前の記事
でBaileyによるnearly-poised${}_5\phi_4$の変換公式
\begin{align}
&\Q54{a,b,c,d,q^{-n}}{aq/b,aq/c,aq/d,a^2q^{-n}/w^2}q\\
&=\frac{(wq/a,w^2q/a;q)_n}{(wq,w^2q/a^2;q)_n}\Q{12}{11}{w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a,\sqrt a,-\sqrt a,\sqrt{aq},-\sqrt{aq},w^2q^{n+1}/a,q^{-n}}{\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,wq/\sqrt a,-wq/\sqrt a,w\sqrt{q/a},-w\sqrt{q/a},aq^{-n}/w,wq^{n+1}}q
\end{align}
を示した. ここで, $w=a^2q/bcd$である.
今回はそれをnon-terminatingに拡張したVerma-Jainの変換公式を示す.
\begin{align}
W(a;b_1,\dots,b_r;x):=\Q{r+3}{r+2}{a,\sqrt aq,-\sqrt aq,b_1,\dots,b_r}{\sqrt a,-\sqrt a,aq/b_1,\dots,aq/b_r}x
\end{align}
とする.
$w=a^2q/bcd,f= a^2e/w^2$とするとき,
\begin{align}
&\Q54{a,b,c,d,e}{aq/b,aq/c,aq/d,f}{q}\\
&\qquad+\frac{(a,b,c,d,e,q/f,aq^2/bf,aq^2/cf,aq^2/df;q)_{\infty}}{(aq/b,aq/c,aq/d,f/q,aq/f,bq/f,cq/f,dq/f,eq/f;q)_{\infty}}\\
&\qquad\cdot \Q54{aq/f,bq/f,cq/f,dq/f,eq/f}{q^2/f,aq^2/bf,aq^2/cf,aq^2/df}q\\
&=\frac{(w q/a,w q/e,w^2q/a,q/f;q)_{\infty}}{(wq,aq/f,eq/f,aq/w f;q)_{\infty}}\\
&\qquad\cdot W(w;\sqrt a,-\sqrt a,\sqrt{aq},-\sqrt{aq},w b/a,w c/a,w d/a,e,aq/f;q)\\
&\qquad+\frac{(a,e,w b/a,w c/a,w d/a,q/f,bcq/f,bdq/f,cdq/f,aq^3/f^2;q)_{\infty}}{(aq/b,aq/c,aq/d,aq/f,bq/f,cq/f,dq/f,eq/f,w f/aq,a^2q^3/w f^2;q)_{\infty}}\\
&\qquad\cdot W(a^2q^2/wf^2;a^{\frac 32}q/w f,-a^{\frac 32}q/w f,(aq)^{\frac 32}/w f,-(aq)^{\frac 32}/w f,w q/a,aq/f,bq/f,cq/f,dq/f;q)
\end{align}
が成り立つ.
以下の方針はGasper-Rahmanの本のExercise 2.24, 2.25に基づいている.
前の記事
と全く同様の議論で,
\begin{align}
&\Q54{a,b,c,d,e}{aq/b,aq/c,aq/d,f}q\\
&=\sum_{0\leq j}\frac{(1-wq^{2j})(w,wb/a,wc/a,wd/a,e;q)_j(a;q)_{2j}}{(1-w)(aq/b,aq/c,aq/d,f,q;q)_j(wq;q)_{2j}}\left(\frac{aq}{w}\right)^j\Q32{aq^{2j},eq^{j},a/w}{wq^{2j+1},fq^j}q
\end{align}
となる. ここで,
non-terminating $q$-Saalschützの和公式
より
\begin{align}
\Q32{aq^{2j},eq^{j},a/w}{wq^{2j+1},fq^j}q&=\frac{(q^{1-j}/f,wq/a,wq^{j+1}/e,w^2q^{2j+1}/a;q)_{\infty}}{(aq^{j+1}/f,eq/f,aq^{1-j}/wf,wq^{2j+1};q)_{\infty}}\\
&\qquad-\frac{(q^{1-j}/f,aq^{2j},eq^{j},a/w,wq^{j+2}/f;q)_{\infty}}{(fq^{j-1},aq^{j+1}/f,eq/f,aq^{1-j}/wf,wq^{2j+1};q)_{\infty}}\Q32{aq^{j+1}/f,eq/f,aq^{1-j}/wf}{wq^{j+2}/f,q^{2-j}/f}q\\
&=\frac{(q/f,wq/a,wq/e,w^2q/a;q)_{\infty}}{(aq/f,eq/f,aq/wf,wq;q)_{\infty}}\frac{(q^{1-j}/f,aq/f;q)_j(wq;q)_{2j}}{(aq^{1-j}/wf,wq/e;q)_j(w^2q/a;q)_{2j}}\\
&\qquad-\frac{(q/f,a,e,a/w,wq^2/f;q)_{\infty}}{(f/q,aq/f,eq/f,aq/wf,wq;q)_{\infty}}\frac{(q^{1-j}/f,f/q,aq/f;q)_j(wq;q)_{2j}}{(aq^{1-j}/wf,e,wq^2/f;q)_j(a;q)_{2j}}\\
&\qquad\cdot\Q32{aq^{j+1}/f,eq/f,aq^{1-j}/wf}{wq^{j+2}/f,q^{2-j}/f}q\\
&=\frac{(q/f,wq/a,wq/e,w^2q/a;q)_{\infty}}{(aq/f,eq/f,aq/wf,wq;q)_{\infty}}\frac{(f,aq/f;q)_j(wq;q)_{2j}}{(wf/a,wq/e;q)_j(w^2q/a;q)_{2j}}\left(\frac wa\right)^j\\
&\qquad-\frac{(q/f,a,e,a/w,wq^2/f;q)_{\infty}}{(f/q,aq/f,eq/f,aq/wf,wq;q)_{\infty}}\frac{(f,f/q,aq/f;q)_j(wq;q)_{2j}}{(wf/a,e,wq^2/f;q)_j(a;q)_{2j}}\left(\frac wa\right)^j\\
&\qquad\cdot\Q32{aq^{j+1}/f,eq/f,aq^{1-j}/wf}{wq^{j+2}/f,q^{2-j}/f}q\\
\end{align}
であるからこれを代入して
\begin{align}
&\sum_{0\leq j}\frac{(1-wq^{2j})(w,wb/a,wc/a,wd/a,e;q)_j(a;q)_{2j}}{(1-w)(aq/b,aq/c,aq/d,f,q;q)_j(wq;q)_{2j}}\left(\frac{aq}{w}\right)^j\Q32{aq^{2j},eq^{j},a/w}{wq^{2j+1},fq^j}q\\
&=\frac{(q/f,wq/a,wq/e,w^2q/a;q)_{\infty}}{(aq/f,eq/f,aq/wf,wq;q)_{\infty}}\sum_{0\leq j}\frac{(1-wq^{2j})(w,wb/a,wc/a,wd/a,e;q)_j(a;q)_{2j}}{(1-w)(aq/b,aq/c,aq/d,f,q;q)_j(wq;q)_{2j}}\left(\frac{aq}{w}\right)^j\\
&\qquad\cdot\frac{(f,aq/f;q)_j(wq;q)_{2j}}{(wf/a,wq/e;q)_j(w^2q/a;q)_{2j}}\left(\frac wa\right)^j\\
&\qquad-\frac{(q/f,a,e,a/w,wq^2/f;q)_{\infty}}{(f/q,aq/f,eq/f,aq/wf,wq;q)_{\infty}}\sum_{0\leq j}\frac{(1-wq^{2j})(w,wb/a,wc/a,wd/a,e;q)_j(a;q)_{2j}}{(1-w)(aq/b,aq/c,aq/d,f,q;q)_j(wq;q)_{2j}}\left(\frac{aq}{w}\right)^j\\
&\qquad\cdot\frac{(f,f/q,aq/f;q)_j(wq;q)_{2j}}{(wf/a,e,wq^2/f;q)_j(a;q)_{2j}}\left(\frac wa\right)^j\Q32{aq^{j+1}/f,eq/f,aq^{1-j}/wf}{wq^{j+2}/f,q^{2-j}/f}q\\
&=\frac{(q/f,wq/a,wq/e,w^2q/a;q)_{\infty}}{(aq/f,eq/f,aq/wf,wq;q)_{\infty}}\sum_{0\leq j}\frac{(1-wq^{2j})(w,wb/a,wc/a,wd/a,e,aq/f;q)_j(a;q)_{2j}}{(1-w)(aq/b,aq/c,aq/d,wq/e,wf/a,q;q)_j(w^2q/a;q)_{2j}}q^j\\
&\qquad-\frac{(q/f,a,e,a/w,wq^2/f;q)_{\infty}}{(f/q,aq/f,eq/f,aq/wf,wq;q)_{\infty}}\sum_{0\leq j}\frac{(1-wq^{2j})(w,wb/a,wc/a,wd/a,f/q,aq/f;q)_j}{(1-w)(aq/b,aq/c,aq/d,wq^2/f,wf/a,q;q)_j}q^j\\
&\qquad\cdot\Q32{aq^{j+1}/f,eq/f,aq^{1-j}/wf}{wq^{j+2}/f,q^{2-j}/f}q\\
&=\frac{(q/f,wq/a,wq/e,w^2q/a;q)_{\infty}}{(aq/f,eq/f,aq/wf,wq;q)_{\infty}} W(w;\sqrt a,-\sqrt a,\sqrt{aq},-\sqrt{aq},w b/a,w c/a,w d/a,e,aq/f;q)\\
&\qquad-\frac{(q/f,a,e,a/w,wq^2/f;q)_{\infty}}{(f/q,aq/f,eq/f,aq/wf,wq;q)_{\infty}}\sum_{0\leq j}\frac{(1-wq^{2j})(w,wb/a,wc/a,wd/a,f/q,aq/f;q)_j}{(1-w)(aq/b,aq/c,aq/d,wq^2/f,wf/a,q;q)_j}q^j\\
&\qquad\cdot\Q32{aq^{j+1}/f,eq/f,aq^{1-j}/wf}{wq^{j+2}/f,q^{2-j}/f}q\\
\end{align}
となる. 第二項は
\begin{align}
&\sum_{0\leq j}\frac{(1-wq^{2j})(w,wb/a,wc/a,wd/a,f/q,aq/f;q)_j}{(1-w)(aq/b,aq/c,aq/d,wq^2/f,wf/a,q;q)_j}q^j\\
&\qquad\cdot\Q32{aq^{j+1}/f,eq/f,aq^{1-j}/wf}{wq^{j+2}/f,q^{2-j}/f}q\\
&=\sum_{0\leq j}\frac{(1-wq^{2j})(w,wb/a,wc/a,wd/a;q)_j}{(1-w)(aq/b,aq/c,aq/d,q;q)_j}q^j\\
&\qquad\cdot\sum_{0\leq k}\frac{(eq/f;q)_k(f/q;q)_{j-k}(aq/f;q)_{j+k}}{(q;q)_k(wf/a;q)_{j-k}(wq^2/f;q)_{j+k}}\left(\frac aw\right)^k\\
&=\sum_{0\leq k}\frac{(eq/f;q)_k}{(q;q)_k}\left(\frac aw\right)^k\sum_{0\leq j}\frac{(1-wq^{2j})(w,wb/a,wc/a,wd/a;q)_j}{(1-w)(aq/b,aq/c,aq/d,q;q)_j}q^j\frac{(f/q;q)_{j-k}(aq/f;q)_{j+k}}{(wf/a;q)_{j-k}(wq^2/f;q)_{j+k}}\\
&=\sum_{0\leq k}\frac{(eq/f,aq/wf,aq/f;q)_k}{(q,q^2/f,wq^2/f;q)_k}q^kW(w;wb/a,wc/a,wd/a,fq^{-k-1},aq^{k+1}/f;q)
\end{align}
となる. ここで,
non-terminating Jacksonの和公式
より
\begin{align}
&W(w;wb/a,wc/a,wd/a,fq^{-k-1},aq^{k+1}/f;q)\\
&=\frac{(wq,b,c,d,aq^{k+2}/bf,aq^{k+2}/cf,aq^{k+2}/df,aq^{k+1}/wf;q)_{\infty}}{(aq/b,aq/c,aq/d,wq^{k+2}/f,bq^{k+1}/f,cq^{k+1}/f,dq^{k+1}/f,a/w;q)_{\infty}}\\
&-\frac{(wq,aq^{k+1}/wf,bcq^{k+1}/f,bdq^{k+1}/f,cdq^{k+1}/f,aq^{2k+3}/f^2,wb/a,wc/a,wd/a,fq^{-k-1};q)_{\infty}}{(aq/b,aq/c,aq/d,wq^{k+2}/f,wfq^{-k-1}/a,bq^{k+1}/f,cq^{k+1}/f,dq^{k+1}/f,a/w,a^2q^{2k+3}/wf^2;q)_{\infty}}\\
&\qquad \cdot W(a^2q^{2k+2}/wf^2;aq^{k+1}/f,bq^{k+1}/f,cq^{k+1}/f,dq^{k+1}/f,a/w;q)
\end{align}
であるからこれを代入すると
\begin{align}
&\sum_{0\leq k}\frac{(eq/f,aq/wf,aq/f;q)_k}{(q,q^2/f,wq^2/f;q)_k}q^kW(w;wb/a,wc/a,wd/a,fq^{-k-1},aq^{k+1}/f;q)\\
&=\sum_{0\leq k}\frac{(eq/f,aq/wf,aq/f;q)_k}{(q,q^2/f,wq^2/f;q)_k}q^k\\
&\qquad\cdot \frac{(wq,b,c,d,aq^{k+2}/bf,aq^{k+2}/cf,aq^{k+2}/df,aq^{k+1}/wf;q)_{\infty}}{(aq/b,aq/c,aq/d,wq^{k+2}/f,bq^{k+1}/f,cq^{k+1}/f,dq^{k+1}/f,a/w;q)_{\infty}}\\
&-\sum_{0\leq k}\frac{(eq/f,aq/wf,aq/f;q)_k}{(q,q^2/f,wq^2/f;q)_k}q^k\\
&\qquad\cdot\frac{(wq,aq^{k+1}/wf,bcq^{k+1}/f,bdq^{k+1}/f,cdq^{k+1}/f,aq^{2k+3}/f^2,wb/a,wc/a,wd/a,fq^{-k-1};q)_{\infty}}{(aq/b,aq/c,aq/d,wq^{k+2}/f,wfq^{-k-1}/a,bq^{k+1}/f,cq^{k+1}/f,dq^{k+1}/f,a/w,a^2q^{2k+3}/wf^2;q)_{\infty}}\\
&\qquad \cdot W(a^2q^{2k+2}/wf^2;aq^{k+1}/f,bq^{k+1}/f,cq^{k+1}/f,dq^{k+1}/f,a/w;q)\\
&= \frac{(wq,b,c,d,aq^{2}/bf,aq^{2}/cf,aq^{2}/df,aq/wf;q)_{\infty}}{(aq/b,aq/c,aq/d,wq^{2}/f,bq/f,cq/f,dq/f,a/w;q)_{\infty}}\Q54{aq/f,bq/f,cq/f,dq/f,eq/f}{aq^2/bf,aq^2/cf,aq^2/df,aq^2/ef}q\\
&-\frac{(wq,aq/wf,bcq/f,bdq/f,cdq/f,aq^{3}/f^2,wb/a,wc/a,wd/a,f/q;q)_{\infty}}{(aq/b,aq/c,aq/d,wq^{2}/f,wf/aq,bq/f,cq/f,dq/f,a/w,a^2q^3/wf^2;q)_{\infty}}\\
&\qquad\cdot\sum_{0\leq k}\frac{(eq/f,aq/f,bq/f,cq/f,dq/f,fq^{-k-1};q)_k(a^2q^3/wf^2;q)_{2k}}{(q,q^2/f,bcq/f,bdq/f,cdq/f,wfq^{-k-1}/a;q)_k(aq^3/f^2;q)_{2k}}q^k\\
&\qquad \cdot W(a^2q^{2k+2}/wf^2;aq^{k+1}/f,bq^{k+1}/f,cq^{k+1}/f,dq^{k+1}/f,a/w;q)\\
&= \frac{(wq,b,c,d,aq^{2}/bf,aq^{2}/cf,aq^{2}/df,aq/wf;q)_{\infty}}{(aq/b,aq/c,aq/d,wq^{2}/f,bq/f,cq/f,dq/f,a/w;q)_{\infty}}\Q54{aq/f,bq/f,cq/f,dq/f,eq/f}{aq^2/bf,aq^2/cf,aq^2/df,aq^2/ef}q\\
&-\frac{(wq,aq^{2}/wf,bcq/f,bdq/f,cdq/f,aq^{3}/f^2,wb/a,wc/a,wd/a,f/q;q)_{\infty}}{(aq/b,aq/c,aq/d,wq^{2}/f,wf/a,bq/f,cq/f,dq/f,a/w,a^2q^3/wf^2;q)_{\infty}}\\
&\qquad\cdot\sum_{0\leq k}\frac{(eq/f,aq/f,bq/f,cq/f,dq/f;q)_k(a^2q^3/wf^2;q)_{2k}}{(q,aq^2/wf,bcq/f,bdq/f,cdq/f;q)_k(aq^3/f^2;q)_{2k}}\left(\frac{aq}w\right)^k\\
&\qquad \cdot W(a^2q^{2k+2}/wf^2;aq^{k+1}/f,bq^{k+1}/f,cq^{k+1}/f,dq^{k+1}/f,a/w;q)
\end{align}
を得る. ここまでをまとめると,
\begin{align}
&\Q54{a,b,c,d,e}{aq/b,aq/c,aq/d,f}q\\
&=\frac{(q/f,wq/a,wq/e,w^2q/a;q)_{\infty}}{(aq/f,eq/f,aq/wf,wq;q)_{\infty}} W(w;\sqrt a,-\sqrt a,\sqrt{aq},-\sqrt{aq},w b/a,w c/a,w d/a,e,aq/f;q)\\
&\qquad-\frac{(q/f,a,e,a/w,wq^2/f;q)_{\infty}}{(f/q,aq/f,eq/f,aq/wf,wq;q)_{\infty}}\\
&\qquad\cdot\frac{(wq,b,c,d,aq^{2}/bf,aq^{2}/cf,aq^{2}/df,aq/wf;q)_{\infty}}{(aq/b,aq/c,aq/d,wq^{2}/f,bq/f,cq/f,dq/f,a/w;q)_{\infty}}\Q54{aq/f,bq/f,cq/f,dq/f,eq/f}{aq^2/bf,aq^2/cf,aq^2/df,aq^2/ef}q\\
&\qquad +\frac{(q/f,a,e,a/w,wq^2/f;q)_{\infty}}{(f/q,aq/f,eq/f,aq/wf,wq;q)_{\infty}}\\
&\qquad\cdot \frac{(wq,aq/wf,bcq/f,bdq/f,cdq/f,aq^{3}/f^2,wb/a,wc/a,wd/a,f/q;q)_{\infty}}{(aq/b,aq/c,aq/d,wq^{2}/f,wf/aq,bq/f,cq/f,dq/f,a/w,a^2q^3/wf^2;q)_{\infty}}\\
&\qquad\cdot\sum_{0\leq k}\frac{(eq/f,aq/f,bq/f,cq/f,dq/f;q)_k(a^2q^3/wf^2;q)_{2k}}{(q,aq^2/wf,bcq/f,bdq/f,cdq/f;q)_k(aq^3/f^2;q)_{2k}}\left(\frac{aq}w\right)^k\\
&\qquad \cdot W(a^2q^{2k+2}/wf^2;aq^{k+1}/f,bq^{k+1}/f,cq^{k+1}/f,dq^{k+1}/f,a/w;q)\\
&=\frac{(q/f,wq/a,wq/e,w^2q/a;q)_{\infty}}{(aq/f,eq/f,aq/wf,wq;q)_{\infty}} W(w;\sqrt a,-\sqrt a,\sqrt{aq},-\sqrt{aq},w b/a,w c/a,w d/a,e,aq/f;q)\\
&\qquad-\frac{(q/f,a,b,c,d,e,aq^{2}/bf,aq^{2}/cf,aq^{2}/df;q)_{\infty}}{(f/q,aq/b,aq/c,aq/d,aq/f,bq/f,cq/f,dq/f,eq/f;q)_{\infty}}\Q54{aq/f,bq/f,cq/f,dq/f,eq/f}{aq^2/bf,aq^2/cf,aq^2/df,aq^2/ef}q\\
&\qquad+\frac{(q/f,a,e,bcq/f,bdq/f,cdq/f,aq^{3}/f^2,wb/a,wc/a,wd/a;q)_{\infty}}{(aq/b,aq/c,aq/d,wf/aq,aq/f,bq/f,cq/f,dq/f,eq/f,a^2q^3/wf^2;q)_{\infty}}\\
&\qquad\cdot\sum_{0\leq k}\frac{(eq/f,aq/f,bq/f,cq/f,dq/f;q)_k(a^2q^3/wf^2;q)_{2k}}{(q,aq^2/wf,bcq/f,bdq/f,cdq/f;q)_k(aq^3/f^2;q)_{2k}}\left(\frac{aq}w\right)^k\\
&\qquad \cdot W(a^2q^{2k+2}/wf^2;aq^{k+1}/f,bq^{k+1}/f,cq^{k+1}/f,dq^{k+1}/f,a/w;q)
\end{align}
ここで, 最後の項は$q$-Saalschützの和公式より
\begin{align}
&\sum_{0\leq k}\frac{(eq/f,aq/f,bq/f,cq/f,dq/f;q)_k(a^2q^3/wf^2;q)_{2k}}{(q,aq^2/wf,bcq/f,bdq/f,cdq/f;q)_k(aq^3/f^2;q)_{2k}}\left(\frac{aq}w\right)^k\\
&\qquad \cdot W(a^2q^{2k+2}/wf^2;aq^{k+1}/f,bq^{k+1}/f,cq^{k+1}/f,dq^{k+1}/f,a/w;q)\\
&=\sum_{0\leq k}\frac{1-a^2q^{2k+2}/wf^2}{1-a^2q^2/wf^2}\frac{(eq/f,aq/f,bq/f,cq/f,dq/f;q)_k(a^2q^2/wf^2;q)_{2k}}{(q,aq^2/wf,bcq/f,bdq/f,cdq/f;q)_k(aq^3/f^2;q)_{2k}}\left(\frac{aq}w\right)^k\\
&\qquad \cdot \sum_{0\leq j}\frac{(1-a^2q^{2j+2k+2}/wf^2)(a^2q^{2k+2}/wf^2,aq^{k+1}/f,bq^{k+1}/f,cq^{k+1}/f,dq^{k+1}/f,a/w;q)_j}{(1-a^2q^{2k+2}/wf^2)(q,aq^{k+2}/wf,bcq^{k+1}/f,bdq^{k+1}/f,cdq^{k+1}/f,aq^{2k+3}/f^2;q)_j}q^j\\
&=\sum_{0\leq k}\frac{(eq/f;q)_k}{(q;q)_k}\left(\frac{aq}w\right)^k\sum_{0\leq j}\frac{(1-a^2q^{2j+2k+2}/wf^2)(a^2q^2/wf^2;q)_{j+2k}(aq/f,bq/f,cq/f,dq/f;q)_{j+k}(a/w;q)_j}{(1-a^2q^{2}/wf^2)(q;q)_j(aq^{2}/wf,bcq/f,bdq/f,cdq/f;q)_{j+k}(aq^{3}/f^2;q)_{j+2k}}q^j\\
&=\sum_{0\leq k}\frac{(eq/f;q)_k}{(q;q)_k}\left(\frac{a}w\right)^k\sum_{k\leq j}\frac{(1-a^2q^{2j+2}/wf^2)(a^2q^2/wf^2;q)_{j+k}(aq/f,bq/f,cq/f,dq/f;q)_{j}(a/w;q)_{j-k}}{(1-a^2q^{2}/wf^2)(q;q)_{j-k}(aq^{2}/wf,bcq/f,bdq/f,cdq/f;q)_{j}(aq^{3}/f^2;q)_{j+k}}q^j\\
&=\sum_{0\leq j}\frac{(1-a^2q^{2j+2}/wf^2)(aq/f,bq/f,cq/f,dq/f;q)_{j}}{(1-a^2q^{2}/wf^2)(aq^{2}/wf,bcq/f,bdq/f,cdq/f;q)_{j}}q^j\sum_{0\leq k}\frac{(eq/f;q)_k}{(q;q)_k}\left(\frac{a}w\right)^k\frac{(a^2q^2/wf^2;q)_{j+k}(a/w;q)_{j-k}}{(q;q)_{j-k}(aq^{3}/f^2;q)_{j+k}}\\
&=\sum_{0\leq j}\frac{(1-a^2q^{2j+2}/wf^2)(a^2q^2/wf^2,aq/f,bq/f,cq/f,dq/f,a/w;q)_{j}}{(1-a^2q^{2}/wf^2)(q,aq^{2}/wf,bcq/f,bdq/f,cdq/f,aq^3/f^2;q)_{j}}q^j\Q32{eq/f,a^2q^{j+2}/wf^2,q^{-j}}{aq^{j+3}/f^2,wq^{1-j}/a}q\\
&=\sum_{0\leq j}\frac{(1-a^2q^{2j+2}/wf^2)(a^2q^2/wf^2,aq/f,bq/f,cq/f,dq/f,a/w;q)_{j}}{(1-a^2q^{2}/wf^2)(q,aq^{2}/wf,bcq/f,bdq/f,cdq/f,aq^3/f^2;q)_{j}}q^j\frac{(a^3q^{j+2}/w^2f^2,wq/a;q)_j}{(aq^{j+3}/f,a/w;q)_j}\\
&=\sum_{0\leq j}\frac{(1-a^2q^{2j+2}/wf^2)(a^2q^2/wf^2,aq/f,bq/f,cq/f,dq/f,wq/a;q)_{j}(a^3q^2/w^2f^2;q)_{2j}}{(1-a^2q^{2}/wf^2)(q,aq^{2}/wf,bcq/f,bdq/f,cdq/f,a^3q^2/w^2f^2;q)_{j}(aq^3/f^2;q)_{2j}}q^j\\
&=W(a^2q^2/wf^2;a^{\frac 32}q/w f,-a^{\frac 32}q/w f,(aq)^{\frac 32}/w f,-(aq)^{\frac 32}/w f,w q/a,aq/f,bq/f,cq/f,dq/f;q)
\end{align}
となる. よって, これを代入して定理を得る.
上の証明は Baileyの4項変換公式 の証明とかなり方針が似ていると思う.
定理1において$d=1$とすると, 左辺は1になるので, $w=a^2q/bc,f= a^2e/w^2$とするとき,
\begin{align}
1&=\frac{(w q/a,w q/e,w^2q/a,q/f;q)_{\infty}}{(wq,aq/f,eq/f,aq/w f;q)_{\infty}}\\
&\qquad W(w;\sqrt a,-\sqrt a,\sqrt{aq},-\sqrt{aq},w/a,e,aq/f;q)\\
&\qquad+\frac{(a,e,w/a,a^2q^2/wf,aq^3/f^2;q)_{\infty}}{(aq,aq/f,eq/f,w f/aq,a^2q^3/w f^2;q)_{\infty}}\\
&\qquad\cdot W(a^2q^2/wf^2;a^{\frac 32}q/w f,-a^{\frac 32}q/w f,(aq)^{\frac 32}/w f,-(aq)^{\frac 32}/w f,w q/a,aq/f,q/f;q)
\end{align}
が成り立つ. ここに$b,c$は現れていないので, 条件$w=a^2q/bc$は不要になる. $f$を消去すると
\begin{align}
1&=\frac{(w q/a,w q/e,w^2q/a,w^2q/a^2e;q)_{\infty}}{(wq,w^2q/ae,w^2q/a^2,wq/ae;q)_{\infty}}\\
&\qquad W(w;\sqrt a,-\sqrt a,\sqrt{aq},-\sqrt{aq},w/a,e,w^2q/a;q)\\
&\qquad+\frac{(a,e,w/a,wq^2/e,w^4q^3/a^3e^2;q)_{\infty}}{(aq,w^2q/ae,w^2q/a^2, ae/wq,w^3q^3/a^2e^2;q)_{\infty}}\\
&\qquad\cdot W(w^3q^2/a^2e^2;wq/\sqrt ae,-wq/\sqrt ae,wq^{\frac 32}/\sqrt ae,-wq^{\frac 32}/\sqrt ae,w q/a,w^2q/ae,w^2q/a^2e;q)
\end{align}
よって,
\begin{align}
&W(w;\sqrt a,-\sqrt a,\sqrt{aq},-\sqrt{aq},e,w/a,w^2q/a;q)\\
&\qquad+\frac{(a,e,wq,w/a,wq^2/e,wq/ae,w^4q^3/a^3e^2;q)_{\infty}}{(aq,w q/a,w q/e,w^2q/a,w^2q/a^2e,ae/wq,w^3q^3/a^2e^2;q)_{\infty}}\\
&\qquad\cdot W(w^3q^2/a^2e^2;wq/\sqrt ae,-wq/\sqrt ae,wq^{\frac 32}/\sqrt ae,-wq^{\frac 32}/\sqrt ae,w q/a,w^2q/ae,w^2q/a^2e;q)\\
&=\frac{(wq,w^2q/a^2,wq/ae,w^2q/ae;q)_{\infty}}{(w q/a,w q/e,w^2q/a,w^2q/a^2e;q)_{\infty}}
\end{align}
文字を置き換えて以下の系を得る.
\begin{align} &W(b;\sqrt a,-\sqrt a,\sqrt{aq},-\sqrt{aq},c,b/a,b^2q/a;q)\\ &\qquad+\frac{(a,c,bq,b/a,bq^2/c,bq/ac,b^4q^3/a^3c^2;q)_{\infty}}{(aq,b q/a,b q/c,b^2q/a,b^2q/a^2c,ac/bq,b^3q^3/a^2c^2;q)_{\infty}}\\ &\qquad\cdot W(b^3q^2/a^2c^2;bq/\sqrt ac,-bq/\sqrt ac,bq^{\frac 32}/\sqrt ac,-bq^{\frac 32}/\sqrt ac,b q/a,b^2q/ac,b^2q/a^2c;q)\\ &=\frac{(bq,b^2q/a^2,bq/ac,b^2q/ac;q)_{\infty}}{(b q/a,b q/c,b^2q/a,b^2q/a^2c;q)_{\infty}} \end{align}
これはRahmanの1986年の論文で$q$-Mellin-Barnes積分に応用されているものである.