 Solution

The Adjoint representation of a Lie algebra acts on the algebra itself: ${\mathrm{ad}}_X(Y)=[X,Y]$. In the basis $T_b$, its matrices areFor any algebra element $X$, the Jacobi identity givesTherefore the $D\times D$ matrices satisfy the same Lie bracket relations:This construction does not require that the algebra be simple or that the adjoint action be faithful.

For the field calculation, write $A_{\mu }=A_{\mu }^a{T}_a$ and $\lambda ={\lambda }^a{T}_a$. In the derivative-plus-connection convention the Yang-Mills field strength isThe sign of the parameter here is the negative of the matter-field parameter in the usual gauge transformation in the derivative-plus-connection convention. Work with the stated $\lambda$ throughout. Expanding the variation givesThe second derivatives cancel by commutativity of partial derivatives, and all remaining derivatives of $\lambda$ cancel by antisymmetry of the Lie bracket. The Jacobi identity makes the last two terms $[[A_{\mu },A_{\nu }],\lambda ]$. HenceUnlike the connection, the curvature has no inhomogeneous derivative term: it transforms in the Adjoint representation.

The quadratic form ${\kappa }_{ab}=\mathrm{tr}(T_a^{\mathrm{ad}}{T}_b^{\mathrm{ad}})$ is the Killing form. Its symmetry follows from cyclicity of the matrix trace. Using the just-proved adjoint commutators,ThusEquivalently, each adjoint generator is skew with respect to this invariant bilinear form. This invariance holds even when the Killing form is degenerate; no inverse form is needed here.

The spacetime metric used to raise Lorentz indices is unaffected by the internal gauge transformation. Since $\delta F=-{\mathrm{ad}}_{\lambda }F$, invariance givesTherefore the Killing-form contraction of the field strengths is gauge invariant, as required for the internal contraction in a Yang-Mills action.