Products of several commutators in a Lie nilpotent associative algebra

Abstract

Let $F$ be a field of characteristic $\ne 2,3$ and let $A$ be a unital associative $F$-algebra. Define a left-normed commutator $[a_1,a_2,\dots ,a_n]$ $(a_i\in A)$ 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 $T^{(n)}(A)$ be the two-sided ideal in $A$ generated by all commutators $[a_1,a_2,\dots ,a_n]$ ($a_i\in A)$. Define $T^{(1)}(A)=A$. Let $k,\ell$ be integers such that $k>0$, $0\le \ell \le k$. Let $m_1,\dots ,m_k$ be positive integers such that $\ell$ of them are odd and $k-\ell$ of them are even. Let $N_{k\ell }={\sum }_{i=1}^k{m}_i-2k+\ell +2$. The aim of the present note is to show that, for any positive integers $m_1,\dots ,m_k$, in general, \[ T^{(m_1)} (A) \dots T^{(m_k)} (A) \nsubseteq T^{(N_{k \ell} +1)} (A). \] It is known that if $\ell <k$ (that is, if at least one of $m_i$ is even) then, for each $A$, \begin{equation*} \label{evenabstr} T^{(m_1)} (A) \dots T^{(m_k)} (A) \subseteq T^{(N_{k \ell} )} (A) \end{equation*} so our result cannot be improved if $\ell <k$. Let $N_k={\sum }_{i=1}^k{m}_i-k+1$. Recently Dangovski has proved that if $m_1,\dots ,m_k$ are any positive integers then, in general, \[ T^{(m_1)} (A) \dots T^{(m_k)} (A) \nsubseteq T^{(N_k+1)} (A) . \] Since $N_{k\ell }=N_k-(k-\ell -1)$, Dangovski's result is stronger than ours if $\ell =k$ and is weaker than ours if $\ell \le k-2$; if $\ell =k-1$ then $N_k=N_{k(k-1)}$ so both results coincide. It is known that if $\ell =k$ (that is, if all $m_i$ are odd) then, for each $A$, \begin{equation*} \label{alloddabstr} T^{(m_1)} (A) \dots T^{(m_k)} (A) \subseteq T^{(N_k)} (A) \end{equation*} so in this case Dangovski's result cannot be improved.

9 pages; v.2: abstract improved, typos fixed; v.3: change of the title, minor changes in notation