Frenkel-Turaevの和公式

$n=0$のときは明らかに成り立つ. $n=m$のときに成り立つと仮定して$n=m+1$の場合を示す. つまり,
$$\begin{array}{rl}& \sum _{k=0}^{m+1}\frac{\theta (a{q}^{2k};p)(a,b,c,d,a^2{q}^{m+2}/bcd,q^{-m-1};q,p{)}_k}{\theta (a;p)(q,aq/b,aq/c,aq/d,bcd{q}^{-m-1}/a,a{q}^{m+2};q,p{)}_k}{q}^k\\ & =\frac{(aq,aq/bc,aq/bd,aq/cd;q,p{)}_{m+1}}{(aq/b,aq/c,aq/d,aq/bcd;q,p{)}_{m+1}}\end{array}$$
を示せば良い. まず,
$$\begin{array}{rl}& \frac{(a^2{q}^{m+2}/bcd,q^{-m-1};q,p{)}_k}{(bcd{q}^{-m-1}/a,a{q}^{m+2};q,p{)}_k}\\ & =\frac{(a^2{q}^{m+1}/bcd,q^{-m};q,p{)}_k}{(bcd{q}^{-m}/a,a{q}^{m+1};q,p{)}_k}\frac{\theta (a^2{q}^{m+k+1}/bcd,bcd{q}^{k-m-1}/a,a{q}^{m+1},q^{-m-1};p)}{\theta (a^2{q}^{m+1}/bcd,bcd{q}^{-m-1}/a,a{q}^{m+k+1},q^{k-m-1};p)}\end{array}$$
である. 定理1より,
$$\begin{array}{rl}& \frac{\theta (a^2{q}^{m+k+1}/bcd,bcd{q}^{k-m-1}/a,a{q}^{m+1},q^{-m-1};p)}{\theta (a^2{q}^{m+1}/bcd,bcd{q}^{-m-1}/a,a{q}^{m+k+1},q^{k-m-1};p)}\\ & =1+\frac{\theta (a^2{q}^{m+k+1}/bcd,bcd{q}^{k-m-1}/a,a{q}^{m+1},q^{-m-1};p)-\theta (a^2{q}^{m+1}/bcd,bcd{q}^{-m-1}/a,a{q}^{m+k+1},q^{k-m-1};p)}{\theta (a^2{q}^{m+1}/bcd,bcd{q}^{-m-1}/a,a{q}^{m+k+1},q^{k-m-1};p)}\\ & =1+\frac{bcd{q}^{-m-1}}{a}\frac{\theta (a{q}^k,q^k,a/bcd,a^2{q}^{2m+2}/bcd;p)}{\theta (a^2{q}^{m+1}/bcd,bcd{q}^{-m-1}/a,a{q}^{m+k+1},q^{k-m-1};p)}\end{array}$$
であるから,
$$\begin{array}{rl}& \sum _{k=0}^{m+1}\frac{\theta (a{q}^{2k};p)(a,b,c,d,a^2{q}^{m+2}/bcd,q^{-m-1};q,p{)}_k}{\theta (a;p)(q,aq/b,aq/c,aq/d,bcd{q}^{-m-1}/a,a{q}^{m+2};q,p{)}_k}{q}^k\\ & =\sum _{k=0}^{m+1}\frac{\theta (a{q}^{2k};p)(a,b,c,d;q,p{)}_k}{\theta (a;p)(q,aq/b,aq/c,aq/d;q,p{)}_k}\\ & \cdot \frac{(a^2{q}^{m+1}/bcd,q^{-m};q,p{)}_k}{(bcd{q}^{-m}/a,a{q}^{m+1};q,p{)}_k}(1+\frac{bcd{q}^{-m-1}}{a}\frac{\theta (a{q}^k,q^k,a/bcd,a^2{q}^{2m+2}/bcd;p)}{\theta (a^2{q}^{m+1}/bcd,bcd{q}^{-m-1}/a,a{q}^{m+k+1},q^{k-m-1};p)})q^k\\ & =\frac{(aq,aq/bc,aq/bd,aq/cd;q,p{)}_m}{(aq/b,aq/c,aq/d,aq/bcd;q,p{)}_m}\\ & +\frac{bcd{q}^{-m-1}}{a}\frac{\theta (a,a/bcd,a^2{q}^{2m+2}/bcd;p)}{\theta (a^2{q}^{m+1}/bcd,bcd{q}^{-m-1}/a,a{q}^{m+1};p)}\sum _{k=1}^{m+1}\frac{\theta (a{q}^{2k};p)(aq,b,c,d;q,p{)}_k}{\theta (a;p)(q;q,p{)}_{k-1}(aq/b,aq/c,aq/d;q,p{)}_k}\\ & \cdot \frac{(a^2{q}^{m+1}/bcd;q,p{)}_k(q^{-m};q,p{)}_{k-1}}{(bcd{q}^{-m}/a,a{q}^{m+2};q,p{)}_k}{q}^k\\ & =\frac{(aq,aq/bc,aq/bd,aq/cd;q,p{)}_m}{(aq/b,aq/c,aq/d,aq/bcd;q,p{)}_m}\\ & +\frac{bcd{q}^{-m}}{a}\frac{\theta (aq,a{q}^2,a/bcd,a^2{q}^{2m+2}/bcd,b,c,d;p)}{\theta (aq/b,aq/c,aq/d,bcd{q}^{-m-1}/a,bcd{q}^{-m}/a,a{q}^{m+1},a{q}^{m+2};p)}\\ & \cdot \sum _{k=0}^m\frac{\theta (a{q}^{2k+2};p)(a{q}^2,bq,cq,dq,a^2{q}^{m+2}/bcd,q^{-m};q,p{)}_k}{\theta (a{q}^2;p)(q,a{q}^2/b,a{q}^2/c,a{q}^2/d,bcd{q}^{1-m}/a,a{q}^{m+3};q,p{)}_k}{q}^k\\ & =\frac{(aq,aq/bc,aq/bd,aq/cd;q,p{)}_m}{(aq/b,aq/c,aq/d,aq/bcd;q,p{)}_m}\\ & +\frac{bcd{q}^{-m}}{a}\frac{\theta (aq,a{q}^2,a/bcd,a^2{q}^{2m+2}/bcd,b,c,d;p)}{\theta (aq/b,aq/c,aq/d,bcd{q}^{-m-1}/a,bcd{q}^{-m}/a,a{q}^{m+1},a{q}^{m+2};p)}\\ & \cdot \frac{(a{q}^3,aq/bc,aq/bd,aq/cd;q,p{)}_m}{(a{q}^2/b,a{q}^2/c,a{q}^2/d,a/bcd;q,p{)}_m}\\ & =\frac{(aq,aq/bc,aq/bd,aq/cd;q,p{)}_m}{(aq/b,aq/c,aq/d,aq/bcd;q,p{)}_m}\\ & +\frac{bcd{q}^{-m}}{a}\frac{\theta (a{q}^m/bcd,a^2{q}^{2m+2}/bcd,b,c,d;p)}{\theta (a{q}^{m+1}/b,a{q}^{m+1}/c,a{q}^{m+1}/d,bcd{q}^{-m-1}/a,bcd{q}^{-m}/a;p)}\\ & \cdot \frac{(aq,aq/bc,aq/bd,aq/cd;q,p{)}_m}{(aq/b,aq/c,aq/d,aq/bcd;q,p{)}_m}\\ & =\frac{(aq,aq/bc,aq/bd,aq/cd;q,p{)}_m}{(aq/b,aq/c,aq/d,aq/bcd;q,p{)}_m}(1-\frac{\theta (a^2{q}^{2m+2}/bcd,b,c,d;p)}{\theta (a{q}^{m+1}/b,a{q}^{m+1}/c,a{q}^{m+1}/d,bcd{q}^{-m-1}/a;p)})\end{array}$$
ここで, 定理1より,
$$\begin{array}{rl}& 1-\frac{\theta (a^2{q}^{2m+2}/bcd,b,c,d;p)}{\theta (a{q}^{m+1}/b,a{q}^{m+1}/c,a{q}^{m+1}/d,bcd{q}^{-m-1}/a;p)}\\ & =\frac{\theta (a{q}^{m+1}/b,a{q}^{m+1}/c,a{q}^{m+1}/d,bcd{q}^{-m-1}/a;p)-\theta (a^2{q}^{2m+2}/bcd,b,c,d;p)}{\theta (a{q}^{m+1}/b,a{q}^{m+1}/c,a{q}^{m+1}/d,bcd{q}^{-m-1}/a;p)}\\ & =\frac{a^2{q}^{2m+2}}{bcd}\frac{\theta (a{q}^{m+1},bc{q}^{-m-1}/a,bd{q}^{-m-1}/a,cd{q}^{-m-1}/a;p)}{\theta (a{q}^{m+1}/b,a{q}^{m+1}/c,a{q}^{m+1}/d,bcd{q}^{-m-1}/a;p)}\\ & =\frac{\theta (a{q}^{m+1},a{q}^{m+1}/bc,a{q}^{m+1}/bd,a{q}^{m+1}/cd;p)}{\theta (a{q}^{m+1}/b,a{q}^{m+1}/c,a{q}^{m+1}/d,a{q}^{m+1}/bcd;p)}\end{array}$$
よって, これを代入すれば,
$$\begin{array}{rl}& \sum _{k=0}^{m+1}\frac{\theta (a{q}^{2k};p)(a,b,c,d,a^2{q}^{m+2}/bcd,q^{-m-1};q,p{)}_k}{\theta (a;p)(q,aq/b,aq/c,aq/d,bcd{q}^{-m-1}/a,a{q}^{m+2};q,p{)}_k}{q}^k\\ & =\frac{(aq,aq/bc,aq/bd,aq/cd;q,p{)}_m}{(aq/b,aq/c,aq/d,aq/bcd;q,p{)}_m}\frac{\theta (a{q}^{m+1},a{q}^{m+1}/bc,a{q}^{m+1}/bd,a{q}^{m+1}/cd;p)}{\theta (a{q}^{m+1}/b,a{q}^{m+1}/c,a{q}^{m+1}/d,a{q}^{m+1}/bcd;p)}\\ & =\frac{(aq,aq/bc,aq/bd,aq/cd;q,p{)}_{m+1}}{(aq/b,aq/c,aq/d,aq/bcd;q,p{)}_{m+1}}\end{array}$$
となって定理を得る.