INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 5 (2005), #A16

RELATIONS AMONG FOURIER COEFFICIENTS OF CERTAIN ETA PRODUCTS

Shaun Cooper

Institute of Information and Mathematical Sciences, Massey University, Auckland 102904, New Zealand

                           s.cooper@massey.ac.nz

Sanoli Gun

Harish-Chandra Research Institute, Jhusi, Allahabad 211019, India

                            jhulan@mri.ernet.in

Michael Hirschhorn

School of Mathematics, University of New South Wales, Sydney 2052, Australia

                          m.hirschhorn@unsw.edu.au

B. Ramakrishnan

Harish-Chandra Research Institute, Jhusi, Allahabad 211019, India

                            ramki@mri.ernet.in

Received: 3/29/05, Accepted: 7/21/05, Published: 7/27/05

Abstract

Certain arithmetic relations for the coefficients in the expansions of (q)_{∞}^{r}, (q)_{∞}^{r}(q^{t})_{∞}^{s}, t = 2, 3, 4, were studied by M. Newman, S. Cooper, M. D. Hirschhorn, R. Lewis, S. Ahlgren and R. Chapman. In this work, we prove similar identities for certain multi-product expansions using an elementary method.

1. Introduction

For an integer r, let

(q)_{∞}^{r} = ∏_{n=1}^{∞}(1 − q^{n})^{r} = ∑_{n≥0} a_{r}(n)q^{n},  (1.1)

where q = e^{2πiz} and Im(z) > 0 and let

f_{j}(q) = (q)_{∞}^{r_{j}}(q^{2})_{∞}^{s_{j}}(q^{4})_{∞}^{t_{j}},  (1.2)

where r_{j}, s_{j}, t_{j} are certain specific integers (see theorem). In this article we consider the following products

f_{i}(q^{ℓ})f_{k}(q^{m}) = ∑_{n=0}^{∞} a(n)q^{n} (l, m > 0),

and prove certain identities involving the Fourier coefficients a(n) by elementary arguments. Similar identities for eta powers and products of two eta functions were earlier obtained by several authors [1, 3, 4, 5].

2. Statement of theorem

Let q be a complex number satisfying |q| < 1. It is readily checked that (−q)_{∞} = \frac{(q^{2})_{∞}^{3}}{(q)_{∞}(q^{4})_{∞}} .

Let

\begin{matrix}f_{1}(q) = (q)_{∞}, & f_{7}(q) = f_{1}(−q) = \frac{(q^{2})_{∞}^{3}}{(q)_{∞}(q^{4})_{∞}} , \\ f_{2}(q) = (q)_{∞}^{3}, & f_{8}(q) = f_{2}(−q) = \frac{(q^{2})_{∞}^{9}}{(q)_{∞}^{3}(q^{4})_{∞}^{3}} , \\ f_{3}(q) = \frac{(q)_{∞}^{2}}{(q^{2})_{∞}} , & f_{9}(q) = f_{3}(−q) = \frac{(q^{2})_{∞}^{5}}{(q)_{∞}^{2}(q^{4})_{∞}^{2}} , \\ f_{4}(q) = \frac{(q^{2})_{∞}^{2}}{(q)_{∞}} , & f_{10}(q) = f_{4}(−q) = \frac{(q)_{∞}(q^{4})_{∞}}{(q^{2})_{∞}} , \\ f_{5}(q) = \frac{(q)_{∞}^{5}}{(q^{2})_{∞}^{2}} , & f_{11}(q) = f_{5}(−q) = \frac{(q^{2})_{∞}^{13}}{(q)_{∞}^{5}(q^{4})_{∞}^{5}} , \\ f_{6}(q) = \frac{(q^{2})_{∞}^{5}}{(q)_{∞}^{2}} , & f_{12}(q) = f_{6}(−q) = \frac{(q)_{∞}^{2}(q^{4})_{∞}^{2}}{(q^{2})_{∞}} . \end{matrix}

Observe that each function f_{i}(q) has the form f_{i}(q) = (q)_{∞}^{r_{i}}(q^{2})_{∞}^{s_{i}}(q^{4})_{∞}^{t_{i}}, for certain integers r_{i}, s_{i}, t_{i}. By the triple product and quintuple product identities, we have [2, pp. 64–65 and 306–307], [4]

f_{1}(q) = ∑_{α≡1 (mod 6)}(−1)^{(α−1)/6}q^{(α^{2}−1)/24},
f_{2}(q) = ∑_{α≡1 (mod 4)} αq^{(α^{2}−1)/8},
f_{3}(q) = ∑_{α}(−1)^{α}q^{α^{2}},
f_{4}(q) = ∑_{α≡1 (mod 4)} q^{(α^{2}−1)/8},
f_{5}(q) = ∑_{α≡1 (mod 6)} αq^{(α^{2}−1)/24},
f_{6}(q) = ∑_{α≡1 (mod 3)}(−1)^{α−1}αq^{(α^{2}−1)/3},  (2.1)

where in each case the sum is over all integers α, positive and negative, satisfying the given congruence.

For 1 ≤ i ≤ 12, let d_{i} = r_{i} + 2s_{i} + 4t_{i} and λ_{i} = ⌈\frac{1}{2}(r_{i} + s_{i} + t_{i})⌉ − 1. Let

(e_{1}, e_{2}, ⋯, e_{12}) = (1, 3, 24, 3, 1, 8, 1, 3, 24, 3, 1, 8),
(n_{1}, n_{2}, ⋯, n_{12}) = (6, 4, 1, 4, 6, 3, 6, 4, 1, 4, 6, 3).

Observe that d_{i} = e_{i} unless i = 3 or 9, in which case d_{3} = d_{9} = 0. For 1 ≤ i ≤ 12 and p an

odd prime, define ϵ_{i}(p) = (\frac{a_{i}}{p} ), where (a_{1}, …, a_{12}) = (3, −1, 1, 1, −3, −3, 6, −2, 1, 2, −6, −3).

The main purpose of this article is to prove the following result.

Theorem. Let ℓ and m be positive integers, and let 1 ≤ j, k ≤ 12. Let p > 3 be any prime satisfying (\frac{−e_{j}e_{k}ℓm}{p}) = −1 and put Δ = \frac{p^{2}−1}{24}. Let f_{j}(q^{ℓ})f_{k}(q^{m}) = ∑_{n=0}^{∞} a(n)q^{n}. Then the coefficients a(n) satisfy

a(pn + (ℓd_{j} + md_{k})Δ) = ϵ_{j}(p)ϵ_{k}(p)p^{λ_{j}+λ_{k}}a (\frac{n}{p}) .

Example. (j = 5, k = 10) We have f_{5}(q) = (q)_{∞}^{5}(q^{2})_{∞}^{−2} and f_{10}(q) = (q)_{∞}(q^{2})_{∞}^{−1}(q^{4})_{∞}, so (r_{5}, s_{5}, t_{5}) = (5, −2, 0), (r_{10}, s_{10}, t_{10}) = (1, −1, 1), d_{5} = r_{5} + 2s_{5} + 4t_{5} = 1, d_{10} = r_{10} + 2s_{10} + 4t_{10} = 3, λ_{5} = ⌈\frac{1}{2}(r_{5} + s_{5} + t_{5})⌉ − 1 = 1, λ_{10} = ⌈\frac{1}{2}(r_{5} + s_{5} + t_{5})⌉ − 1 = 0.e_{5} = 1, e_{10} =

3, ϵ_{5}(p) = (\frac{−3}{p}) , and ϵ_{10}(p) = (\frac{2}{p}) . Let p be any prime satisfying (\frac{−e_{5}e_{10}ℓm}{p}) = −1, i.e.,

(\frac{−3ℓm}{p}) = −1. Let f_{5}(q^{ℓ})f_{10}(q^{m}) = ∑_{n=0}^{∞} a(n)q^{n}. Then the Theorem implies

a(pn + (ℓ + 3m)Δ) = (\frac{−3}{p}) (\frac{2}{p}) p a (\frac{n}{p}) ,

i.e.,

a(pn + (ℓ + 3m)Δ) = (\frac{−6}{p}) p a (\frac{n}{p}) .

3. Proofs

We shall require the following elementary lemma, which we state without further comment.

Lemma. Let ℓ and m be positive integers and let p be an odd prime satisfying (\frac{−ℓm}{p}) = −1.

Let α and β be integers satisfying ℓα^{2} + mβ^{2} ≡ 0 (mod p). Then α ≡ 0 (mod p) and β ≡ 0 (mod p).

In order to illustrate the technique, we first prove the example, before proving the general statement of the theorem.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 5 (2005), #A16 4

Proof of example. We have

f_{5}(q^{ℓ})f_{10}(q^{m}) = ∑_{α≡1 (mod 6)} αq^{ℓ(α^{2}−1)/24} ∑_{β≡1 (mod 4)}(−q^{m})^{(β^{2}−1)/8},

so

a(n) = ∑_{{α≡1 (mod 6), β≡1 (mod 4) \atop ℓ(α^{2}−1)/24+m(β^{2}−1)/8=n}} α(−1)^{(β^{2}−1)/8}
= ∑_{{α≡1 (mod 6), β≡1 (mod 4) \atop ℓα^{2}+3mβ^{2}=24n+ℓ+3m}} α(−1)^{(β^{2}−1)/8}.

Therefore

a(pn + (ℓ + 3m)Δ) = ∑_{{α≡1 (mod 6), β≡1 (mod 4) \atop ℓα^{2}+3mβ^{2}=24pn+(ℓ+3m)p^{2}}} α(−1)^{(β^{2}−1)/8}. (3.1)

Now ℓα^{2} + 3mβ^{2} ≡ 0 (mod p), and the lemma implies p | α, p | β. Let

α = (\frac{−3}{p}) pα′, β = (\frac{−1}{p}) pβ′.  (3.2)

Then α′ ≡ 1 (mod 6) and β′ ≡ 1 (mod 4). Also, modulo 2,

\frac{β^{2} − 1}{8} − \frac{β^{′2} − 1}{8} = \frac{β^{2} − β′^{2}}{8}
= \frac{(p^{2} − 1)β′^{2}}{8}
≡ \frac{p^{2} − 1}{8}
≡ \{ \begin{matrix}0 & if p ≡ 1 or 7 (mod 8) \\ 1 & if p ≡ 3 or 5 (mod 8) \end{matrix}

Therefore

(−1)^{(β^{2}−1)/8} = (\frac{2}{p}) (−1)^{(β′^{2}−1)/8}.  (3.3)

Substituting (3.2) and (3.3) into (3.1) we get

a(pn + (ℓ + 3m)Δ) = ∑_{{α′≡1 (mod 6), β′≡1 (mod 4) \atop ℓα′^{2}+3mβ′^{2}=24n/p+ℓ+3m}} (\frac{−3}{p}) pα′ (\frac{2}{p}) (−1)^{(β′^{2}−1)/8}
= (\frac{−6}{p}) p a (\frac{n}{p}) .

This completes the proof of the example. Proof of Theorem. Writing

f_{j}(q^{ℓ})f_{k}(q^{m}) = ∑_{n=0}^{∞} a(n)q^{n},

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 5 (2005), #A16 5

we have, using (2.1),

a(n) = ∑_{{α≡1 (mod n_{j}), β≡1 (mod n_{k}) \atop e_{j} ℓα^{2}+e_{k}mβ^{2}=24n+d_{j} ℓ+d_{k}m}} ϕ_{j}(α)ϕ_{k}(β),

where

\begin{matrix}ϕ_{1}(α) = (−1)^{(α−1)/6}, & ϕ_{7}(α) = (−1)^{(α−1)/6+(α^{2}−1)/24}, \\ ϕ_{2}(α) = α, & ϕ_{8}(α) = α(−1)^{(α^{2}−1)/8}, \\ ϕ_{3}(α) = (−1)^{α}, & ϕ_{9}(α) = 1, \\ ϕ_{4}(α) = 1, & ϕ_{10}(α) = (−1)^{(α^{2}−1)/8}, \\ ϕ_{5}(α) = α, & ϕ_{11}(α) = α(−1)^{(α^{2}−1)/24}, \\ ϕ_{6}(α) = (−1)^{α−1}α, & ϕ_{12}(α) = (−1)^{α−1+(α^{2}−1)/3}α. \end{matrix}

Therefore

a(pn + (ℓd_{j} + md_{k})Δ) = ∑_{{α≡1 (mod n_{j}), β≡1 (mod n_{k}) \atop e_{j} ℓα^{2}+e_{k}mβ^{2}=24pn+p^{2}(d_{j} ℓ+d_{k}m)}} ϕ_{j}(α)ϕ_{k}(β).

Observe that e_{j}ℓα^{2} + e_{k}mβ^{2} ≡ 0 (mod p). The Lemma implies p | α, p | β. Let

α = \{ \begin{matrix}(\frac{−3}{p}) pα′ & if j = 1, 5, 6, 7, 11 or 12 \\ (\frac{−1}{p}) pα′ & if j = 2, 4, 8 or 10 \\ pα′ & if j = 3 or 9, \end{matrix}

β = \{ \begin{matrix}(\frac{−3}{p}) pβ′ & if k = 1, 5, 6, 7, 11 or 12 \\ (\frac{−1}{p}) pβ′ & if k = 2, 4, 8 or 10 \\ pβ′ & if k = 3 or 9. \end{matrix}

Then it is easily verified that α′ ≡ 1 (mod n_{j}) and β′ ≡ 1 (mod n_{k}), and that ϕ_{j}(α) = ϵ_{j}(p) p^{λ_{j}} ϕ_{j}(α′) and ϕ_{k}(β) = ϵ_{k}(p) p^{λ_{k}} ϕ_{k}(β′). Consequently,

\begin{aligned}a(pn + (ℓd_{j} + md_{k})Δ) & = & ∑_{{α′≡1 (mod n_{j}), β′≡1 (mod n_{k}) \atop e_{j} ℓα′^{2}+e_{k}mβ′^{2}=24n/p+d_{j} ℓ+d_{k}m}} ϵ_{j}(p)ϵ_{k}(p)p^{λ_{j}+λ_{k}}ϕ_{j}(α′)ϕ_{k}(β′) \\ & = & ϵ_{j}(p)ϵ_{k}(p)p^{λ_{j}+λ_{k}} a (\frac{n}{p}) . \end{aligned}

This completes the proof of the theorem.

Remark. Though our theorem can be proved using the theory of lacunary modular forms, we prefer to present an elementary proof for its simplicity. Acknowledgement The first author thanks the staff and students at Harish-Chandra Research Institute for warm hospitality during his visit.

References

[1] S. Ahlgren, Multiplicative relations in powers of Euler’s product, J. Number Theory 89 (2001), 222-233.

[2] J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, New York, 1987.

[3] R. Chapman, Coefficients of Products of Powers of Eta Functions, Ramanujan J. 5 (2001), 271–279.

[4] S. Cooper, M. Hirschhorn and R. Lewis, Powers of Euler’s product and related identities, Ramanujan J. 4, (2000), 137–155.

[5] M. Newman, An identity for the coefficients of certain modular forms, J. London Math. Soc. 30 (1955), 488–493.