Quantized mixed tensor space and Schur-Weyl duality

Abstract

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