1
$\begingroup$

It's a very basic question. If I'm not wrong the Lie theorem says that any solvable sub-algebra of $\mathfrak{gl}\left(V\right)$ over complex numbers with $V$ finite dimensional is isomorphic to a sub-algebra of the algebra of upper triangular matrices $\mathfrak{b}(n)$ to some $n$.

Isn't Ado theorem + Lie theorem implying that every solvable finitedimensional Lie algebra is isomorphic to a sub-algebra of $\mathfrak{b}(n)$?

I suppose that should be the case, but I couldn't find an explicit reference and wanted to be sure that I'm not missing something...

  • 0
    You mean *a subalgebra of*... – 2017-02-19
  • 0
    which part are you refering to? – 2017-02-19
  • 0
    Both. An algebra of dimension $2$ (which is soluble), say, is definitely not isomorphic to an algebra of upper triangular matrices. – 2017-02-19
  • 0
    Why not? matrices of the form $\left(\begin{array}{cc} a & 0\\ 0 & b \end{array}\right)$ don't they form a dimension 2 abelian algebra? – 2017-02-19
  • 0
    But you were talking of ***the*** algebra of upper triangular matrices. – 2017-02-19
  • 0
    Please be more explicit. Ok it was just a copy and paste error I edited it, thank you – 2017-02-19
  • 1
    The answer to your question is yes. – 2017-02-19

0 Answers 0