Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 102 / 1

An Ideal of a Lie algebra is a vector subspace $I\subseteq \mathfrak{g}$ such that $[\mathfrak{g},I]\subseteq I$. The derived series of a Lie algebra is defined by ${\mathfrak{g}}^{(0)}=\mathfrak{g}$ and ${\mathfrak{g}}^{(r+1)}=[{\mathfrak{g}}^{(r)},{\mathfrak{g}}^{(r)}]$.

Suppose that $I$ is an ideal and $x\in \mathfrak{g}$, $u,v\in I$. The Jacobi identity givesBoth terms on the right lie in $[I,I]$, because $[x,u],[x,v]\in I$. Thus $[I,I]$ is again an ideal. Starting from the ideal $\mathfrak{g}$ and applying this observation inductively proves that every term ${\mathfrak{g}}^{(r)}$ of the derived series is an ideal of $\mathfrak{g}$.

A Simple Lie algebra is a nonabelian Lie algebra whose only Ideals of a Lie algebra are $0$ and the whole algebra.

Use the standard basis $e,f,h$ of the sl2 Lie algebra, withLet $I\ne 0$ be an ideal and choose $0\ne x=ae+bf+ch\in I$. Since $I$ is invariant under the Adjoint representation, it is invariant under the linear operator $\mathrm{ad}h$. The three basis vectors are eigenvectors of $\mathrm{ad}h$ with distinct eigenvalues $2,-2,0$. Applying the corresponding polynomial spectral projections to $x$ shows that $I$ contains at least one nonzero multiple of $e$, $f$, or $h$.

If $e\in I$, then $[f,e]=-h\in I$ and $[f,h]=2f\in I$; the cases $f\in I$ and $h\in I$ are identical after taking brackets with the other basis vectors. Hence $e,f,h\in I$, so $I={\mathfrak{sl}}_2(\mathbb{C})$. Therefore ${\mathfrak{sl}}_2(\mathbb{C})$ is simple.

The Killing form of a finite-dimensional Lie algebra $\mathfrak{g}$ is the symmetric bilinear formThe cyclicity of the trace makes it invariant:Consequently its radical of a bilinear form $R=\{x:\kappa (x,\mathfrak{g})=0\}$ is an ideal, since $x\in R$ implies $\kappa ([y,x],z)=\kappa (x,[z,y])=0$ for all $y,z\in \mathfrak{g}$.

If $\mathfrak{g}$ is simple, then $R$ is either $0$ or $\mathfrak{g}$. In the second case the Killing form vanishes identically, so the stated solvability criterion makes $\mathfrak{g}$ a Solvable Lie algebra. A nonabelian simple Lie algebra cannot be solvable: its first derived algebra is a nonzero ideal and hence equals $\mathfrak{g}$, after which the derived series never reaches zero. Thus $R=0$, and the Killing form of a simple Lie algebra is nondegenerate.