---- The following result provides a link between [[semisimple Lie algebra|semisimplicity]], (lack of) [[derived and central series of a Lie algebra|solvability]], and the [[adjoint representation]] for [[Lie algebra|Lie algebras]]. > [!theorem] Theorem. ([[The Cartan-Killing Criterion]]) > For a nonzero [[Lie algebra]] $\mathfrak{g}$ over $\mathbb{C}$, the following are equivalent: > >1. $\mathfrak{g}$ is [[semisimple Lie algebra|semisimple]], i.e., a [[direct sum of Lie algebras|direct sum]] of [[simple Lie algebra|simple]] [[ideal of a Lie algebra|ideals]]; >2. The [[radical of a Lie algebra|radical]] of $\mathfrak{g}$ is zero ($\mathfrak{g}$ has no nonzero [[derived and central series of a Lie algebra|solvable]] ideals); >3. The [[killing form]] $\kappa:\mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}$ is a [[nondegenerate bilinear form|nondegenerate]]. ^theorem > [!proof]- Proof. ([[The Cartan-Killing Criterion]]) > recall: [[the killing form includes into the ambient Lie algebra]] > > We will show $1 \implies 2 \implies 3 \implies 1$. > > > > **$1 \implies 2$.** > Summary: > 1. Define $I_{i} \subset \mathfrak{g}_{i}$ projection of $\text{Rad }\mathfrak{g}$ onto $\mathfrak{g}_{i}$ > 2. $I_{i} \subset \text{Rad }\mathfrak{g}$ is solvable (subalgebra of solvable). $\mathfrak{g}_i$ is not (by simplicity). so $I_{i} \neq \mathfrak{g}_{i}$. > 3. But simplicity means $I_{i} \subsetneq \mathfrak{g}_{i}$. So $I_{i}$ has to be zero. > 4. Now justify/recall why that means $\text{Rad }\mathfrak{g}$ is zero. > > > Suppose that $\mathfrak{g}$ is [[semisimple Lie algebra|semisimple]], $\mathfrak{g}=\mathfrak{g}_{1} \oplus \dots \oplus \mathfrak{g}_{k}$ for $\mathfrak{g}_{1},\dots, \mathfrak{g}_{k}$ all [[simple Lie algebra|simple]]. Write $I_{i}$ for the projection of $\text{Rad }\mathfrak{g}$ onto $\mathfrak{g}_{i}$. Then $I_{i} \subset \mathfrak{g}_{i}$ is a [[derived and central series of a Lie algebra|solvable]] [[ideal of a Lie algebra|ideal]], as a subalgebra of a solvable ideal. Since $\mathfrak{g}_{i}$ is [[simple Lie algebra|simple]], $[\mathfrak{g}_{i}, \mathfrak{g}_{i}]=\mathfrak{g}_{i}$, and so $\mathfrak{g}_{i}$ is not solvable. So, if $I_{i}$ is nonzero, we must not have $I_{i} = \mathfrak{g}_{i}$ (LHS is solvable, RHS isn't) nor $I_{i} \subsetneq \mathfrak{g}_{i}$ (because $\mathfrak{g}_{i}$ is simple). The conclusion is that $I_{i}=0$ for all $i$. Hence $\text{Rad }\mathfrak{g}= \bigoplus_{i} I_{i}=0$. > > > > **$2 \implies 3$.** > Summary: > - Given $\text{Rad }\mathfrak{g}=0$, WTS $\mathfrak{g}^{\perp}=\operatorname{ker }( x \mapsto \kappa(x,-))=0$. Enough to show $\mathfrak{g}^{\perp}$ is a solvable ideal. > - First have to check it *is* an ideal. This follows from the lie bracket + trace form interaction. > - To see solvable, use Cartan's trace theorem to show $\text{ad}(\mathfrak{g}^{\perp})$ is solvable. $\operatorname{ker }\text{ad}=Z(\mathfrak{g})$, so have $\frac{\mathfrak{g}^{\perp}}{Z(\mathfrak{g})} \cong \text{ad}(\mathfrak{g^{\perp}})$ is solvable. > - But $\mathfrak{g}$ is semisimple, so $\text{ad}$ is faithful: $Z(\mathfrak{g})=0$. Thus, $\mathfrak{g}^{\perp}$ is solvable. > > > Assume $\text{Rad}(\mathfrak{g})=0$. Want to show the [[killing form]] $\kappa$ is a [[nondegenerate bilinear form]]. [[matrix of a bilinear form|To do this]] [[linear map is injective iff kernel is trivial|it suffices]] to show that the [[linear map]] $x \mapsto \kappa(x, -)$ has trivial [[kernel of a linear map|kernel]], which is to say it suffices to show that the set $\mathfrak{g}^{\perp}=\{ x \in \mathfrak{g}: \kappa(x,y)=0 \text{ for all }y \in \mathfrak{g} \}$ > is trivial. The assumption that $\text{Rad}(\mathfrak{g})=0$ tells us that if we can show $\mathfrak{g}^{\perp}$ is a [[derived and central series of a Lie algebra|solvable]] [[ideal of a Lie algebra|ideal]] we are done. It is *an* ideal by the [[trace form|identity]] $\kappa([x,y], z)=\kappa(x, [y,z])$: > $x \in \mathfrak{g}^{\perp}, z \in \mathfrak{g} \implies \kappa([x,y], z)=\kappa(x, [y,z])=0 \text{ for all }y \in \mathfrak{g}.$ > To see it is solvable, consider the [[adjoint representation]] $\text{ad}:\mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})$. It follows from [[Cartan's Trace Theorem]] that $\text{ad}(\mathfrak{g}^{\perp}) \subset \mathfrak{gl}(\mathfrak{g})$ is solvable. Essentially by definition, $\operatorname{ker }\text{ad}=Z(\mathfrak{g})$, and the [[center of a Lie algebra]] is always solvable (it is [[abelian Lie algebra|abelian]]). [[first isomorphism theorem|Now]], $\frac{\mathfrak{g}^{\perp}}{Z(\mathfrak{g})} \cong \text{ad }(\mathfrak{g}^{\perp})$ , hence is solvable. We are now done by [[if quotient by a solvable ideal is solvable, then the whole Lie algebra is solvable|Lie algebra is solvable iff ideal and quotient are]], > > > $(3) \implies (1)$ is nonexaminable, for there was an error in the lecture. --- #### ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```