2
$\begingroup$

Let $k$ be a field with $\operatorname{char} k = 0$. Let $L$ be a Lie $k$-algebra. Then the universal envelope $U(L)$ is a PI-algebra iff $L$ is abelian.

Remark: PI-algebra means polynomial identity algebra, an algebra that satisfies a nonzero polynomial.

Obviously, if $L$ is abelian, then $U(L)$ is commutative. But the reverse direction is difficult for me. I also know that this is a result in an article "Two remarks on PI-algebras, V.N.Latysev". But I can not find the article.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

The article is available here, it seems that it hasn't been translated. The result you're interested in is Теорема 2; the proof is to notice that a non-abelian Lie algebra in characteristic zero contains as a subalgebra either a two-dimensional non-abelian Lie algebra or a three-dimensional Heisenberg Lie algebra, and their universal envelopings are not PI.

UPD: This only works for finite-dimensional Lie algebras.

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ But this fact (existence of a non-abelian 2-dimensional or 3-dimensional nonabelian subalgebra) is true for finite-dimensional non-abelian Lie algebras, but not in general. For instance, it fails for free Lie algebras. $\endgroup$
    – YCor
    Commented May 1, 2022 at 5:44
  • 1
    $\begingroup$ @YCor That is true; the article only claims the result for finite-dimensional Lie algebras. $\endgroup$
    – Grisha Papayanov
    Commented May 1, 2022 at 5:56
  • $\begingroup$ Actually, the result holds for arbitrary Lie algebras over fields of characteristic zero. This was noticed by Bahturin in his book "Identities on Lie algebras". $\endgroup$
    – Salvatore Siciliano
    Commented Jun 23, 2022 at 15:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .