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Let ${\hat{T}}_a={\mathrm{ad}}_{T_a}$ on the Lie algebra itself. The Jacobi identity says, for every $X$,Therefore $[{\hat{T}}_a,{\hat{T}}_b]=c_{ab}^c{\hat{T}}_c$, proving that the displayed structure-constant matrices form the Adjoint representation of a Lie algebra.

The bilinear form ${\kappa }_{ab}=\mathrm{tr}({\hat{T}}_a{\hat{T}}_b)$ is the Killing form. It is symmetric by cyclicity of trace. Its invariance follows by expanding the two commutators:In components this is preciselyEquivalently, the adjoint action preserves the bilinear form, so it is an invariant tensor.

The invariant abelian subalgebra is an abelian ideal of a Lie algebra $\mathfrak{a}$. Assume it is nonzero, as required for the conclusion. For $X\in \mathfrak{a}$ and $Y\in \mathfrak{g}$, the map ${\mathrm{ad}}_X{\mathrm{ad}}_Y$ sends $\mathfrak{g}$ into $\mathfrak{a}$, because it is an ideal. It vanishes on $\mathfrak{a}$, because ${\mathrm{ad}}_Y$ preserves the ideal and ${\mathrm{ad}}_X$ kills it. Relative to $\mathfrak{a}$ and a vector-space complement, this map has zero diagonal blocks, so its trace is zero. ThusThis proves that abelian ideals lie in the radical of the Killing form. Choose a nonzero $X=x^a{T}_a$ in the ideal. Then ${\kappa }_{ba}{x}^a=0$ for every $b$, so the coordinate vector $x$ is a nonzero null eigenvector of the Killing matrix. The nonzero-ideal qualification cannot be dropped: the zero ideal is abelian even in a semisimple algebra with nondegenerate Killing form.

Now suppose $\kappa$ is invertible and let $C_{\kappa }={\kappa }^{bc}{t}_b{t}_c$ in any Lie algebra representation. In matrix notation invariance is $A_a^T\kappa +\kappa A_a=0$, where $(A_a{)}^d{}_b=c_{ab}^d$. Multiplying by the inverse form gives $A_a{\kappa }^{-1}+{\kappa }^{-1}{A}_a^T=0$. HenceNo commutativity of the representation matrices has been assumed. This proves the centrality of the quadratic Casimir operator formed with the inverse Killing form. Schur lemma makes it a scalar on every finite-dimensional complex irreducible representation.

For SU(2), make the normalization explicit. In a Hermitian angular-momentum basis with $[J_a,J_b]=i{ϵ}_{abc}{J}_c$, the structure-constant matrices obeyThus $C_{\kappa }=\frac{1}{2}{\sum }_a{J}_a^2$. In the compact anti-Hermitian basis $T_a=-i{J}_a$, the brackets have real structure constants ${ϵ}_{abc}$ and the Killing form of the SU(2) Lie algebra is instead $-2{\delta }_{ab}$; the inverse contraction gives the same $C_{\kappa }$. This sign change is a basis convention, not a different spectrum.

In the complex highest-weight basis $(H,E^{+},E^{-})$ used earlier, the same Killing form and its inverse areConsequentlyOn a highest vector of weight $m=2j$, the ladder calculation gives $E^{+}{E}^{-}v=mv$ and $E^{-}{E}^{+}v=0$. Its Casimir eigenvalue is thereforeCentrality gives the same value throughout the irreducible multiplet. This is the Killing-normalized SU(2) quadratic Casimir; the familiar angular-momentum operator ${\sum }_a{J}_a^2$ has eigenvalue $j(j+1)$ and is twice this inverse-Killing-form contraction.