Quantized mixed tensor space and Schur–Weyl duality

Abstract

Let $R$ be a commutative ring with $1$ and $q$ an invertible element of $R$. The (specialized) quantum group $U=U_q\left({\mathfrak{g}\mathfrak{l}}_n\right)$ over $R$ of the general linear group acts on mixed tensor space $V^{\otimes r}\otimes {V^{\ast }}^{\otimes s}$, where $V$ denotes the natural $U$-module $R^n$, $r$ and $s$ are nonnegative integers and $V^{\ast }$ is the dual $U$-module to $V$. The image of $U$ in ${End}_R\left(V^{\otimes r}\otimes {V^{\ast }}^{\otimes s}\right)$ is called the rational $q$-Schur algebra $S_q\left(n;r,s\right)$. We construct a bideterminant basis of $S_q\left(n;r,s\right)$. There is an action of a $q$-deformation ${\mathfrak{B}}_{r,s}^n\left(q\right)$ of the walled Brauer algebra on mixed tensor space centralizing the action of $U$. We show that ${End}_{{\mathfrak{B}}_{r,s}^n\left(q\right)}\left(V^{\otimes r}\otimes {V^{\ast }}^{\otimes s}\right)=S_q\left(n;r,s\right)$. By a previous result, the image of ${\mathfrak{B}}_{r,s}^n\left(q\right)$ in ${End}_R\left(V^{\otimes r}\otimes {V^{\ast }}^{\otimes s}\right)$ is ${End}_U\left(V^{\otimes r}\otimes {V^{\ast }}^{\otimes s}\right)$. Thus, a mixed tensor space as $\left(U,{\mathfrak{B}}_{r,s}^n\left(q\right)\right)$-bimodule satisfies Schur–Weyl duality.