"Pieri rule"의 두 판 사이의 차이
imported>Pythagoras0 |
imported>Pythagoras0 |
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==introduction== | ==introduction== | ||
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| − | expressed in generating function form | + | |
| + | ==Pieri rules for Schur polynomials== | ||
| + | * $S_{\lambda}$ denotes a Schur polynomial of $k$-variables | ||
| + | |||
| + | $$ | ||
| + | S_{\lambda}S_{(m,0\cdots, 0)}=S_{\nu} | ||
| + | $$ | ||
| + | where the sum is over all $\nu$ such that $\nu_1\geq \lambda_1\geq \nu_2 \geq \lambda_2\cdots \nu_k\geq \lambda_k\geq 0$ and $\sum \nu_j=m+\sum \lambda_j$ | ||
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| + | ==Pieri rules for Macdonal polynomials== | ||
| + | * $g$- and $e$-Pieri rules for Macdonald polynomials expressed in generating function form | ||
| + | ===$g$-Pieri case=== | ||
\begin{equation} | \begin{equation} | ||
P_{\mu}(q,t) \prod_{i\geq 1} \frac{(atx_i;q)_{\infty}}{(ax_i;q)_{\infty}} | P_{\mu}(q,t) \prod_{i\geq 1} \frac{(atx_i;q)_{\infty}}{(ax_i;q)_{\infty}} | ||
| 19번째 줄: | 31번째 줄: | ||
\frac{(t^{j-i+1};q)_{\mu_i-\lambda_{j+1}}}{(qt^{j-i};q)_{\mu_i-\lambda_{j+1}}}. | \frac{(t^{j-i+1};q)_{\mu_i-\lambda_{j+1}}}{(qt^{j-i};q)_{\mu_i-\lambda_{j+1}}}. | ||
\end{multline} | \end{multline} | ||
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| + | |||
| + | ===$e$-Pieri case=== | ||
Similarly, the $e$-Pieri rule is given by | Similarly, the $e$-Pieri rule is given by | ||
\begin{equation} | \begin{equation} | ||
2015년 10월 27일 (화) 00:34 판
introduction
Pieri rules for Schur polynomials
- $S_{\lambda}$ denotes a Schur polynomial of $k$-variables
$$ S_{\lambda}S_{(m,0\cdots, 0)}=S_{\nu} $$ where the sum is over all $\nu$ such that $\nu_1\geq \lambda_1\geq \nu_2 \geq \lambda_2\cdots \nu_k\geq \lambda_k\geq 0$ and $\sum \nu_j=m+\sum \lambda_j$
Pieri rules for Macdonal polynomials
- $g$- and $e$-Pieri rules for Macdonald polynomials expressed in generating function form
$g$-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}
$e$-Pieri case
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 $\lambda'_j>\mu_j'$.