Using a $q$-shuffle algebra to describe the basic module $V(\Lambda_0)$ for the quantized enveloping algebra $U_q(\widehat{\mathfrak{sl}}_2)$

TL;DR

This paper describes the basic module of the quantum affine algebra U_q(sl_2) using a q-shuffle algebra, establishing an explicit algebraic and module-theoretic correspondence.

Contribution

It introduces a novel realization of the basic module V(Λ₀) via a q-shuffle algebra structure, connecting it to Rosso's algebra and providing explicit module isomorphisms.

Findings

The subalgebra generated by x,y in the q-shuffle algebra is isomorphic to the positive part of U_q(𝔰𝔩̂₂).

The module generated by the empty word is isomorphic to V(Λ₀).

The subspace of words not starting with y or xx equals the intersection of the algebra with a specific subspace.

Abstract

We consider the quantized enveloping algebra $U_q(\widehat{\mathfrak{sl}}_2)$ and its basic module $V(\Lambda_0)$. This module is infinite-dimensional, irreducible, integrable, and highest-weight. We describe $V(\Lambda_0)$ using a $q$-shuffle algebra in the following way. Start with the free associative algebra $\mathbb V$ on two generators $x,y$. The standard basis for $\mathbb V$ consists of the words in $x,y$. In 1995 M. Rosso introduced an associative algebra structure on $\mathbb V$, called a $q$-shuffle algebra. For $u,v\in \lbrace x,y\rbrace$ their $q$-shuffle product is $u\star v = uv+q^{(u,v)}vu$, where $( u,v) =2$ (resp. $(u,v) =-2$) if $u=v$ (resp. $u\not=v$). Let $\mathbb U$ denote the subalgebra of the $q$-shuffle algebra $\mathbb V$ that is generated by $x, y$. Rosso showed that the algebra $\mathbb U$ is isomorphic to the positive part of $U_q(\widehat{\mathfrak{sl}}_2)$. In our first main result, we turn $\mathbb U$ into a $U_q(\widehat{\mathfrak{sl}}_2)$-module. Let $\bf U$ denote the $U_q(\widehat{\mathfrak{sl}}_2)$-submodule of $\mathbb U$ generated by the empty word. In our second main result, we show that the $U_q(\widehat{\mathfrak{sl}}_2)$-modules $\bf U$ and $V(\Lambda_0)$ are isomorphic. Let $\bf V$ denote the subspace of $\mathbb V$ spanned by the words that do not begin with $y$ or $xx$. In our third main result, we show that $\bf U = \mathbb U \cap {\bf V}$.