Let $A$ be an associative algebra over a field. Then $A$ can be regarded as a Lie algebra via the Lie bracket defined by $[a,b]=ab-ba$ for every $a,b\in A$. The algebra $A$ is called Lie locally nilpotent if it is locally nilpotent as a Lie algebra. Also, $A$ is said to be locally Lie nilpotent if every finitely generated associative subalgebra of $A$ is nilpotent as a Lie algebra. Clearly, if $A$ is locally Lie nilpotent then it is Lie locally nilpotent. Is the converse true?
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asked Aug 4, 2011 at 16:41
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$\begingroup$ Of course, my question is equivalent to the following. Suppose that a finitely generated associative algebra $A$ is locally nilpotent as a Lie algebra. Is $A$ necessary nilpotent as a Lie algebra? $\endgroup$– Salvatore SicilianoCommented Aug 5, 2011 at 15:27
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$\begingroup$ It maybe that the following paper: P. Etingof, J. Kim, X. Ma, On universal Lie nilpotent associative algebras, J. Algebra 321 (2009), N2, 697-703 DOI: 10.1016/j.jalgebra.2008.09.042; arXiv:0805.1909, and/or references therein, are relevant. $\endgroup$– Pasha ZusmanovichCommented Aug 20, 2011 at 16:12
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$\begingroup$ Hi Pasha, thank you for pointing out this reference. I took a look at that paper, however a satisfactory answer to my question is not present there. $\endgroup$– Salvatore SicilianoCommented Aug 20, 2011 at 16:59
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I have to mention that for associative algebras over fields of characteristic not 3 the question has now a positive answer. This is a consequence of a very recent paper by Dias and Krasilnikov: https://arxiv.org/pdf/1709.05728.pdf
answered Sep 19, 2017 at 10:02