Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 102 / 2 / c

Bilinearity and alternatingness ofare immediate. For three elements, the $\mathfrak{g}$ component of the Jacobi sum vanishes by the Jacobi identity in $\mathfrak{g}$. In the $V$ component, the coefficient of a vector such as $u$ isbecause the action is a Lie algebra representation; the other terms cancel cyclically in the same way. Hence the bracket satisfies Jacobi and defines the semidirect product of a Lie algebra and a module $\mathfrak{g}⋉V$.

Let $\mathfrak{g}={\mathfrak{sl}}_2(\mathbb{C})$ and let $V={\mathbb{C}}^2$ be its defining irreducible representation. SetSince $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$ and $\mathfrak{g}V=V$,If $(x,v)$ is central, commuting with every $(0,w)$ gives $xw=0$ for all $w$, so faithfulness of the defining representation gives $x=0$. Commuting with every $(y,0)$ then gives $yv=0$ for all $y$; irreducibility and nontriviality give $v=0$. Thus $Z(\mathfrak{h})=0$.

The nonzero abelian subspace $V$ is a proper ideal of a Lie algebra, so $\mathfrak{h}$ is not simple. It is not a direct product of simple Lie algebras either, because such a product is semisimple and has no nonzero solvable ideal, whereas $V$ is one.

Relative to $\mathfrak{g}\oplus V$, the adjoint action has block formMultiplying two such block-triangular matrices and taking the trace givesthe Killing form of a semidirect product with a module. It does not depend on $v$ or $w$, so $V$ lies in its radical. Therefore $K$ can be nondegenerate only if $V=0$. In that case $K=\kappa$, which is nondegenerate exactly when $\mathfrak{g}$ is semisimple by the Cartan criterion for semisimplicity. Thus