Lie polynomials in $q$-deformed Heisenberg algebras

TL;DR

This paper investigates the structure of Lie polynomials within the $q$-deformed Heisenberg algebra, revealing how the algebra's relations influence its Lie subalgebra, especially when $q$ is not a root of unity.

Contribution

It characterizes the Lie subalgebra generated by $A,B$ in the $q$-deformed Heisenberg algebra, especially showing that for non-root of unity $q$, it forms a Lie ideal with a specific quotient structure.

Findings

If $q$ is not a root of unity, $ ext{Lie}(q)$ is a Lie ideal of $ ext{Heisenberg}(q)$.

The quotient Lie algebra is infinite-dimensional and one-step nilpotent.

The relation $AB - qBA = I$ influences the Lie algebra structure despite not being expressible solely in Lie algebra operations.

Abstract

Let $\mathbb{F}$ be a field, and let $q\in\mathbb{F}$. The $q$-deformed Heisenberg algebra is the unital associative $\mathbb{F}$-algebra $\mathcal{H}(q)$ with generators $A,B$ and relation $AB-qBA=I$, where $I$ is the multiplicative identity in $\mathcal{H}(q)$. The set of all Lie polynomials in $A,B$ is the Lie subalgebra $\mathcal{L}(q)$ of $\mathcal{H}(q)$ generated by $A,B$. If $q\neq 1$ or the characteristic of $\mathbb{F}$ is not $2$, then the equation $AB-qBA=I$ cannot be expressed in terms of Lie algebra operations only, yet this equation still has consequences on the Lie algebra structure of $\mathcal{L}(q)$, which we investigate. We show that if $q$ is not a root of unity, then $\mathcal{L}(q)$ is a Lie ideal of $\mathcal{H}(q)$, and the resulting quotient Lie algebra is infinite-dimensional and one-step nilpotent.