What is the relationship between the radical of the Killing form and the radical of a Lie algebra?

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 $$\mathfrak{g}^\perp = \{x \in \mathfrak{g} : \kappa(x,y) = 0, \forall y \in \mathfrak{g} \}.$$

Clearly it is true that $$\mathfrak{g}^\perp \subset R(\mathfrak{g})$$, as by the invariance of the Klling form $$\mathfrak{g}^\perp$$ is an ideal, and so restricting $$\kappa$$ to $$\mathfrak{g}^\perp$$ we find that $$\kappa$$ is trivial and so $$\mathfrak{g}^\perp$$ is solvable by Cartan's criterion.

I thought I had proved also $$R(g) \subset \mathfrak{g}^\perp$$ but this does not appear to be true.

Where does it appear that I am making an error in the below attempted proof?

Since $$\mathfrak{g}^\perp$$ ideal by invariance, consider $$\mathfrak{g} / \mathfrak{g}^\perp$$. Since $$\mathfrak{g} / \mathfrak{g}^\perp$$ has non-degenerate Killing form (as we have quotiented by its kernel), by the Cartan criterion $$\mathfrak{g} / \mathfrak{g}^\perp$$ is semisimple, and so $$R(\mathfrak{g}) \subset \mathfrak{g}^\perp$$.

• Can you elaborate on the very last implication? Namely, "$\mathfrak g /\mathfrak h$ semisimple $\implies$ $R(\mathfrak g) \subseteq \mathfrak h$? It does not immediately sound wrong to me, but I also don't think it's straightforward if it's true. May 28 '19 at 13:49
• Consider the projection map $p: \mathfrak{g} \to \mathfrak{g} / \mathfrak{g}^\perp$. Consider the image $p(R(\mathfrak{g}))$, which must be an ideal in $\mathfrak{g} / \mathfrak{g}^\perp$ because $p$ is surjective, and must be solvable (just play around with the definition of homomorphism). Since we are given $\mathfrak{g} / \mathfrak{g}^\perp$ semisimple, we must have $p(R(\mathfrak{g})) = 0 \implies R(\mathfrak{g}) \subset \mathfrak{g}^\perp$. This is a copy of math.stackexchange.com/questions/3242646/… May 28 '19 at 13:57
• Then I think the only thing that can possibly fail is that the canonical Killing form of $\mathfrak{g}/\mathfrak{g}^\perp$ does not agree with the reduction of the Killing form of $\mathfrak{g}$ to $\mathfrak{g}/\mathfrak{g}^\perp$. Writing out the two expressions for them in a basis of $\mathfrak{g}$, I also couldn't make out why those should be equal, so maybe that's indeed the problem. May 29 '19 at 8:59

Like I mentioned in my comment, the Killing form of the quotient Lie algebra $$\mathfrak{g}/\mathfrak{g}^\perp$$ does not necessarily equal the reduction of the Killing form of $$\mathfrak{g}$$ to $$\mathfrak{g}/\mathfrak{g}^\perp$$.
Elaborating: As mentioned in the comment of Torsten Schoeneberger here, a good example is a 3-dimensional Lie algebra so that the corresponding Killing form has a two-dimensional radical like here (whenever $$\lambda^2 \neq -1$$). The Lie algebra $$\mathfrak{g}/\mathfrak{g}^\perp$$ must be one-dimensional, thus abelian, thus the canonical Killing form on $$\mathfrak{g}/\mathfrak{g}^\perp$$ vanishes. However, the reduction of the Killing form of $$\mathfrak{g}$$ does not vanish on the one-dimensional quotient, as we constructed it to be non-degenerate.