Assuming $L$ to be a semisimple Lie algebra and $I$ an arbitrary ideal of $L$, define
$$I^\perp = \{x\in L \mid \kappa(x,y) =0 \text{ for }y\in I\},$$
where $\kappa$ is the Killing-form. I'd like to show that $L=I\oplus I^\perp$.
First one can show that $I^\perp$ is an ideal (follows directly from the associativity of $\kappa$). The next step is, I think, to deduce from Cartan's criterion that the ideal $I\cap I^\perp$ of $L$ is solvable and zero. Together with the fact that $\mathrm{dim}I + \mathrm{dim}I^\perp = \mathrm{dim}L$, we then have $L=I\oplus I^\perp.$
My problem here is that I don't really understand how exactly one applies Cartan's criterion here and why from the solvability we can directly deduce that the intersection is zero.
Cartan's criterion for solvability (applied to this setting) is:
The Lie algebra $I$ is solvable if and only if its Killing form $\kappa|_{I\times I}$ satisfies $\kappa(X,Y)=0$ for all $X\in I$ and $Y\in [II]$.
How exactly can we deduce from this that $I\cap I^\perp$ is solvable and zero?
Note: This question came up when reading Theorem 5.2 in Humphreys' book.
Best Answer
Let $x\in I, y\in I^{\perp}$, and $k$ the killing form, we have $k(x,y)=0$. Let $z$ be any element of $L$, we have $k([x,y],z)+k([z,x],y)+k([y,z],x)=0$.
Since $I, I^{\perp}$ are ideals, we deduce that $[z,x]\in I$ and $k([z,x],y)=0$ since $y\in I^{\perp}$. $[y,z]\in I^{\perp}$ and $k([y,z],x)=0$ since $x\in I$.
This implies that $k([x,y],z)=0$ since $k$ is not degenerated, we deduce that $[x,y]=0$.
We have $[I\cap I^{\perp},I\cap I^{\perp}]\subset [I,I^{\perp}]=0$ implies that $I\cap I^{perp}$ is commutative and zero.