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Let $\eta$ be the Dedekind eta function. Show that $\dfrac{\eta(q^9)^3}{\eta(q^3)}=\displaystyle\sum_{a,b\in \mathbb{Z}^2}q^{3(a^2+b^2+ab+a+b)+1}$.

I'm pretty sure the RHS is equal to $\theta_2(q^3)\psi_6(q^9)+\theta_3(q^3)\psi_3(q^9)$, but I'm not sure how to show this is equal to the LHS.

whetham
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    One should have $a,b\in \mathbb{Z}$ instead. Also, you need to define the functions $\psi_6(q)$ and $\psi_3(q)$ and/or provide a reference. – T.A.Tarbox Apr 13 '17 at 02:16

2 Answers2

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The equation is precisely equivalent to the cubic theta function identity (equation 2.1)

$$c(q^3) = \frac{(a(q)-b(q))}{3}$$

The proof appears on page 3 of "SOME CUBIC MODULAR IDENTITIES OF RAMANUJAN", J. M. Borwein, P. B. Borwein and F. G. Garvan, Trans. Amer. Math. Soc. 343 (1994), 35-47.

T.A.Tarbox
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  • How is the identity you mention "precisely equivalent" to the identity in the question? – Somos Sep 08 '18 at 18:18
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The two sides of the equation are equal up to a factor of $3$. That is, the left side is $\, q + q^4 + 2q^7 + \dots, \,$ the generating function of OEIS sequence A033687, and the right side is the generating function of OEIS sequence A005882 which is $3$ times that. It is also the right side of equation $(63)$ on page $111$ of Conway and Sloane "Sphere Packings, Lattices and Groups". On page $103$ is equation $(11)$ with the definition $\, \psi_k(z) = e^{\pi i/ z^2} \, \theta_3(\pi z/k|z) = \sum_{m=-\infty}^\infty q^{(m+1/k)^2}. \,$ The left side is equal to $$\, \frac{q}3(2\, \psi(q^6)\, f(q^6, q^{12}) + \phi(q^3)\, f(q^3, q^{15}))$$ where $\phi(), \psi()$ are Ramanujan theta functions and $f(, )$ is Ramanujan's general theta function. The right side can be written as $$ \theta_2(0, q^3)\, q^{1/4} f(q^6, q^{12}) + \theta_3(0, q^3)\, qf(q^3, q^{15}). $$ This can be shown by splitting the infinite sum into two infinite sums depending on the parity of $\,a\,$ and then identifying each sum with one of the two products of a theta null function and a Ramanujan general theta function.

In more detail, the infinite sum exponent of $q$ function $$ e(a,b) := 3 (a^2 + b^2 + a b + a + b) + 1 $$ regarded as a matrix $\, \{e(i,j)\} \,$ has rank $3$ while the matrices $\, \{e(2i,j-i)\} \,$ and $\, \{e(2i+1,j-i)\} \,$ have rank $2$. Note that the matrix $\, \{q^{e(i,j)}\} \,$ has infinite rank, while the two matrices $\, \{q^{e(2i,j-i)}\} \,$ and $\, \{q^{e(2i+1,j-i)}\} \,$ have rank $1$, and thus the two infinite sums of their entries factor into products of two Ramanujan theta functions each.

Somos
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