is a theta function. We now turn to the proof of Theorem 1., assuming that fz = ηdz r d is in Θm for some m whose only prime factors are and 3. Under the assumption that fz is monic, we must consider two cases: either the Fourier series of fz has the form 1 + Oq or it has the form q + Oq 4. In the first case, the Fourier expansion has the form 1 + Oq and we must have that m = 1 since fz is a linear combination of theta functions and the only theta function whose Fourier expansions begins in this fashion is θ 1 z. Lemma.1 then tells us that we are limited to factors ηdz with d 144. Unfortunately, we must now split into further cases, depending on the location of the first non-zero coefficient after the constant term.

14 ROBERT J. LEMKE OLIVER If the Fourier series begins 1 aq + Oq with a 0, then we must have that ηz a is a factor of fz. We have that η 0 z a = 1 aq + aa 3 q + Oq 3, and consequently, we must have that ηz a divides fz, where a = aa 3. We then see that η3z a 3 must also divide fz, where a 3 = a3 4a. The coefficient of q 4 can be arbitrary, 3 however, so we let η4z b denote the power of η4z dividing fz. We then compute the coefficient of q 5 in η 0 z a η 0 z a η 0 3z a 3 η 0 4z b to be a b 1 0 a4 + 3 4 a 1 a 6. 5 This coefficient must be 0, since we cannot fix it with some ηdz with d 144. Since a was assumed to be non-zero, we see that b is in fact not allowed to be arbitrary. We let a 4 be the required value, namely 1 0 a4 3 4 a + 1a + 6. We then continue as before, seeing that we 5 must have a factor of η6z a 6, where a 6 = 1 30 a6 1 a3 + 8 15 a + a. The coefficient of q 7 is then 35 aa 1a + 1a a + a + 5, and, again since a 0, we must have that a = ±1, ±. ηz If a =, the above yields that fz is divisible by 5, which is exactly the etaquotient representation of θ 1 z we found earlier. In particular, the exponents in its Fourier ηz η4z series are supported on the squares, and any additional factor ηdz r d of fz must then have the property that d is a square. But because of the q term in the Fourier series of fz, this would introduce a term of order q d+1 in the Fourier series which we cannot change. Since d + 1 is not a perfect square, we have found the only possible fz with a =. Similarly, we find that there is exactly one form in Θ 1 for each of a = 1, a = 1, and a = : ηz η3z ηzη6z, ηzη4zη6z ηz, and ηzη3zη1z ηz. If the first non-zero coefficient of fz after the constant term is aq 4, the argument above translates almost exactly to this case, with the modification that q is replaced by q 4. One must argue that there can be no η9z r 9 factor, but this is clear, as it would introduce an irreparable q 13 in the Fourier expansion. We thus obtain in this case the previous etaquotients with z replaced by 4z. Two of these are in Θ 1, namely η8z 5 η4zη16zη4z and η4z η16z η8zη1zη48z. If the Fourier series starts 1 aq 9 + Oq 16, we proceed similarly, obtaining two elements of Θ 0 1, namely η9z and η18z 5. Throughout all of the remaining cases, corresponding η18z η9z η36z to the first non-constant term being q 16, q 36, and q 144, we obtain exactly two forms whose coefficients are supported on the squares: η36z, which is not in Θ η7z η7z 1, and 5, which η36z η144z is in Θ 1. We now consider the case when the Fourier expansion of fz has the form q + Oq 4. By Lemma.5, the factor ηdz r d of fz with rd 0 and d minimal has d {3, 8, 4, 48, 88}. We consider each of these cases in turn.

ETA-QUOTIENTS AND THETA FUNCTIONS 15 If a := r 3 0, in order for the Fourier coefficients to cancel properly, we must also have that η6z a η9z a 3 η1z a 4 η18z a 5 divides fz, where a = aa 3, a 3 = aa a+, a 3 4 = aa+a 1, and a 4 5 = aa a+a3 +4a+15. The coefficient of q 1 in the Fourier expansion 30 of the corresponding fz is 35 aa 1a + 1a a + a + 5, from which we see that a must be one of ±1, ±. These give rise to η3z η1z, η6z η3zη18z η6zη9z, η6z η9zη36z η6z5, and η3zη1zη18z η3z. As before, one verifies that each of these is in some Θ m, and no additional factors ηdz r d can be added to each of these eta-quotients while maintaining the property that the Fourier coefficients are supported on the squares. If r 3 = 0 and a := r 8 0, then we must also have that η16z a η4z b η3z a 4 divides fz, where a = aa 3, b is arbitrary, and a 4 = ab 1 1 a4 + 7 1 a + 1 a. Requiring the coefficient of q 40 to be 0 as we are not permitted to change it, we must have that b = a4 0a + 18. 15a Since b is an integer, we have that a 18, and one observes that the above is an integer only for a 6. If a = ±1, we find the two forms η8zη3z η16z and η16z η8z, both of which are theta functions. Any factors of η48z or η88z would introduce irreparable coefficients, so these are all the forms arising from a = ±1. If a = ±, we find η8z η48z η16zη4z and η16z5 η4zη96z η8z η3z η48z, the first of which is in Θ 0, the second of which is also a form of some interest, namely θ χ z + 3θ χ 9z where χn = n, but the Serre-Stark basis theorem [13] implies that it cannot lie in any Θ m. If a = ±3, we obtain the forms η8z 3 and η16z 9 η8z 3 η3z 3, both of which are theta functions. Lastly, if a = ±6, we obtain no forms, eventually running into a non-zero coefficient of q 88. We now suppose that η4z a η48z b is the smallest divisor of fz. We consider these two variables, since either coefficient can be arbitrary. Consequently, we permit one, but not both, of a and b to be 0. We proceed as before, eventually finding potentially problematic coefficients of q 40 and q 64, say A 1 and A, respectively. Both A 1 and A are polynomials in a and b, whose degrees in a are 10 and 11, respectively, and whose degrees in b are both 5. A 1 is irreducible, whereas A has a factor of a and is otherwise irreducible. Substituting a = 0 into A 1, we find that we must have bb 4 6 = 0, which cannot hold since we assumed that not both a and b are 0. Now, we compute the resultant in b of A 1 and A /a. This yields a degree 50 polynomial in a whose only rational roots are a = 1, a = 1, a = 5, and a = 5,

16 ROBERT J. LEMKE OLIVER which yield that b = 0 or, b = 1 or 3, b =, and b = 13, respectively. These yield the η4zη96zη144z forms η4z, 5, η48zη7z, η48z 3, η4z5 η48z, and 13. η48z η7z η88z η4zη144z η4zη96z η48z η4z 5 η96z 5 Finally, we suppose that η88z a is the smallest divisor of fz. We then must also have that η576z a η864z a 3 η115z a 4 divides fz, where a = aa 3, a 3 = aa a+ 3, and a 4 = aa+a 1 4. The coefficient of q 1440 is 1 5 aa4 6, which is non-zero, and so there are no such fz. This finishes the proof of Theorem 1.. References [1] V. Blomer and G. Harcos. Hybrid bounds for twisted L-functions. J. Reine Angew. Math., 61:53 79, 008. [] K. Bringmann and K. Ono. Coefficients of harmonic weak maass forms. Proc. of the 008 Univ. of Flor. Conf., accepted for publication. [3] J. H. Bruinier, G. van der Geer, G. Harder, and D. Zagier. The 1--3 of modular forms. Universitext. Springer-Verlag, Berlin, 008. Lectures from the Summer School on Modular Forms and their Applications held in Nordfjordeid, June 004, Edited by Kristian Ranestad. [4] F. Diamond and J. Shurman. A first course in modular forms, volume 8 of Graduate Texts in Mathematics. Springer-Verlag, New York, 005. [5] D. Dummit, H. Kisilevsky, and J. McKay. Multiplicative products of η-functions. In Finite groups coming of age Montreal, Que., 198, volume 45 of Contemp. Math., pages 89 98. Amer. Math. Soc., Providence, RI, 1985. [6] T. Jagathesan and M. Manickam. On Shimura correspondence for non-cusp forms of half-integral weight. J. Ramanujan Math. Soc., 33:11, 008. [7] Y. Martin. Multiplicative η-quotients. Trans. Amer. Math. Soc., 3481:485 4856, 1996. [8] G. Mersmann. Holomorphe η-produkte und nichtverschwindende ganze modulformen für γ 0 n. Master s Thesis, Univ. of Bonn, 1991. [9] P. Mihăilescu. Primary cyclotomic units and a proof of Catalan s conjecture. J. Reine Angew. Math., 57:167 195, 004. [10] K. Ono. The web of modularity: arithmetic of the coefficients of modular forms and q-series, volume 10 of CBMS Regional Conference Series in Mathematics. Published for the Conference Board of the Mathematical Sciences, Washington, DC, 004. [11] H. Rademacher. On the expansion of the partition function in a series. Ann. of Math., 44:416 4, 1943. [1] H. Rademacher and H. S. Zuckerman. On the Fourier coefficients of certain modular forms of positive dimension. Ann. of Math., 39:433 46, 1938. [13] J.-P. Serre and H. M. Stark. Modular forms of weight 1/. In Modular functions of one variable, VI Proc. Second Internat. Conf., Univ. Bonn, Bonn, 1976, pages 7 67. Lecture Notes in Math., Vol. 67. Springer, Berlin, 1977. [14] H. S. Zuckerman. On the coefficients of certain modular forms belonging to subgroups of the modular group. Trans. Amer. Math. Soc., 45:98 31, 1939. [15] H. S. Zuckerman. On the expansions of certain modular forms of positive dimension. Amer. J. Math., 6:17 15, 1940. Emory University, Department of Mathematics and Computer Science, 400 Dowman Dr, Atlanta, GA 303