Milneの二項定理の楕円類似
前の記事
の記法を用いる.
\begin{align}
\Delta(z_1,\dots,z_n;p):=\prod_{1\leq j< k\leq n}z_j\theta(z_k/z_j;p)
\end{align}
とする. これは$p=0$のとき,
\begin{align}
\Delta(z_1,\dots,z_n;0)&=\prod_{1\leq j< k\leq n}(z_j-z_k)\\
&=:\Delta(z_1,\dots,z_n)
\end{align}
となって通常の差積と一致する. 以下は部分分数分解の楕円類似である.
$n\geq 1$に対し,
\begin{align}
\prod_{j=1}^n\frac{\theta(t/b_j;p)}{\theta(t/a_j;p)}=-\sum_{k=1}^n\frac{\theta(b_1\cdots b_na_k/a_1\cdots a_nt;p)\prod_{j=1}^n\theta(a_k/b_j;p)}{\theta(a_k/t;p)\theta(b_1\cdots b_n/a_1\cdots a_n;p)\prod_{\substack{1\leq j\leq n\\j\neq k}}\theta(a_k/a_j;p)}
\end{align}
が成り立つ.
以下の証明において$\theta(x;p)=-x\theta(1/x;p)$であることは特に断りなく用いる.
まず, $n=2$の場合を考える.
\begin{align}
\frac{\theta(t/b_1,t/b_2;p)}{\theta(t/a_1,t/a_2;p)}=-\frac{\theta(b_1b_2/a_2t,a_1/b_1,a_1/b_2;p)}{\theta(a_1/t,b_1b_1/a_1a_2,a_1/a_2;p)}-\frac{\theta(b_1b_2/a_1t,a_2/b_1,a_2/b_2;p)}{\theta(a_2/t,b_1b_2/a_1a_2,a_2/a_1;p)}
\end{align}
を示せばよい. 両辺に$\theta(t/a_1,t/a_2,b_1b_2/a_1a_2,a_1/a_2)$を掛けると, これは
\begin{align}
&\theta(t/b_1,t/b_2,b_1b_1/a_1a_2,a_1/a_2;p)\\
&=\frac{t}{a_1}\theta(b_1b_2/a_2t,a_1/b_1,a_1/b_2,t/a_2;p)-\frac{a_1t}{a_2^2}\theta(b_1b_2/a_1t,a_2/b_1,a_2/b_2,t/a_1;p)
\end{align}
と同値である. 無限積の三項関係式(
前の記事
の定理1)において, $b,c,d,e$を$t/b_1,t/b_2,b_1b_2/a_1a_2,a_2/a_1$とすると,
\begin{align}
&\theta(b_1/a_1,b_2/a_1,a_2t/b_1b_2,t/a_2;p)-\theta(t/b_1,t/b_2,b_1b_2/a_1a_2,a_2/a_1;p)\\
&=\frac t{b_1}\theta(t/a_1,b_1b_2/a_1t,a_1/b_2,b_1/a_2;p)
\end{align}
となるからこれを整理することによって示すべき等式が得られる. よって, $n=2$の場合が示せた. 一般の場合は$n=2$の場合を用いると,
\begin{align}
&\frac{\theta(b_1\cdots b_na_k/a_1\cdots a_nt;p)}{\theta(a_k/t)}\frac{\theta(t/b_{n+1})}{\theta(t/a_{n+1})}\\
&=\frac{b_1\cdots b_n}{a_1\cdots a_n}\frac{\theta(a_1\cdots a_nt/a_kb_1\cdots b_n;p)}{\theta(t/a_k)}\frac{\theta(t/b_{n+1})}{\theta(t/a_{n+1})}
\end{align}
が$t$の関数として
\begin{align}
\frac{\theta(b_1\cdots b_{n+1}a_{n+1}/a_1\cdots a_{n+1}t)}{\theta(a_k/t)}, \frac{\theta(b_1\cdots b_{n+1}/a_1\cdots a_{n+1}t)}{\theta(a_{n+1}/t)}
\end{align}
の線形和で表されることが分かる. よって, 帰納的に
\begin{align}
\prod_{j=1}^n\frac{\theta(t/b_j;p)}{\theta(t/a_j;p)}=\sum_{k=1}^nC_k\frac{\theta(b_1\cdots b_na_k/a_1\cdots a_nt;p)}{\theta(a_k/t;p)}
\end{align}
と展開できることが分かる. 両辺に$\theta(t/a_k;p)$を掛けてから$t\to a_k$とすると,
\begin{align}
\frac{\prod_{j=1}^n\theta(a_k/b_j;p)}{\prod_{\substack{1\leq j\leq n\\j\neq k}}\theta(a_k/a_j;p)}=-C_k\theta(b_1\cdots b_n/a_1\cdots a_n;p)
\end{align}
となるので,
\begin{align}
C_k=-\frac{\prod_{j=1}^n\theta(a_k/b_j;p)}{\theta(b_1\cdots b_n/a_1\cdots a_n;p)\prod_{\substack{1\leq j\leq n\\j\neq k}}\theta(a_k/a_j;p)}
\end{align}
を得る.
$n\geq 1, a_1\cdots a_n=b_1\cdots b_n$のとき,
\begin{align}
\sum_{k=1}^n\frac{\prod_{j=1}^n\theta(a_k/b_j;p)}{\prod_{\substack{1\leq j\leq n\\j\neq k}}\theta(a_k/a_j;p)}=0
\end{align}
が成り立つ.
$n=1$のときは明らか. $n\geq 1$とする. 補題1において, $t=a_{n+1}$とすると,
\begin{align}
\prod_{j=1}^n\frac{\theta(a_{n+1}/b_j;p)}{\theta(a_{n+1}/a_j;p)}=-\sum_{k=1}^n\frac{\theta(b_1\cdots b_na_k/a_1\cdots a_{n+1};p)\prod_{j=1}^n\theta(a_k/b_j;p)}{\theta(a_k/a_{n+1};p)\theta(b_1\cdots b_n/a_1\cdots a_n;p)\prod_{\substack{1\leq j\leq n\\j\neq k}}\theta(a_k/a_j;p)}
\end{align}
となる. これは$b_{n+1}=a_1\cdots a_{n+1}/b_1\cdots b_n$とすると,
\begin{align}
\prod_{j=1}^n\frac{\theta(a_{n+1}/b_j;p)}{\theta(a_{n+1}/a_j;p)}=-\sum_{k=1}^n\frac{\prod_{j=1}^{n+1}\theta(a_k/b_j;p)}{\theta(a_{n+1}/b_{n+1};p)\prod_{\substack{1\leq j\leq n+1\\j\neq k}}\theta(a_k/a_j;p)}
\end{align}
となる. よって両辺に$\theta(a_{n+1}/b_{n+1};p)$を掛けて移項すると,
\begin{align}
\sum_{k=1}^{n+1}\frac{\prod_{j=1}^{n+1}\theta(a_k/b_j;p)}{\prod_{\substack{1\leq j\leq n+1\\j\neq k}}\theta(a_k/a_j;p)}=0
\end{align}
となって示すべきことが得られる.
$n\geq 1,b=a_1\cdots a_{n+1}z_1\cdots z_n$であるとき,
\begin{align}
&\sum_{\substack{0\leq y_1,\dots,y_n\\y_1+\cdots+y_n=N}}\frac{\Delta(z_1q^{y_1},\dots,z_nq^{y_n};p)}{\Delta(z_1,\dots,z_n;p)}\prod_{k=1}^n\frac{(a_1z_k,\dots,a_{n+1}z_k;q,p)_{y_k}}{(bz_k,z_kq/z_1,\dots,z_kq/z_n;q,p)_{y_k}}\\
&=\frac{(b/a_1,\dots,b/a_{n+1};q,p)_N}{(q,bz_1,\dots,bz_n;q,p)_N}
\end{align}
が成り立つ.
$N=1$のとき, $y_1,\dots,y_n$のどれか1つだけが$1$である. $y_j=1$で他の$y_i$が$0$のとき,
\begin{align}
&\frac{\Delta(z_1q^{y_1},\dots,z_nq^{y_n};p)}{\Delta(z_1,\dots,z_n;p)}\prod_{k=1}^n\frac{(a_1z_k,\dots,a_{n+1}z_k;q,p)_{y_k}}{(bz_k,z_kq/z_1,\dots,z_kq/z_n;q,p)_{y_k}}\\
&=\frac{\Delta(z_1,\dots,z_jq,\dots,z_n;p)}{\Delta(z_1,\dots,z_n;p)}\frac{\theta(a_1z_j,\dots,a_{n+1}z_j;p)}{\theta(bz_j,z_jq/z_1,\dots,z_jq/z_n;p)}\\
&=\left(\prod_{k=1}^{j-1}\frac{z_k\theta(z_jq/z_k;p)}{z_k\theta(z_j/z_k;p)}\right)\left(\prod_{k=j+1}^{n}\frac{z_jq\theta(z_k/z_jq;p)}{z_j\theta(z_k/z_j;p)}\right)\frac{\theta(a_1z_j,\dots,a_{n+1}z_j;p)}{\theta(bz_j,z_jq/z_1,\dots,z_jq/z_n;p)}\\
&=\frac{\theta(a_1z_j,\dots,a_{n+1}z_j;p)}{\theta(q,bz_j;p)\prod_{\substack{1\leq k\leq n\\k\neq j}}\theta(z_j/z_k;p)}
\end{align}
となる. ここで, 系1において$n$を$n+1$に置き換えて, $a_1,\dots,a_{n+1}$を$z_1,\dots,z_n,1/b$として, $b_1,\dots,b_{n+1}$を$1/a_1,\dots,1/a_{n+1}$とすれば,
\begin{align}
\sum_{j=1}^{n}\frac{\theta(a_1z_j,\dots,a_{n+1}z_j;p)}{\theta(bz_j)\prod_{\substack{1\leq k\leq n\\k\neq j}}\theta(z_j/z_k;p)}&=-\frac{\theta(a_1/b,\dots,a_{n+1}/b;p)}{\theta(1/bz_1,\dots,1/bz_{n};p)}\\
&=\frac{\theta(b/a_1,\dots,b/a_{n+1};p)}{\theta(bz_1,\dots,bz_n;p)}
\end{align}
を得る. これらより, 定理の$N=1$の場合
\begin{align}
&\sum_{\substack{0\leq y_1,\dots,y_n\\y_1+\cdots+y_n=1}}\frac{\Delta(z_1q^{y_1},\dots,z_nq^{y_n};p)}{\Delta(z_1,\dots,z_n;p)}\prod_{k=1}^n\frac{(a_1z_k,\dots,a_{n+1}z_k;q,p)_{y_k}}{(bz_k,z_kq/z_1,\dots,z_kq/z_n;q,p)_{y_k}}\\
&=\frac{\theta(b/a_1,\dots,b/a_{n+1};p)}{\theta(q,bz_1,\dots,bz_n;p)}
\end{align}
が示される. 次に, $N$のときに定理が成り立つとすると, 定理の右辺を$R_N$として,
\begin{align}
R_{N+1}&=\frac{\theta(bq^N/a_1,\dots,bq^N/a_{n+1};p)}{\theta(q^{N+1},bz_1q^N,\dots,bz_{n}q^N;p)}R_N\\
&=\frac{\theta(bq^N/a_1,\dots,bq^N/a_{n+1};p)}{\theta(q^{N+1},bz_1q^N,\dots,bz_{n}q^N;p)}\\
&\qquad\cdot\sum_{\substack{0\leq y_1,\dots,y_n\\y_1+\cdots+y_n=N}}\frac{\Delta(z_1q^{y_1},\dots,z_nq^{y_n};p)}{\Delta(z_1,\dots,z_n;p)}\prod_{k=1}^n\frac{(a_1z_k,\dots,a_{n+1}z_k;q,p)_{y_k}}{(bz_k,z_kq/z_1,\dots,z_kq/z_n;q,p)_{y_k}}\\
&=\frac{\theta(q;p)}{\theta(q^{N+1};p)}\sum_{\substack{0\leq y_1,\dots,y_n\\y_1+\cdots+y_n=N}}\frac{\Delta(z_1q^{y_1},\dots,z_nq^{y_n};p)}{\Delta(z_1,\dots,z_n;p)}\prod_{k=1}^n\frac{\theta(bz_kq^{N+y_k};p)(a_1z_k,\dots,a_{n+1}z_k;q,p)_{y_k}}{\theta(bz_kq^N;p)(bz_k,z_kq/z_1,\dots,z_kq/z_n;q,p)_{y_k}}\\
&\qquad\cdot\frac{\theta(bq^N/a_1,\dots,bq^N/a_{n+1};p)}{\theta(q,bz_1q^{N+y_1},\dots,bz_nq^{N+y_n};p)}
\end{align}
となる. ここで, 定理1の$N=1$の場合を用いると,
\begin{align}
&\frac{\theta(bq^N/a_1,\dots,bq^N/a_{n+1};p)}{\theta(q,bz_1q^{N+y_1},\dots,bz_nq^{N+y_n};p)}\\
&=\sum_{\substack{0\leq x_1,\dots,x_n\\x_1+\cdots+x_n=1}}\frac{\Delta(z_1q^{y_1+x_1},\dots,z_nq^{y_n+x_n};p)}{\Delta(z_1q^{y_1},\dots,z_nq^{y_n};p)}\prod_{k=1}^n\frac{(a_1z_kq^{y_k},\dots,a_{n+1}z_kq^{y_k};q,p)_{x_k}}{(bz_kq^{N+y_k},z_kq^{1+y_k-y_1}/z_1,\dots,z_kq^{1+y_k-y_n}/z_n;q,p)_{x_k}}
\end{align}
と展開できるので, これを代入して,
\begin{align}
R_{N+1}&=\frac{\theta(q;p)}{\theta(q^{N+1};p)}\sum_{\substack{0\leq y_1,\dots,y_n\\y_1+\cdots+y_n=N}}\frac{\Delta(z_1q^{y_1},\dots,z_nq^{y_n};p)}{\Delta(z_1,\dots,z_n;p)}\prod_{k=1}^n\frac{\theta(bz_kq^{N+y_k};p)(a_1z_k,\dots,a_{n+1}z_k;q,p)_{y_k}}{\theta(bz_kq^N;p)(bz_k,z_kq/z_1,\dots,z_kq/z_n;q,p)_{y_k}}\\
&\qquad\cdot\sum_{\substack{0\leq x_1,\dots,x_n\\x_1+\cdots+x_n=1}}\frac{\Delta(z_1q^{y_1+x_1},\dots,z_nq^{y_n+x_n};p)}{\Delta(z_1q^{y_1},\dots,z_nq^{y_n};p)}\prod_{k=1}^n\frac{(a_1z_kq^{y_k},\dots,a_{n+1}z_kq^{y_k};q,p)_{x_k}}{(bz_kq^{N+y_k},z_kq^{1+y_k-y_1}/z_1,\dots,z_kq^{1+y_k-y_n}/z_n;q,p)_{x_k}}\\
&=\frac{\theta(q;p)}{\theta(q^{N+1};p)}\sum_{\substack{0\leq y_1,\dots,y_n,x_1,\dots,x_n\\y_1+\cdots+y_n=N\\x_1+\cdots+x_n=1}}\frac{\Delta(z_1q^{y_1+x_1},\dots,z_nq^{y_n+x_n};p)}{\Delta(z_1,\dots,z_n;p)}\prod_{k=1}^n\frac{\theta(bz_kq^{N+y_k};p)(a_1z_k,\dots,a_{n+1}z_k;q,p)_{y_k+x_k}}{\theta(bz_kq^N;p)(bz_k,z_kq/z_1,\dots,z_kq/z_n;q,p)_{y_k}}\\
&\qquad\cdot\prod_{k=1}^n\frac{1}{(bz_kq^{N+y_k},z_kq^{1+y_k-y_1}/z_1,\dots,z_kq^{1+y_k-y_n}/z_n;q,p)_{x_k}}\\
&=\frac{\theta(q;p)}{\theta(q^{N+1};p)}\sum_{\substack{0\leq y_1,\dots,y_n,x_1,\dots,x_n\\y_1+\cdots+y_n=N+1\\x_1+\cdots+x_n=1}}\frac{\Delta(z_1q^{y_1},\dots,z_nq^{y_n};p)}{\Delta(z_1,\dots,z_n;p)}\prod_{k=1}^n\frac{\theta(bz_kq^{N+y_k-x_k};p)(a_1z_k,\dots,a_{n+1}z_k;q,p)_{y_k}}{\theta(bz_kq^N;p)(bz_k,z_kq/z_1,\dots,z_kq/z_n;q,p)_{y_k-x_k}}\\
&\qquad\cdot\prod_{k=1}^n\frac{1}{(bz_kq^{N+y_k-x_k},z_kq^{1+y_k-y_1-x_k+x_1}/z_1,\dots,z_kq^{1+y_k-y_n-x_k+x_n}/z_n;q,p)_{x_k}}\qquad(y_k\mapsto y_k-x_k)
\end{align}
ここで, $x_1+\cdots+x_n=1$であることから,
\begin{align}
\frac{\theta(bz_kq^{N+y_k-x_k};p)}{(bz_kq^{N+y_k-x_k};q,p)_{x_k}}&=\frac{\theta(bz_kq^{N+y_k};p)}{(bz_kq^{N+y_k};q,p)_{x_k}}\\
(A;q,p)_{y_k-x_k}&=\frac{(A;q,p)_{y_k}}{(Aq^{y_k-1};q,p)_{x_k}}
\end{align}
となることと,
\begin{align}
&\prod_{k=1}^n\frac{1}{(z_kq^{1+y_k-y_1-x_k+x_1}/z_1,\dots,z_kq^{1+y_k-y_n-x_k+x_n}/z_n;q,p)_{x_k}}\\
&=\frac 1{\theta(q;p)\prod_{\substack{1\leq j,k\leq n\\j\neq k}}(z_kq^{y_k-y_j}/z_j;q,p)_{x_k}}
\end{align}
となることを用いると,
\begin{align}
R_{N+1}&=\frac{1}{\theta(q^{N+1};p)}\sum_{\substack{0\leq y_1,\dots,y_n\\y_1+\cdots+y_n=N+1}}\frac{\Delta(z_1q^{y_1},\dots,z_nq^{y_n};p)}{\Delta(z_1,\dots,z_n;p)}\prod_{k=1}^n\frac{\theta(bz_kq^{N+y_k};p)(a_1z_k,\dots,a_{n+1}z_k;q,p)_{y_k}}{\theta(bz_kq^N;p)(bz_k,z_kq/z_1,\dots,z_kq/z_n;q,p)_{y_k}}\\
&\qquad\cdot\sum_{\substack{0\leq x_1,\dots,x_n\\x_1+\cdots+x_n=1}}\prod_{k=1}^n\frac{(bz_kq^{y_k-1},z_kq^{y_k}/z_1,\dots,z_kq^{y_k}/z_n;q,p)_{x_k}}{(bz_kq^{N+y_k};q,p)_{x_k}\prod_{\substack{1\leq j\leq n\\j\neq k}}(z_kq^{y_k-y_j}/z_j;q,p)_{x_k}}
\end{align}
となる. ここで,
\begin{align}
&\sum_{\substack{0\leq x_1,\dots,x_n\\x_1+\cdots+x_n=1}}\prod_{k=1}^n\frac{(bz_kq^{y_k-1},z_kq^{y_k}/z_1,\dots,z_kq^{y_k}/z_n;q,p)_{x_k}}{(bz_kq^{N+y_k};q,p)_{x_k}\prod_{\substack{1\leq j\leq n\\j\neq k}}(z_kq^{y_k-y_j}/z_j;q,p)_{x_k}}\\
&=\sum_{k=1}^n\frac{\theta(bz_kq^{y_k-1},z_kq^{y_k}/z_1,\dots,z_kq^{y_k}/z_n;p)}{\theta(bz_kq^{N+y_k};p)\prod_{\substack{1\leq j\leq n\\j\neq k}}\theta(z_kq^{y_k-y_j}/z_j;p)}
\end{align}
となることから, 先ほど示した式
\begin{align}
\sum_{j=1}^{n}\frac{\theta(a_1z_j,\dots,a_{n+1}z_j;p)}{\theta(bz_j)\prod_{\substack{1\leq k\leq n\\k\neq j}}\theta(z_j/z_k;p)}&=\frac{\theta(b/a_1,\dots,b/a_{n+1};p)}{\theta(bz_1,\dots,bz_n;p)}
\end{align}
において, $z_k\mapsto z_kq^{y_k},b\mapsto bq^N$として, $a_1,\dots,a_{n+1}$を$1/z_1,\dots,1/z_n,b/q$とすると,
\begin{align}
R_{N+1}&=\frac{1}{\theta(q^{N+1};p)}\sum_{\substack{0\leq y_1,\dots,y_n\\y_1+\cdots+y_n=N+1}}\frac{\Delta(z_1q^{y_1},\dots,z_nq^{y_n};p)}{\Delta(z_1,\dots,z_n;p)}\prod_{k=1}^n\frac{\theta(bz_kq^{N+y_k};p)(a_1z_k,\dots,a_{n+1}z_k;q,p)_{y_k}}{\theta(bz_kq^N;p)(bz_k,z_kq/z_1,\dots,z_kq/z_n;q,p)_{y_k}}\\
&\qquad\cdot\frac{\theta(bz_1q^N,\dots,bz_nq^N,q^{N+1};p)}{\theta(bz_1q^{N+y_1},\dots,bz_nq^{N+y_n};p)}\\
&=\sum_{\substack{0\leq y_1,\dots,y_n\\y_1+\cdots+y_n=N+1}}\frac{\Delta(z_1q^{y_1},\dots,z_nq^{y_n};p)}{\Delta(z_1,\dots,z_n;p)}\prod_{k=1}^n\frac{(a_1z_k,\dots,a_{n+1}z_k;q,p)_{y_k}}{(bz_k,z_kq/z_1,\dots,z_kq/z_n;q,p)_{y_k}}
\end{align}
となるので, $N+1$の場合が示せた. よって$N$に関する帰納法により, 示すべき定理を得る.
特に, $p=0$とすると以下の系を得る.
$n\geq 1,b=a_1\cdots a_{n+1}z_1\cdots z_n$であるとき,
\begin{align}
&\sum_{\substack{0\leq y_1,\dots,y_n\\y_1+\cdots+y_n=N}}\frac{\Delta(z_1q^{y_1},\dots,z_nq^{y_n})}{\Delta(z_1,\dots,z_n)}\prod_{k=1}^n\frac{(a_1z_k,\dots,a_{n+1}z_k;q)_{y_k}}{(bz_k,z_kq/z_1,\dots,z_kq/z_n;q)_{y_k}}\\
&=\frac{(b/a_1,\dots,b/a_{n+1};q)_N}{(q,bz_1,\dots,bz_n;q)_N}
\end{align}
が成り立つ.
さらに, $b,a_{n+1}$以外を固定して$a_{n+1}\to 0$とすると, 以下の系を得る.
$n\geq 1$とするとき,
\begin{align}
&\sum_{\substack{0\leq y_1,\dots,y_n\\y_1+\cdots+y_n=N}}\frac{\Delta(z_1q^{y_1},\dots,z_nq^{y_n})}{\Delta(z_1,\dots,z_n)}\prod_{k=1}^n\frac{(a_1z_k,\dots,a_{n}z_k;q)_{y_k}}{(z_kq/z_1,\dots,z_kq/z_n;q)_{y_k}}\\
&=\frac{(a_1\cdots a_nz_1\cdots z_n;q)_N}{(q;q)_N}
\end{align}
が成り立つ.
これは 前の記事 で示したMilneの二項定理である.
