Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 302 / 2

A Lie bracket is bilinear, antisymmetric, and satisfies the Jacobi identity. For a basis $T_a$,The structure constant of a Lie algebra coefficients satisfy $f_{ab}{}^c=-f_{ba}{}^c$ and $f_{ab}{}^d{f}_{dc}{}^e+f_{bc}{}^d{f}_{da}{}^e+f_{ca}{}^d{f}_{db}{}^e=0$.

Conjugation differentiates to the Adjoint representationBecause conjugations compose, ${\mathrm{Ad}}_{gh}={\mathrm{Ad}}_g{\mathrm{Ad}}_h$ and ${\mathrm{Ad}}_e=I$, so this is a group representation.

Differentiating ${\mathrm{Ad}}_{\exp (tX)}Y$ at zero gives the Adjoint representation of a Lie algebraThe Jacobi identity implies $[{\mathrm{ad}}_X,{\mathrm{ad}}_Y]={\mathrm{ad}}_{[X,Y]}$, proving that it is a Lie-algebra representation.

Since $({\mathrm{ad}}_{T_a}{)}^c{}_b=f_{ab}{}^c$, the Killing form has componentsThis is the matrix trace of ${\mathrm{ad}}_{T_a}{\mathrm{ad}}_{T_b}$.

The displayed basis obeys $[{\tau }_a,{\tau }_b]={ϵ}_{abc}{\tau }_c$. ThereforeThe basis is adapted to the Killing form in the usual orthogonal-basis sense, since its Gram matrix is diagonal; rescaling by $1/\sqrt{2}$ makes it orthonormal for $-\kappa$.

The complexification of a Lie algebra is $\mathfrak{su}(2{)}_{\mathbb{C}}=\mathfrak{su}(2){\otimes }_{\mathbb{R}}\mathbb{C}\cong {\mathfrak{sl}}_2(\mathbb{C})$. With $[H,E_{\pm }]=\pm 2E_{\pm }$ and $[E_{+},E_{-}]=H$, the Killing matrix in the ordered basis $(E_{+},E_{-},H)$ isThis Cartan-Weyl basis is adapted to the root decomposition, but it is not Killing-orthogonal because $\kappa (E_{+},E_{-})=4$.

Using the matrix-unit commutator and summing the adjoint indices givesEquivalently, in components this is $2n{X}^i{}_j{Y}^j{}_i-2X^i{}_i{Y}^j{}_j$. It vanishes on the scalar center, as expected because ${\mathfrak{gl}}_n$ is not semisimple.