Decomposition of Lie algebras using Weyl's Reducibility Suppose I have a semisimple ideal $\mathfrak{g}$ of a Lie algebra $\mathfrak{l}$, is it possible to uniquely write $\mathfrak{l}=\mathfrak{g}\oplus\mathfrak{i}$ where $\mathfrak{i}\subset\mathfrak{l}$ is an ideal? I was told to use Weyl's complete reducibility but no clues on how to proceed so far. Can anyone help me with it? Thank you in advance.
 A: We can apply Weyl's reducibility to the adjoint representation. Then we have to assume that $\mathfrak{g}$ is a semisimple $\mathfrak{g}$-module via the adjoint representation, i.e., that $\mathfrak{g}$ is semisimple. The reducibility gives that $\mathfrak{g}$ is the direct sum of simple ideals (which are $\mathfrak{g}$-submodules), i.e.,
$$
\mathfrak{g}=\mathfrak{s_1}\oplus \cdots \oplus \mathfrak{s_r}
$$
for simple ideals $\mathfrak{s_i}$. If $\mathfrak{l}$ is a Lie algebra which is not semisimple, we cannot apply Weyl's theorem to $\mathfrak{l}$, but only to a semisimple subalgebra.
A: Weyl gives you a complement of ${\mathfrak g}$ in ${\mathfrak l}$ as a ${\mathfrak g}$-module, but I don't see why it would be an ideal in ${\mathfrak l}$.
Instead, you might take ${\mathfrak i} := {\mathfrak g}^{\perp}$ the orthogonal of ${\mathfrak g}$ with respect to the Killing form of ${\mathfrak l}$. Then ${\mathfrak i}\lhd {\mathfrak l}$ and $${\mathfrak i}\cap{\mathfrak g} = {\mathfrak g}^{\perp}\cap{\mathfrak g}\stackrel{(\dagger)}{=}\text{rad}(\kappa_{\mathfrak g})\stackrel{(\ddagger)}{=}\{0\},$$ where in $(\dagger)$ we used that the restriction of the Killing form of a Lie algebra to an ideal is the Killing form of the ideal, and in $(\ddagger)$ that the Killing form $\kappa_{\mathfrak g}$ of ${\mathfrak g}$ is non-degenerate due to the semi-simplicity of ${\mathfrak g}$. Since $\text{dim}({\mathfrak g}^{\perp}) \geq\text{dim}{\mathfrak l}-\text{dim}{\mathfrak g}$, it follows that $${\mathfrak l} = {\mathfrak g}^{\perp}\oplus{\mathfrak g}.$$
Conversely, in any decomposition ${\mathfrak l} = {\mathfrak g} \oplus {\mathfrak i}$ you have ${\mathfrak i}\subseteq{\mathfrak g}^{\perp}$, hence ${\mathfrak i} = {\mathfrak g}^{\perp}$. 
