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*}


  • $\begingroup$ I know it should be easy to prove it, but I am trying to pin down the simplest way to do it. It is mentioned on pg 28 of arxiv.org/pdf/hep-th/9308079.pdf. Since I am not an expert in the field, I was hoping I could get a hint from some of you... $\endgroup$ – Andrew Jun 26 '12 at 20:01
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    $\begingroup$ You might want to start with 2.3.2 in my survey: math.ucla.edu/~pak/papers/psurvey.pdf $\endgroup$ – Igor Pak Jun 27 '12 at 0:02

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.

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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).

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  • $\begingroup$ I don't see how: e.g. r+s=1 and s+t = 1 gives r=0, s=1, t=0, and r = 1, s=0, t=1, which leads to \frac{q}{(q)} + \frac{q^2}{(q)^2}, which is the same when you take a=1 and b=1 and calculate the corresponding coefficient $\endgroup$ – Andrew Jun 27 '12 at 11:58
  • $\begingroup$ I agree with your calculation for the coefficient on the LHS (which simplifies to just $\frac{q}{(q)} + \frac{q^2}{(q)^2} = 1 + 1 = 2$). But as I said, the RHS coefficient (for $a=b=1$) is $q^{1-1+1}/(q)^2 = q/q^2 = 1/q$. There's a fine chance that I'm doing something quite stupid, in which case, let me know! $\endgroup$ – Pat Devlin Jun 27 '12 at 14:08
  • $\begingroup$ OK, i understand now - in my notation $(q)_a$ means $\prod_{i=1}^{a}(1−q^i)$. Sorry for not clarifying... – Andrew 0 secs ago $\endgroup$ – Andrew Jun 27 '12 at 14:16
  • $\begingroup$ Ah! I had guessed it was the Pochhammer symbol. $\endgroup$ – Pat Devlin Jun 27 '12 at 14:26
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    $\begingroup$ Andrew: perhaps edit your original question too to clarify the notation, so that other readers aren't confused. $\endgroup$ – Zsbán Ambrus Sep 5 '12 at 16:54

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