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Levi decomposition theorem states that any finite-dimensional real Lie algebra $L$ is the semidirect product of a solvable ideal and a semisimple subalgebra. I don't understand this since to me it looks trivial after looking at the exact sequence $ 0 \rightarrow \textrm{rad}(L) \rightarrow L \rightarrow L/\textrm{rad}(L) \rightarrow 0 $. As we are dealing with vector spaces the sequence splits and we have that the Lie algebra $L$ is a direct sum of a solvable ideal and a semisimple subalgebra. I guess that the problem is that it is a direct sum a vector spaces but not as Lie algebras, if this is the case can someone provide an example? Thanks

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As you write, it is true that without using semisimplicity of $L/\mathrm{rad}(L)$, the sequence of vector spaces splits. It is not true that such a sequence of Lie algebras is necessarily split: this is where semisimplicity is crucial. Consider the example in which $L$ is $3$ by $3$ strictly upper triangular matrices (that is, with zeros on the diagonal) and $I$ is the (central) subalgebra consisting of those matrices with a non-zero entry only in the upper right-hand corner. The quotient is a two-dimensional abelian Lie algebra. However, there is no two dimensional abelian subalgebra of $L$ that does not contain $I$, so the sequence $$0 \rightarrow I \rightarrow L \rightarrow L/I \rightarrow 0$$ is not split as Lie algebras.

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