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If two finite dimensional matrix Lie algebras are isomorphic, is it always possible to see the isomorphism as a similarity transformation $g \mapsto M^{-1} g M$ ?

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Are you asking: given two subalgebras $g$ and $g'$ of the matrx algebra $M_n(k)$, is there a matrix $M$ such that $g'=MgM^{-1}$? – Mariano Suárez-Alvarez Mar 27 '13 at 19:19

If the question is

given two isomorphic Lie subalgebras $\mathfrak g$, $\mathfrak g'$ of $M_n(k)$, does there exist a matrix $M$ such that $\mathfrak g'=M\mathfrak gM^{-1}$?

then the answer is no. Just take $\mathfrak g$ and $\mathfrak g'$ to be two $1$-dimensional Lie algebras spanned by matrices which are not conjugated.

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In fact, in some sense, the whole theory of representations of Lie algebras gives a ton of examples: Any two non-equivalent faithful representations of the same dimension give a counter example. – Jason DeVito Mar 27 '13 at 19:22
Is it possible that two non-equivalent faithful representations of the same dimension that are as well irreducible are non equivalent? Is there probably a difference between semisimple or general Lie algebras? – user9784 Mar 31 '13 at 17:19
@user9784 I don't understand the question. – Mariano Suárez-Alvarez Mar 31 '13 at 21:11

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