Let $L$ be a Lie algebra over a field $F$ and denote by $U(L)$ the universal enveloping algebra of $L$. Regard $U(L)$ as a Lie algebra with respect to the Lie bracket $[a,b]=ab-ba$ for any $a,b\in U(L)$. If $U(L)$ is solvable as a Lie algebra, is $L$ necessarily abelian?
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
If $F$ has characteristic different from 2 the answer is yes. This follows from Corollary 6.1 in the paper by D. Riley - A.Shalev: The Lie structure of enveloping algebras, J. Algebra 162, 46-61 (1993).
On the other hand, in characteristic 2 this conclusion is false. For instance, if $L$ is a 2-dimensional nonabelian Lie algebra or a 3-dimensional Heisenberg algebra, then one can see by explicite calculations that $U(L)$ is Lie solvable of derived length 3.
answered May 13, 2012 at 20:43
-
$\begingroup$ is there some background for this ? or it is just standing alone paper ? In this case I would be very much admired and surprised how can any one (except the authors) know about it :) $\endgroup$ Commented May 14, 2012 at 5:50
-
1$\begingroup$ Polynomial identities of (ordinary and restricted) enveloping algebras have been studied by a lot of people (Latysev, Bahturin, Passman, Petrogradski, Riley, Shalev, etc.) and this topic is one of my interest research. In particular, restricted enveloping algebras which are PI were characterized by Passman and, indipendently, by Petrogradski in 1991, and in the paper mentioned in my answer the authors established when a restricted enveloping algebra is Lie nilpotent, bounded Lie Engel or Lie solvable (in odd characteristic). $\endgroup$ Commented May 14, 2012 at 8:19