Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 302 / 3

The Adjoint representation of a Lie algebra isIt is linear. The Jacobi identity givesso it is a Lie algebra representation.

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The Killing form is $\kappa (X,Y)=\mathrm{Tr}({\mathrm{ad}}_X{\mathrm{ad}}_Y)$. From part (a),where the middle equality uses cyclicity of the trace.

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Each structure constant of a Lie algebra satisfies $f_{ab}{}^d=-f_{ba}{}^d$, so $f_{abc}=f_{ab}{}^d{\kappa }_{dc}$ is antisymmetric in $a,b$. Invariance of the Killing form givesThis cyclic symmetry together with antisymmetry in the first pair implies antisymmetry under every transposition. Thus $f_{abc}$ is totally antisymmetric.

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Let $W\subseteq \mathfrak{g}$ be invariant under the Adjoint representation of a Lie algebra. Then $[\mathfrak{g},W]\subseteq W$, so $W$ is an ideal. If $\mathfrak{g}$ is a Simple Lie algebra, its only ideals are $0$ and $\mathfrak{g}$. Hence the adjoint representation is irreducible. Compact type is compatible with the anti-Hermitian realization used below, although simplicity alone proves this adjoint irreducibility of a simple Lie algebra.

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The Trace form of a Lie algebra representation is invariant becauseChoose a basis orthonormal for the positive-definite form $-\kappa$ of the compact simple algebra. Any invariant bilinear form determines an endomorphism commuting with the irreducible adjoint action; Schur lemma makes it scalar. Thus $H(T_a,T_b)=c{\delta }_{ab}$. Since $d(T_a)$ is anti-Hermitian,The inequality is strict because the kernel of the nontrivial irreducible representation is an ideal and hence zero. Therefore $c=-\mu$ with $\mu >0$.

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For the trace trilinear form of a Lie algebra representation, the trace of a commutator vanishes:Taking $X=T_b$, $Y=T_a$, $Z=T_c$, $W=T_d$ and expanding each Lie bracket givesSince $f_a{}^{bc}$ is antisymmetric in $b,c$,Raising indices with the inverse Killing form gives $f_a{}^{bc}{f}_{bc}{}^e={\delta }_a{}^e$ in the stated normalization. Using $H(T_e,T_d)=-\mu {\delta }_{ed}$ proves

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