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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 $R(\mathfrak{g}) \subset \mathfrak{g}^\perp$.

seeker_after_truth
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  • 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. – Lukas Miaskiwskyi May 28 '19 at 13:49
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    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 https://math.stackexchange.com/questions/3242646/if-mathfrakg-i-is-semisimple-then-is-mathrmrad-mathfrakg-subset – seeker_after_truth May 28 '19 at 13:57
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    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. – Lukas Miaskiwskyi May 29 '19 at 08:59

1 Answers1

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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.

Lukas Miaskiwskyi
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