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WarnaarによるJacobiの三重積の一般化

Jacobiの三重積は
$\begin{aligned} \sum_{n \in Z} (- 1)^{n} q^{\frac{n (n - 1)}{2}} a^{n} & = (a, q / a, q; q)_{\infty} \end{aligned}$
によって与えられる公式である. 2003年にWarnaarによって以下の美しい一般化が得られている.

Warnaar(2003)

$\begin{aligned} 1 + \sum_{0 < n} (- 1)^{n} q^{\frac{n (n - 1)}{2}} (a^{n} + b^{n}) & = (a, b, q; q)_{\infty} \sum_{0 \leq n} \frac{(a b / q; q)_{2 n}}{(a, b, a b, q; q)_{n}} q^{n} \end{aligned}$
が成り立つ.

$b = q / a$ とすれば, Jacobiの三重積が得られることが分かる.

示すべき等式は
$\begin{aligned} \sum_{0 \leq n} \frac{(a q^{n}, b q^{n}; q)_{\infty} (a b q^{n - 1}; q)_{n}}{(1 - a b q^{n - 1}) (q; q)_{n}} q^{n} & = \frac{1}{(1 - \frac{a b}{q}) (q; q)_{\infty}} (1 + \sum_{0 < n} (- 1)^{n} q^{(\binom{n}{2})} (a^{n} + b^{n})) \end{aligned}$
と書き換えられる. 右辺を $q$ 二項定理, 等比級数で展開すると,
$\begin{aligned} \sum_{0 \leq n} \frac{(a q^{n}, b q^{n}; q)_{\infty} (a b q^{n - 1}; q)_{n}}{(1 - a b q^{n - 1}) (q; q)_{n}} q^{n} \\ = \sum_{0 \leq n} \frac{q^{n}}{(q; q)_{n}} \sum_{0 \leq i, j, k, l} \frac{(- a q^{n})^{i} q^{(\binom{i}{2})}}{(q; q)_{i}} \frac{(- b q^{n})^{j} q^{(\binom{j}{2})}}{(q; q)_{j}} \frac{(q; q)_{n}}{(q; q)_{k} (q; q)_{n - k}} q^{(\binom{k}{2})} (- a b q^{n - 1})^{k} (a b q^{n - 1})^{l} \\ = \sum_{0 \leq n, i, j, k, l} \frac{(- 1)^{i + j + k} a^{i + k + l} b^{j + k + l} q^{(\binom{i}{2}) + (\binom{j}{2}) + (\binom{k}{2}) + n (i + j + k + l + 1) - k - l}}{(q; q)_{i} (q; q)_{j} (q; q)_{k} (q; q)_{n - k}} \\ = \sum_{0 \leq n, i, j, k, l} \frac{(- 1)^{i + j + k} a^{i + k + l} b^{j + k + l} q^{(\binom{i}{2}) + (\binom{j}{2}) + (\binom{k}{2}) + (n + k) (i + j + k + l + 1) - k - l}}{(q; q)_{i} (q; q)_{j} (q; q)_{k} (q; q)_{n}} \\ = \sum_{0 \leq i, j, k, l} \frac{(- 1)^{i + j + k} a^{i + k + l} b^{j + k + l} q^{(\binom{i}{2}) + (\binom{j}{2}) + (\binom{k}{2}) + k (i + j + k + l + 1) - k - l}}{(q; q)_{i} (q; q)_{j} (q; q)_{k}} \sum_{0 \leq n} \frac{q^{n (i + j + k + l + 1)}}{(q; q)_{n}} \\ = \sum_{0 \leq i, j, k, l} \frac{(- 1)^{i + j + k} a^{i + k + l} b^{j + k + l} q^{(\binom{i}{2}) + (\binom{j}{2}) + (\binom{k}{2}) + k (i + j + k + l + 1) - k - l}}{(q; q)_{i} (q; q)_{j} (q; q)_{k}} \frac{1}{(q^{i + j + k + l + 1}; q)_{\infty}} \\ = \frac{1}{(q; q)_{\infty}} \sum_{0 \leq i, j, k, l} \frac{(- 1)^{i + j + k} a^{i + k + l} b^{j + k + l} q^{(\binom{i}{2}) + (\binom{j}{2}) + (\binom{k}{2}) + k (i + j + k + l + 1) - k - l} (q; q)_{i + j + k + l}}{(q; q)_{i} (q; q)_{j} (q; q)_{k}} \\ = \frac{1}{(q; q)_{\infty}} \sum_{0 \leq i, j, k, l} \frac{(- 1)^{i + j - k} a^{i + l} b^{j + l} q^{(\binom{i - k}{2}) + (\binom{j - k}{2}) + (\binom{k}{2}) + k (i + j - k + l + 1) - k - l} (q; q)_{i + j + l - k}}{(q; q)_{i - k} (q; q)_{j - k} (q; q)_{k}} \\ = \frac{1}{(q; q)_{\infty}} \sum_{0 \leq i, j, l} (- 1)^{i + j} a^{i + l} b^{j + l} q^{(\binom{i}{2}) + (\binom{j}{2}) - l} \sum_{0 \leq k} \frac{(- 1)^{k} q^{(\binom{k + 1}{2}) + k l} (q; q)_{i + j + l - k}}{(q; q)_{i - k} (q; q)_{j - k} (q; q)_{k}} \end{aligned}$
ここで, $q$ -Vandermondeの恒等式より,
$\begin{aligned} \sum_{0 \leq k} \frac{(- 1)^{k} q^{(\binom{k + 1}{2}) + k l} (q; q)_{i + j + l - k}}{(q; q)_{i - k} (q; q)_{j - k} (q; q)_{k}} \\ = \frac{(q; q)_{i + j + l}}{(q; q)_{i} (q; q)_{j}}_{2} ϕ_{1} [\begin{array}{c} q^{- i}, q^{- j} \\ q^{- i - j - l} \end{array}; q] \\ = \frac{(q; q)_{i + j + l}}{(q; q)_{i} (q; q)_{j}} \frac{(q^{- i - l}; q)_{i}}{(q^{- i - j - l}; q)_{i}} q^{- i j} \\ = \frac{(q; q)_{i + j + l}}{(q; q)_{i} (q; q)_{j}} \frac{(q^{l + 1}; q)_{i}}{(q^{j + l + 1}; q)_{i}} \\ = \frac{(q; q)_{i + l} (q; q)_{j + l}}{(q; q)_{i} (q; q)_{j} (q; q)_{l}} \end{aligned}$
であるから, これを代入して,
$\begin{aligned} \frac{1}{(q; q)_{\infty}} \sum_{0 \leq i, j, l} (- 1)^{i + j} a^{i + l} b^{j + l} q^{(\binom{i}{2}) + (\binom{j}{2}) - l} \sum_{0 \leq k} \frac{(- 1)^{k} q^{(\binom{k + 1}{2}) + k l} (q; q)_{i + j + l - k}}{(q; q)_{i - k} (q; q)_{j - k} (q; q)_{k}} \\ = \frac{1}{(q; q)_{\infty}} \sum_{0 \leq i, j, l} (- 1)^{i + j} a^{i + l} b^{j + l} q^{(\binom{i}{2}) + (\binom{j}{2}) - l} \frac{(q; q)_{i + l} (q; q)_{j + l}}{(q; q)_{i} (q; q)_{j} (q; q)_{l}} \\ = \frac{1}{(q; q)_{\infty}} \sum_{0 \leq i, j, l} (- 1)^{i + j} a^{i} b^{j} q^{(\binom{i - l}{2}) + (\binom{j - l}{2}) - l} \frac{(q; q)_{i} (q; q)_{j}}{(q; q)_{i - l} (q; q)_{j - l} (q; q)_{l}} \\ = \frac{1}{(q; q)_{\infty}} \sum_{0 \leq i, j} (- 1)^{i + j} a^{i} b^{j} q^{(\binom{i}{2}) + (\binom{j}{2})} (q; q)_{i} (q; q)_{j} \sum_{0 \leq l} \frac{q^{l (l - i - j)}}{(q; q)_{i - l} (q; q)_{j - l} (q; q)_{l}} \end{aligned}$
ここで, 再び $q$ -Vandermondeの恒等式より,
$\begin{aligned} \sum_{0 \leq l} \frac{q^{l (l - i - j)}}{(q; q)_{i - l} (q; q)_{j - l} (q; q)_{l}} \\ = \frac{1}{(q; q)_{i} (q; q)_{j}}_{2} ϕ_{1} [\begin{array}{c} q^{- i}, q^{- j} \\ 0 \end{array}; q] \\ = \frac{1}{(q; q)_{i} (q; q)_{j}} q^{- i j} \end{aligned}$
であるから, これを代入して,
$\begin{aligned} \frac{1}{(q; q)_{\infty}} \sum_{0 \leq i, j} (- 1)^{i + j} a^{i} b^{j} q^{(\binom{i}{2}) + (\binom{j}{2})} (q; q)_{i} (q; q)_{j} \sum_{0 \leq l} \frac{q^{l (l - i - j)}}{(q; q)_{i - l} (q; q)_{j - l} (q; q)_{l}} \\ = \frac{1}{(q; q)_{\infty}} \sum_{0 \leq i, j} (- 1)^{i + j} a^{i} b^{j} q^{(\binom{i}{2}) + (\binom{j}{2}) - i j} \\ = \frac{1}{(q; q)_{\infty}} \sum_{\begin{array}{c} 0 \leq j \\ - j \leq i \end{array}} (- 1)^{i} a^{i + j} b^{j} q^{(\binom{i + j}{2}) + (\binom{j}{2}) - i j - j^{2}} \\ = \frac{1}{(q; q)_{\infty}} \sum_{\begin{array}{c} 0 \leq j \\ - j \leq i \end{array}} (- 1)^{i} a^{i + j} b^{j} q^{(\binom{i}{2}) - j} \\ = \frac{1}{(q; q)_{\infty}} \sum_{\begin{array}{c} i \in Z \\ \max {0, - i} \leq j \end{array}} (- 1)^{i} a^{i + j} b^{j} q^{(\binom{i}{2}) - j} \\ = \frac{1}{(1 - \frac{a b}{q}) (q; q)_{\infty}} \sum_{\begin{array}{c} i \in Z \end{array}} (- 1)^{i} a^{i} (a b / q)^{\max {0, - i}} q^{(\binom{i}{2})} \\ = \frac{1}{(1 - \frac{a b}{q}) (q; q)_{\infty}} (1 + \sum_{0 < n} (- a)^{n} q^{(\binom{n}{2})} + \sum_{n < 0} (- 1)^{n} b^{- n} q^{(\binom{n + 1}{2})}) \\ = \frac{1}{(1 - \frac{a b}{q}) (q; q)_{\infty}} (1 + \sum_{0 < n} (- a)^{n} q^{(\binom{n}{2})} + \sum_{0 < n} (- 1)^{n} b^{n} q^{(\binom{n}{2})}) \end{aligned}$
となって示すべき等式が得られた.

$b = - a$ とした後, $a \mapsto \sqrt{a q}$ として, 以下の系を得る.

$\begin{aligned} 1 + 2 \sum_{0 < n} a^{n} q^{2 n^{2}} & = (q; q)_{\infty} (a q; q^{2})_{\infty} \sum_{0 \leq n} \frac{(- a; q)_{2 n} q^{n}}{(q, - a q; q)_{n} (a q; q^{2})_{n}} \end{aligned}$