I was reading Classification of complex and real semisimple Lie Algebras by Florian Wisser on real semisimple Lie algebras and came across this result early in the notes.
The proof makes no sense to me. An inner product $\langle X, Y \rangle$ is defined which may or may not be related to the Killing form, they don’t say. And what they show is that with respect to this form, the orthogonal complement of any ideal in $\mathfrak{g}$ is itself an ideal. Then the proof ends abuptly. What am I missing here?
1.4. Proposition. Let $\mathfrak{g}$ be a real Lie algebra of matrices over $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$. If $\mathfrak{g}$ is closed under conjugate transpose (i.e, $X^* = (\overline{X})^t \in \mathfrak{g}$ for all $X \in \mathfrak{g}$), then $\mathfrak{g}$ is reductive.
Proof. Define an inner product $\langle X, Y \rangle = \operatorname{Re} \mathrm{Tr}(X Y^*)$ for $X, Y$ in $\mathfrak{g}$. Let $\mathfrak{a}$ be an ideal in $\mathfrak{g}$ and denote by $\mathfrak{a}^\perp$ the orthogonal complement of $\mathfrak{a}$. Then $\mathfrak{g} = \mathfrak{a} \oplus \mathfrak{a}^\perp$ as vector spaces. To see that $\mathfrak{a}^\perp$ is an idael in $\mathfrak{g}$ we choose arbitrary elements $X \in \mathfrak{a}^\perp$, $Y \in \mathfrak{g}$ and $Z \in \mathfrak{a}$ and compute \begin{align*} \langle [X, Y], Z \rangle &= \operatorname{Re} \mathrm{Tr}(X Y Z^* - Y X Z^*) \\ &= - \operatorname{Re} \mathrm{Tr}(X Z^* Y - X Y Z^*) \\ &= - \operatorname{Re} \mathrm{Tr}(X (Y^* Z)^* - X (Z Y^*)^*) \\ &= - \langle X, [Y^*, Z] \rangle \,. \end{align*} Since $Y^*$ is in $\mathfrak{g}$, $[Y^*, Z]$ is in $\mathfrak{a}$. Thus the right hand side is $0$ for all $Z$ and hence $[X, Y]$ is in $\mathfrak{a}^\perp$ and $\mathfrak{a}^\perp$ is an ideal and $\mathfrak{g}$ is reductive. $\square$
Consider $\mathfrak{g} = \mathfrak{h} + \mathfrak{s}$, where $\mathfrak{s}$ is the radical of $\mathfrak{g}$ and $\mathfrak{h} = \mathfrak{s}^{\perp}$. Then we see that $\mathfrak{h}$ and $\mathfrak{s}$ are both ideals of $\mathfrak{g}$. Moreover, it is easy to see that $\mathfrak{h}$ is semisimple (otherwise the radical of $\mathfrak{g}$ will be strictly larger than $\mathfrak{s}$). On the other hand, it is not hard to prove that $\mathfrak{s}$ lies in the center of $\mathfrak{g}$ (by considering the derived series). This shows that $\mathfrak{g}$ is reductive.