Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 102 / 2 / Solution

A Lie algebra is semisimple when its soluble radical is zero. Its Killing form isThe radical of this invariant symmetric form is an ideal. The solvability result behind the Cartan criterion for semisimplicity applied to that ideal shows that it is soluble; semisimplicity therefore makes it zero. Hence $B_L$ is nondegenerate.

An abelian subalgebra $H$ is a Cartan subalgebra when its elements are semisimple and it is maximal toral, equivalently when $C_L(H)=H$. An arbitrary abelian subalgebra need not lie in one: in ${\mathfrak{sl}}_2$, the line spanned by the nilpotent matrix $e$ is abelian, whereas every element of a Cartan subalgebra is semisimple.

Let $H=C_L(h_0)$ for a regular $h_0\in H$, as allowed. Generalized eigenspaces of $\mathrm{ad}{h}_0$ giveIf $x\in H$ is orthogonal to $H$, invariance givesThus $x$ is orthogonal to all of $L$, and nondegeneracy gives $x=0$. Therefore $B_L{|}_H$ is nondegenerate.

The commuting semisimple maps $\{\mathrm{ad}h:h\in H\}$ can be simultaneously diagonalized. ConsequentlyHere $L_0=C_L(H)=H$, the nonzero weights $\Phi \subseteq H^{\ast }$ are the roots, and the Jacobi identity gives $[L_{\alpha },L_{\beta }]\subseteq L_{\alpha +\beta }$.