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...

  • $\begingroup$ You mean a subalgebra of... $\endgroup$ – Andreas Caranti Feb 19 '17 at 14:38
  • $\begingroup$ which part are you refering to? $\endgroup$ – Dac0 Feb 19 '17 at 14:39
  • $\begingroup$ Both. An algebra of dimension $2$ (which is soluble), say, is definitely not isomorphic to an algebra of upper triangular matrices. $\endgroup$ – Andreas Caranti Feb 19 '17 at 14:40
  • $\begingroup$ 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? $\endgroup$ – Dac0 Feb 19 '17 at 14:43
  • 1
    $\begingroup$ The answer to your question is yes. $\endgroup$ – Dustan Levenstein Feb 19 '17 at 14:54

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