"Pieri rule"의 두 판 사이의 차이

수학노트
둘러보기로 가기 검색하러 가기
imported>Pythagoras0
imported>Pythagoras0
1번째 줄: 1번째 줄:
 +
==introduction==
 +
* special case of [[Littlewood-Richardson rule]]
 +
* expansion of the product of a Schur function with a complete homogeneous symmetric polynomial or an elementary symmetric polynomial
 +
* representation theoretically, it corresponds to a tensor product in which one of the factors is a symmetric or exterior
 +
power of the defining representation
 +
* $g$-Pieri is related to complete homogeneous symmetric polynomial
 +
* $e$-Pieri is dual to $g$-pieri and is related to complete elementary symmetric polynomial
 +
* in more geometric setting, let $G$ be a classical Lie group and $P$ a parabolic subgroup. The Pieri rule is a combinatorial formula describing the multiplication by a special Schubert class in the (quantum) cohomology ring of the homogeneous space $X=G/P$.
  
 +
==Pieri rules for Schur polynomials==
 +
* $S_{\lambda}$ denotes a Schur polynomial of $k$-variables
 +
$$
 +
S_{\lambda}S_{(m,0\cdots, 0)}=\sum_{\nu} S_{\nu}
 +
$$
 +
where the sum is over all $\nu$ such that $\nu_1\geq \lambda_1\geq \nu_2 \geq \lambda_2\cdots \geq \nu_k\geq \lambda_k\geq 0$ and $\sum \nu_j=m+\sum \lambda_j$
 +
 +
 +
===example===
 +
* $S_{(2,1)}S_{(2)}=S_{(4,1)}+S_{(3,2)}+S_{(3,1,1)}+S_{(2,2,1)}$
 +
 +
 +
===generating function form===
 +
* recall that $S_{(m,0\cdots, 0)}=H_m$ and
 +
$$
 +
\prod_{i\geq 1}^k \frac{1}{1-a x_i } = \sum_{j=0}^{\infty}H_ja^j
 +
$$
 +
* thus
 +
$$
 +
S_{\mu}\prod_{i\geq 1} \frac{1}{1-a x_i}=\sum_{\lambda\supseteq\mu} a^{\lvert \lambda/\mu \rvert} \varphi_{\lambda/\mu}S_{\lambda}
 +
$$
 +
where $\varphi_{\lambda/\mu}=1$ only when $\lambda/\mu$ is a horizontal strip and zero otherwise
 +
 +
==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 $\lambda'_j>\mu_j'$.
 +
 +
==related items==
 +
* [[Branching rules for Macdonald polynomials]]
 +
 +
 +
==computational resource==
 +
* https://drive.google.com/file/d/0B8XXo8Tve1cxekhvdXh2bFctX3M/view
 +
 +
[[분류:symmetric polynomials]]
 +
 +
== articles ==
 +
 +
* Soichi Okada, Pieri rules for classical groups and equinumeration between generalized oscillating tableaux and semistandard tableaux, arXiv:1606.02375 [math.CO], June 08 2016, http://arxiv.org/abs/1606.02375
 +
[[분류:migrate]]

2020년 11월 13일 (금) 02:19 판

introduction

  • special case of Littlewood-Richardson rule
  • expansion of the product of a Schur function with a complete homogeneous symmetric polynomial or an elementary symmetric polynomial
  • representation theoretically, it corresponds to a tensor product in which one of the factors is a symmetric or exterior

power of the defining representation

  • $g$-Pieri is related to complete homogeneous symmetric polynomial
  • $e$-Pieri is dual to $g$-pieri and is related to complete elementary symmetric polynomial
  • in more geometric setting, let $G$ be a classical Lie group and $P$ a parabolic subgroup. The Pieri rule is a combinatorial formula describing the multiplication by a special Schubert class in the (quantum) cohomology ring of the homogeneous space $X=G/P$.

Pieri rules for Schur polynomials

  • $S_{\lambda}$ denotes a Schur polynomial of $k$-variables

$$ S_{\lambda}S_{(m,0\cdots, 0)}=\sum_{\nu} S_{\nu} $$ where the sum is over all $\nu$ such that $\nu_1\geq \lambda_1\geq \nu_2 \geq \lambda_2\cdots \geq \nu_k\geq \lambda_k\geq 0$ and $\sum \nu_j=m+\sum \lambda_j$


example

  • $S_{(2,1)}S_{(2)}=S_{(4,1)}+S_{(3,2)}+S_{(3,1,1)}+S_{(2,2,1)}$


generating function form

  • recall that $S_{(m,0\cdots, 0)}=H_m$ and

$$ \prod_{i\geq 1}^k \frac{1}{1-a x_i } = \sum_{j=0}^{\infty}H_ja^j $$

  • thus

$$ S_{\mu}\prod_{i\geq 1} \frac{1}{1-a x_i}=\sum_{\lambda\supseteq\mu} a^{\lvert \lambda/\mu \rvert} \varphi_{\lambda/\mu}S_{\lambda} $$ where $\varphi_{\lambda/\mu}=1$ only when $\lambda/\mu$ is a horizontal strip and zero otherwise

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 $\lambda'_j>\mu_j'$.

related items


computational resource

articles

  • Soichi Okada, Pieri rules for classical groups and equinumeration between generalized oscillating tableaux and semistandard tableaux, arXiv:1606.02375 [math.CO], June 08 2016, http://arxiv.org/abs/1606.02375