Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 102 / 1

The Adjoint representation of a Lie algebra isThe Killing form is the symmetric invariant bilinear form

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A vector subspace $I\subseteq \mathfrak{g}$ is an ideal of a Lie algebra when $[\mathfrak{g},I]\subseteq I$. A Nilpotent Lie algebra is one whose lower central serieseventually vanishes.

By the Engel theorem, the operators ${\mathrm{ad}}_x$ for a nilpotent complex Lie algebra can be represented simultaneously by strictly upper triangular matrices. Their products are strictly upper triangular and have zero trace. Hence

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A Solvable Lie algebra is one whose derived serieseventually vanishes. By the Lie theorem, the adjoint operators of a solvable complex Lie algebra are simultaneously upper triangular. If $x\in [\mathfrak{g},\mathfrak{g}]$, then ${\mathrm{ad}}_x$ is a sum of commutators of upper triangular matrices and is therefore strictly upper triangular. For every $y\in \mathfrak{g}$, the product ${\mathrm{ad}}_x{\mathrm{ad}}_y$ is strictly upper triangular, soThus

For a nonzero example, let $\mathfrak{g}$ have basis $h,e$ with $[h,e]=e$. It is solvable because $[\mathfrak{g},\mathfrak{g}]=\mathbb{C}e$ is abelian, but in the ordered basis $(h,e)$,

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Invariance of the Killing form givesIf $x\in {\mathfrak{g}}^{\perp }$, the right-hand side vanishes for every $y$, so $[a,x]\in {\mathfrak{g}}^{\perp }$. Thus ${\mathfrak{g}}^{\perp }$ is an ideal.

Let $\mathfrak{h}=\mathrm{ad}({\mathfrak{g}}^{\perp })\subseteq \mathfrak{gl}(\mathfrak{g})$. For $x\in [{\mathfrak{g}}^{\perp },{\mathfrak{g}}^{\perp }]$ and $y\in {\mathfrak{g}}^{\perp }$,The Cartan solvability criterion makes $\mathfrak{h}$ solvable. The kernel of the adjoint map on ${\mathfrak{g}}^{\perp }$ lies in its center and is abelian, so ${\mathfrak{g}}^{\perp }$ is itself solvable. This proves the Solvability of the radical of the Killing form.

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The Killing form of a complex Simple Lie algebra is nondegenerate. Since $B$ is also nondegenerate, there is a unique endomorphism $T$ of $\mathfrak{g}$ satisfyingInvariance of both forms givesso $T$ intertwines the adjoint representation. That representation is irreducible because its invariant subspaces are ideals. The Schur lemma therefore gives $T=cI$. Since $\kappa$ is nondegenerate, $c\ne 0$, andThis is the uniqueness of an invariant bilinear form on a simple Lie algebra.

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For ${\mathfrak{sl}}_3(\mathbb{C})$, the Killing form of the special linear Lie algebra isFor $H_1=E_{11}-E_{22}$ and $H_2=E_{22}-E_{33}$, its matrix on the Cartan part iswhose determinant is $108$. Each $E_{ij}$ pairs only with $E_{ji}$, with value $6$. In the stated ordering, the remaining block iswhose determinant is $-6^6$. Therefore

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