"BCn interpolation polynomials"의 두 판 사이의 차이
imported>Pythagoras0 |
imported>Pythagoras0 |
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| 1번째 줄: | 1번째 줄: | ||
==introduction== | ==introduction== | ||
===notation=== | ===notation=== | ||
| − | We | + | * We define relations $\prec$ and $\succ$ such that $\kappa\prec\lambda$ (equivalently $\lambda\succ\kappa$) for two partitions iff $\lambda/\kappa$ is a vertical strip; that is, $\kappa_i\le \lambda_i\le \kappa_i+1$ for all $i$. |
| − | $\prec$ and $\succ$ such that $\kappa\prec\lambda$ (equivalently | + | * we frequently use the product of the form |
| − | $\lambda\succ\kappa$) for two partitions iff $\lambda/\kappa$ is a vertical | + | \[ |
| − | strip; that is, $\kappa_i\le \lambda_i\le \kappa_i+1$ for all $i$. | + | \prod_{(i,j)\in \lambda} f(i,j), |
| − | + | \] | |
| + | where $(i,j)\in \lambda$ means that $1\le i$ and $1\le j\le | ||
| + | \lambda'_i$ | ||
| + | * let | ||
\begin{align} | \begin{align} | ||
C^+_\lambda(x;q,t)&:=\prod_{(i,j)\in \lambda} (1-q^{\lambda_i+j-1} | C^+_\lambda(x;q,t)&:=\prod_{(i,j)\in \lambda} (1-q^{\lambda_i+j-1} | ||
2015년 8월 24일 (월) 23:29 판
introduction
notation
- We define relations $\prec$ and $\succ$ such that $\kappa\prec\lambda$ (equivalently $\lambda\succ\kappa$) for two partitions iff $\lambda/\kappa$ is a vertical strip; that is, $\kappa_i\le \lambda_i\le \kappa_i+1$ for all $i$.
- we frequently use the product of the form
\[ \prod_{(i,j)\in \lambda} f(i,j), \] where $(i,j)\in \lambda$ means that $1\le i$ and $1\le j\le \lambda'_i$
- let
\begin{align} C^+_\lambda(x;q,t)&:=\prod_{(i,j)\in \lambda} (1-q^{\lambda_i+j-1} t^{2-\lambda'_j-i} x)\\ &\phantom{:}= \prod_{1\le i\le l} \frac{(q^{\lambda_i} t^{2-l-i} x;q)} {(q^{2\lambda_i} t^{2-2i} x;q)} \prod_{1\le i<j\le l} \frac{(q^{\lambda_i+\lambda_j} t^{3-i-j} x;q)} {(q^{\lambda_i+\lambda_j} t^{2-i-j} x;q)},\\ C^-_\lambda(x;q,t)&:=\prod_{(i,j)\in \lambda} (1-q^{\lambda_i-j} t^{\lambda'_j-i} x)\\ &\phantom{:}= \prod_{1\le i\le l} \frac{(x;q)} {(q^{\lambda_i} t^{l-i} x;q)} \prod_{1\le i<j\le l} \frac{(q^{\lambda_i-\lambda_j} t^{j-i} x;q)} {(q^{\lambda_i-\lambda_j} t^{j-i-1} x;q)},\\ C^0_\lambda(x;q,t)&:=\prod_{(i,j)\in \lambda} (1-q^{j-1} t^{1-i} x)\\ &\phantom{:}= \prod_{1\le i\le l} (t^{1-i} x;q)_{\lambda_i}. \end{align}
branching rule
- thm (3.9)
We have \[ \bar{P}^{*(n+m)}_\lambda(x_1,x_2,\dots x_n,t^{m-1} v,t^{m-2} v,\dots v;q,t,s) = \sum_{\substack{\mu\subset\lambda\\\ell(\mu)\le n}} \psi^{(B)}_{\lambda/\mu}(v,vt^m;q,t,s t^n) \bar{P}^{*(n)}_\mu(x_1,x_2,\dots x_n;q,t,s), \] where \[ \psi^{(B)}_{\lambda/\mu}(v,v';q,t,s) = \frac{ C^0_\lambda(s/v;q,t) C^0_\lambda(t/sv';1/q,1/t)} { C^0_\mu(s/v;q,t) C^0_\mu(t/sv';1/q,1/t)} P_{\lambda/\mu}(\left[\frac{v^k-v^{\prime k}}{1-t^k}\right];q,t) \]
- cor (3.10)
We have \[ \bar{P}^{*(n+1)}_\lambda(x_1,x_2,\dots x_n,v;q,t,s) = \sum_{\substack{\mu'\prec\lambda'\\\mu_{n+1}=0}} \psi^{(b)}_{\lambda/\mu}(v;q,t,s t^n) \bar{P}^{*(n)}_\mu(x_1,x_2,\dots x_n;q,t,s), \] where \[ \psi^{(b)}_{\lambda/\mu}(v;q,t,s) = \psi_{\lambda/\mu}(q,t) \prod_{(i,j)\in \lambda/\mu} (v+1/v-q^{j-1} t^{1-i}s-q^{1-j}t^{i-1}/s) \]
articles
- Rains, Eric M. “BCn-Symmetric Polynomials.” Transformation Groups 10, no. 1 (March 2005): 63–132. doi:10.1007/s00031-005-1003-y. http://arxiv.org/abs/math/0112035.
- Okounkov, A. “BC-Type Interpolation Macdonald Polynomials and Binomial Formula for Koornwinder Polynomials.” Transformation Groups 3, no. 2 (June 1998): 181–207. doi:10.1007/BF01236432.