PARTITION IDENTITIES AND A THEOREM OF ZAGIER

Jayce Getz and Karl Mahlburg

July 30, 2001

Abstract. In partition theory and q-series, one often seeks identities between series and infinite products. Using a recent result of Zagier, we obtain such identities for every positive integer m. For example if m = 1, then we obtain the classical Eisenstein series identity

∑_{λ≥1 odd} \frac{(−1)^{(λ−1)/2}q^{λ}}{(1 − q^{2λ})} = q ∏_{n=1}^{∞} \frac{(1 − q^{8n})^{4}}{(1 − q^{4n})^{2}} .

If m = 2 and (\frac{·}{3}) denotes the usual Legendre symbol modulo 3, then we obtain

∑_{λ≥1} \frac{(\frac{λ}{3})q^{λ}}{(1 − q^{2λ})} = q ∏_{n=1}^{∞} \frac{(1 − q^{n})(1 − q^{6n})^{6}}{(1 − q^{2n})^{2}(1 − q^{3n})^{3}} .

We describe some of the partition theoretic consequences of these identities. For instance, we obtain simple formulas for the number of representations of an integer as a sum of an arbitrary number of triangular numbers.

1. Introduction and Statement of Results

We begin by recalling the following identity of Euler:

(Euler)  ∏_{n=1}^{∞}(1 − q^{n}) = ∑_{k=−∞}^{∞}(−1)^{k}q^{(3k^{2}+k)/2}.

Note that the left hand side of Euler’s identity is related to partitions of integers into distinct parts. More precisely, each partition of n into an odd number of distinct parts adds −1 to the coefficient of q^{n} and each partition of n into an even number of distinct parts adds +1.

2000 Mathematics Subject Classification. 11P83, 11E25 The authors thank the National Science Foundation for their support. The first author also thanks the Clay Mathematical Institute for their help.

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Therefore, this identity, known as Euler’s pentagonal number theorem, shows us that the number of partitions of n into an odd number of distinct parts is equal to the number of partitions of n into an even number of distinct parts except when n is a pentagonal number, that is, a number of the form (3k^{2} + k)/2.

Other identities of this sort exist, perhaps the most famous being the Rogers-Ramanujan identities. The following is one of these celebrated identities:

(Rogers-Ramanujan) ∏_{n=0}^{∞} \frac{1}{(1 − q^{5n+2})(1 − q^{5n+3})} = 1 + ∑_{n=1}^{∞} \frac{q^{n^{2}+n}}{(1 − q)(1 − q^{2}) ⋯ (1 − q^{n})}.

Combinatorially, this identity establishes that the number of partitions of an integer n into parts that are congruent to 2, 3 (mod 5) equals the number of partitions of n in which any two summands differ by at least 2 and all summands exceed 1.

Such results are interesting both intrinsically and for their combinatorial consequences. Unfortunately, the methods used to prove them generate only finitely many natural identities. The purpose of this paper is to use a master theorem of Zagier [Z] previously conjectured by Kac and Wakimoto [K-W] to prove a natural infinite family of similar simple q-series identities relating infinite products to natural partition generating functions (perhaps well known to the experts). Through a combinatorial examination of these identities, we obtain closed explicit formulas for the number of representations of integers as a sum of an arbitrary number of triangular numbers. This problem is closely related to the discovery of formulas for the number of representations of n as a sum of squares and triangular numbers, a subject that has been examined recently by Milne, Ono and Zagier (see [M], [O], [Z]).

Before we state the main theorem we must offer some notation. Throughout, if m is a positive integer, then let s(m) denote the integer

(1.1)  s(m) := [\frac{m + 1}{2}].

For convenience, define the following set of vectors which we will see determines the summands in the partitions we consider.

Definition 1.1. If m is a positive integer, then let S(m) denote the set of integral vectors Λ = (λ_{1}, λ_{2}, …, λ_{s(m)}) for which the following hold:

(i) If j > i, then λ_{i} > λ_{j} > 0.

(ii) For every i we have λ_{i} ≡ m (mod 2).

(iii) For every i we have λ_{i} ≢ 0 (mod 2m + 2).

(iv) For every i ≠ j we have λ_{i} ≢ ±λ_{j} (mod 2m + 2).

Further, define subsets S_{±}(m) of S(m) in the following manner:

(1.2)  S_{+}(m) := {Λ ∈ S(m) : the number of λ_{i} (mod 2m + 2) > m + 1 is even.} ,
(1.3)  S_{−}(m) := {Λ ∈ S(m) : the number of λ_{i} (mod 2m + 2) > m + 1 is odd.} .

In what follows, we will see that S_{±}(m) determine two different partition functions, one whose parts contain an even number of λ_{i} such that the λ_{i} are in the larger half of the residue classes modulo 2m + 2, and the other whose parts contain an odd number of λ_{i} such that the λ_{i} are in the larger half of the residue classes modulo 2m + 2. These partition functions arise naturally from the following theorems.

Theorem 1. If m ≥ 1 is odd, then

∑_{Λ∈S_{+}(m)} \frac{q^{(λ_{1}+···+λ_{s(m)})/2}}{(1 − q^{λ_{1}}) ⋯ (1 − q^{λ_{s(m)}})} − ∑_{Λ∈S_{−}(m)} \frac{q^{(λ_{1}+···+λ_{s(m)})/2}}{(1 − q^{λ_{1}}) ⋯ (1 − q^{λ_{s(m)}})}

= q^{(m+1)^{2}/8} ∏_{n=1}^{∞} \frac{(1 − q^{2(m+1)n})^{2m+2}}{(1 − q^{(m+1)n})^{m+1}} .

Theorem 2. If m ≥ 1 is even, then

∑_{Λ∈S_{+}(m)} \frac{q^{(λ_{1}+···+λ_{s(m)})/2}}{(1 − q^{λ_{1}}) ⋯ (1 − q^{λ_{s(m)}})} − ∑_{Λ∈S_{−}(m)} \frac{q^{(λ_{1}+···+λ_{s(m)})/2}}{(1 − q^{λ_{1}}) ⋯ (1 − q^{λ_{s(m)}})}

= q^{(m^{2}+2m)/8} ∏_{n=1}^{∞} \frac{(1 − q^{n})}{(1 − q^{2n})^{2}} \frac{(1 − q^{2(m+1)n})^{2m+2}}{(1 − q^{(m+1)n})^{m+1}} .

Before we state the some partition theoretic interpretations of this set of identities, we must offer some more definitions.

Definition 1.2. If m is a positive integer, then let P_{±}(n, m) denote the number of partitions of n of the form

n = ∑_{i=1}^{s(m)}(2n_{i} + 1)λ_{i},

where Λ = (λ_{1}, …, λ_{s(m)}) ∈ S_{±}(m) and each n_{i} ≥ 0.

In simple terms, the function P_{±}(n, m) counts the number of partitions of n into parts, occuring with odd multiplicity, that are elements of a vector Λ ∈ S_{±}(m). Obviously,

(1.4)  ∑_{Λ∈S_{±}(m)} \frac{q^{(λ_{1}+···+λ_{s(m)})}}{(1 − q^{2λ_{1}}) ⋯ (1 − q^{2λ_{s(m)}})} = ∑_{n=1}^{∞} P_{±}(n, m)q^{n}.

These generating functions are closely related to the problem of counting the number of representations of an integer as a sum of triangular numbers, numbers of the form (k^{2} + k)/2 with k ≥ 0. If k is a positive integer, then let T (n, k) denote the number of representations of n as a sum of k triangular numbers. A well known identity due to Jacobi implies that

(1.5)  ∑_{n=0}^{∞} T (n, k)q^{n} = (∏_{n=1}^{∞} \frac{(1 − q^{2n})^{2}}{(1 − q^{n})} )^{k} .

In view of (1.4) and (1.5), it is simple to verify the following two corollaries which give partition theoretic formulas for T (n, k).

Corollary 3. If k ≥ 2 is even, then

T (n, k) = P_{+} (2kn + \frac{k^{2}}{4} , k − 1) − P_{−} (2kn + \frac{k^{2}}{4} , k − 1) .

Corollary 4. If k ≥ 1 is odd, then

T (n, k) = ∑_{j=0}^{∞} (P_{+} (2kn + \frac{k^{2} − 1}{4} − j^{2} − j, k − 1) − P_{−} (2kn + \frac{k^{2} − 1}{4} − j^{2} − j, k − 1)) .

Remark. Although the sum in Corollary 4 appears to be infinite, it is easy to see that it is indeed finite. After all, there are no partitions of negative integers into positive parts. More importantly, observe that Corollaries 3 and 4 completely characterize the number of representations of every integer n as a sum of an arbitrary number of triangular numbers in terms of the partition functions P_{±}(n, k). One should compare these results with those appearing in [M] where explicit formulas of a different type are obtained for T (n, k) for those k of the form 4s^{2} and 4s^{2} + 4s. It would be very interesting to obtain a combinatorial proof of these results.

In view of (1.4), Theorems 1 and 2 immediately imply the following corollaries.

Corollary 5. If k ≥ 2 is even, then for every non-negative integer n we have

P_{+}(n, k − 1) ≥ P_{−}(n, k − 1).

Furthermore, if n ≢ k^{2}/4 (mod 2k), then

P_{+}(n, k − 1) = P_{−}(n, k − 1).

Corollary 6. If k ≥ 2 is even and n is a positive odd integer, then

P_{+}(n, k) = P_{−}(n, k).

Remark. If k ≥ 4 (mod 2k) is even and n ≡ k^{2}/4 (mod 2k), then P_{+}(n, k −1) > P_{−}(n, k −1) by Gauss’ Eureka Theorem (i.e. T (n, 3) > 0 for all n). If k = 2, then an easy analysis shows that P_{+}(n, 1) = P_{−}(n, 1) for almost all n.

One may also make the observation that the right hand side of the identities of Theorems 1 and 2 are quotients of powers of Dedekind eta-functions. In Section 3 we use a well known theorem on such eta-products (see [N1], [N2]) to show that the generating function for P_{+}(n, m) − P_{−}(n, m) is a holomorphic integer weight modular form. Then, using a powerful result of Serre [S], we prove the following corollary:

Corollary 7. If k is a positive integer and M is any integer, then

P_{+}(n, k) ≡ P_{−}(n, k) (mod M)

for a set of positive integers n with arithmetic density 1.

Notice that Corollary 7 is an analog of Euler’s pentagonal number theorem mentioned above. Though the result is not nearly as precise as Euler’s, it does tell us that P_{+}(n, k) and P_{−}(n, k) are congruent modulo M for almost all positive n.

2. Proof of Theorems 1 and 2

Our proof of Theorems 1 and 2 relies on the Kac-Wamimoto Conjecture, proved by Zagier. The following two functions appear in the result:

Θ_{0}(x) = ∏_{n=1}^{∞}(1 − q^{2n})(1 − q^{2n−1}x)(1 − q^{2n−1}x^{−1})

Θ_{1}(x) = ∏_{n=1}^{∞}(1 − q^{2n})(1 − q^{2n−2}x)(1 − q^{2n}x^{−1}).

Alternative forms of these functions will be derived shortly. To prove our theorems, we use the following form of the conjecture, which Zagier derived as an intermediate step in the proof (see [Z]):

Theorem 2.1. (Zagier) If m ∈ ℤ^{+}, s(m) = [\frac{m+1}{2}] and x_{1}, x_{2}, …, x_{m+1} ∈ ℂ are distinct complex numbers such that |q| < |x_{i}/x_{j}| < |q|^{−1}, then

(2.1) ∑_{\begin{matrix}λ_{1}>λ_{2}>···>λ_{s(m)}>0 \\ λ_{i}≡m (mod 2) \end{matrix}} \frac{q^{(λ_{1}+···+λ_{s(m)})/2}}{(1 − q^{λ_{1}})(1 − q^{λ_{2}}) … (1 − q^{λ_{s(m)}})} × (Alt)_{m+1} (∏_{i=1}^{s} (\frac{x_{i}}{x_{m+2−i}} )^{λ_{i}/2})

= θ(τ)^{2s(m)} ∏_{1≤i<j≤m+1} F (x_{j}/x_{i}),

where

F (x) := q^{1/4}x^{−1/2} \frac{Θ_{1}(x)}{Θ_{0}(x)}

and Alt_{m+1} denotes the alternating sum over all permutations of x_{1}, …, x_{m+1}, and

θ(τ) = q^{1/8}Δ(q) = q^{1/8} ∏_{n=1}^{∞}(1 − q^{2n})^{2}/(1 − q^{n}).

The identities given in Theorems 1 and 2 are derived directly from Theorem 2.1. To prove this, we first simplify the infinite products on the right side of the equation by substituting roots of cyclotomic polynomials for the x_{i}, and then use combinatorial arguments to reduce the complicated alternating sum to known values.

Proof of Theorems 1 and 2. Our first task is to simplify the right side of (2.1). The functions Θ_{0} and Θ_{1} are rewritten so that the exponents on all of the q terms are similar:

(2.2)  Θ_{0}(x) = ∏_{n=1}^{∞}(1 − q^{2n−1})(1 − q^{2n−1}x)(1 − q^{2n−1}x^{−1}) · ∏_{n=1}^{∞} \frac{(1 − q^{2n})}{(1 − q^{2n−1})}

= ∏_{n=1}^{∞}(1 − q^{2n−1})(1 − q^{2n−1}x)(1 − q^{2n−1}x^{−1}) · ∏_{n=1}^{∞} \frac{(1 − q^{2n})^{2}}{(1 − q^{n})} ,

(2.3)  Θ_{1}(x) = (1 − x) ∏_{n=1}^{∞}(1 − q^{2n})(1 − q^{2n}x)(1 − q^{2n}x^{−1}).

Before fully expanding the product of F (x_{j}/x_{i}), we consider the terms (x_{j}/x_{i})^{−1/2}. Since we will be substituting complex values for the x_{i}, the branching of the square root function necessitates a careful treatment of these terms. Fortunately, when all such terms are moved to the left side and factored into the expanded alternating sum, the exponents all become integral.

Consider the product of all of the Θ_{1},

(2.4) ∏_{1≤i<j≤m+1} Θ_{1}(x_{j}/x_{i}) =

∏_{1≤i<j≤m+1} ((1 − x_{j}/x_{i}) ∏_{n=1}^{∞}(1 − q^{2n})(1 − q^{2n}(x_{j}/x_{i}))(1 − q^{2n}(x_{j}/x_{i})^{−1})) .

The expansion of the Θ_{0} are similar, and all of the x_{i} appear with integral exponents. Thus we can make the substitution x_{i} = ζ_{m+1}^{i−1}, 1 ≤ i ≤ m+1, where ζ_{m+1} is a primitive (m+1)−st root of unity. The fraction (x_{j}/x_{i}) now reduces to ζ_{m+1}^{j−i}, and since the set {1, ζ_{m+1}, ζ_{m+1}^{2}, …, ζ_{m+1}^{m}} contains all (m + 1)-st roots of unity, the product

(1 − q^{2n})(1 − q^{2n}ζ_{m+1}) ⋯ (1 − q^{2n}ζ_{m+1}^{m})

reduces to (1−q^{2n(m+1)}). Note that in the product of the Θ_{1} there are m+1 copies of the term ∏_{n=1}^{∞}(1 − q^{2n}ζ_{m+1}^{k}) for all 1 ≤ k ≤ m. This is because x_{j′}/x_{i′} = ζ_{m+1}^{k} whenever j′ ≡ i′ + k (mod m + 1), and there are m + 1 choices for i′, ignoring the restrictions on the sum. If j′ > i′, then let j = j′ and i = i′; the term occurs as (x_{j}/x_{i}). When j′ < i′, let j = i′ and i = j′; the term occurs as (x_{i}/x_{j})^{−1} since the higher subscript is always the numerator. Thus (2.4) becomes

∏_{1≤i<j≤m+1} Θ_{1}(ζ_{m+1}^{j−i})

= ∏_{1≤i<j≤m+1}(1 − ζ_{m+1}^{j−i}) ∏_{n=1}^{∞}(1 − q^{2n})^{(m+1)m/2}(1 − q^{2n}ζ_{m+1})^{m+1} ⋯ (1 − q^{2n}ζ_{m+1}^{m})^{m+1}

= ∏_{k=1}^{m}(1 − ζ_{m+1}^{k})^{m+1−k} ∏_{n=1}^{∞}(1 − q^{2n})^{(m+1)(m−2)/2}(1 − q^{2n(m+1)})^{m+1}.

PARTITION IDENTITIES 7

Similarly, with powers of ζ_{m+1} substituted in for the x_{i}, the expansion of the Θ_{0} product is

∏_{1≤i<j≤m+1} Θ_{0}(ζ_{m+1}^{j−i})

= ∏_{n=1}^{∞}(1 − q^{2n−1})^{(m+1)m/2}(1 − q^{2n−1}ζ_{m+1})^{m+1} ⋯ (1 − q^{2n−1}ζ_{m+1}^{m})^{m+1}

× ∏_{n=1}^{∞} \frac{(1 − q^{2n})^{(m+1)m}}{(1 − q^{n})^{(m+1)m/2}}

= ∏_{n=1}^{∞}(1 − q^{2n−1})^{(m+1)(m−2)/2}(1 − q^{(2n−1)(m+1)})^{m+1} ∏_{n=1}^{∞} \frac{(1 − q^{2n})^{(m+1)m}}{(1 − q^{n})^{(m+1)m/2}} .

After all of these simplifications (including the removal of the fractional powers), the right side of (2.1) becomes

θ(τ)^{2s(m)} ∏_{1≤i<j≤m+1} q^{1/4} \frac{Θ_{1}(x_{j}/x_{i})}{Θ_{0}(x_{j}/x_{i})}

= q^{(m^{2}+m+2s(m))/8} ∏_{k=1}^{m}(1 − ζ_{m+1}^{k})^{m+1−k} ∏_{n=1}^{∞} \frac{(1 − q^{2n})^{4s(m)}}{(1 − q^{n})^{2s(m)}}

× ∏_{n=1}^{∞} \frac{(1 − q^{2n})^{(m+1)(m−2)/2}(1 − q^{2n(m+1)})^{m+1}}{(1 − q^{2n−1})^{(m+1)(m−2)/2}(1 − q^{(2n−1)(m+1)})^{m+1} ∏_{n=1}^{∞} \frac{(1−q^{2n})^{(m+1)m}}{(1−q^{n})^{(m+1)m/2}}}

= q^{(m^{2}+m+2s(m))/8} ∏_{k=1}^{m}(1 − ζ_{m+1}^{k})^{m+1−k}

× ∏_{n=1}^{∞} \frac{(1 − q^{n})^{(m+1)m/2−2s(m)}}{(1 − q^{2n})^{(m+1)(m+2)/2−4s(m)}(1 − q^{2n−1})^{(m+1)(m−2)/2}} ∏_{n=1}^{∞} \frac{(1 − q^{(m+1)2n})^{2m+2}}{(1 − q^{(m+1)n})^{m+1}} .

If m is odd, then 2s(m) = m + 1, so the above product is equal to

(2.5) q^{(m+1)^{2}/8} ∏_{k=1}^{m}(1 − ζ_{m+1}^{k})^{m+1−k} ∏_{n=1}^{∞} (\frac{(1 − q^{n})}{(1 − q^{2n})(1 − q^{2n−1})})^{(m+1)(m−2)/2}

× ∏_{n=1}^{∞} \frac{(1 − q^{(m+1)2n})^{2m+2}}{(1 − q^{(m+1)n})^{m+1}}

= q^{(m+1)^{2}/8} ∏_{k=1}^{m}(1 − ζ_{m+1}^{k})^{m+1−k} ∏_{n=1}^{∞} \frac{(1 − q^{(m+1)2n})^{2m+2}}{(1 − q^{(m+1)n})^{m+1}} ,

which is nearly the form seen in the right side Theorem 1. And when m is even, 2s(m) = m, so the product is equal to

(2.6) q^{(m^{2}+2m)/8} ∏_{k=1}^{m}(1 − ζ_{m+1}^{k})^{m+1−k} ∏_{n=1}^{∞} \frac{(1 − q^{n})}{(1 − q^{2n})^{2}} (\frac{(1 − q^{n})}{(1 − q^{2n})(1 − q^{2n−1})})^{(m+1)(m−2)/2}

× ∏_{n=1}^{∞} \frac{(1 − q^{(m+1)2n})^{2m+2}}{(1 − q^{(m+1)n})^{m+1}}

= q^{(m^{2}+2m)/8} ∏_{k=1}^{m}(1 − ζ_{m+1}^{k})^{m+1−k} ∏_{n=1}^{∞} \frac{(1 − q^{n})}{(1 − q^{2n})^{2}} \frac{(1 − q^{(m+1)2n})^{2m+2}}{(1 − q^{(m+1)n})^{m+1}} .

This resembles the right side of Theorem 2.

Note that when (2.5) and (2.4) are expanded into an infinite sum, the first nonzero term is q^{(m+1)^{2}/8} when m is odd, q^{(m^{2}+2m)/8} when m is even, and has coefficient

∏_{k=1}^{m}(1 − ζ_{m+1}^{k})^{m+1−k}

in either case.

Now consider the left side of (2.1),

∑_{\begin{matrix}λ_{1}>λ_{2}>···>λ_{s(m)}>0 \\ λ_{i}≡m (mod 2) \end{matrix}} \frac{q^{(λ_{1}+···+λ_{s(m)})/2}}{(1 − q^{λ_{1}})(1 − q^{λ_{2}}) … (1 − q^{λ_{s(m)}})} × (Alt)_{m+1} (∏_{i=1}^{s} (\frac{x_{i}}{x_{m+2−i}} )^{λ_{i}/2}) X̄,

where X̄ = ∏_{1≤i<j≤m+1}(x_{j}/x_{i})^{1/2} is the product of all of the radical terms transferred from the right side. Before substituting cyclotomic roots for the x_{i}, we ensure that all of the exponents are integral. Note that X̄ = ∏_{k=1}^{m+1}(x_{k})^{(2k−m−2)/2}. If m is even, then each of the λ_{i} are even so all of the powers are already integral. If m is odd, then all of the λ_{i} are also odd. Each of the m + 1 variables appears in each term of the alternating sum, so once again all of the exponents simplify to integers. Thus we can substitute x_{k} = ζ_{m+1}^{k−1}.

For all but the simplest cases, the alternating sum becomes too unwieldy to calculate directly. Fortunately, we can still evaluate it by comparing the series expansion of the left and right sides of (2.1). Consider the term of minimum degree in the left side sum. Since the λ_{i} form a decreasing, nonnegative sequence, there is a unique vector of minimum sum; namely, Λ′ = (m, m − 2, …, 1) when m is odd and Λ′ = (m, m − 2, …, 2) when m is even. Therefore the minimum exponent is (λ′_{1} + ⋯ + λ′_{s(m)})/2, where Λ′ = (λ′_{1}, …, λ′_{s(m)}). This sum evaluates

to \frac{1}{2}(\frac{m+1}{2})^{2} = \frac{(m+1)^{2}}{8} when m is odd and \frac{1}{2}(2(\frac{m}{2}(\frac{m}{2} + 1)/2)) = \frac{m^{2}+2m}{8}. These are precisely the exponents of least degree on the right hand side of (2.5) and (2.6).

The series expansion of the left side must correspond to the expansion of the right side, so the coefficient on the first terms are equal. Thus,

(Alt)_{m+1} (∏_{i=1}^{s} (\frac{x_{i}}{x_{m+2−i}} )^{λ′_{i}/2}) X̄ = ∏_{k=1}^{m}(1 − ζ_{m+1}^{k})^{m+1−k}.

Now we show that if the value of the alternating sum for some Λ is nonzero, then it can be easily related to the value for Λ′.

We need only consider Λ from the residue classes modulo 2m+2, for if λ_{i} ≡ λ_{j} (mod 2m+2), and thus λ_{i} = λ_{j} + r(2m + 2) for some integer r, then

(ζ_{m+1}^{k})^{λ_{i}/2} = (ζ_{m+1}^{k})^{r(2m+2)/2}(ζ_{m+1}^{k})^{λ_{j}/2} = (ζ_{m+1}^{k})^{λ_{j}/2},

for ζ_{m+1}^{k} is an (m + 1)-th root of unity. Thus we can replace λ_{i} by λ_{j} in the alternating sum without affecting its value.

Consider the case where λ_{i} ≡ 0 (mod 2m + 2) for some 1 ≤ i ≤ m + 1. Under the identity permutation, λ_{i}/2 is the exponent on the term (x_{i}/x_{m+2−i}). If the exponent is zero, then this term is always 1. Thus it is irrelevant which variable is the numerator and which is the denominator, so the sum is unaffected when we transpose x_{i} and x_{m+2−i}. However, adding this single transposition also changes the parity of every permutation in the sum. Therefore every summand in the alternating sum has a complementary term that is equal in magnitude and opposite in sign. Thus the sum is zero for Λ of this form.

Now consider the case λ_{i} ≡ λ_{j} (mod 2m + 2) for some 1 ≤ i, j ≤ m + 1. Under the identity permutation, these two values appear in the terms (x_{i}/x_{m+2−i})^{λ_{i}/2} and (x_{j}/x_{m+2−j})^{λ_{j}/2}. Since the exponents are equal, these two terms are combined to give ((x_{i}x_{j})/(x_{m+2−i}x_{m+2−j}))^{λ_{i}/2}. Clearly the x_{i} and x_{j} can be interchanged without affecting the sum. Thus, as before, every summand has a complement. Therefore the sum is zero for these Λ as well.

Note that if we have an alternating sum where the identity permutation gives the term

(x_{1}/x_{m+1})^{λ_{1}/2}(x_{2}/x_{m})^{λ_{2}/2} ⋯ (x_{i}/x_{m+2−i})^{λ_{i}/2} ⋯ (x_{s(m)}/x_{m+2−s(m)})^{λ_{s(m)}/2},

then we can construct an equivalent alternating sum in which the identity permutation gives

(x_{1}/x_{m+1})^{λ_{1}/2}(x_{2}/x_{m})^{λ_{2}/2} ⋯ (x_{m+2−i}/x_{i})^{(2m+2−λ_{i})/2} ⋯ (x_{s(m)}/x_{m+2−s(m)})^{λ_{s(m)}/2}.

This allows us to show that if λ_{i} ≡ −λ_{j} (mod 2m+2), then the alternating sum is zero. These two exponents appear on the terms (x_{i}/x_{m+2−i})^{λ_{i}/2} and (x_{j}/x_{m+2−j})^{λ_{j}/2} under the identity permutation. As discussed, we can replace the second term by (x_{m+2−j}/x_{j})^{(2m+2−λ_{j})/2} = (x_{m+2−j}/x_{j})^{λ_{i}/2}. As in the case where two exponents were equal, we combine the terms and interchange x_{i} and x_{m+2−j}. In this manner we again find a complement of opposite sign and equal magnitude for every summand. Thus, the alternating sum is zero for Λ of this form.

The only remaining cases are those in which there is exactly one λ_{i} contained in each pair of residue classes {k, 2m + 2 − k}. Recall the flipping process described above. If the identity permutation gives us the term

(x_{1}/x_{m+1})^{λ_{1}/2}(x_{2}/x_{m})^{λ_{2}/2} ⋯ (x_{i}/x_{m+2−i})^{λ_{i}/2} ⋯ (x_{s(m)}/x_{m+2−s(m)})^{λ_{s(m)}/2},

then there is an equivalent alternating sum whose identity term is

(x_{1}/x_{m+1})^{λ_{1}/2}(x_{2}/x_{m})^{λ_{2}/2} ⋯ (x_{m+2−i}/x_{i})^{(2m+2−λ_{i})/2} ⋯ (x_{s(m)}/x_{m+2−s(m)})^{λ_{s(m)}/2}.

Now if we transpose x_{i} and x_{m+2−i} we obtain an alternating sum equal in magnitude and opposite in sign with identity permutation

(x_{1}/x_{m+1})^{λ_{1}/2}(x_{2}/x_{m})^{λ_{2}/2} ⋯ (x_{i}/x_{m+2−i})^{−λ_{i}/2} ⋯ (x_{s(m)}/x_{m+2−s(m)})^{λ_{s(m)}/2}.

For a given m, each λ_{i} has the same parity as m, and thus all of the residue classes modulo 2m + 2 also have the same parity as m, and so at most half of the 2m + 2 residue classes can be achieved. Since we need not consider Λ in which additive inverses are present, we can again halve the number of distinct classes that can occur. But our known alternating sum for Λ′ has [\frac{m+1}{2}] distinct residue classes, none of which are additive inverses. Thus we can determine the value of the alternating sum for all Λ by multiplying a factor of −1 for each inverse exponent. Note that in Λ′, 1 ≤ λ_{i} ≤ m for every 1 ≤ i ≤ s(m). Thus the inverses modulo 2m + 2 satisfy m + 2 ≤ −λ_{i} ≤ 2m + 2.

Hence our definition of S_{±}(m). If Λ ∉ S_{±}(m), then

(Alt)_{m+1} (∏_{i=1}^{s} (\frac{x_{i}}{x_{m+2−i}} )^{λ_{i}/2}) X̄ = 0,

and thus is ignored in the sum.

If Λ ∈ S_{±}(m), and there are r inverse exponents, then

(Alt)_{m+1} (∏_{i=1}^{s} (\frac{x_{i}}{x_{m+2−i}} )^{λ_{i}/2}) X̄ = (−1)^{r} ∏_{k=1}^{m}(1 − ζ_{m+1}^{k})^{m+1−k}.

So if there are an even number of inverses in Λ, then the sum is the same as that for Λ′, but if there are an odd number, the sum is negated. Dividing both sides of our identity by ∏_{k=1}^{m}(1 − ζ_{m+1}^{k})^{m+1−k} gives Theorems 1 and 2.

Q.E.D.

Remark. It is also possible to let x_{1} = ζ_{m+1} … x_{m+1} = ζ_{m+1}^{m} in (2.1) and obtain identities similar to those in Theorem 1 and 2. The infinite product in these other identities is essentially the same, but the definition of the restricted partition functions is slightly different.

3. Modular forms and the proof of Corollary 7

For a positive integer N, define the subgroup Γ_{0}(N) of SL_{2}(ℤ) as follows:

(3.1)  Γ_{0}(N) := \{( \begin{matrix}a & b \\ c & d \end{matrix} ) : ad − bc = 1, c ≡ 0 (mod N)\} .

Now let f(z) be a meromorphic function on the upper half of the complex plane. We say that f(z) is a modular function of weight k with respect to Γ_{0}(N) if

f (\frac{az + b}{cz + d}) = (cz + d)^{k}f(z)

for all z in the upper half of the complex plane and all ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ_{0}(N). Further, if f(z) is

holomorphic on the upper half of the complex plane and at the cusps of Γ_{0}(N), it is a modular form of weight k with respect to Γ_{0}(N). Finally, let χ be a Dirichlet character modulo N. We say that f(z) is a modular form with character if

(3.2)  f (\frac{az + b}{cz + d}) = χ(d)(cz + d)^{k}f(z),

again, for z in the upper half of the complex plane with ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ_{0}(N). The finite dimen-

sional space of such modular forms is denoted by M_{k}(Γ_{0}(N), χ).

If we let q = e^{2πiz}, we may construct a Fourier expansion for such a holomorphic modular form:

f(z) = ∑_{n=0}^{∞} a(n)q^{n}

In fact, we may identify any form with its Fourier expansion.

In this paper we will be studying modular forms that are products of Dedekind eta-functions, defined by the infinite product

(3.3)  η(z) = q^{1/24} ∏_{n=1}^{∞}(1 − q^{n}).

A function f(z) is called an eta-product if it can be expressed as a finite product of the form

(3.4)  f(z) = ∏_{δ|N} η^{r_{δ}}(δz).

where N and each r_{δ} are integers. Such functions have many unique properties. In the following result ([N1] and [N2]), Newman proves that certain eta-products fulfill the functional equation (3.2) for modular forms with character:

Theorem 3.1. (Newman) If f(z) = ∏_{δ|N} η^{r_{δ}}(δz) is an eta-product for which

(3.5)  ∑_{δ|N} δr_{δ} ≡ 0 (mod 24)

and

(3.6)  ∑_{δ|N} \frac{N}{δ} r_{δ} ≡ 0 (mod 24)

then f(z) satisfies the functional equation (3.2) for all ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ_{0}(N), where k = \frac{1}{2} ∑_{δ|N} r_{δ}.

Here χ is the character defined by:

χ(d) = (\frac{(−1)^{k}s}{d}) and s = ∏_{δ|N} δ^{r_{δ}}.

If an eta-product satisfies the functional equation (3.4), we still must demonstrate holomorphicity at the cusps of Γ_{0}(N) to prove it is a holomorphic modular form. The following observation is well known (for example, see Ligozat [L]):

Proposition 3.2. Let c, d, and N be positive integers with d|N and (c, d) = 1. With the notation as above, if the eta-product f(z) satisfies (3.5) and (3.6), then the order of vanishing of f(z) at the cusp \frac{c}{d} is

\frac{1}{24} ∑_{δ|N} \frac{N(d, δ)^{2}r_{δ}}{(d, \frac{N}{d} ) dδ} .

Using Theorem 3.1 and Proposition 3.2 we obtain the following:

Theorem 3.3. If m is even, then

(3.7)  ∑_{n=0}^{∞} ((P_{+}(n, m) − P_{+}(n, m)) q^{n/2} ∈ M_{m/2}(Γ_{0}(2(m + 1)), χ)

where χ(d) = (\frac{(−1)^{m/2}(m+1)}{d}).

If m is odd, then

(3.8)  ∑_{n=0}^{∞} ((P_{+}(n, m) − P_{+}(n, m)) q^{n} ∈ M_{(m+1)/2}(Γ_{0}(4(m + 1)), χ)

where χ(d) = (\frac{(−1)^{(m+1)/2}(m+1)}{d}).

Proof of Theorem 3.3. We have, for even m,

(3.9) ∑_{n=0}^{∞} (P_{+}(n, m) − P_{−}(n, m)) q^{n/2} = q^{(m^{2}+2m)/8} ∏_{n=1}^{∞} \frac{(1 − q^{n})}{(1 − q^{2n})^{2}} \frac{(1 − q^{2(m+1)n})^{2m+2}}{(1 − q^{(m+1)n})^{m+1}}

= ∏_{n=1}^{∞} \frac{q^{1/24}(1 − q^{n})}{q^{4/24}(1 − q^{2n})^{2}} \frac{q^{4(m+1)^{2}/24}(1 − q^{2(m+1)n})^{2m+2}}{q^{(m+1)^{2}/24}(1 − q^{(m+1)n})^{m+1}}

= \frac{η(z)}{η^{2}(2z)} \frac{η^{2m+2}(2(m + 1)z)}{η^{m+1}((m + 1)z)}.

Let f(z) = \frac{η(z)}{η^{2}(2z)} \frac{η^{2m+2}(2(m+1)z)}{η^{m+1}((m+1)z)} and write f(z) = ∏_{δ|N} η^{r_{δ}}(δz) as in Theorem 3.1 and Proposition 3.2. We have

∑_{δ|N} δr_{δ} = 1(1) − 2(2) + 4(m + 1)^{2} − (m + 1)^{2} = −3 + 3(m + 1)^{2}.

Note that 3(m + 1)^{2} ≡ 3 (mod 24) if and only if (m + 1)^{2} ≡ 1 (mod 8). One can easily check that this congruence is true for all even m.

∑_{δ|N} \frac{N}{δ} r_{δ} = 2(m + 1) (\frac{1}{1} − \frac{2}{2} + \frac{2(m + 1)}{2(m + 1)} − \frac{(m + 1)}{(m + 1)}) = 0 ≡ 0 (mod 24).

So f(z) satisfies (3.5) and (3.6). Therefore, by Theorem 3.1, f(z) satisfies (3.2) with

( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ_{0}(2(m + 1)), k = \frac{m}{2}, and χ(d) = (\frac{(−1)^{m/2+}(m+1)^{m+1}}{d} ).

Now we must check the holomorphicity of f(z) at the cusps of Γ_{0}(2(m+1)). By Proposition 3.2 the order of vanishing of f(z) at the cusp \frac{c}{d} is

\frac{1}{24} ∑_{δ|N} \frac{N(d, δ)^{2}r_{δ}}{(d, \frac{N}{d} ) dδ} .

Therefore, it suffices for our purposes to show that the following is non-negative:

\frac{(d, 1)^{2}(1)}{1} − \frac{(d, 2)^{2}(2)}{2} + \frac{(d, 2(m + 1))^{2}(2(m + 1))}{2(m + 1)} − \frac{(d, (m + 1))^{2}((m + 1))}{(m + 1)}.

If 2 |̸d, then

(d, 1)^{2} − (d, 2)^{2} + (d, 2(m + 1))^{2} − (d, (m + 1))^{2} = 0.

If 2|d, then

(d, 1)^{2} − (d, 2)^{2} + (d, 2(m + 1))^{2} − (d, (m + 1))^{2} = 1 − 4 + 4(d, (m + 1))^{2} − (d, (m + 1))^{2} ≥ 0.

Therefore, the order of vanishing of f(z) at the cusps of Γ_{0}(2(m + 1)) is nonnegative, so f(z) is holomorphic on the entire upper half of the complex plane.

We have, for odd m,

∑_{n=0}^{∞} (P_{+}(n, m) − P_{−}(n, m)) q^{n} = q^{(m+1)^{2}/4} ∏_{n=1}^{∞} \frac{(1 − q^{4(m+1)n})^{2m+2}}{(1 − q^{2(m+1)n})^{m+1}}

= ∏_{n=1}^{∞} \frac{q^{8(m+1)^{2}/24}(1 − q^{4(m+1)n})^{2(m+1)}}{q^{2(m+1)^{2}/24}(1 − q^{2(m+1)n})^{m+1}}

= \frac{η^{2(m+1)}(2(m + 1)z)}{η^{m+1}((m + 1)z)}.

Let g(z) = \frac{η^{2(m+1)}(2(m+1)z)}{η^{m+1}((m+1)z)} and write g(z) = ∏_{δ|N} η^{r_{δ}}(δz). We have

∑_{δ|N} δr_{δ} = 8(m + 1)^{2} − 2(m + 1)^{2} = 6(m + 1)^{2} ≡ 0 (mod 24)

for odd m.

∑_{δ|N} \frac{N}{δ} r_{δ} = 4(m + 1) (\frac{4(m + 1)}{2(m + 1)} − \frac{2(m + 1)}{(m + 1)}) = 0 ≡ 0 (mod 24).

So g(z) satisfies (3.5) and (3.6), with ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ_{0}(4(m + 1)), k = \frac{m+1}{2}, and χ(d) =

(\frac{(−1)^{(m+1)/2}(m+1)^{m+1}}{d} ).

Now we must check the holomorphicity of g(z) at the cusps of Γ_{0}(4(m + 1)). Arguing as above, it suffices for our purposes to show that the following is non-negative:

\frac{(d, 4(m + 1))^{2}(2(m + 1))}{4(m + 1)} − \frac{(d, 2(m + 1))^{2}((m + 1))}{2(m + 1)}.

Note that (\frac{(d,4(m+1))^{2}}{2} − \frac{(d,2(m+1))^{2}}{2} ) ≥ 0, so the order of vanishing of g(z) is non-negative for all the cusps of Γ_{0}(4(m + 1)). Therefore, g(z) is holomorphic on the entire upper half of the complex plane.

Q.E.D.

To prove Corollary 7, we recall a theorem of Serre [S].

Theorem 3.4. (Serre) Let f(z) be a holomorphic modular form of positive integer weight k with character χ on Γ_{0}(N) with Fourier expansion

(3.10)  f(z) = ∑_{n=0}^{∞} a(n)q^{n}

where a(n) are algebraic integers in some number field and N is a positive number. If M is a positive integer, then there exists a constant α such that there are O (\frac{x}{log^{α}x}) integers n ≤ x such that a(n) is not divisible by M.

Proof of Corollary 7. By Theorem 3.3, for every positive integer k the generating function

∑_{n=0}^{∞} (P_{+}(n, k) − P_{−}(n, k)) q^{n}

is a holomorphic integer weight modular form. Therefore, Theorem 3.4 immediately implies Corollary 7.

Q.E.D.

5. Acknowledgements

The authors thank Ken Ono for suggesting this project and for help during the preparation of this manuscript. The first author thanks Ken Ono especially for organizing this work as an undergraduate summer research opportunity.

References

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[Ko] N. Koblitz, Introduction to elliptic curves and modular forms, Springer-Verlag, 1984.

[L] G. Ligozat, Courbes modulaires de genre 1, Bull. Soc. Math. France 43 (1972), 1-80.

[M] S. Milne, Infinite families of exact sums of squares formulas, Jacobi elliptic functions, continued fractions, and Schur functions, to appear, Ramanujan J..

[N1] M. Newman, Construction and application of a certain class of modular functions, Proc. London Math. Soc. (3) 9 (1956), 334-350.

[N2] M. Newman, Construction and application of a certain class of modular functions II, Proc. London Math. Soc. (3) 9 (1959), 373-387.

[O] K. Ono, Representations of integers as sums of squares, preprint.

[S] J.-P. Serre, Divisibilité des coefficients des formes modulaires de poids entier, C.R. Acad. Sci. Paris (A) 279 (1974), 679-682.

[Z] D. Zagier, A proof of the Kac-Wakimoto affine denominator formula for the strange series, Math. Res. Lettr. 7 (2000), 597-604.

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