Pieri rule

introduction

We also need the $g$- and $e$-Pieri rules for Macdonald polynomials expressed in generating function form. First, in the $g$-Pieri case, \begin{equation} P_{\mu}(q,t) \prod_{i\geq 1} \frac{(atx_i;q)_{\infty}}{(ax_i;q)_{\infty}} =\sum_{\lambda\supseteq\mu} a^{\lvert \lambda/\mu \rvert} \varphi_{\lambda/\mu}(q,t) P_{\lambda}(q,t). \end{equation} Here the Pieri coefficient $\varphi_{\lambda/\mu}(q,t)=0$ unless $\lambda/\mu$ is a horizontal strip, in which case \begin{multline}\label{Eq_varphi} \varphi_{\lambda/\mu}(q,t)= \prod_{1\leq i\leq j\leq l(\lambda)} \frac{(qt^{j-i};q)_{\lambda_i-\lambda_j}}{(t^{j-i+1};q)_{\lambda_i-\lambda_j}}\cdot \frac{(qt^{j-i};q)_{\mu_i-\mu_{j+1}}}{(t^{j-i+1};q)_{\mu_i-\mu_{j+1}}} \\ \times \frac{(t^{j-i+1};q)_{\lambda_i-\mu_j}}{(qt^{j-i};q)_{\lambda_i-\mu_j}}\cdot \frac{(t^{j-i+1};q)_{\mu_i-\lambda_{j+1}}}{(qt^{j-i};q)_{\mu_i-\lambda_{j+1}}}. \end{multline} Similarly, the $e$-Pieri rule is given by \begin{equation} P_{\mu}(x;q,t)\prod_{i\geq 1} (1+ax_i)= \sum_{\lambda\supseteq\mu} a^{\lvert \lambda/\mu \rvert} \psi'_{{\lambda}/{\mu}}q,t) P_{\lambda}(x;q,t), \end{equation} where $\psi'_{{\lambda}/{\mu}}(q,t)$ is zero unless $\lambda/\mu$ is a vertical strip, in which case \cite[page 336]{Macdonald95} \begin{equation}\label{Eq_psip} \psi'_{{\lambda}/{\mu}}(q,t) = \prod \frac{1-q^{\mu_i-\mu_j}t^{j-i-1}}{1-q^{\mu_i-\mu_j}t^{j-i}}\cdot \frac{1-q^{\lambda_i-\lambda_j}t^{j-i+1}}{1-q^{\lambda_i-\lambda_j}t^{j-i}}. \end{equation} The product in the above is over all $i<j$ such that $\lambda_i=\mu_i$ and $\lambda_j>\mu_j$. An alternative expression for $\psi'_{{\lambda}/{\mu}}(q,t)$ is given by \cite[page 340]{Macdonald95} \begin{equation}\label{Eq_psip340} \psi'_{{\lambda}/{\mu}}(q,t)=\prod \frac{b_{\lambda}(s;q,t)}{b_{\mu}(s;q,t)} \end{equation} where the product is over all squares $s=(i,j)\in\mu\subseteq\lambda$ such that $i<j$, $\mu_i=\lambda_i$ and $\lambda'_j>\mu_j'$.