Products of commutators in a Lie nilpotent associative algebra

TL;DR

This paper proves a conjecture about the product of certain ideals in a free associative algebra, showing that the product is contained in a larger ideal only when at least one of the indices is odd.

Contribution

It confirms the remaining part of a conjecture regarding the conditions under which the product of two specific ideals is contained in a larger ideal in a free associative algebra.

Findings

Proves that if both indices are even, the product of the ideals is not contained in the larger ideal.

Validates the conjecture over any field, not just characteristic zero.

Completes the characterization of ideal products in the algebra setting.

Abstract

Let $F$ be a field and let $F \langle X \rangle$ be the free unital associative algebra over $F$ freely generated by an infinite countable set $X = \{x_1, x_2, \dots \}$. Define a left-normed commutator $[a_1, a_2, \dots, a_n]$ recursively by $[a_1, a_2] = a_1 a_2 - a_2 a_1$, $[a_1, \dots, a_{n-1}, a_n] = [[a_1, \dots, a_{n-1}], a_n]$ $(n \ge 3)$. For $n \ge 2$, let $T^{(n)}$ be the two-sided ideal in $F \langle X \rangle$ generated by all commutators $[a_1, a_2, \dots, a_n]$ ($a_i \in F \langle X \rangle)$. Let $F$ be a field of characteristic $0$. In 2008 Etingof, Kim and Ma conjectured that $T^{(m)} T^{(n)} \subset T^{(m+n -1)}$ if and only if $m$ or $n$ is odd. In 2010 Bapat and Jordan confirmed the "if" direction of the conjecture: if at least one of the numbers $m$, $n$ is odd then $T^{(m)} T^{(n)} \subset T^{(m + n -1)}.$ The aim of the present note is to confirm the "only if" direction of the conjecture. We prove that if $m = 2 m'$ and $n = 2 n'$ are even then $T^{(m)} T^{(n)} \nsubseteq T^{(m +n -1)}.$ Our result is valid over any field $F$.