WarnaarによるC_n型Jacksonの和公式
楕円超幾何級数
の記法を用いる. また,
\begin{align}
\sum_{k_1,\dots,k_n=0}^N:=\sum_{k_1=0}^N\cdots\sum_{k_n=0}^N
\end{align}
のような記法を用いる. 今回はWarnaarによる楕円$C_n$型Jacksonの和公式
\begin{align}
&\sum_{k_1,\dots,k_n=0}^N\prod_{1\leq i< j\leq n}\frac{\theta(x_iq^{k_i-k_j}/x_j;p)q^{k_j}}{\theta(x_i/x_j;p)}\frac{\theta(ax_ix_jq^{k_i+k_j};p)}{\theta(ax_ix_jq^N;p)}\\
&\qquad\cdot\prod_{i=1}^n\frac{\theta(ax_i^2q^{2k_i};p)(ax_i^2,bx_i,cx_i,dx_i,ex_i,q^{-N};q,p)_{k_i}q^{k_i}}{\theta(ax_i^2;p)(q,ax_iq/b,ax_iq/c,ax_iq/d,ax_iq/e,ax_i^2q^{N+1};q,p)_{k_i}}\\
&=\prod_{i=1}^n\frac{(ax_i^2q,aq^{2-i}/bc,aq^{2-i}/bd,aq^{2-i}/cd;q,p)_N}{(aq^{2-n}/bcdx_i,ax_iq/b,ax_iq/c,ax_iq/d;q,p)_N}\qquad a^2q^{N-n+2}=bcde
\end{align}
を示す.
行列式の補題
まず, 補題を用意する.
$X_1,\dots,X_n,A_2,\dots,A_n,C$を不定元とする. $j=0,\dots,n-1$に対し, $P_j$は$0<|x|$で解析的で, $P_j(px)=(C/px^2)^jP_j(x)$を満たし, $P_j(C/X)=P_j(X)$を満たすとする. このとき,
\begin{align}
&\det\left(P_{j-1}(X_i)\prod_{k=j+1}^n\theta(A_kX_i,CA_k/X_i;p)\right)_{1\leq i,j\leq n}\\
&=\prod_{1\leq i< j\leq n} A_jX_j\theta(X_i/X_j,C/X_iX_j;p)\prod_{i=1}^nP_{i-1}(1/A_i)
\end{align}
が成り立つ.
左辺, 右辺をそれぞれ$X_i$の関数と見たものを$L(X_i),R(X_i)$と表すとする. このとき, 左辺の行列の$i$行目は
\begin{align}
&P_{j-1}(pX_i)\prod_{k=j+1}^n\theta(pA_kX_i,CA_k/pX_i;p)\\
&=(C/pX_i^2)^{j-1}P_{j-1}(X_i)\prod_{k=j+1}^n\frac{C}{pX_i^2}\theta(A_kX_i,CA_k/X_i;p)\\
&=(C/pX_i^2)^{n-1}P_{j-1}(X_i)\prod_{k=j+1}^n\theta(A_kX_i,CA_k/X_i;p)
\end{align}
となるので,
\begin{align}
L(pX_i)=(C/pX_i^2)^{n-1}L(X_i)
\end{align}
を満たす. また, 右辺は$X_i$に関する部分は
\begin{align}
&\prod_{i< j\leq n}X_j\theta(pX_i/X_j,C/pX_iX_j;p)\prod_{1\leq j< i}pX_i\theta(X_j/pX_i,C/pX_iX_j;p)\\
&=\prod_{i< j\leq n}\frac{C}{pX_i^2}X_j\theta(X_i/X_j,C/X_iX_j;p)\prod_{1\leq j< i}\frac{C}{pX_i^2}X_i\theta(X_j/pX_i,C/pX_iX_j;p)\\
&=(C/pX_i^2)^{n-1}\prod_{i< j\leq n}X_j\theta(X_i/X_j,C/X_iX_j;p)\prod_{1\leq j< i}X_i\theta(X_j/X_i,C/X_iX_j;p)
\end{align}
となるので,
\begin{align}
R(pX_i)=(C/pX_i^2)^{n-1}R(X_i)
\end{align}
となる. よって, $f(X_i):=L(X_i)/R(X_i)$とすると
\begin{align}
f(pX_i)=f(X_i)
\end{align}
が成り立つ. 次に, $f(X_i)$の極となりうるのは$R(X_i)$の$1$位の零点$X_i=p^kX_j, C/p^kX_j,k\in\ZZ$である. $X_i=p^kX_j$のとき, 左辺の$i$行目を$\ell(X_i)$と書いたとき, 先ほど見たように,
\begin{align}
\ell(pX_i)=(C/pX_i^2)^{n-1}\ell(X_i)
\end{align}
であるから, これを繰り返し用いれば, ある定数$c$があって
\begin{align}
\ell(p^kX_j)=c\ell(X_j)
\end{align}
と書ける. よって, $X_i=p^kX_j$のときは$i$行目が$j$行目の定数倍になるから$L=0$である. $\ell$は$X_i\mapsto C/X_i$に関して不変であるから$X_i=C/p^kX_j$の場合も同様に$L=0$になること分かる. よって, これらは極にならず, $f(X_i)$は$X_i$に関して$0<|X_i|$で正則であることが分かる. 全く同様に$f$は$0<|X_1|,\dots,|X_n|$において正則であることが分かる. 特にコンパクト集合$|p|\leq |X_1|,\dots,|X_n|\leq 1$の上で正則であり, $f(pX_i)=f(X_i)$を満たすことから, $X_1,\dots,X_n$について$\CC^{\times}$で有界である. よってLiouvilleの定理より, $f$は$X_1,\dots,X_n$に依存しない定数である. $X_i=1/A_i,1\leq i\leq n$とすると, 左辺の行列は上三角行列になり,
\begin{align}
&\det\left(P_{j-1}(1/A_i)\prod_{k=j+1}^n\theta(A_k/A_i,CA_iA_k;p)\right)_{1\leq i,j\leq n}\\
&=\prod_{i=1}^n\left(P_{i-1}(1/A_i)\prod_{k=i+1}^n\theta(A_k/A_i,CA_iA_k;p)\right)
\end{align}
と求められる. これは右辺において$X_i=1/A_i,1\leq i\leq n$とした値と等しい. よって, $f=1$であることが分かり, 示すべき等式が得られる.
特に, $A_i=Aq^{n-i}, P_i(X)=(BXq^{n-i-1},BCq^{n-i-1}/X;q,p)_i$とすると, 左辺は
\begin{align}
&\det\left((BX_iq^{n-j},BCq^{n-j}/X_i;q,p)_{j-1}\prod_{k=j+1}^n\theta(AX_iq^{n-k},CAq^{n-k}/X_i;p)\right)_{1\leq i,j\leq n}\\
&=\det\left(\frac{(AX_i,AC/X_i;q,p)_{n-j}}{(BX_i,BC/X_i;q,p)_{n-j}}\right)_{1\leq i,j\leq n}\prod_{i=1}^n(BX_i,BC/X_i;q,p)_{n-1}\\
\end{align}
となり, 右辺は
\begin{align}
&\prod_{1\leq i< j\leq n} AX_jq^{n-j}\theta(X_i/X_j,C/X_iX_j;p)\prod_{i=1}^n(B/A,ABCq^{2n-2i};q,p)_{i-1}\\
&=\prod_{1\leq i< j\leq n} AX_jq^{i-1}\theta(X_i/X_j,C/X_iX_j;p)\prod_{i=1}^n(B/A,ABCq^{2n-2i};q,p)_{i-1}
\end{align}
となる. つまり, 以下の系を得る.
$X_1,\dots,X_n,A,B,C$を不定元として,
\begin{align}
&\det\left(\frac{(AX_i,AC/X_i;q,p)_{n-j}}{(BX_i,BC/X_i;q,p)_{n-j}}\right)_{1\leq i,j\leq n}\\
&=\left(\prod_{1\leq i< j\leq n} AX_jq^{i-1}\theta(X_i/X_j,C/X_iX_j;p)\right)\prod_{i=1}^n\frac{(B/A,ABCq^{2n-2i};q,p)_{i-1}}{(BX_i,BC/X_i;q,p)_{n-1}}
\end{align}
が成り立つ.
$C_n$型Jacksonの和公式
$N$を非負整数として, $a^2q^{N-n+2}=bcde$が成り立つとする. このとき,
\begin{align}
&\sum_{k_1,\dots,k_n=0}^N\prod_{1\leq i< j\leq n}\frac{\theta(x_iq^{k_i-k_j}/x_j;p)q^{k_j}}{\theta(x_i/x_j;p)}\frac{\theta(ax_ix_jq^{k_i+k_j};p)}{\theta(ax_ix_jq^N;p)}\\
&\qquad\cdot\prod_{i=1}^n\frac{\theta(ax_i^2q^{2k_i};p)(ax_i^2,bx_i,cx_i,dx_i,ex_i,q^{-N};q,p)_{k_i}q^{k_i}}{\theta(ax_i^2;p)(q,ax_iq/b,ax_iq/c,ax_iq/d,ax_iq/e,ax_i^2q^{N+1};q,p)_{k_i}}\\
&=\prod_{i=1}^n\frac{(ax_i^2q,aq^{2-i}/bc,aq^{2-i}/bd,aq^{2-i}/cd;q,p)_N}{(aq^{2-n}/bcdx_i,ax_iq/b,ax_iq/c,ax_iq/d;q,p)_N}
\end{align}
が成り立つ.
系1において, $X_i=q^{-k_i}/x_i, C=a$とすると, $\theta(x;p)=-x\theta(1/x;p)$であることも用いて,
\begin{align}
&\det\left(\frac{(Aq^{-k_i}/x_i,Aax_iq^{k_i};q,p)_{n-j}}{(Bq^{-k_i}/x_i,Bax_iq^{k_i};q,p)_{n-j}}\right)_{1\leq i,j\leq n}\\
&=\left(\prod_{1\leq i< j\leq n} (-Aq^{i-k_i-1}/x_i)\theta(x_iq^{k_i-k_j}/x_j,ax_ix_jq^{k_i+k_j};p)\right)\prod_{i=1}^n\frac{(B/A,ABaq^{2n-2i};q,p)_{i-1}}{(Bq^{-k_i}/x_i,Bax_iq^{k_i};q,p)_{n-1}}
\end{align}
となる. さらに$A=b/a, B=q^{2-n}/c$とすると
\begin{align}
&\det\left(\frac{(bq^{-k_i}/ax_i,bx_iq^{k_i};q,p)_{n-j}}{(q^{2-n-k_i}/cx_i,ax_iq^{2-n+k_i}/c;q,p)_{n-j}}\right)_{1\leq i,j\leq n}\\
&=\left(\prod_{1\leq i< j\leq n} (-bq^{i-k_i-1}/ax_i)\theta(x_iq^{k_i-k_j}/x_j,ax_ix_jq^{k_i+k_j};p)\right)\prod_{i=1}^n\frac{(aq^{2-n}/bc,bq^{2+n-2i}/c;q,p)_{i-1}}{(q^{2-n-k_i}/cx_i,ax_iq^{2-n+k_i}/c;q,p)_{n-1}}
\end{align}
となる. これより,
\begin{align}
&\prod_{1\leq i< j\leq n} \theta(x_iq^{k_i-k_j}/x_j,ax_ix_jq^{k_i+k_j};p)q^{k_j}\\
&=\det\left(\frac{(bq^{-k_i}/ax_i,bx_iq^{k_i};q,p)_{n-j}}{(q^{2-n-k_i}/cx_i,ax_iq^{2-n+k_i}/c;q,p)_{n-j}}\right)_{1\leq i,j\leq n}\frac 1{\prod_{1\leq i< j\leq n} (-bq^{i-k_i-k_j-1}/ax_i)}\prod_{i=1}^n\frac{(q^{2-n-k_i}/cx_i,ax_iq^{2-n+k_i}/c;q,p)_{n-1}}{(aq^{2-n}/bc,bq^{2+n-2i}/c;q,p)_{i-1}}\\
&=\det\left(\frac{(bq^{-k_i}/ax_i,bx_iq^{k_i};q,p)_{n-j}(q^{2-n-k_i}/cx_i,ax_iq^{2-n+k_i}/c;q,p)_{n-1}}{(q^{2-n-k_i}/cx_i,ax_iq^{2-n+k_i}/c;q,p)_{n-j}}\right)_{1\leq i,j\leq n}\\
&\qquad\cdot\frac 1{\prod_{1\leq i< j\leq n} (-bq^{i-k_i-k_j-1}/ax_i)}\prod_{i=1}^n\frac{1}{(aq^{2-n}/bc,bq^{2+n-2i}/c;q,p)_{i-1}}
\end{align}
となる. ここで,
\begin{align}
&\frac{(bq^{-k_i}/ax_i,bx_iq^{k_i};q,p)_{n-j}(q^{2-n-k_i}/cx_i,ax_iq^{2-n+k_i}/c;q,p)_{n-1}}{(q^{2-n-k_i}/cx_i,ax_iq^{2-n+k_i}/c;q,p)_{n-j}}\\
&=\frac{(b/ax_i;q,p)_{n-j-k_i}(bx_i;q,p)_{n-j+k_i}(q^{2-n}/cx_i;q,p)_{n-1-k_i}(ax_iq^{2-n}/c;q,p)_{n-1+k_i}}{(b/ax_i;q,p)_{-k_i}(bx_i;q,p)_{k_i}(q^{2-n}/cx_i;q,p)_{n-j-k_i}(ax_iq^{2-n}/c;q,p)_{n-j+k_i}}\\
&=\frac{(b/ax_i,bx_i;q,p)_{n-j}(q^{2-n}/cx_i,ax_iq^{2-n}/c;q,p)_{n-1}}{(q^{2-n}/cx_i,ax_iq^{2-n}/c;q,p)_{n-j}}\\
&\qquad\cdot\frac{(bq^{n-j}/ax_i;q,p)_{-k_i}(bx_iq^{n-j};q,p)_{k_i}(q/cx_i;q,p)_{-k_i}(ax_iq/c;q,p)_{k_i}}{(b/ax_i;q,p)_{-k_i}(bx_i;q,p)_{k_i}(q^{2-j}/cx_i;q,p)_{-k_i}(ax_iq^{2-j}/c;q,p)_{k_i}}\\
&=\frac{(b/ax_i,bx_i;q,p)_{n-j}}{(q/cx_i,ax_iq/c;q,p)_{1-j}}q^{(1-n)k_i}\frac{(ax_iq/b,bx_iq^{n-j},cx_iq^{j-1},ax_iq/c;q,p)_{k_i}}{(ax_iq^{1+j-n}/b,bx_i,cx_i,ax_iq^{2-j}/c;q,p)_{k_i}}\\
&=(aq^2/c^2)^{j-1}q^{-2\binom j2}(cx_i,c/ax_i;q,p)_{j-1}(bx_i,b/ax_i;q,p)_{n-j}q^{(1-n)k_i}\frac{(ax_iq/b,bx_iq^{n-j},cx_iq^{j-1},ax_iq/c;q,p)_{k_i}}{(ax_iq^{1+j-n}/b,bx_i,cx_i,ax_iq^{2-j}/c;q,p)_{k_i}}
\end{align}
であるから,
\begin{align}
&\prod_{1\leq i< j\leq n} \theta(x_jq^{k_j-k_i}/x_i,ax_ix_jq^{k_i+k_j};p)q^{k_j}\\
&=\det\left((aq^2/c^2)^{j-1}q^{-2\binom j2}(cx_i,c/ax_i;q,p)_{j-1}(bx_i,b/ax_i;q,p)_{n-j}q^{(1-n)k_i}\frac{(ax_iq/b,bx_iq^{n-j},cx_iq^{j-1},ax_iq/c;q,p)_{k_i}}{(ax_iq^{1+j-n}/b,bx_i,cx_i,ax_iq^{2-j}/c;q,p)_{k_i}}\right)_{1\leq i,j\leq n}\\
&\qquad\cdot\frac 1{\prod_{1\leq i< j\leq n} (-bq^{i-k_i-k_j-1}/ax_i)}\prod_{i=1}^n\frac{1}{(aq^{2-n}/bc,bq^{2+n-2i}/c;q,p)_{i-1}}\\
&=\det\left((cx_i,c/ax_i;q,p)_{j-1}(bx_i,b/ax_i;q,p)_{n-j}\frac{(ax_iq/b,bx_iq^{n-j},cx_iq^{j-1},ax_iq/c;q,p)_{k_i}}{(ax_iq^{1+j-n}/b,bx_i,cx_i,ax_iq^{2-j}/c;q,p)_{k_i}}\right)_{1\leq i,j\leq n}\\
&\qquad\cdot\left(\prod_{1\leq i< j\leq n} \frac {-a^2x_i}{bc^2q^{3i-3}}\right)\prod_{i=1}^n\frac{1}{(aq^{2-n}/bc,bq^{2+n-2i}/c;q,p)_{i-1}}
\end{align}
となる. よって,
楕円Jacksonの和公式
を用いて,
\begin{align}
&\sum_{k_1,\dots,k_n=0}^N\prod_{1\leq i< j\leq n}\frac{\theta(x_iq^{k_i-k_j}/x_j;p)q^{k_j}}{\theta(x_i/x_j;p)}\frac{\theta(ax_ix_jq^{k_i+k_j};p)}{\theta(ax_ix_jq^N;p)}\\
&\qquad\cdot\prod_{i=1}^n\frac{\theta(ax_i^2q^{2k_i};p)(ax_i^2,bx_i,cx_i,dx_i,ex_i,q^{-N};q,p)_{k_i}q^{k_i}}{\theta(ax_i^2;p)(q,ax_iq/b,ax_iq/c,ax_iq/d,ax_iq/e,ax_i^2q^{N+1};q,p)_{k_i}}\\
&=\sum_{k_1,\dots,k_n=0}^N\prod_{i=1}^n\frac{\theta(ax_i^2q^{2k_i};p)(ax_i^2,bx_i,cx_i,dx_i,ex_i,q^{-N};q,p)_{k_i}}{\theta(ax_i^2;p)(q,ax_iq/b,ax_iq/c,ax_iq/d,ax_iq/e,ax_i^2q^{N+1};q,p)_{k_i}}q^{k_i}\\
&\qquad\cdot\det\left((cx_i,c/ax_i;q,p)_{j-1}(bx_i,b/ax_i;q,p)_{n-j}\frac{(ax_iq/b,bx_iq^{n-j},cx_iq^{j-1},ax_iq/c;q,p)_{k_i}}{(ax_iq^{1+j-n}/b,bx_i,cx_i,ax_iq^{2-j}/c;q,p)_{k_i}}\right)_{1\leq i,j\leq n}\\
&\qquad\cdot\left(\prod_{1\leq i< j\leq n} \frac {-a^2x_i}{bc^2q^{3i-3}\theta(x_i/x_j,ax_ix_jq^N;p)}\right)\prod_{i=1}^n\frac{1}{(aq^{2-n}/bc,bq^{2+n-2i}/c;q,p)_{i-1}}\\
&=\det\left((cx_i,c/ax_i;q,p)_{j-1}(bx_i,b/ax_i;q,p)_{n-j}\sum_{0\leq k}\frac{\theta(ax_i^2q^{2k};p)(ax_i^2,bx_iq^{n-j},cx_iq^{j-1},dx_i,ex_i,q^{-N};q,p)_{k}}{\theta(ax_i^2;p)(q,ax_iq^{1+j-n}/b,ax_iq^{2-j}/c,ax_iq/d,ax_iq/e,ax_i^2q^{N+1};q,p)_{k}}q^k\right)_{1\leq i,j\leq n}\\
&\qquad\cdot\left(\prod_{1\leq i< j\leq n} \frac {a^2x_j}{bc^2q^{3i-3}\theta(x_j/x_i,ax_ix_jq^N;p)}\right)\prod_{i=1}^n\frac{1}{(aq^{2-n}/bc,bq^{2+n-2i}/c;q,p)_{i-1}}\\
&=\det\left((cx_i,c/ax_i;q,p)_{j-1}(bx_i,b/ax_i;q,p)_{n-j}\frac{(ax_i^2q,aq^{2-n}/bc,aq^{1+j-n}/bd,aq^{2-j}/cd;q,p)_N}{(ax_iq^{1+j-n}/b,ax_iq^{2-j}/c,ax_iq/d,aq^{2-n}/bcdx_i;q,p)_N}\right)_{1\leq i,j\leq n}\\
&\qquad\cdot\left(\prod_{1\leq i< j\leq n} \frac {a^2x_j}{bc^2q^{3i-3}\theta(x_j/x_i,ax_ix_jq^N;p)}\right)\prod_{i=1}^n\frac{1}{(aq^{2-n}/bc,bq^{2+n-2i}/c;q,p)_{i-1}}\\
&=\det\left((cx_i,c/ax_i;q,p)_{j-1}(bx_i,b/ax_i;q,p)_{n-j}\frac{(ax_iq/b;q,p)_{j-n}(ax_iq/c;q,p)_{1-j}}{(ax_iq/b;q,p)_{N+j-n}(ax_iq/c;q,p)_{N+1-j}}\right)_{1\leq i,j\leq n}\\
&\qquad\cdot\left(\prod_{1\leq i< j\leq n} \frac {a^2x_j}{bc^2q^{3i-3}\theta(x_j/x_i,ax_ix_jq^N;p)}\right)\prod_{i=1}^n\frac{(ax_i^2q,aq^{2-n}/bc,aq^{2-i}/bd,aq^{2-i}/cd;q,p)_N}{(aq^{2-n}/bc,bq^{2+n-2i}/c;q,p)_{i-1}(ax_iq/d,aq^{2-n}/bcdx_i;q,p)_N}\\
&=\det\left((cx_i,cq^{-N}/ax_i;q,p)_{j-1}(bx_i,bq^{-N}/ax_i;q,p)_{n-j}q^{(n-1)N}\right)_{1\leq i,j\leq n}\\
&\qquad\cdot\left(\prod_{1\leq i< j\leq n} \frac {a^2x_j}{bc^2q^{3i-3}\theta(x_j/x_i,ax_ix_jq^N;p)}\right)\prod_{i=1}^n\frac{(ax_i^2q,aq^{2-i}/bc,aq^{2-i}/bd,aq^{2-i}/cd;q,p)_N}{(aq^{2+N-n}/bc,bq^{2+n-2i}/c;q,p)_{i-1}(ax_iq/b,ax_iq/c,ax_iq/d,aq^{2-n}/bcdx_i;q,p)_N}\\
\end{align}
ここで, 系1において, $X_i=1/x_i,A=bq^{-N}/a,B=q^{2-n}/c,C=aq^N$としたものを用いると,
\begin{align}
&\det\left((cx_i,cq^{-N}/ax_i;q,p)_{j-1}(bx_i,bq^{-N}/ax_i;q,p)_{n-j}\right)_{1\leq i,j\leq n}\\
&=\det\left((cx_i,cq^{-N}/ax_i;q,p)_{n-1}(c^2q^{-N-2}/a)^{j-n}q^{2\binom{j}2-2\binom{n}2}\frac{(bx_i,bq^{-N}/ax_i;q,p)_{n-j}}{(q^{2-n}/cx_i,ax_iq^{2+N-n}/c;q,p)_{n-j}}\right)_{1\leq i,j\leq n}\\
&=\det\left(\frac{(bx_i,bq^{-N}/ax_i;q,p)_{n-j}}{(q^{2-n}/cx_i,ax_iq^{2+N-n}/c;q,p)_{n-j}}\right)_{1\leq i,j\leq n}\prod_{i=1}^n(cx_i,cq^{-N}/ax_i;q,p)_{n-1}(c^2q^{-N-2}/a)^{i-n}q^{2\binom{i}2-2\binom{n}2}\\
&=\left(\prod_{1\leq i< j\leq n}(bq^{i-N-1}/ax_j)\theta(x_j/x_i,ax_ix_jq^N;p)\right)\prod_{i=1}^n\frac{(aq^{2+N-n}/bc,bq^{2+n-2i}/c;q,p)_{i-1}(cx_i,cq^{-N}/ax_i;q,p)_{n-1}(c^2q^{-N-2}/a)^{i-n}q^{2\binom{i}2-2\binom{n}2}}{(q^{2-n}/cx_i,ax_iq^{2+N-n}/c;q,p)_{n-1}}\\
&=\left(\prod_{1\leq i< j\leq n}(bq^{i-N-1}/ax_j)\theta(x_j/x_i,ax_ix_jq^N;p)\right)\prod_{i=1}^n{(aq^{2+N-n}/bc,bq^{2+n-2i}/c;q,p)_{i-1}(c^2q^{-N-2}/a)^{i-1}q^{2\binom{i}2}}
\end{align}
となるから, これを代入すると,
\begin{align}
&\sum_{k_1,\dots,k_n=0}^N\prod_{1\leq i< j\leq n}\frac{\theta(x_iq^{k_i-k_j}/x_j;p)q^{k_j}}{\theta(x_i/x_j;p)}\frac{\theta(ax_ix_jq^{k_i+k_j};p)}{\theta(ax_ix_jq^N;p)}\\
&\qquad\cdot\prod_{i=1}^n\frac{\theta(ax_i^2q^{2k_i};p)(ax_i^2,bx_i,cx_i,dx_i,ex_i,q^{-N};q,p)_{k_i}q^{k_i}}{\theta(ax_i^2;p)(q,ax_iq/b,ax_iq/c,ax_iq/d,ax_iq/e,ax_i^2q^{N+1};q,p)_{k_i}}\\
&=q^{n(n-1)N}\left(\prod_{1\leq i< j\leq n}(bq^{i-N-1}/ax_j)\theta(x_j/x_i,ax_ix_jq^N;p)\right)\prod_{i=1}^n{(aq^{2+N-n}/bc,bq^{2+n-2i}/c;q,p)_{i-1}(c^2q^{-N-2}/a)^{i-1}q^{2\binom{i}2}}\\
&\qquad\cdot\left(\prod_{1\leq i< j\leq n} \frac {a^2x_j}{bc^2q^{3i-3}\theta(x_j/x_i,ax_ix_jq^N;p)}\right)\prod_{i=1}^n\frac{(ax_i^2q,aq^{2-i}/bc,aq^{2-i}/bd,aq^{2-i}/cd;q,p)_N}{(aq^{2+N-n}/bc,bq^{2+n-2i}/c;q,p)_{i-1}(ax_iq/b,ax_iq/c,ax_iq/d,aq^{2-n}/bcdx_i;q,p)_N}\\
&=q^{n(n-1)N}\left(\prod_{1\leq i< j\leq n}\frac {a}{c^2q^{N+2i-2}}\right)\prod_{i=1}^n(c^2q^{-N-2}/a)^{i-1}q^{2\binom{i}2}\\
&\qquad\cdot\prod_{i=1}^n\frac{(ax_i^2q,aq^{2-i}/bc,aq^{2-i}/bd,aq^{2-i}/cd;q,p)_N}{(ax_iq/b,ax_iq/c,ax_iq/d,aq^{2-n}/bcdx_i;q,p)_N}\\
&=q^{n(n-1)N}\left(\prod_{1\leq i< j\leq n}\frac {1}{q^{2N}}\right)\prod_{i=1}^n\frac{(ax_i^2q,aq^{2-i}/bc,aq^{2-i}/bd,aq^{2-i}/cd;q,p)_N}{(ax_iq/b,ax_iq/c,ax_iq/d,aq^{2-n}/bcdx_i;q,p)_N}\\
&=\prod_{i=1}^n\frac{(ax_i^2q,aq^{2-i}/bc,aq^{2-i}/bd,aq^{2-i}/cd;q,p)_N}{(ax_iq/b,ax_iq/c,ax_iq/d,aq^{2-n}/bcdx_i;q,p)_N}
\end{align}
となって示すべき等式を得る.
