A Lie algebra $\mathfrak{g}$ is semi-simple if the maximal solvable ideal ${\rm rad}(\mathfrak{g})$ is trivial (let's take this as the definition for this question). Cartan's Second Criterion says that $\mathfrak{g}$ is semi-simple if and only if the Killing form $\kappa_{\mathfrak{g}}$ is non-degenerate, which is equivalent to saying that the radical ${\rm rad}(\kappa_{\mathfrak{g}})=\mathfrak{g}^\perp$ (as in the radical of a symmetric bilinear form) is trivial.
So I am asking myself if we have that ${\rm rad}(\mathfrak{g})={\rm rad}(\kappa_{\mathfrak{g}})$ in general.
We always have that ${\rm rad}(\kappa_{\mathfrak{g}})$ is solvable by Cartan's First Criterion, for if $X \in {\rm rad}(\kappa_{\mathfrak{g}})$ and $Y \in {\rm rad}(\kappa_{\mathfrak{g}})'$, then in particular $Y$ is in ${\rm rad}(\kappa_{\mathfrak{g}})$ and so $\kappa_{\mathfrak{g}^\perp}(X,Y) = \kappa_{\mathfrak{g}}(X,Y)=0$. This implies that ${\rm rad}(\kappa_{\mathfrak{g}}) \subseteq {\rm rad}(\mathfrak{g})$.
I am struggling with the reverse inclusion. Namely, I would like to show that if $\mathfrak{h}$ is an arbitrary solvable ideal of $\mathfrak{g}$, then $\mathfrak{h} \subseteq{\rm rad}(\kappa_{\mathfrak{g}})$. I could not immediately see how to use Cartan's First Criterion again here (in other words, I couldn't see how $\kappa_{\mathfrak{h}}(\mathfrak{h},\mathfrak{h}')=0$ helps). Is this only true if $\mathfrak{g}$ is assumed to be semi-simple so one can use $\mathfrak{g} = \mathfrak{h}\oplus \mathfrak{h}^\perp$ somehow?
Best Answer
For the two-dimensional solvable but non-nilpotent Lie algebra $\mathfrak g$ with basis vectors $x,y$ and bracket $[x,y]=y$ we have $\kappa(x,x)= 1$ so $\mathrm{rad}(\kappa) \neq \mathfrak g = \mathrm{rad}(\mathfrak g)$.