WarnaarによるJacobiの三重積の一般化

示すべき等式は
\begin{align} \sum_{0\leq n}\frac{(aq^n,bq^n;q)_{\infty}(abq^{n-1};q)_n}{(1-abq^{n-1})(q;q)_n}q^n&=\frac 1{\left(1-\frac{ab}q\right)(q;q)_{\infty}}\left(1+\sum_{0< n}(-1)^nq^{\binom n2}(a^n+b^n)\right) \end{align}
と書き換えられる. 右辺を$q$二項定理, 等比級数で展開すると,
\begin{align} &\sum_{0\leq n}\frac{(aq^n,bq^n;q)_{\infty}(abq^{n-1};q)_n}{(1-abq^{n-1})(q;q)_n}q^n\\ &=\sum_{0\leq n}\frac{q^n}{(q;q)_n}\sum_{0\leq i,j,k,l}\frac{(-aq^n)^iq^{\binom i2}}{(q;q)_i}\frac{(-bq^n)^jq^{\binom j2}}{(q;q)_j}\frac{(q;q)_n}{(q;q)_k(q;q)_{n-k}}q^{\binom k2}(-abq^{n-1})^k(abq^{n-1})^l\\ &=\sum_{0\leq n,i,j,k,l}\frac{(-1)^{i+j+k}a^{i+k+l}b^{j+k+l}q^{\binom i2+\binom j2+\binom k2+n(i+j+k+l+1)-k-l}}{(q;q)_i(q;q)_j(q;q)_k(q;q)_{n-k}}\\ &=\sum_{0\leq n,i,j,k,l}\frac{(-1)^{i+j+k}a^{i+k+l}b^{j+k+l}q^{\binom i2+\binom j2+\binom k2+(n+k)(i+j+k+l+1)-k-l}}{(q;q)_i(q;q)_j(q;q)_k(q;q)_n}\\ &=\sum_{0\leq i,j,k,l}\frac{(-1)^{i+j+k}a^{i+k+l}b^{j+k+l}q^{\binom i2+\binom j2+\binom k2+k(i+j+k+l+1)-k-l}}{(q;q)_i(q;q)_j(q;q)_k}\sum_{0\leq n}\frac{q^{n(i+j+k+l+1)}}{(q;q)_n}\\ &=\sum_{0\leq i,j,k,l}\frac{(-1)^{i+j+k}a^{i+k+l}b^{j+k+l}q^{\binom i2+\binom j2+\binom k2+k(i+j+k+l+1)-k-l}}{(q;q)_i(q;q)_j(q;q)_k}\frac 1{(q^{i+j+k+l+1};q)_{\infty}}\\ &=\frac 1{(q;q)_{\infty}}\sum_{0\leq i,j,k,l}\frac{(-1)^{i+j+k}a^{i+k+l}b^{j+k+l}q^{\binom i2+\binom j2+\binom k2+k(i+j+k+l+1)-k-l}(q;q)_{i+j+k+l}}{(q;q)_i(q;q)_j(q;q)_k}\\ &=\frac 1{(q;q)_{\infty}}\sum_{0\leq i,j,k,l}\frac{(-1)^{i+j-k}a^{i+l}b^{j+l}q^{\binom{i-k}2+\binom{j-k}2+\binom k2+k(i+j-k+l+1)-k-l}(q;q)_{i+j+l-k}}{(q;q)_{i-k}(q;q)_{j-k}(q;q)_k}\\ &=\frac 1{(q;q)_{\infty}}\sum_{0\leq i,j,l}(-1)^{i+j}a^{i+l}b^{j+l}q^{\binom{i}2+\binom{j}2-l}\sum_{0\leq k}\frac{(-1)^kq^{\binom{k+1}2+kl}(q;q)_{i+j+l-k}}{(q;q)_{i-k}(q;q)_{j-k}(q;q)_k} \end{align}
ここで, $q$-Vandermondeの恒等式 より,
\begin{align} &\sum_{0\leq k}\frac{(-1)^kq^{\binom{k+1}2+kl}(q;q)_{i+j+l-k}}{(q;q)_{i-k}(q;q)_{j-k}(q;q)_k}\\ &=\frac{(q;q)_{i+j+l}}{(q;q)_i(q;q)_j}\Q21{q^{-i},q^{-j}}{q^{-i-j-l}}{q}\\ &=\frac{(q;q)_{i+j+l}}{(q;q)_i(q;q)_j}\frac{(q^{-i-l};q)_i}{(q^{-i-j-l};q)_i}q^{-ij}\\ &=\frac{(q;q)_{i+j+l}}{(q;q)_i(q;q)_j}\frac{(q^{l+1};q)_i}{(q^{j+l+1};q)_i}\\ &=\frac{(q;q)_{i+l}(q;q)_{j+l}}{(q;q)_i(q;q)_j(q;q)_l} \end{align}
であるから, これを代入して,
\begin{align} &\frac 1{(q;q)_{\infty}}\sum_{0\leq i,j,l}(-1)^{i+j}a^{i+l}b^{j+l}q^{\binom{i}2+\binom{j}2-l}\sum_{0\leq k}\frac{(-1)^kq^{\binom{k+1}2+kl}(q;q)_{i+j+l-k}}{(q;q)_{i-k}(q;q)_{j-k}(q;q)_k}\\ &=\frac 1{(q;q)_{\infty}}\sum_{0\leq i,j,l}(-1)^{i+j}a^{i+l}b^{j+l}q^{\binom{i}2+\binom{j}2-l}\frac{(q;q)_{i+l}(q;q)_{j+l}}{(q;q)_i(q;q)_j(q;q)_l}\\ &=\frac 1{(q;q)_{\infty}}\sum_{0\leq i,j,l}(-1)^{i+j}a^{i}b^{j}q^{\binom{i-l}2+\binom{j-l}2-l}\frac{(q;q)_{i}(q;q)_{j}}{(q;q)_{i-l}(q;q)_{j-l}(q;q)_l}\\ &=\frac 1{(q;q)_{\infty}}\sum_{0\leq i,j}(-1)^{i+j}a^{i}b^{j}q^{\binom{i}2+\binom{j}2}(q;q)_{i}(q;q)_{j}\sum_{0\leq l}\frac{q^{l(l-i-j)}}{(q;q)_{i-l}(q;q)_{j-l}(q;q)_l} \end{align}
ここで, 再び $q$-Vandermondeの恒等式 より,
\begin{align} &\sum_{0\leq l}\frac{q^{l(l-i-j)}}{(q;q)_{i-l}(q;q)_{j-l}(q;q)_l}\\ &=\frac 1{(q;q)_i(q;q)_j}\Q21{q^{-i},q^{-j}}{0}{q}\\ &=\frac 1{(q;q)_i(q;q)_j}q^{-ij} \end{align}
であるから, これを代入して,
\begin{align} &\frac 1{(q;q)_{\infty}}\sum_{0\leq i,j}(-1)^{i+j}a^{i}b^{j}q^{\binom{i}2+\binom{j}2}(q;q)_{i}(q;q)_{j}\sum_{0\leq l}\frac{q^{l(l-i-j)}}{(q;q)_{i-l}(q;q)_{j-l}(q;q)_l}\\ &=\frac 1{(q;q)_{\infty}}\sum_{0\leq i,j}(-1)^{i+j}a^{i}b^{j}q^{\binom{i}2+\binom{j}2-ij}\\ &=\frac 1{(q;q)_{\infty}}\sum_{\substack{0\leq j\\-j\leq i}}(-1)^{i}a^{i+j}b^{j}q^{\binom{i+j}2+\binom{j}2-ij-j^2}\\ &=\frac 1{(q;q)_{\infty}}\sum_{\substack{0\leq j\\ -j\leq i}}(-1)^{i}a^{i+j}b^{j}q^{\binom{i}2-j}\\ &=\frac 1{(q;q)_{\infty}}\sum_{\substack{i\in\ZZ\\\mathrm{max}\{0,-i\}\leq j}}(-1)^{i}a^{i+j}b^{j}q^{\binom{i}2-j}\\ &=\frac 1{\left(1-\frac{ab}{q}\right)(q;q)_{\infty}}\sum_{\substack{i\in\ZZ}}(-1)^{i}a^{i}(ab/q)^{\mathrm{max}\{0,-i\}}q^{\binom{i}2}\\ &=\frac 1{\left(1-\frac{ab}{q}\right)(q;q)_{\infty}}\left(1+\sum_{0< n}(-a)^nq^{\binom n2}+\sum_{n<0}(-1)^nb^{-n}q^{\binom{n+1}2}\right)\\ &=\frac 1{\left(1-\frac{ab}{q}\right)(q;q)_{\infty}}\left(1+\sum_{0< n}(-a)^nq^{\binom n2}+\sum_{0< n}(-1)^nb^{n}q^{\binom{n}2}\right) \end{align}
となって示すべき等式が得られた.