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

For a finite-dimensional Lie algebra representation $\varphi :\mathfrak{g}\to \mathfrak{gl}(V)$, the Trace form of a Lie algebra representation isThe Killing form is the trace form of the Adjoint representation of a Lie algebra:A bilinear form $B$ is $\mathfrak{g}$-invariant whenequivalently $B([x,y],z)=B(x,[y,z])$. For a trace form this follows from cyclicity of trace:

Use the nondegenerate restriction of the Killing form to $\mathfrak{t}$ to define $t_{\alpha }\in \mathfrak{t}$ byKilling-form invariance and the root-space decomposition show that ${\mathfrak{g}}_{\alpha }$ pairs nondegenerately with ${\mathfrak{g}}_{-\alpha }$ and orthogonally with every other root space. Choose nonzero $e\in {\mathfrak{g}}_{\alpha }$ and $f\in {\mathfrak{g}}_{-\alpha }$ with $\kappa (e,f)\ne 0$. For $t\in \mathfrak{t}$,so

We need $\alpha (t_{\alpha })\ne 0$. If it were zero, the span of $e,f,t_{\alpha }$ would be a solvable Heisenberg-type Lie algebra with central commutator $[e,f]$. By Lie theorem, its adjoint action on $\mathfrak{g}$ can be upper triangularized, so $\mathrm{ad}[e,f]$ is nilpotent. But $[e,f]\in \mathfrak{t}$, and elements of the Cartan subalgebra act semisimply; hence $\mathrm{ad}[e,f]=0$. A semisimple Lie algebra has zero center, contradicting $[e,f]\ne 0$.

SetRescale $f$ so that $[e_{\alpha },f_{\alpha }]=h_{\alpha }$. Since $e_{\alpha }$ and $f_{\alpha }$ lie in the $\alpha$ and $-\alpha$ root spaces,Thus ${\mathfrak{m}}_{\alpha }=⟨e_{\alpha },h_{\alpha },f_{\alpha }⟩$ is the sl2 subalgebra associated with a root.

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