6
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 Siciliano Aug 5 '11 at 15:27
  • $\begingroup$ 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 Zusmanovich Aug 20 '11 at 16:12
  • $\begingroup$ 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 Siciliano Aug 20 '11 at 16:59
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.