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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – T.A.Tarbox
    Apr 13, 2017 at 2:16

2 Answers 2


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.

  • $\begingroup$ How is the identity you mention "precisely equivalent" to the identity in the question? $\endgroup$
    – Somos
    Sep 8, 2018 at 18:18

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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .