1

Can one give me some examples of three dimensional Non Nilpotent Leibniz algebras? Any references to the classification of three dimensional Non Nilpotent Leibniz or Lie algebras will also be helpful.

user64494
  • 3,309
Ripan Saha
  • 586
  • Are you familiar with J. Milnor, "Curvatures of left-invariant metrics on Lie groups", Advances in Math. 3 (21), p 293-329, 1976? Three-dimensional Lie algebras are classified in Section 4. – Nate Eldredge Mar 16 '19 at 13:39
  • Milnor's account (1976) is very sensitive to working with the real field. In Jacobson's Lie algebra book (1961), there's an account of 3-dim Lie algebras over an arbitrary field, in the first chapter. Also I guess that there are many close questions already on MathSE in the Lie algebra case. – YCor Mar 16 '19 at 13:45
  • As regards the Leibniz case (every 3-dim non-Lie Leibniz algebra is solvable), see https://arxiv.org/abs/1301.7665 and references therein. – YCor Mar 16 '19 at 14:13
  • For your first question, I'm sure it's not what you mean, but it seems inappropriate not even to mention $\mathfrak{sl}_2$ in the comments. (Presumably you mean complex Lie algebras?) – LSpice Mar 16 '19 at 14:28
  • 1
    @LSpice since my previous comments were inappropriate :), the classification of 3-dimensional perfect (= simple, = non-solvable, in dim 3) Lie algebras (over an arbitrary field of char $\neq 2,3$) is the following: these are the $\mathfrak{so}(q)$ for $q$ 3-dim non-degenerate quadratic form; this is immediate using Killing form. At least in char 0, two such $\mathfrak{so}(q)$ and $\mathfrak{so}(q')$ are isomorphic iff $q$ and $tq'$ are equivalent for some $t$ (reference: https://mathoverflow.net/a/200947/14094). $\mathfrak{sl}_2$ corresponds to isotropic $q$. – YCor Mar 16 '19 at 14:35
  • PS: no need to exclude char 3 in my previous comment. The point is that a 3-dim simple Lie algebra has a nonzero Killing form. Indeed, otherwise, given $x$, both $ad(x)$ and $ad(x)^2$ have trace zero, and since 0 is eigenvalue and $2\neq 0$, this forces $ad(x)$ to be nilpotent, and since this is for all $x$ this forces the Lie algebra to be nilpotent, contradiction (see Jacobson's book for the proof, in every characteristic, that a non-nilpotent finite-dim Lie algebra has some $x$ with $ad(x)$ not nilpotent). – YCor Mar 16 '19 at 16:56

0 Answers0