Let L be a Lie algebra with a one-dimensional maximal subalgebra. Is the following true? Over a perfect field of characteristic 0 or p > 3, every such finite-dimensional Lie algebra is either 2-dimensional, or is a form of sl(2). General structure theory seems to indicate this.

  • $\begingroup$ As the answer by Yves suggests, it's risky here to work over a field which isn't algebraically closed. Probably that will cause problems also in prime characteristic, but I'm not sure. $\endgroup$ – Jim Humphreys Jul 29 '13 at 20:05
  • $\begingroup$ @Jim: I think the main issue is certainly the characteristic: issues such that I mention are not a big deal. Also the obvious 3-dim examples disappear when the field is algebraically closed, at least in char 0. $\endgroup$ – YCor Jul 29 '13 at 20:53
  • $\begingroup$ The only possibilities over an algebraically closed field are two dimensional, so the question then is no longer interesting. $\endgroup$ – David Towers Jul 30 '13 at 10:34

If $\mathfrak{a}$ is 1-dimensional and $\mathfrak{v}$ is an irreducible $\mathfrak{a}$-module then $\mathfrak{a}$ is maximal in $\mathfrak{a}\ltimes\mathfrak{v}$. Such $\mathfrak{v}$ of dimension $\ge 3$ indeed exists if the ground field has extensions of degree $\ge 3$, and then this Lie algebra has dimension $\ge 4$ with a maximal 1-dimensional subalgebra. (I guess this are the only finite-dimensional counterexamples, at least in characteristic zero).

  • $\begingroup$ Of course. Thank you Yves. That probably does sort out the characteristic zero case. $\endgroup$ – David Towers Jul 29 '13 at 17:15
  • 1
    $\begingroup$ Yes: assume $\dim(\mathfrak{g})\ge 3$, $\mathfrak{a}$ is 1-dim and maximal. If $\mathfrak{g}$ is not simple and $\mathfrak{s}$ is a minimal ideal, then $\mathfrak{g}=\mathfrak{a}\ltimes\mathfrak{s}$ by maximality. If $\mathfrak{s}$ is abelian it has to be irreducible as in my example. Otherwise it's simple and since derivations are inner the semidirect we see that $\mathfrak{a}$ is not maximal, contradiction. Otherwise, $\mathfrak{g}$ is simple. [...] $\endgroup$ – YCor Jul 29 '13 at 20:58
  • 1
    $\begingroup$ [...] if $\mathfrak{g}$ is simple, its rank (in the sense of Bourbaki Lie Algebras Chap 7) is 1. This passes to the "complexification" (which is semisimple) and thus $\mathfrak{g}$ is 3-dimensional by the classification, since the Bourbaki rank is the usual (absolute) rank of semisimple Lie algebras. [If $\mathfrak{g}$ is a $n$-dimensional Lie algebra over an infinite field, its Bourbaki rank is the minimal dimension of $Ker(ad(x)^n)$ when $x$ ranges over the Lie algebra.] $\endgroup$ – YCor Jul 29 '13 at 21:04
  • $\begingroup$ Yes, if the characteristic of F is zero and g is simple then g is three-dimensional and \sqrt{F} \not \subseteq F. When char F > 7 then the results of Benkart and Osborn apply, but I'm not sure about characteristics 5 and 7. $\endgroup$ – David Towers Jul 30 '13 at 9:15
  • $\begingroup$ @David: but, well, I'm not sure what's going on with this argument in positive characteristic: if you have a simple Lie algebra $\mathfrak{g}$ and $K$ is the algebraic closure of the ground field, what can be said of $\mathfrak{g}\otimes K$? is it a product of simple Lie algebras? I can't even prove that having a trivial solvable radical passes to non-separable extensions, but I'm not a specialist. $\endgroup$ – YCor Jul 30 '13 at 22:06

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.