連続q-Jacobi多項式の線形化公式4: 結果の整理
1つ目の記事
で$x:=\cos\theta$
\begin{align}
r_n(x)=r_n(x;a,b,c,d)&:=\Q43{q^{-n},abcdq^{n-1},ae^{i\theta},ae^{-i\theta}}{ab,ac,ad}q\\
w(x)=w(x;a,b,c,d)&:=\frac 1{\sqrt{1-x^2}}\frac{(e^{2i\theta},e^{-2i\theta};q)_{\infty}}{(ae^{i\theta},ae^{-i\theta},be^{i\theta},be^{-i\theta},ce^{i\theta},ce^{-i\theta},de^{i\theta},de^{-i\theta};q)_{\infty}}\\
g_k(m,n;a,b,c,d)&:=\int_{-1}^1r_m(x)r_n(x)r_k(x)w(x)\,dx\\
h_0&:=\frac{2\pi(abcd;q)_{\infty}}{(q,ab,ac,ad,bc,bd,cd;q)_{\infty}}
\end{align}
として, $0\leq s\leq n,\quad0\leq j\leq 2n-2s$のとき,
\begin{align}
g_{s+j}&=g_{s+j}(n-s,n;a,b,c,d)\\
&=h_0\frac{q^{-js}(abcdq^{n-s-1};q)_{n-s}(q^{j+1},abcdq^{s+j-1};q)_{s}(q,bc,bd,cd;q)_n}{(ab,ac,ad;q)_{n-s}(q,ab,ac,ad;q)_{s}(abcd;q)_{2n}}a^{2n}F_j\\
F_j&:=\sum_{\lambda=0}^{j}\frac{(q^{-j},abcdq^{2s+j-1},q^{s-n},q^{1+s-n}/bc,q^{1+s-n}/bd,q^{1+s-n}/cd;q)_{\lambda}}{(q,abq^s,acq^s,adq^s,q^{2+2s-2n}/abcd,q^{1+s-2n}/abcd;q)_{\lambda}}q^{\lambda}\\
&\qquad\cdot\Q{10}9{abcdq^{2n-\lambda-s-1},q\sqrt{abcdq^{2n-\lambda-s-1}},-q\sqrt{abcdq^{2n-\lambda-s-1}},abq^n,acq^n,adq^n,abcdq^{n-1},q^{-\lambda-s},abcdq^{2n-2s-1-\lambda},q^{-\lambda}}{\sqrt{abcdq^{2n-\lambda-s-1}},-\sqrt{abcdq^{2n-\lambda-s-1}},bcq^{n-\lambda-s},bdq^{n-\lambda-s},cdq^{n-s-\lambda},q^{1+n-\lambda-s},abcdq^{2n},q^{s+1},abcdq^{2n-s}}{\frac{q^2}{a^2}}
\end{align}
と与えられることを示した. ここで, $a=q^{\frac 12},d=-q^{\frac 12}$とすると,
\begin{align}
g_{s+j}&=g_{s+j}(n-s,n;q^{\frac 12},b,c,-q^{\frac 12})\\
&=h_0\frac{q^{-js}(-bcq^{n-s};q)_{n-s}(q^{j+1},-bcq^{s+j};q)_{s}(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_n}{(bq^{\frac 12},cq^{\frac 12},-q;q)_{n-s}(q,bq^{\frac 12},cq^{\frac 12},-q;q)_{s}(-bcq;q)_{2n}}q^{n}F_j\\
h_0&=\frac{2\pi(-bcq;q)_{\infty}}{(q,bq^{\frac 12},cq^{\frac 12},-q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_{\infty}}\\
&=\frac{2\pi(-bcq;q)_{\infty}}{(bc;q)_{\infty}(q^2,b^2q,c^2q;q^2)_{\infty}}
\end{align}
3つ目の記事
で
\begin{align}
F_j&=\frac{(-cq^{s+\frac 12},q^{2s-2n},b^2c^2q^{2n},-b/c;q)_j}{(bq^{s+\frac 12},-bcq^{2n+1},-q^{1+2s-2n}/bc,c^2q^{2s+1};q)_j}\left(\frac{q^{\frac 32+s}}{b}\right)^{j}\\
&\qquad\cdot\Q{10}9{c^2q^{2s},cq^{s+1},-cq^{s+1},c^2,c^2q^{2n+1},-bcq^{2s+j},-bcq^{2s+j+1},q^{1+2s-2n}/b^2,q^{-j},q^{1-j}}{cq^s,-cq^s,q^{2s+2},q^{1+2s-2n},-cq^{2-j}/b,-cq^{1-j}/b,b^2c^2q^{2n+1},c^2q^{2s+2+j},c^2q^{2s+1+j}}{q^2;q^2}
\end{align}
であることを示したので, これを代入すると,
\begin{align}
g_{s+j}&=h_0\frac{q^{-js}(-bcq^{n-s};q)_{n-s}(q^{j+1},-bcq^{s+j};q)_{s}(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_n}{(bq^{\frac 12},cq^{\frac 12},-q;q)_{n-s}(q,bq^{\frac 12},cq^{\frac 12},-q;q)_{s}(-bcq;q)_{2n}}q^{n}\\
&\qquad\cdot\frac{(-cq^{s+\frac 12},q^{2s-2n},b^2c^2q^{2n},-b/c;q)_j}{(bq^{s+\frac 12},-bcq^{2n+1},-q^{1+2s-2n}/bc,c^2q^{2s+1};q)_j}\left(\frac{q^{\frac 32+s}}{b}\right)^{j}\\
&\qquad\cdot\Q{10}9{c^2q^{2s},cq^{s+1},-cq^{s+1},c^2,c^2q^{2n+1},-bcq^{2s+j},-bcq^{2s+j+1},q^{1+2s-2n}/b^2,q^{-j},q^{1-j}}{cq^s,-cq^s,q^{2s+2},q^{1+2s-2n},-cq^{2-j}/b,-cq^{1-j}/b,b^2c^2q^{2n+1},c^2q^{2s+2+j},c^2q^{2s+1+j}}{q^2;q^2}
\end{align}
ここで, 係数は
\begin{align}
&\frac{q^{-js}(-bcq^{n-s};q)_{n-s}(q^{j+1},-bcq^{s+j};q)_{s}(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_n}{(bq^{\frac 12},cq^{\frac 12},-q;q)_{n-s}(q,bq^{\frac 12},cq^{\frac 12},-q;q)_{s}(-bcq;q)_{2n}}q^{n}\\
&\qquad\cdot\frac{(-cq^{s+\frac 12},q^{2s-2n},b^2c^2q^{2n},-b/c;q)_j}{(bq^{s+\frac 12},-bcq^{2n+1},-q^{1+2s-2n}/bc,c^2q^{2s+1};q)_j}\left(\frac{q^{\frac 32+s}}{b}\right)^{j}\\
&=\frac{(-bc;q)_{2n-2s}(q^{j+1},-bcq^{s+j};q)_{s}(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_n}{(-bc,bq^{\frac 12},cq^{\frac 12},-q;q)_{n-s}(q,cq^{\frac 12},-q;q)_{s}(bq^{\frac 12};q)_{s+j}(-bcq;q)_{2n+j}}q^{n}\\
&\qquad\cdot\frac{(-cq^{\frac 12};q)_{s+j}(q^{2s-2n},b^2c^2q^{2n},-b/c;q)_j}{(-cq^{\frac 12};q)_s(-q^{1+2s-2n}/bc,c^2q^{2s+1};q)_j}\left(\frac{q^{\frac 32}}{b}\right)^{j}\\
&=\frac{(-bc;q)_{2n-2s}(q,-cq^{\frac 12};q)_{s+j}(-bc;q)_{2s+j}(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_n(b^2c^2;q)_{2n+j}}{(-bc,bq^{\frac 12},cq^{\frac 12},-q;q)_{n-s}(q^2;q^2)_s(c^2q;q^2)_{s}(-bc,bq^{\frac 12};q)_{s+j}(-bcq;q)_{2n+j}(b^2c^2;q)_{2n}}q^{n}\\
&\qquad\cdot\frac{(q^{2s-2n},-b/c;q)_j(c^2q;q)_{2s}}{(q,-q^{1+2s-2n}/bc;q)_j(c^2q;q)_{2s+j}}\left(\frac{q^{\frac 32}}{b}\right)^{j}\\
&=\frac{(-bc;q)_{2n-2s-j}(q,-cq^{\frac 12};q)_{s+j}(-bc;q)_{2s+j}(c^2q^2;q^2)_{s}(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_n(b^2c^2;q)_{2n+j}(q;q)_{2n-2s}(-b/c;q)_j}{(q;q)_{2n-2s-j}(-bc,bq^{\frac 12},cq^{\frac 12},-q;q)_{n-s}(c^2q;q)_{2s+j}(q^2;q^2)_s(-bc,bq^{\frac 12};q)_{s+j}(-bcq;q)_{2n+j}(b^2c^2;q)_{2n}(q;q)_j}q^{n}(-cq^{\frac 12})^j
\end{align}
となる. よって,
\begin{align}
g_{s+j}(n-s,n)&=g_{s+j}(n-s,n;q^{\frac 12},b,c,-q^{\frac 12})\\
&=h_0\frac{(-bc;q)_{2n-2s-j}(q,-cq^{\frac 12};q)_{s+j}(-bc;q)_{2s+j}(c^2q^2;q^2)_{s}(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_n(b^2c^2;q)_{2n+j}(q;q)_{2n-2s}(-b/c;q)_j}{(q;q)_{2n-2s-j}(-bc,bq^{\frac 12};q)_{s+j}(-bc,bq^{\frac 12},cq^{\frac 12},-q;q)_{n-s}(c^2q;q)_{2s+j}(q^2;q^2)_s(-bcq;q)_{2n+j}(b^2c^2;q)_{2n}(q;q)_j}q^{n}(-cq^{\frac 12})^j\\
&\qquad\cdot\Q{10}9{c^2q^{2s},cq^{s+1},-cq^{s+1},c^2,c^2q^{2n+1},-bcq^{2s+j},-bcq^{2s+j+1},q^{1+2s-2n}/b^2,q^{-j},q^{1-j}}{cq^s,-cq^s,q^{2s+2},q^{1+2s-2n},-cq^{2-j}/b,-cq^{1-j}/b,b^2c^2q^{2n+1},c^2q^{2s+2+j},c^2q^{2s+1+j}}{q^2;q^2}
\end{align}
を得る.
Askey-Wilson多項式の直交性
\begin{align}
\int_{-1}^1r_m(x;a,b,c,d)r_n(x;a,b,c,d)w(x;a,b,c,d)\,dx&=h_0\frac{a^{2n}(q,bc,bd,cd;q)_n}{(abcd/q,ab,ac,ad;q)_n}\frac{1-abcd/q}{1-abcdq^{2n-1}}
\end{align}
を特殊化すると, 連続$q$-Jacobi多項式の直交性
\begin{align}
\int_{-1}^1r_m(x;q^{\frac 12},b,c,-q^{\frac 12})r_n(x;q^{\frac 12},b,c,-q^{\frac 12})w(x;q^{\frac 12},b,c,-q^{\frac 12})\,dx&=h_0\frac{q^n(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_n}{(-q,-bc,bq^{\frac 12},cq^{\frac 12};q)_n}\frac{1+bc}{1+bcq^{2n}}
\end{align}
を得る. これを用いると, 線形化公式
\begin{align}
r_{n-s}(x;q^{\frac 12},b,c,-q^{\frac 12})r_n(x;q^{\frac 12},b,c,-q^{\frac 12})&=\sum_{j=0}^{2n-2s}b_jp_{s+j}(x;q^{\frac 12},b,c,-q^{\frac 12})
\end{align}
の係数$b_j$を
\begin{align}
b_j&=\left(h_0\frac{q^{s+j}(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_{s+j}}{(-q,-bc,bq^{\frac 12},cq^{\frac 12};q)_{s+j}}\frac{1+bc}{1+bcq^{2s+2j}}\right)^{-1}g_{s+j}(n-s,n)\\
&=\frac{1+bcq^{2s+2j}}{1+bc}\frac{(-bc;q)_{2n-2s-j}(-q,cq^{\frac 12};q)_{s+j}(-bc;q)_{2s+j}(c^2q^2;q^2)_{s}(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_n(b^2c^2;q)_{2n+j}(q;q)_{2n-2s}(-b/c;q)_j}{(q;q)_{2n-2s-j}(bc,-bq^{\frac 12};q)_{s+j}(-bc,bq^{\frac 12},cq^{\frac 12},-q;q)_{n-s}(c^2q;q)_{2s+j}(q^2;q^2)_s(-bcq;q)_{2n+j}(b^2c^2;q)_{2n}(q;q)_j}q^{n-s}(-cq^{-\frac 12})^j\\
&\qquad\cdot\Q{10}9{c^2q^{2s},cq^{s+1},-cq^{s+1},c^2,c^2q^{2n+1},-bcq^{2s+j},-bcq^{2s+j+1},q^{1+2s-2n}/b^2,q^{-j},q^{1-j}}{cq^s,-cq^s,q^{2s+2},q^{1+2s-2n},-cq^{2-j}/b,-cq^{1-j}/b,b^2c^2q^{2n+1},c^2q^{2s+2+j},c^2q^{2s+1+j}}{q^2;q^2}
\end{align}
と表すことができる. $b_j$の係数は
\begin{align}
&\frac{1+bcq^{2s+2j}}{1+bc}\frac{(-bc;q)_{2n-2s-j}(-q,cq^{\frac 12};q)_{s+j}(-bc;q)_{2s+j}(c^2q^2;q^2)_{s}(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_n(b^2c^2;q)_{2n+j}(q;q)_{2n-2s}(-b/c;q)_j}{(q;q)_{2n-2s-j}(bc,-bq^{\frac 12};q)_{s+j}(-bc,bq^{\frac 12},cq^{\frac 12},-q;q)_{n-s}(c^2q;q)_{2s+j}(q^2;q^2)_s(-bcq;q)_{2n+j}(b^2c^2;q)_{2n}(q;q)_j}q^{n-s}(-cq^{-\frac 12})^j\\
&=\frac{(-bc;q)_{2n-2s}(-q,cq^{\frac 12};q)_{s}(-bc;q)_{2s}(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_n}{(bc,-bq^{\frac 12};q)_{s}(-bc,bq^{\frac 12},cq^{\frac 12},-q;q)_{n-s}(q^2,c^2q;q^2)_s(-bcq;q)_{2n}}q^{n-s}\\
&\qquad\cdot\frac{(q^{2s-2n},-q^{s+1},cq^{s+\frac 12},-bcq^{2s},b^2c^2q^{2n},-b/c;q)_j}{(-q^{1+2s-2n}/bc,bcq^s,-bq^{s+\frac 12},c^2q^{2s+1},-bcq^{2n+1},q;q)_j}\left(\frac{q^{\frac 12}}b\right)^j\frac{1+bcq^{2s+2j}}{1+bc}\\
&=\frac{(-bc;q)_{2s}(-bc;q)_{2n-2s}(q,-bq^{\frac 12},-cq^{\frac 12},bc;q)_n}{(-bcq;q)_{2n}(q,-bq^{\frac 12},-cq^{\frac 12},bc;q)_{s}(-q,bq^{\frac 12},cq^{\frac 12},-bc;q)_{n-s}}\\
&\qquad\cdot\frac{(1+bcq^{2s+2j})(-bcq^{2s},-q^{s+1},cq^{s+\frac 12},-b/c,b^2c^2q^{2n},q^{2s-2n};q)_j}{(1+bc)(q,bcq^s,-bq^{s+\frac 12},c^2q^{2s+1},-q^{1+2s-2n}/bc,-bcq^{2n+1};q)_j}q^{n-s+j/2}b^{-j}
\end{align}
と表すこともできる. つまり,
\begin{align}
b_j&=\frac{(-bc;q)_{2s}(-bc;q)_{2n-2s}(q,-bq^{\frac 12},-cq^{\frac 12},bc;q)_n}{(-bcq;q)_{2n}(q,-bq^{\frac 12},-cq^{\frac 12},bc;q)_{s}(-q,bq^{\frac 12},cq^{\frac 12},-bc;q)_{n-s}}\\
&\qquad\cdot\frac{(1+bcq^{2s+2j})(-bcq^{2s},-q^{s+1},cq^{s+\frac 12},-b/c,b^2c^2q^{2n},q^{2s-2n};q)_j}{(1+bc)(q,bcq^s,-bq^{s+\frac 12},c^2q^{2s+1},-q^{1+2s-2n}/bc,-bcq^{2n+1};q)_j}q^{n-s+j/2}b^{-j}\\
&\qquad\cdot\Q{10}9{c^2q^{2s},cq^{s+1},-cq^{s+1},c^2,c^2q^{2n+1},-bcq^{2s+j},-bcq^{2s+j+1},q^{1+2s-2n}/b^2,q^{-j},q^{1-j}}{cq^s,-cq^s,q^{2s+2},q^{1+2s-2n},-cq^{2-j}/b,-cq^{1-j}/b,b^2c^2q^{2n+1},c^2q^{2s+2+j},c^2q^{2s+1+j}}{q^2;q^2}
\end{align}
を得る. $b,c,s,j$を$a,-b,m,k$に置き換えたものがGasper-Rahmanの本のEx.8.24に載っているものである. 途中の$K_{\lambda}, F_j$の計算において, Rahmanの論文に書かれている式とは微妙に異なる形になったが, 同じ方針で計算を進めることがで正しい結果が得られたことになる.
対称的な表示
ここからはRahmanの論文には書かれていないことであるが, 先ほどの$g_{s+j}(n-s,n)$の表示
\begin{align}
g_{s+j}(n-s,n)&=h_0\frac{(-bc;q)_{2n-2s-j}(q,-cq^{\frac 12};q)_{s+j}(-bc;q)_{2s+j}(c^2q^2;q^2)_{s}(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_n(b^2c^2;q)_{2n+j}(q;q)_{2n-2s}(-b/c;q)_j}{(q;q)_{2n-2s-j}(-bc,bq^{\frac 12};q)_{s+j}(-bc,bq^{\frac 12},cq^{\frac 12},-q;q)_{n-s}(c^2q;q)_{2s+j}(q^2;q^2)_s(-bcq;q)_{2n+j}(b^2c^2;q)_{2n}(q;q)_j}q^{n}(-cq^{\frac 12})^j\\
&\qquad\cdot\Q{10}9{c^2q^{2s},cq^{s+1},-cq^{s+1},c^2,c^2q^{2n+1},-bcq^{2s+j},-bcq^{2s+j+1},q^{1+2s-2n}/b^2,q^{-j},q^{1-j}}{cq^s,-cq^s,q^{2s+2},q^{1+2s-2n},-cq^{2-j}/b,-cq^{1-j}/b,b^2c^2q^{2n+1},c^2q^{2s+2+j},c^2q^{2s+1+j}}{q^2;q^2}
\end{align}
において, $k=s+j, m=n-s$として書き換えると,
\begin{align}
g_k(m,n)&=h_0\frac{(-bc;q)_{m+n-k}(q,-cq^{\frac 12};q)_k(-bc;q)_{n+k-m}(c^2q^2;q^2)_{n-m}(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_n(b^2c^2;q)_{m+n+k}(q;q)_{2m}(-b/c;q)_{m+k-n}}{(q;q)_{m+n-k}(-bc,bq^{\frac 12};q)_{k}(-bc,bq^{\frac 12},cq^{\frac 12},-q;q)_{m}(c^2q;q)_{n+k-m}(q^2;q^2)_{n-m}(-bcq;q)_{m+n+k}(b^2c^2;q)_{2n}(q;q)_{m+k-n}}q^{n}(-cq^{\frac 12})^{m+k-n}\\
&\qquad\cdot\Q{10}9{c^2q^{2n-2m},cq^{n-m+1},-cq^{n-m+1},c^2,c^2q^{2n+1},-bcq^{n+k-m},-bcq^{n+k-m+1},q^{1-2m}/b^2,q^{n-m-k},q^{1+n-m-k}}{cq^{n-m},-cq^{n-m},q^{2n-2m+2},q^{1-2m},-cq^{2+n-m-k}/b,-cq^{1+n-m-k}/b,b^2c^2q^{2n+1},c^2q^{2+n+k-m},c^2q^{1+n+k-m}}{q^2;q^2}
\end{align}
を得る. この左辺は$k,m,n$に関して対称なので, 右辺の${}_{10}\phi_9$も対称な形に変換したいところである.
Baileyの変換公式
より, $m+k-n=2r$を偶数として,
\begin{align}
&\Q{10}9{c^2q^{2n-2m},cq^{n-m+1},-cq^{n-m+1},c^2,c^2q^{2n+1},-bcq^{n+k-m},-bcq^{n+k-m+1},q^{1-2m}/b^2,q^{n-m-k},q^{1+n-m-k}}{cq^{n-m},-cq^{n-m},q^{2n-2m+2},q^{1-2m},-cq^{2+n-m-k}/b,-cq^{1+n-m-k}/b,b^2c^2q^{2n+1},c^2q^{2+n+k-m},c^2q^{1+n+k-m}}{q^2;q^2}\\
&=\frac{(c^2q^{2n-2m+2},b^2c^2q^{2n+r},q^{2+2n-2m-2k+r}/b^2,q^{2+2n-2k};q^2)_r}{(c^2q^{1+2n-2m+r},b^2c^2q^{1+2n},q^{3+2n-2m-2k}/b^2,q^{1+2n-2k+r};q^2)_r}\\
&\qquad\cdot\Q{10}9{q^{1+2n-2m-2k}/b^2,q\sqrt{q^{1+2n-2m-2k}/b^2},-q\sqrt{q^{1+2n-2m-2k}/b^2},q^{1-2k}/b^2,c^2q^{2n+1},-q^{1+n-m-k}/bc,-q^{2+n-m-k}/bc,q^{1-2m}/b^2,q^{n-m-k},q^{1+n-m-k}}{\sqrt{q^{1+2n-2m-2k}/b^2},-\sqrt{q^{1+2n-2m-2k}/b^2},q^{2n-2m+2},q^{2-2m-2k}/b^2c^2,-cq^{2+n-m-k}/b,-cq^{1+n-m-k}/b,q^{2n-2k+2},q^{3+n-m-k}/b^2,q^{2+n-m-k}/b^2}{q^2;q^2}
\end{align}
となる. ここで, 係数は
\begin{align}
&\frac{(c^2q^{2n-2m+2},b^2c^2q^{n+k-m+r},q^{2+2n-2m-2k+r}/b^2,q^{2+2n-2k};q^2)_r}{(c^2q^{1+2n-2m+r},b^2c^2q^{1+n+k-m},q^{3+2n-2m-2k}/b^2,q^{1+2n-2k+r};q^2)_r}\\
&=\frac{(c^2q^{2n-2m+1},q^{1+2n-2k};q)_{m+k-n}(b^2c^2q^{2n},q^{2+2n-2m-2k}/b^2;q^2)_{m+k-n}}{(c^2q^{2n-2m+1},q^{1+2n-2k};q^2)_{m+k-n}(b^2c^2q^{2n},q^{2+2n-2m-2k}/b^2;q)_{m+k-n}}\\
&=\frac{(c^2q;q)_{n+k-m}(c^2q;q^2)_{n-m}(q;q)_{m+n-k}(q;q^2)_{n-k}(b^2c^2;q^2)_{m+k}(b^2c^2;q)_{2n}(q^2/b^2;q)_{2n-2m-2k}}{(c^2q;q)_{2n-2m}(c^2q;q^2)_k(q;q)_{2n-2k}(q;q^2)_{m}(b^2c^2;q^2)_n(b^2c^2;q)_{m+n+k}(q^2/b^2;q)_{n-m-k}(q^2/b^2;q^2)_{n-m-k}}\\
&=\frac{(c^2q;q)_{n+k-m}(q;q)_{m+n-k}(b^2c^2;q^2)_{m+k}(b^2c^2q;q^2)_{n}(q^3/b^2;q^2)_{n-m-k}}{(c^2q^2;q^2)_{n-m}(c^2q;q^2)_k(q^2;q^2)_{n-k}(q;q^2)_{m}(b^2c^2;q)_{m+n+k}(q^2/b^2;q)_{n-m-k}}\\
&=\frac{(c^2q;q)_{n+k-m}(q;q)_{m+n-k}(b^2c^2;q^2)_{m+k}(b^2c^2q;q^2)_{n}(b^2/q;q)_{m+k-n}}{(c^2q^2;q^2)_{n-m}(c^2q;q^2)_k(q^2;q^2)_{n-k}(q;q^2)_{m}(b^2c^2;q)_{m+n+k}(b^2/q;q^2)_{m+k-n}}q^{\binom{m+k-n}2}
\end{align}
となる. これは$m+k-n$が奇数の場合も全く同様である. よって,
\begin{align}
g_k(m,n)&=h_0\frac{(-bc;q)_{m+n-k}(q,-cq^{\frac 12};q)_k(-bc;q)_{n+k-m}(c^2q^2;q^2)_{n-m}(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_n(b^2c^2;q)_{m+n+k}(q;q)_{2m}(-b/c;q)_{m+k-n}}{(q;q)_{m+n-k}(-bc,bq^{\frac 12};q)_{k}(-bc,bq^{\frac 12},cq^{\frac 12},-q;q)_{m}(c^2q;q)_{n+k-m}(q^2;q^2)_{n-m}(-bcq;q)_{m+n+k}(b^2c^2;q)_{2n}(q;q)_{m+k-n}}q^{n}(-cq^{\frac 12})^{m+k-n}\\
&\qquad\cdot\frac{(c^2q;q)_{n+k-m}(q;q)_{m+n-k}(b^2c^2;q^2)_{m+k}(b^2c^2q;q^2)_{n}(b^2/q;q)_{m+k-n}}{(c^2q^2;q^2)_{n-m}(c^2q;q^2)_k(q^2;q^2)_{n-k}(q;q^2)_{m}(b^2c^2;q)_{m+n+k}(b^2/q;q^2)_{m+k-n}}q^{\binom{m+k-n}2}\\
&\qquad\cdot\Q{10}9{q^{1+2n-2m-2k}/b^2,q\sqrt{q^{1+2n-2m-2k}/b^2},-q\sqrt{q^{1+2n-2m-2k}/b^2},q^{1-2k}/b^2,c^2q^{2n+1},-q^{1+n-m-k}/bc,-q^{2+n-m-k}/bc,q^{1-2m}/b^2,q^{n-m-k},q^{1+n-m-k}}{\sqrt{q^{1+2n-2m-2k}/b^2},-\sqrt{q^{1+2n-2m-2k}/b^2},q^{2n-2m+2},q^{2-2m-2k}/b^2c^2,-cq^{2+n-m-k}/b,-cq^{1+n-m-k}/b,q^{2n-2k+2},q^{3+n-m-k}/b^2,q^{2+n-m-k}/b^2}{q^2;q^2}\\
&=h_0\frac{(-bc;q)_{m+n-k}(q;q)_k(-bc;q)_{n+k-m}(q;q)_{m}(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_n(-b/c;q)_{m+k-n}(b^2c^2;q^2)_{m+k}(b^2/q;q)_{m+k-n}}{(-bc,bq^{\frac 12},cq^{\frac 12};q)_{k}(q^2;q^2)_{n-k}(-bc,bq^{\frac 12},cq^{\frac 12};q)_{m}(q^2;q^2)_{n-m}(-bcq;q)_{m+n+k}(b^2c^2;q^2)_{n}(q;q)_{m+k-n}(b^2/q;q^2)_{m+k-n}}(-cq^{\frac 12})^{m+k-n}q^{n+\binom{m+k-n}2}\\
&\qquad\cdot\Q{10}9{q^{1+2n-2m-2k}/b^2,q\sqrt{q^{1+2n-2m-2k}/b^2},-q\sqrt{q^{1+2n-2m-2k}/b^2},q^{1-2k}/b^2,c^2q^{2n+1},-q^{1+n-m-k}/bc,-q^{2+n-m-k}/bc,q^{1-2m}/b^2,q^{n-m-k},q^{1+n-m-k}}{\sqrt{q^{1+2n-2m-2k}/b^2},-\sqrt{q^{1+2n-2m-2k}/b^2},q^{2n-2m+2},q^{2-2m-2k}/b^2c^2,-cq^{2+n-m-k}/b,-cq^{1+n-m-k}/b,q^{2n-2k+2},q^{3+n-m-k}/b^2,q^{2+n-m-k}/b^2}{q^2;q^2}
\end{align}
を得る. これは$m,k$に関して対称的な表示である. さらに
\begin{align}
&\Q{10}9{q^{1+2n-2m-2k}/b^2,q\sqrt{q^{1+2n-2m-2k}/b^2},-q\sqrt{q^{1+2n-2m-2k}/b^2},q^{1-2k}/b^2,c^2q^{2n+1},-q^{1+n-m-k}/bc,-q^{2+n-m-k}/bc,q^{1-2m}/b^2,q^{n-m-k},q^{1+n-m-k}}{\sqrt{q^{1+2n-2m-2k}/b^2},-\sqrt{q^{1+2n-2m-2k}/b^2},q^{2n-2m+2},q^{2-2m-2k}/b^2c^2,-cq^{2+n-m-k}/b,-cq^{1+n-m-k}/b,q^{2n-2k+2},q^{3+n-m-k}/b^2,q^{2+n-m-k}/b^2}{q^2;q^2}\\
&=\sum_{0\leq j}\frac{1-q^{4j+1+2n-2m-2k}/b^2}{1-q^{1+2n-2m-2k}/b^2}\frac{(q^{1+2n-2m-2k}/b^2,q^{1-2k}/b^2,c^2q^{2n+1},-q^{1+n-m-k}/bc,-q^{2+n-m-k}/bc,q^{1-2m}/b^2,q^{n-m-k},q^{1+n-m-k};q^2)_j}{(q^2,q^{2n-2m+2},q^{2-2m-2k}/b^2c^2,-cq^{2+n-m-k}/b,-cq^{1+n-m-k}/b,q^{2n-2k+2},q^{3+n-m-k}/b^2,q^{2+n-m-k}/b^2;q^2)_j}q^{2j}\\
&=\frac{(q^2;q^2)_{n-m}(q^2;q^2)_{n-k}(q^{2-2m-2n-2k}/b^2c^2;q^2)_n(-cq^{1-m-n-k}/b,q^{2-m-n-k}/b^2;q)_{2n}}{(q^{1-2m-2k}/b^2,q^{1-2n-2k}/b^2,q^{1-2n-2m}/b^2,c^2q;q^2)_{n}(-q^{1-m-n-k}/bc,q^{-m-n-k};q)_{2n}}\\
&\qquad\cdot\sum_{0\leq j}\frac{1-q^{4j+1+2n-2m-2k}/b^2}{1-q^{1+2n-2m-2k}/b^2}\frac{(q^{1-2m-2k}/b^2,q^{1-2n-2k}/b^2,q^{1-2n-2m}/b^2,c^2q;q^2)_{n+j}(-q^{1-m-n-k}/bc,q^{-m-n-k};q)_{2n+2j}}{(q^2;q^2)_{n+j-m}(q^2;q^2)_{n+j-k}(q^2;q^2)_j(q^{2-2m-2n-2k}/b^2c^2;q^2)_{n+j}(-cq^{1-m-n-k}/b,q^{2-m-n-k}/b^2;q)_{2n+2j}}q^{2j}\\
&=\frac{(q^2;q^2)_{n-m}(q^2;q^2)_{n-k}(q^{2-2m-2n-2k}/b^2c^2;q^2)_n(-cq^{1-m-n-k}/b,q^{2-m-n-k}/b^2;q)_{2n}}{(1-q^{1+2n-2m-2k}/b^2)(q^{1-2m-2k}/b^2,q^{1-2n-2k}/b^2,q^{1-2n-2m}/b^2,c^2q;q^2)_{n}(-q^{1-m-n-k}/bc,q^{-m-n-k};q)_{2n}}q^{-2n}\\
&\qquad\cdot\sum_{0\leq j}\frac{(1-q^{4j+1-2n-2m-2k}/b^2)(q^{1-2m-2k}/b^2,q^{1-2n-2k}/b^2,q^{1-2n-2m}/b^2,c^2q;q^2)_{j}(-q^{1-m-n-k}/bc,q^{-m-n-k};q)_{2j}}{(q^2;q^2)_{j-m}(q^2;q^2)_{j-n}(q^2;q^2)_{j-k}(q^{2-2m-2n-2k}/b^2c^2;q^2)_{j}(-cq^{1-m-n-k}/b,q^{2-m-n-k}/b^2;q)_{2j}}q^{2j}
\end{align}
この$j$に関する和は$m,n,k$に関して対称である. 係数は
\begin{align}
&\frac{(q^2;q^2)_{n-m}(q^2;q^2)_{n-k}(q^{2-2m-2n-2k}/b^2c^2;q^2)_n(-cq^{1-m-n-k}/b,q^{2-m-n-k}/b^2;q)_{2n}}{(1-q^{1+2n-2m-2k}/b^2)(q^{1-2m-2k}/b^2,q^{1-2n-2k}/b^2,q^{1-2n-2m}/b^2,c^2q;q^2)_{n}(-q^{1-m-n-k}/bc,q^{-m-n-k};q)_{2n}}q^{-2n}\\
&=\frac{(q^2;q^2)_{n-m}(q^2;q^2)_{n-k}(b^2c^2;q^2)_{m+n+k}(-b/c,b^2/q;q)_{m+n+k}(b^2q;q^2)_{m+k-n}(b^2q;q^2)_m(b^2q;q^2)_k(-bc,q;q)_{m+k-n}}{(1-q^{1+2n-2m-2k}/b^2)(b^2c^2;q^2)_{m+k}(-b/c,b^2/q;q)_{m+k-n}(b^2q;q^2)_{m+k}(b^2q;q^2)_{n+m}(b^2q;q^2)_{n+k}(c^2q;q^2)_{n}(-bc,q;q)_{m+n+k}}\\
&\qquad\cdot\left(-\frac{q^{2-2m-2n-2k}}{b^2c^2}\right)^n\left(\frac{cq^{1-m-n-k}}b\cdot\frac{q^{2-m-n-k}}{b^2}\right)^{2n}\left(-\frac{q^{1-2m-2k}}{b^2}\cdot\frac{q^{1-2n-2k}}{b^2}\cdot\frac{q^{1-2n-2m}}{b^2}\right)^{-n}\left(\frac{q^{1-m-n-k}}{bc}\cdot q^{-m-n-k}\right)^{-2n}q^{-4\binom{n}2-2n}\\
&=\frac{(q^2;q^2)_{n-m}(q^2;q^2)_{n-k}(b^2c^2;q^2)_{m+n+k}(-b/c,b^2/q;q)_{m+n+k}(b^2q;q^2)_{m+k-n}(b^2q;q^2)_m(b^2q;q^2)_k(-bc,q;q)_{m+k-n}}{(1-q^{1+2n-2m-2k}/b^2)(b^2c^2;q^2)_{m+k}(-b/c,b^2/q;q)_{m+k-n}(b^2q;q^2)_{m+k}(b^2q;q^2)_{n+m}(b^2q;q^2)_{n+k}(c^2q;q^2)_{n}(-bc,q;q)_{m+n+k}}\left(c^2q^{2m+2n+2k+1}\right)^nq^{-4\binom{n}2}
\end{align}
となる. これを代入すると
\begin{align}
g_k(m,n)&=h_0\frac{(-bc;q)_{m+n-k}(q;q)_k(-bc;q)_{n+k-m}(q;q)_{m}(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_n(-b/c;q)_{m+k-n}(b^2c^2;q^2)_{m+k}(b^2/q;q)_{m+k-n}}{(-bc,bq^{\frac 12},cq^{\frac 12};q)_{k}(q^2;q^2)_{n-k}(-bc,bq^{\frac 12},cq^{\frac 12};q)_{m}(q^2;q^2)_{n-m}(-bcq;q)_{m+n+k}(b^2c^2;q^2)_{n}(q;q)_{m+k-n}(b^2/q;q^2)_{m+k-n}}(-cq^{\frac 12})^{m+k-n}q^{n+\binom{m+k-n}2}\\
&\qquad\cdot\frac{(q^2;q^2)_{n-m}(q^2;q^2)_{n-k}(b^2c^2;q^2)_{m+n+k}(-b/c,b^2/q;q)_{m+n+k}(b^2q;q^2)_{m+k-n}(b^2q;q^2)_m(b^2q;q^2)_k(-bc,q;q)_{m+k-n}}{(1-q^{1+2n-2m-2k}/b^2)(b^2c^2;q^2)_{m+k}(-b/c,b^2/q;q)_{m+k-n}(b^2q;q^2)_{m+k}(b^2q;q^2)_{n+m}(b^2q;q^2)_{n+k}(c^2q;q^2)_{n}(-bc,q;q)_{m+n+k}}\left(c^2q^{2m+2n+2k+1}\right)^nq^{-4\binom{n}2}\\
&\qquad\cdot\sum_{0\leq j}\frac{(1-q^{4j+1-2m-2n-2k}/b^2)(q^{1-2m-2k}/b^2,q^{1-2n-2k}/b^2,q^{1-2n-2m}/b^2,c^2q;q^2)_{j}(-q^{1-m-n-k}/bc,q^{-m-n-k};q)_{2j}}{(q^2;q^2)_{j-m}(q^2;q^2)_{j-n}(q^2;q^2)_{j-k}(q^{2-2m-2n-2k}/b^2c^2;q^2)_{j}(-cq^{1-m-n-k}/b,q^{2-m-n-k}/b^2;q)_{2j}}q^{2j}\\
&=h_0\frac{(-bc;q)_{m+n-k}(-bq^{\frac 12},q;q)_k(-bc;q)_{n+k-m}(-bq^{\frac 12},q;q)_{m}(-bc;q)_{m+k-n}(-bq^{\frac 12},q;q)_n}{(-bc,cq^{\frac 12};q)_{k}(-bc,cq^{\frac 12};q)_{m}(-bc,cq^{\frac 12};q)_{n}(-bcq;q)_{m+n+k}}(-c)^{m+n+k}q^{n+\binom{m+k-n}2+\frac 12(m+k-n)+(2m+2n+2k+1)n-4\binom n2}\\
&\qquad\cdot\frac{(1-b^2q^{2m+2k-2n-1})(b^2c^2;q^2)_{m+n+k}(-b/c,b^2/q;q)_{m+n+k}}{(1-b^2/q)(1-q^{1+2n-2m-2k}/b^2)(b^2q;q^2)_{m+k}(b^2q;q^2)_{n+m}(b^2q;q^2)_{n+k}(-bc,q;q)_{m+n+k}}\\
&\qquad\cdot\sum_{0\leq j}\frac{(1-q^{4j+1-2m-2n-2k}/b^2)(q^{1-2m-2k}/b^2,q^{1-2n-2k}/b^2,q^{1-2n-2m}/b^2,c^2q;q^2)_{j}(-q^{1-m-n-k}/bc,q^{-m-n-k};q)_{2j}}{(q^2;q^2)_{j-m}(q^2;q^2)_{j-n}(q^2;q^2)_{j-k}(q^{2-2m-2n-2k}/b^2c^2;q^2)_{j}(-cq^{1-m-n-k}/b,q^{2-m-n-k}/b^2;q)_{2j}}q^{2j}\\
&=h_0\frac{(-bq^{\frac 12},q;q)_k(-bq^{\frac 12},q;q)_{m}(-bq^{\frac 12},q;q)_n}{(-bc,cq^{\frac 12};q)_{k}(-bc,cq^{\frac 12};q)_{m}(-bc,cq^{\frac 12};q)_{n}}(-c)^{m+n+k}q^{n-\binom{m+k-n}2+\frac 12(m+k-n)+(2m+2n+2k+1)n-4\binom n2+2m+2k-2n}\\
&\qquad\cdot\frac{(-bc;q)_{m+n-k}(-bc;q)_{m+k-n}(-bc;q)_{n+k-m}(bc,-b/c,b^2/q;q)_{m+n+k}}{(b^2q;q^2)_{m+k}(b^2q;q^2)_{m+n}(b^2q;q^2)_{n+k}(-bcq,q;q)_{m+n+k}}\\
&\qquad\cdot\sum_{0\leq j}\frac{(1-q^{4j+1-2n-2m+2k}/b^2)(q^{1-2m-2k}/b^2,q^{1-2n-2k}/b^2,q^{1-2n-2m}/b^2,c^2q;q^2)_{j}(-q^{1-m-n-k}/bc,q^{-m-n-k};q)_{2j}}{(1-q/b^2)(q^2;q^2)_{j-m}(q^2;q^2)_{j-n}(q^2;q^2)_{j-k}(q^{2-2m-2n-2k}/b^2c^2;q^2)_{j}(-cq^{1-m-n-k}/b,q^{2-m-n-k}/b^2;q)_{2j}}q^{2j}
\end{align}
ここで, $q$の指数の部分は
\begin{align}
&n+\binom{m+k-n}2+\frac 12(m+k-n)+(2m+2n+2k+1)n-4\binom n2+2m+2k-2n\\
&=\binom{m+k}2-n(m+k)+\binom n2+\frac 12(m+k-n)+(2m+2n+2k+1)n-4\binom n2+2m+2k\\
&=\binom{m+k}2+n(m+k)+\binom n2+\frac 12(m+k-n)+3n+2m+2k\\
&=\binom{m+n+k}2+\frac 52(m+n+k)
\end{align}
と表される. よって,
\begin{align}
g_k(m,n)&=h_0\frac{(-bq^{\frac 12},q;q)_k(-bq^{\frac 12},q;q)_{m}(-bq^{\frac 12},q;q)_n}{(-bc,cq^{\frac 12};q)_{k}(-bc,cq^{\frac 12};q)_{m}(-bc,cq^{\frac 12};q)_{n}}(-cq^{\frac 52})^{m+n+k}q^{\binom{m+n+k}2}\\
&\qquad\cdot\frac{(-bc;q)_{m+n-k}(-bc;q)_{m+k-n}(-bc;q)_{n+k-m}(bc,-b/c,b^2/q;q)_{m+n+k}}{(b^2q;q^2)_{m+k}(b^2q;q^2)_{m+n}(b^2q;q^2)_{n+k}(-bcq,q;q)_{m+n+k}}\\
&\qquad\cdot\sum_{0\leq j}\frac{(1-q^{4j+1-2n-2m+2k}/b^2)(q^{1-2m-2k}/b^2,q^{1-2n-2k}/b^2,q^{1-2n-2m}/b^2,c^2q;q^2)_{j}(-q^{1-m-n-k}/bc,q^{-m-n-k};q)_{2j}}{(1-q/b^2)(q^2;q^2)_{j-m}(q^2;q^2)_{j-n}(q^2;q^2)_{j-k}(q^{2-2m-2n-2k}/b^2c^2;q^2)_{j}(-cq^{1-m-n-k}/b,q^{2-m-n-k}/b^2;q)_{2j}}q^{2j}
\end{align}
が得られた. これは$m,n,k$に関して対称的な表示である. これは
\begin{align}
&\frac{(q^{1-2m-2k}/b^2,q^{1-2n-2k}/b^2,q^{1-2n-2m}/b^2;q^2)_{j}(-q^{1-m-n-k}/bc,q^{-m-n-k};q)_{2j}}{(q^{2-2m-2n-2k}/b^2c^2;q^2)_{j}(-cq^{1-m-n-k}/b,q^{2-m-n-k}/b^2;q)_{2j}}\\
&=\frac{(b^2q;q^2)_{m+n}(b^2q;q^2)_{m+k}(b^2q;q^2)_{n+k}(-bc,q;q)_{m+n+k}}{(b^2c^2;q^2)_{m+n+k}(-b/c,b^2/q;q)_{m+n+k}}\\
&\qquad\cdot\frac{(b^2c^2;q^2)_{m+n+k-j}(-b/c,b^2/q;q)_{m+n+k-2j}}{(b^2q;q^2)_{m+n-j}(b^2q;q^2)_{m+k-j}(b^2q;q^2)_{n+k-j}(-bc,q;q)_{m+n+k-2j}}\left(\frac{c^2q^{1-2m-2n-2k}}{b^4}\right)^j\left(\frac{b^2}{c^2q^2}\right)^{2j}q^{4\binom j2}\\
&=\frac{(b^2q;q^2)_{m+n}(b^2q;q^2)_{m+k}(b^2q;q^2)_{n+k}(q;q)_{m+n+k}}{(bc,-b/c,b^2/q;q)_{m+n+k}}\\
&\qquad\cdot\frac{(b^2c^2;q^2)_{m+n+k-j}(-b/c,b^2/q;q)_{m+n+k-2j}}{(b^2q;q^2)_{m+n-j}(b^2q;q^2)_{m+k-j}(b^2q;q^2)_{n+k-j}(-bc,q;q)_{m+n+k-2j}}\left(\frac{q^{-2m-2n-2k-3}}{c^2}\right)^jq^{4\binom j2}
\end{align}
となることから,
\begin{align}
g_k(m,n)&=h_0\frac{(-bq^{\frac 12},q;q)_k(-bq^{\frac 12},q;q)_{m}(-bq^{\frac 12},q;q)_n(-bc;q)_{m+n-k}(-bc;q)_{m+k-n}(-bc;q)_{n+k-m}}{(-bc,cq^{\frac 12};q)_{k}(-bc,cq^{\frac 12};q)_{m}(-bc,cq^{\frac 12};q)_{n}(-bcq;q)_{m+n+k}}(-cq^{\frac 52})^{m+n+k}q^{\binom{m+n+k}2}\\
&\qquad\cdot\sum_{0\leq j}\frac{(1-q^{4j+1-2m-2n-2k}/b^2)(c^2q;q^2)_j(b^2c^2;q^2)_{m+n+k-j}(-b/c,b^2/q;q)_{m+n+k-2j}}{(1-q/b^2)(q^2;q^2)_{j-m}(q^2;q^2)_{j-n}(q^2;q^2)_{j-k}(b^2q;q^2)_{m+n-j}(b^2q;q^2)_{m+k-j}(b^2q;q^2)_{n+k-j}(-bc,q;q)_{m+n+k-2j}}\left(\frac{q^{-2m-2n-2k-1}}{c^2}\right)^jq^{4\binom j2}
\end{align}
と書き換えることもできる.
