Question

What do you use to prove the following equality (and possibly more general ones of the kind)?

\begin{align*}\sum_{r,s,t} \frac{q^{r^2+rs+s^2+st+t^2}}{(q)_r (q)_s (q)_t} z_1^{r+s} z_2^{s+t} = \sum_{a,b} \frac{q^{a^2-ab+b^2}}{(q)_a (q)_b} z_1^a z_2^b \end{align*}

Thanks!

Answer 1

Actually, I believe the generating functions are in fact not equal. They differ on the coefficient for $z_1 z_2$. The left-hand side is $1+z_1+z_2+2z_1 z_2 + \cdots$, and the right-hand side is $1+z_1+z_2+z_1 z_2 /q + \cdots$ (this is assuming you start the running variables all at 0; starting some of them at 1 doesn't fix this inequality).

Answer 2

I keep the notation $(q)_s=\prod_{i=1}^s(1-q^i)$, and in addition set \begin{equation}\binom{a}{s}_q=\frac{(q)_a}{(q)_s(q)_{a-s}},\end{equation}the Gaussian binomial coefficient.

Comparing coefficients, the question amounts to showing that \begin{equation} \sum_s\binom{a}{s}_q\binom{b}{s}_q(q)_sq^{(a-s)(b-s)}=1. \end{equation}

This, however follows immediately from induction over $a$: The assertion is obvious for $a=0$. Now suppose $a\ge1$. Use \begin{equation} \binom{a}{s}_q=q^s\binom{a-1}{s}_q+\binom{a-1}{s-1}_q \end{equation} to replace $\binom{a}{s}_q$. The contribution from the first summand of the right hand side gives $q^b$ (I omit the trivial calculation), by applying the induction hypothesis to $a-1$. In order to handle the contribution from the second summand, we note that \begin{equation} \binom{b}{s}_q(q)_s=(1-q^b)(q)_{s-1}\binom{b-1}{s-1}_q. \end{equation} So, applying the induction hypothesis for $a-1$ once again, we see that this other contribution is $1-q^b$. Now $q^b+(1-q^b)=1$ yields the result.