# Concrete example of Lie algebra $\mathfrak{g}$ with $x$ s.t. $x$ is in the radical of $\mathfrak{g}$ but not the radical of the Killing form?

Let $$\mathfrak{g}$$ be a Lie algebra over algebraically closed field $$k$$ of characteristic $$0$$.

The radical $$R(\mathfrak{g})$$ is the largest solvable ideal of $$\mathfrak{g}$$.

The Killing form $$\kappa$$ is the trace form of the adjoint representation of $$\mathfrak{g}$$.

The radical of the Killing form on $$\mathfrak{g}$$ is $$\mathfrak{g}^\perp = \{x \in \mathfrak{g} : \kappa(x,y) = 0, \forall y \in \mathfrak{g} \}.$$

Could anyone help me find a concrete example of a Lie algebra $$\mathfrak{g}$$ and and an element $$x \in \mathfrak{g}$$ such that $$x \in R(\mathfrak{g})$$ but $$x \not\in \mathfrak{g}^\perp$$?