Rogers多項式の積公式

$x=\cos\theta,y=\cos\phi, w=a^2ce^{i\phi}/\sqrt q$とする. 前の記事 における系1
\begin{align} &p_n(x;a,a\sqrt q,c,c\sqrt q|q)p_n(y;a,a\sqrt q,c,c\sqrt q|q)\\ &=(a^2\sqrt q,ac,ac\sqrt q,ac\sqrt q,acq,c^2\sqrt q;q)_n(ac\sqrt q)^{-n}\\ &\qquad\cdot\sum_{m=0}^n\frac{(q^{-n},a^2c^2q^n,c\sqrt qe^{-i\phi},ace^{i(\phi-\theta)}\sqrt q,ace^{i(\phi+\theta)}\sqrt q;q)_m}{(q,ac\sqrt q,acq,c^2\sqrt q,a^2ce^{i\phi}\sqrt q;q)_m}q^m\\ &\qquad\cdot \Q{10}9{w,\sqrt wq,-\sqrt wq,ce^{i\phi},ae^{i\phi},a\sqrt qe^{i\phi},ae^{i\theta},ae^{-i\theta},a^2c^2q^m,q^{-m}}{\sqrt w,-\sqrt w,a^2\sqrt q,ac\sqrt q,ac,ace^{i(\phi-\theta)}\sqrt q,ace^{i(\phi+\theta)}\sqrt q,e^{i\phi}q^{\frac 12-m}/c,a^2ce^{i\phi}q^{\frac 12+m}}q \end{align}
より,
\begin{align} r_n(x;a,b,c,d|q):=\Q43{q^{-n},abcdq^{n-1},ae^{i\theta},ae^{-i\theta}}{ab,ac,ad}q \end{align}
とすると,
\begin{align} &r_n(x;a,a\sqrt q,c,c\sqrt q|q)r_n(y;a,a\sqrt q,c,c\sqrt q|q)\\ &=\frac{(acq,c^2\sqrt q;q)_n}{(ac,a^2\sqrt q;q)_n}(c\sqrt q/a)^{-n}\sum_{m=0}^n\frac{(q^{-n},a^2c^2q^n,c\sqrt qe^{-i\phi},ace^{i(\phi-\theta)}\sqrt q,ace^{i(\phi+\theta)}\sqrt q;q)_m}{(q,ac\sqrt q,acq,c^2\sqrt q,a^2ce^{i\phi}\sqrt q;q)_m}q^m\\ &\qquad\cdot \Q{10}9{w,\sqrt wq,-\sqrt wq,ce^{i\phi},ae^{i\phi},a\sqrt qe^{i\phi},ae^{i\theta},ae^{-i\theta},a^2c^2q^m,q^{-m}}{\sqrt w,-\sqrt w,a^2\sqrt q,ac\sqrt q,ac,ace^{i(\phi-\theta)}\sqrt q,ace^{i(\phi+\theta)}\sqrt q,e^{i\phi}q^{\frac 12-m}/c,a^2ce^{i\phi}q^{\frac 12+m}}q \end{align}
となる. ここで, $c=-a$とすると,
\begin{align} &r_n(x;a,a\sqrt q,-a,-a\sqrt q|q)r_n(y;a,a\sqrt q,-a,-a\sqrt q|q)\\ &=\frac{1+a^2q^n}{1+a^2}(-\sqrt q)^{-n}\sum_{m=0}^n\frac{(q^{-n},a^4q^n,-a\sqrt qe^{-i\phi},-a^2e^{i(\phi-\theta)}\sqrt q,-a^2e^{i(\phi+\theta)}\sqrt q;q)_m}{(q,-a^2\sqrt q,-a^2q,a^2\sqrt q,-a^3e^{i\phi}\sqrt q;q)_m}q^m\\ &\qquad\cdot W(-a^3e^{i\phi}/\sqrt q;ae^{i\phi},a\sqrt qe^{i\phi},-ae^{i\phi},ae^{i\theta},ae^{-i\theta},a^4q^m,q^{-m};q) \end{align}
を得る. ここで,
\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}
である. $w=a^2q/bcd$として, Baileyによるnearly-poised${}_5\phi_4$の変換公式
\begin{align} &\frac{(wq,w^2q/a^2;q)_m}{(wq/a,w^2q/a;q)_m}\Q54{a,b,c,d,q^{-m}}{aq/b,aq/c,aq/d,a^2q^{-m}/w^2}q\\ &=W(w;wb/a,wc/a,wd/a,\sqrt a,-\sqrt a,\sqrt{aq},-\sqrt{aq},w^2q^{m+1}/a,q^{-m};q) \end{align}
において, $d=-\sqrt{aq}$とすると,
\begin{align} &\frac{(wq,w^2q/a^2;q)_m}{(wq/a,w^2q/a;q)_m}\Q43{a,b,c,q^{-m}}{aq/b,aq/c,a^2q^{-m}/w^2}q\\ &=W(w;-\sqrt{aq}/b,-\sqrt{aq}/c,\sqrt a,-\sqrt a,\sqrt{aq},w^2q^{m+1}/a,q^{-m};q)\qquad w=-a^{\frac 32}\sqrt q/bcd \end{align}
となる. $a\mapsto a^2e^{2i\phi},b\mapsto -\sqrt q e^{i(\phi-\theta)},c\mapsto -\sqrt qe^{i(\phi+\theta)}$とすると, $w=-a^3e^{i\phi}/\sqrt q$となり,
\begin{align} &W(-a^3e^{i\phi}/\sqrt q;ae^{i\phi},a\sqrt qe^{i\phi},-ae^{i\phi},ae^{i\theta},ae^{-i\theta},a^4q^m,q^{-m};q)\\ &=\frac{(-a^3e^{i\phi}\sqrt q,a^2e^{-2i\phi};q)_m}{(-ae^{-i\phi}\sqrt q,a^4;q)_m}\Q43{a^2e^{2i\phi},-\sqrt qe^{i(\phi-\theta)},-\sqrt qe^{i(\phi+\theta)},q^{-m}}{-a^2\sqrt qe^{i(\phi+\theta)},-a^2\sqrt qe^{i(\phi-\theta)},e^{2i\phi}q^{1-m}/a^2}q \end{align}
を得る. ここで, Searsの変換公式 より,
\begin{align} &\frac{(-a^3e^{i\phi}\sqrt q,a^2e^{-2i\phi};q)_m}{(-ae^{-i\phi}\sqrt q,a^4;q)_m}\Q43{a^2e^{2i\phi},-\sqrt qe^{i(\phi-\theta)},-\sqrt qe^{i(\phi+\theta)},q^{-m}}{-a^2\sqrt qe^{i(\phi+\theta)},-a^2\sqrt qe^{i(\phi-\theta)},e^{2i\phi}q^{1-m}/a^2}q\\ &=\frac{(-a^3e^{i\phi}\sqrt q,a^2e^{-2i\phi};q)_m}{(-ae^{-i\phi}\sqrt q,a^4;q)_m}\frac{(-\sqrt q e^{i(\theta-\phi)},q^{1-m}/a^4;q)_m}{(-a^2\sqrt qe^{i(\phi+\theta)},e^{2i\phi}q^{1-m}/a^2;q)_m}(a^2e^{2i\phi})^{m}\\ &\qquad\cdot\Q43{a^2e^{2i\phi},a^2,a^2e^{-2i\theta},q^{-m}}{-a^2\sqrt qe^{i(\phi-\theta)},a^4,-e^{i(\phi-\theta)}q^{\frac 12-m}}q\\ &=\frac{(-a^3e^{i\phi}\sqrt q,-\sqrt q e^{i(\theta-\phi)};q)_m}{(-ae^{-i\phi}\sqrt q,-a^2\sqrt qe^{i(\phi+\theta)};q)_m}\Q43{a^2e^{2i\phi},a^2,a^2e^{-2i\theta},q^{-m}}{-a^2\sqrt qe^{i(\phi-\theta)},a^4,-e^{i(\phi-\theta)}q^{\frac 12-m}}q \end{align}
であるから, これを代入すると,
\begin{align} &r_n(x;a,a\sqrt q,-a,-a\sqrt q|q)r_n(y;a,a\sqrt q,-a,-a\sqrt q|q)\\ &=\frac{1+a^2q^n}{1+a^2}(-\sqrt q)^{-n}\sum_{m=0}^n\frac{(q^{-n},a^4q^n,-a\sqrt qe^{-i\phi},-a^2e^{i(\phi-\theta)}\sqrt q,-a^2e^{i(\phi+\theta)}\sqrt q;q)_m}{(q,-a^2\sqrt q,-a^2q,a^2\sqrt q,-a^3e^{i\phi}\sqrt q;q)_m}q^m\\ &\qquad\cdot \frac{(-a^3e^{i\phi}\sqrt q,-\sqrt q e^{i(\theta-\phi)};q)_m}{(-ae^{-i\phi}\sqrt q,-a^2\sqrt qe^{i(\phi+\theta)};q)_m}\Q43{a^2e^{2i\phi},a^2,a^2e^{-2i\theta},q^{-m}}{-a^2\sqrt qe^{i(\phi-\theta)},a^4,-e^{i(\phi-\theta)}q^{\frac 12-m}}q\\ &=\frac{1+a^2q^n}{1+a^2}(-\sqrt q)^{-n}\sum_{m=0}^n\frac{(q^{-n},a^4q^n,-a^2e^{i(\phi-\theta)}\sqrt q,-\sqrt q e^{i(\theta-\phi)};q)_m}{(q,-a^2\sqrt q,-a^2q,a^2\sqrt q;q)_m}q^m\\ &\qquad\cdot \Q43{a^2e^{2i\phi},a^2,a^2e^{-2i\theta},q^{-m}}{-a^2\sqrt qe^{i(\phi-\theta)},a^4,-e^{i(\phi-\theta)}q^{\frac 12-m}}q \end{align}
となる. ここで, Askey-Wilson積分 より, $z=\cos\psi$として,
\begin{align} &\int_{-1}^1w(z;ae^{i(\phi-\theta)},ae^{i(\theta-\phi)},ae^{i(\theta+\phi)},ae^{-i(\theta+\phi)})(ae^{i(\phi-\theta+\psi)},ae^{i(\phi-\theta-\psi)};q)_j\,dz\\ &=\int_{-1}^1w(z;ae^{i(\phi-\theta)}q^j,ae^{i(\theta-\phi)},ae^{i(\theta+\phi)},ae^{-i(\theta+\phi)})\,dz\\ &=\frac{2\pi(a^4q^j;q)_{\infty}}{(q,a^2q^j,a^2e^{2i\phi}q^j,a^2e^{-2i\theta}q^j,a^2e^{2i\theta},a^2e^{-2i\phi},a^2;q)_{\infty}}\\ &=\frac{2\pi(a^4;q)_{\infty}}{(q,a^2,a^2;q)_{\infty}|(a^2e^{2i\theta},a^2e^{2i\phi};q)_{\infty}|^2}\frac{(a^2,a^2e^{2i\phi},a^2e^{-2i\theta};q)_j}{(a^4;q)_j} \end{align}
であるから,
\begin{align} &\Q43{a^2e^{2i\phi},a^2,a^2e^{-2i\theta},q^{-m}}{-a^2\sqrt qe^{i(\phi-\theta)},a^4,-e^{i(\phi-\theta)}q^{\frac 12-m}}q\\ &=\sum_{0\leq j}\frac{(a^2e^{2i\phi},a^2,a^2e^{-2i\theta},q^{-m};q)_j}{(q,-a^2\sqrt qe^{i(\phi-\theta)},a^4,-e^{i(\phi-\theta)}q^{\frac 12-m};q)_j}q^j\\ &=\frac{(q,a^2,a^2;q)_{\infty}|(a^2e^{2i\theta},a^2e^{2i\phi};q)_{\infty}|^2}{2\pi(a^4;q)_{\infty}}\sum_{0\leq j}\frac{(q^{-m};q)_j}{(q,-a^2\sqrt qe^{i(\phi-\theta)},-e^{i(\phi-\theta)}q^{\frac 12-m};q)_j}q^j\\ &\qquad\cdot\int_{-1}^1w(z;ae^{i(\phi-\theta)},ae^{i(\theta-\phi)},ae^{i(\theta+\phi)},ae^{-i(\theta+\phi)})(ae^{i(\phi-\theta+\psi)},ae^{i(\phi-\theta-\psi)};q)_j\,dz\\ &=\frac{(q,a^2,a^2;q)_{\infty}|(a^2e^{2i\theta},a^2e^{2i\phi};q)_{\infty}|^2}{2\pi(a^4;q)_{\infty}}\\ &\qquad\cdot\int_{-1}^1w(z;ae^{i(\phi-\theta)},ae^{i(\theta-\phi)},ae^{i(\theta+\phi)},ae^{-i(\theta+\phi)})\\ &\qquad\cdot\sum_{0\leq j}\frac{(ae^{i(\phi-\theta+\psi)},ae^{i(\phi-\theta-\psi)},q^{-m};q)_j}{(q,-a^2\sqrt qe^{i(\phi-\theta)},-e^{i(\phi-\theta)}q^{\frac 12-m};q)_j}q^j\,dz\\ &=\frac{(q,a^2,a^2;q)_{\infty}|(a^2e^{2i\theta},a^2e^{2i\phi};q)_{\infty}|^2}{2\pi(a^4;q)_{\infty}}\\ &\qquad\cdot\int_{-1}^1w(z;ae^{i(\phi-\theta)},ae^{i(\theta-\phi)},ae^{i(\theta+\phi)},ae^{-i(\theta+\phi)})\frac{(-a\sqrt qe^{i\psi},-a\sqrt qe^{-i\psi};q)_m}{(-a^2\sqrt qe^{i(\phi-\theta)},-\sqrt qe^{i(\theta-\phi)};q)_m}\,dz \end{align}
となるので, これを代入すると,
\begin{align} &r_n(x;a,a\sqrt q,-a,-a\sqrt q|q)r_n(y;a,a\sqrt q,-a,-a\sqrt q|q)\\ &=\frac{(q,a^2,a^2;q)_{\infty}|(a^2e^{2i\theta},a^2e^{2i\phi};q)_{\infty}|^2}{2\pi(a^4;q)_{\infty}}\frac{1+a^2q^n}{1+a^2}(-\sqrt q)^{-n}\sum_{m=0}^n\frac{(q^{-n},a^4q^n,-a^2e^{i(\phi-\theta)}\sqrt q,-\sqrt q e^{i(\theta-\phi)};q)_m}{(q,-a^2\sqrt q,-a^2q,a^2\sqrt q;q)_m}q^m\\ &\qquad\cdot\int_{-1}^1w(z;ae^{i(\phi-\theta)},ae^{i(\theta-\phi)},ae^{i(\theta+\phi)},ae^{-i(\theta+\phi)})\frac{(-a\sqrt qe^{i\psi},-a\sqrt qe^{-i\psi};q)_m}{(-a^2\sqrt qe^{i(\phi-\theta)},-\sqrt qe^{i(\theta-\phi)};q)_m}\,dz\\ &=\frac{(q,a^2,a^2;q)_{\infty}|(a^2e^{2i\theta},a^2e^{2i\phi};q)_{\infty}|^2}{2\pi(a^4;q)_{\infty}}\frac{1+a^2q^n}{1+a^2}(-\sqrt q)^{-n}\\ &\qquad\cdot\int_{-1}^1w(z;ae^{i(\phi-\theta)},ae^{i(\theta-\phi)},ae^{i(\theta+\phi)},ae^{-i(\theta+\phi)})\sum_{m=0}^n\frac{(q^{-n},a^4q^n,-a\sqrt qe^{i\psi},-a\sqrt qe^{-i\psi};q)_m}{(q,-a^2\sqrt q,-a^2q,a^2\sqrt q;q)_m}q^m\,dz\\ &=\frac{(q,a^2,a^2;q)_{\infty}|(a^2e^{2i\theta},a^2e^{2i\phi};q)_{\infty}|^2}{2\pi(a^4;q)_{\infty}}\frac{1+a^2q^n}{1+a^2}(-\sqrt q)^{-n}\\ &\qquad\cdot\int_{-1}^1w(z;ae^{i(\phi-\theta)},ae^{i(\theta-\phi)},ae^{i(\theta+\phi)},ae^{-i(\theta+\phi)})\frac{(-a\sqrt q)^np_n(z;a,a\sqrt q,-a,-a\sqrt q)}{(a^2\sqrt q,-a^2\sqrt q,-a^2q;q)_n}\,dz\\ &=\frac{(q,a^2,a^2;q)_{\infty}|(a^2e^{2i\theta},a^2e^{2i\phi};q)_{\infty}|^2}{2\pi(a^4;q)_{\infty}}\\ &\qquad\cdot\int_{-1}^1w(z;ae^{i(\phi-\theta)},ae^{i(\theta-\phi)},ae^{i(\theta+\phi)},ae^{-i(\theta+\phi)})r_n(z;a,a\sqrt q,-a,-a\sqrt q)\,dz \end{align}
つまり,
\begin{align} &r_n(x;a,a\sqrt q,-a,-a\sqrt q|q)r_n(y;a,a\sqrt q,-a,-a\sqrt q|q)\\ &=\int_{-1}^1K(x,y,z;a^2|q)r_n(z;a,a\sqrt q,-a,-a\sqrt q)\,dz \end{align}
を得る. ここで, 前の記事 で示した表示
\begin{align} C_n(x;a^2|q)=\frac{(a^4;q)_n}{(q;q)_n}a^{-n}r_n(x;a,a\sqrt q,-a,-a\sqrt q|q) \end{align}
を代入すると,
\begin{align} C_n(x;a^2|q)C_n(y;a^2|q)&=\frac{(a^4;q)_n}{(q;q)_n}a^{-n}\int_{-1}^1K(x,y,z;a^2|q)C_n(z;a^2|q)\,dz \end{align}
となるから, $a\mapsto \sqrt a$とすると定理を得る.