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I have some difficulties with expressing the following series $\sum\limits_{b=-\infty}^{+\infty}\sum\limits_{a=-\infty}^{+\infty}q^{1 + 3 a + 3 a^2 - 3 b - 3 a b + 3 b^2}$ using standart theta functions. I've tried to get the degree for the sum of squares: $1 + 3 a + 3 a^2 - 3 b - 3 a b + 3 b^2=3t^2+(\frac{3b-1}{2})^2$, where $a=t+\frac{b-1}{2}$ so as that the initial sum splits in $\sum\limits_{b=-\infty}^{+\infty}\sum\limits_{a=-\infty}^{+\infty}q^{1 + 3 a + 3 a^2 - 3 b - 3 a b + 3 b^2}$ with $b$ even and $\sum\limits_{t=-\infty}^{+\infty}\sum\limits_{b=-\infty}^{+\infty}q^{3t^2+(\frac{3b-1}{2})^2}$ , where $b$ is odd and then I tried analyzing cases with $\frac{3b-1}{2}\equiv 1 (mod3)$ and $\frac{3b-1}{2}\equiv 2 (mod3)$, but I were unsucceded.

It is also clear how to express $\sum\limits_{b=-\infty}^{+\infty}\sum\limits_{a=-\infty}^{+\infty}q^{3 a^2 - 3 a b + 3 b^2}$, so that we may decide it as being standart.

I will be really grateful for any suggestions.

Some useful definitions:


$\theta_2(z)=\sum\limits_{a=-\infty}^{+\infty}q^{(a+1/2)^2}$, where $q=e^{\pi i z}$;
$\theta_3(z)=\sum\limits_{a=-\infty}^{+\infty}q^{a^2}$;

$\theta_4(z)=\sum\limits_{a=-\infty}^{+\infty}(-q)^{a^2}$.

Some extra comments:

Let $\phi(z)=\sum\limits_{b=-\infty}^{+\infty}\sum\limits_{a=-\infty}^{+\infty}q^{a^2 - a b + b^2}$, and assume that $a^2 - a b + b^2=(a-\frac{b}{2})^2+\frac{3b^2}{4}$, then

$\sum\limits_{b=-\infty}^{+\infty}\sum\limits_{a=-\infty}^{+\infty}q^{(a-\frac{b}{2})^2+\frac{3b^2}{4}}=\sum\limits_{a=-\infty}^{+\infty}\sum\limits_{\substack{b=-\infty \\ \text{b even}}}^{+\infty}q^{(a-\frac{b}{2})^2+\frac{3b^2}{4}}+ \sum\limits_{a=-\infty}^{+\infty}\sum\limits_{\substack{b=-\infty \\ \text{b odd}}}^{+\infty}q^{(a-\frac{b}{2})^2+\frac{3b^2}{4}}= \sum\limits_{a=-\infty}^{+\infty}\sum\limits_{k=-\infty}^{+\infty}q^{(a-k)^2+3k^2} +\sum\limits_{a=-\infty}^{+\infty}\sum\limits_{k=-\infty}^{+\infty}q^{(a-k-1/2)^2+3(k+1/2)^2}= \sum\limits_{t=-\infty}^{+\infty}\sum\limits_{t=-\infty}^{+\infty}q^{t^2+3k^2}+ \sum\limits_{a=-\infty}^{+\infty}\sum\limits_{t=-\infty}^{+\infty}q^{(t+1/2)^2+3(k+1/2)^2}=\theta_2(z)\theta_2(3z)+\theta_3(z)\theta_3(3z)$.

And then $\sum\limits_{b=-\infty}^{+\infty}\sum\limits_{a=-\infty}^{+\infty}q^{3(a^2 - a b + b^2)}=\phi(3z)$.

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  • $\begingroup$ Could you please write how to express the "standard" sum in terms of thetas explicitly? It might serve as a hint for the answer. $\endgroup$
    – Heterotic
    Jun 28, 2016 at 15:42
  • $\begingroup$ I think it will be more useful to rewrite the exponent as $$\frac34 (a+b)^2+\frac94(a-b)^2+3(a-b)+1$$ $\endgroup$ Jun 29, 2016 at 20:06
  • $\begingroup$ and what will be the next step in this case? $\endgroup$ Jun 30, 2016 at 7:44
  • $\begingroup$ it's all about finding the most suitable presentation. And the main problem is how to deal with fractions, because the only pleasant alternative is having $1/2$ or $z+1/2$, where $z \in \mathbb{Z}$ in expressions under squares $\endgroup$ Jun 30, 2016 at 7:48
  • 2
    $\begingroup$ You proved a well known equation for the Borwein-Borwein-Garvan cubic theta function $a(q)$, viz. $\phi(z)=a(q)=\theta_2(z)\theta_2(3z)+\theta_3(z)\theta_3(3z)$. $\endgroup$
    – T.A.Tarbox
    Mar 16, 2017 at 2:21

1 Answer 1

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Complete the square with respect to $k$ to get\begin{align}\sum_{k,n\in\mathbb{Z}}q^{3\big(k+\tfrac{1-n}2\big)^2+\tfrac14+\tfrac{9n^2}4-\tfrac{3n}2}&=\sum_{k,n\in\mathbb{Z}}q^{3\big(k+\tfrac{1-n}2\big)^2+\tfrac14(3n-1)^2}\\ &=\sum_{n\in\mathbb{Z}}q^{\tfrac14(3n-1)^2}\sum_{k\in\mathbb{Z}}q^{3\big(k+\tfrac{1-n}2\big)^2} \end{align}then split the sum by parity of $n$:\begin{align}\sum_{n\in\mathbb{Z}}q^{\tfrac14(3n-1)^2}\sum_{k\in\mathbb{Z}}q^{3\big(k+\tfrac{1-n}2\big)^2}&=\sum_{n\in\mathbb{Z}}q^{\tfrac14(6n-1)^2}\sum_{k\in\mathbb{Z}}q^{3\big(k+\tfrac{1-2n}2\big)^2}+\sum_{n\in\mathbb{Z}}q^{\tfrac14(6n-2-1)^2}\sum_{k\in\mathbb{Z}}q^{3\big(k+\tfrac{1-2n+1}2\big)^2}\\ &=\sum_{n\in\mathbb{Z}}q^{\tfrac14(6n-1)^2}\sum_{k\in\mathbb{Z}}q^{3\big(k-n+\tfrac12\big)^2}+\sum_{n\in\mathbb{Z}}q^{\tfrac94(2n-1)^2}\sum_{k\in\mathbb{Z}}q^{3\big(k-n+1\big)^2}\\ &=\sum_{n\in\mathbb{Z}}q^{9(n-\frac12)^2}\sum_{k\in\mathbb{Z}}q^{3\big(k+1\big)^2}+\sum_{k\in\mathbb {Z}}q^{3(k+\frac12)^2}\left[\sum_{n\ge1}+\sum_{n\le0}\right]q^{\tfrac14(6n-1)^2}\\ &=\theta_2(9z)\theta_3(3z)+\theta_2(3z)\left[q^{5^2/4}+q^{11^2/4}+q^{17^2/4}+\cdots+q^{(-1)^2/4}+q^{(-7)^2/4}+q^{(-13)^2/4}+\cdots\right]\\ &=\theta_2(9z)\theta_3(3z)+\theta_2(3z)\sum_{\substack{k\ge1\\3\nmid {2k-1}}}q^{\tfrac14(2k-1)^2}\\ &=\theta_2(9z)\theta_3(3z)+\theta_2(3z)\left[\sum_{k\ge1}q^{(k-1/2)^2}-\sum_{k\ge1}q^{9(k-1/2)^2}\right]\\ &=\theta_2(9z)\theta_3(3z)+\theta_2(3z)\left[\theta_2(z)-\theta_2(9z)\right]\\ &=\theta_2(z)\theta_2(3z)+\theta_2(9z)\theta_4(3z/4) \end{align} (The sums don't change when you add an integer to their indices, and you can split them into odd and even parts.)

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