1
$\begingroup$

I am a bit confused about semisimple Lie algebras.

For the sake of simplicity, let's take $\mathfrak{g}=M_n(k)$ where $k=\bar{k}$. According to Wiki, $M_n(k)$ is solvable if the radical of $M_n(k)$ is zero. But the set of all upper triangular matrices $\mathfrak{b}_n$ is a subalgebra of $M_n(k)$, which is solvable. Why would that not imply that $M_n(k)$ is solvable?

$\mathit{Edit}$: Here is a better question. Suppose $\mathfrak{h}$ is a subalgebra of a Lie algebra $\mathfrak{g}$. Suppose $\mathfrak{h}$ is solvable.

$\mathbf{first \; question}$: When is it the case that solvability cannot be induced to $\mathfrak{g}$?

$\mathbf{second \; question}$: When is it the case that solvability can be induced to $\mathfrak{g}$?

$$ $$

  • 3
    A Lie algebra is not solvable if the radical is 0, it is semisimple. You need a solvable ideal with solvable quotient for the Lie algebra to be solvable. Any non-trivial Lie algebra will contain a non-trivial solvable (in fact abelian) subalgebra. It is the existence of solvable ideals that has some interest.2012-06-28
  • 0
    Thanks Tobias! When you say solvable quotient, do you mean such a quotient in $\mathfrak{g}/\mathfrak{h}$? And what do you mean by nontrivial Lie algebra?2012-06-28
  • 2
    By non-trivial, I just mean not equal to 0. By quotient I mean the Lie algebra you get by taking the quotient with the ideal, which then needs to be solvable.2012-06-28
  • 0
    Oh, I see. =) Thanks again!2012-06-28
  • 2
    You can guess at everything Tobias says by analogy with groups. Every group has solvable subgroups (in fact the subgroup generated by an element is abelian) but that doesn't imply that every group is solvable; we want solvable _normal_ subgroups with solvable quotient.2012-06-28
  • 0
    Thanks for the additional comment, Qiaochu!2012-06-28

0 Answers 0