Macdonald identities and multidimensional theta-functions (Q1966343)

Let \(S_n\) be the symmetric group of order \(n\) and \(\# \sigma\) be the sign of the permutation \(\sigma\in S_n\). When \(\sigma=(p_0\quad p_1\cdots p_{n-1})\), define \(I(\sigma)\) by \[ I(\sigma) :=\{l\in\mathbb Z: l_j\equiv j-p_j\pmod n, j=1,\cdots, n-1\}. \] Set \[ F_n:=F_n(y_1,\cdots, y_{n-1},q):= \sum_{\sigma\in S_n} (-1)^{\# \sigma}\sum_{l\in I(\sigma)} y_1^{l_1}y_2^{l_2}\cdots y_{n-1}^{l_{n-1}}q^{A(\mathbf{l})}, \] where \[ A(\mathbf{l})= A(l_1,\cdots,l_{n-1}) =\frac{1}{2}\left(\sum_{r=1}^{n-1} l_r^2+\left(\sum_{r=1}^{n-1} l_r\right)^2\right), \] and \[ \begin{aligned} G_n:= G_n(y_1,\cdots, y_{n-1},q)&:=\prod_{m=1}^\infty\{\prod_{1\leq j\leq n-1} [(1-y_jq^{mn-n+j})(1-y_j^{-1}q^{mn-j})]\\ &\times \prod_{1\leq j<k\leq n-1}[(1-y_ky_j^{-1}q^{mn-n+k-j})(1-y_k^{-1}q^{mn-k+j})]\}.\end{aligned} \] In this paper, the author gives a new proof of the well-known Macdonald identity \[ F_n(y_1,\cdots, y_{n-1},q)= \prod_{m=1}^\infty (1-q^{mn})^{n-1}G_n(y_1,\cdots, y_{n-1},q). \] By setting \(y_j = e^{2\pi i z_j}\), \(z_j\in \mathbb C\) and \(q=e^{\pi i\tau}\), \(\tau\in\mathbb H\), the author first shows that the functions \(F_n \) and \(G_n \) are analytic functions in \(\mathbb C^{n-1}\times \mathbb H\) invariant under the translations \((z_1,\cdots,z_{n-1}) \leftarrow (z_1,\cdots, z_{n-1})+ (m_1\cdots m_{n-1})\) and \((z_1,\cdots,z_{n-1}) \leftarrow (z_1,\cdots, z_{n-1})+ (m_1\cdots m_{n-1})\Omega\), where \(m_j\in\mathbb Z, j=1\cdots n-1\) and \(\Omega = nA\tau\), with \[ A=\frac{1}{2}\left(\begin{matrix} 2 & 1 &\cdots & 1 & 1 \\ 1 & 2 &\cdots & 1 & 1\\ \vdots &\vdots &\vdots & \vdots &\vdots \\ 1 &1 &\cdots &1 &2\end{matrix}\right). \] The author then compares the zeros of \(F_n\) and \(G_n\) and conclude that \(F_n/G_n\) is a function of \(q\). The proof of Macdonald's identity is then completed by determining the quotient \(F_n/G_n\) using the differential operator \(\mathcal L\) defined by \[ \mathcal L=\frac{\partial}{\partial \tau}-\frac{1}{4\pi i}\left(\sum_{j=1}^{n-1} \frac{\partial^2}{\partial z_j^2} +\sum_{1\leq k<j\leq n-1}\frac{\partial^2}{\partial z_k\partial z_j}\right). \]