The universal DAHA of type $(C_1^\vee,C_1)$ and Leonard triples

Hau-wen Huang
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引用次数: 3

Abstract

Assume that $\mathbb F$ is an algebraically closed field and $q$ is a nonzero scalar in $\mathbb F$ that is not a root of unity. The universal Askey--Wilson algebra $\triangle_q$ is a unital associative $\mathbb F$-algebra generated by $A,B, C$ and the relations state that each of $$ A+\frac{q BC-q^{-1} CB}{q^2-q^{-2}}, \qquad B+\frac{q CA-q^{-1} AC}{q^2-q^{-2}}, \qquad C+\frac{q AB-q^{-1} BA}{q^2-q^{-2}} $$ is central in $\triangle_q$. The universal DAHA $\mathfrak H_q$ of type $(C_1^\vee,C_1)$ is a unital associative $\mathbb F$-algebra generated by $\{t_i^{\pm 1}\}_{i=0}^3$ and the relations state that \begin{gather*} t_it_i^{-1}=t_i^{-1} t_i=1 \quad \hbox{for all $i=0,1,2,3$}; \\ \hbox{$t_i+t_i^{-1}$ is central} \quad \hbox{for all $i=0,1,2,3$}; \\ t_0t_1t_2t_3=q^{-1}. \end{gather*} It was given an $\mathbb F$-algebra homomorphism $\triangle_q\to \mathfrak H_q$ that sends \begin{eqnarray*} A &\mapsto & t_1 t_0+(t_1 t_0)^{-1}, \\ B &\mapsto & t_3 t_0+(t_3 t_0)^{-1}, \\ C &\mapsto & t_2 t_0+(t_2 t_0)^{-1}. \end{eqnarray*} Therefore any $\mathfrak H_q$-module can be considered as a $\triangle_q$-module. Let $V$ denote a finite-dimensional irreducible $\mathfrak H_q$-module. In this paper we show that $A,B,C$ are diagonalizable on $V$ if and only if $A,B,C$ act as Leonard triples on all composition factors of the $\triangle_q$-module $V$.
类型$(C_1^\vee,C_1)$和伦纳德三元组的通用DAHA
假设 $\mathbb F$ 代数闭场是和吗 $q$ 一个非零的标量在里面吗 $\mathbb F$ 这不是团结的根源。通用的Askey- Wilson代数 $\triangle_q$ 是单位结合律吗 $\mathbb F$-代数由 $A,B, C$ 关系式表明每一个 $$ A+\frac{q BC-q^{-1} CB}{q^2-q^{-2}}, \qquad B+\frac{q CA-q^{-1} AC}{q^2-q^{-2}}, \qquad C+\frac{q AB-q^{-1} BA}{q^2-q^{-2}} $$ 是市中心 $\triangle_q$. 普遍的DAHA $\mathfrak H_q$ 类型 $(C_1^\vee,C_1)$ 是单位结合律吗 $\mathbb F$-代数由 $\{t_i^{\pm 1}\}_{i=0}^3$ 关系式表明 \begin{gather*} t_it_i^{-1}=t_i^{-1} t_i=1 \quad \hbox{for all $i=0,1,2,3$}; \\ \hbox{$t_i+t_i^{-1}$ is central} \quad \hbox{for all $i=0,1,2,3$}; \\ t_0t_1t_2t_3=q^{-1}. \end{gather*} 它被赋予了 $\mathbb F$-代数同态 $\triangle_q\to \mathfrak H_q$ 这就把 \begin{eqnarray*} A &\mapsto & t_1 t_0+(t_1 t_0)^{-1}, \\ B &\mapsto & t_3 t_0+(t_3 t_0)^{-1}, \\ C &\mapsto & t_2 t_0+(t_2 t_0)^{-1}. \end{eqnarray*} 因此任何 $\mathfrak H_q$-module可以看作是 $\triangle_q$-module。让 $V$ 表示有限维不可约 $\mathfrak H_q$-module。在本文中,我们证明了这一点 $A,B,C$ 是对角化的吗 $V$ 当且仅当 $A,B,C$ 作为伦纳德三倍的所有组成因子 $\triangle_q$-模块 $V$.
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