 Solution

The Adjoint representation of a Lie algebra acts on the Lie algebra itself by ${\mathrm{ad}}_X(Y)=[X,Y]$. The Jacobi identity givesThus it is a Lie algebra representation. In the basis $T_a$ with $[T_a,T_b]=c^c{}_{ab}{T}_c$, its matrices areThe column index labels the basis vector being acted on; this convention fixes the matrix signs.

Work in finite dimension over characteristic zero, as in the usual real or complex gauge algebras. A simple Lie algebra is nonabelian and has no nonzero proper Lie algebra ideal. A semisimple Lie algebra has no nonzero solvable ideal; equivalently it is a direct sum of simple ideals. The nondegenerate Killing form is $\kappa (X,Y)=\mathrm{tr}({\mathrm{ad}}_X{\mathrm{ad}}_Y)$. Its invariance follows from the representation identity and trace cyclicity:

For the reduction, take a nonzero proper ideal $\mathfrak{a}$, if one exists, and define its Killing-orthogonal complement ${\mathfrak{a}}^{\perp }$. Invariance makes it an ideal: for $y\in {\mathfrak{a}}^{\perp }$, $x\in L$, $z\in \mathfrak{a}$, $\kappa ([x,y],z)=-\kappa (y,[x,z])=0$. Moreover $[\mathfrak{a},{\mathfrak{a}}^{\perp }]=0$, because for $u\in \mathfrak{a}$, $v\in {\mathfrak{a}}^{\perp }$, and any $x\in L$,and the global Killing form is nondegenerate. Their intersection is consequently an abelian ideal: its elements lie in both commuting ideals. Semisimplicity forces that intersection to be zero. The dimension formula for a nondegenerate bilinear form then givesThe summands commute and are semisimple, since a solvable ideal in either summand would also be an ideal of $L$. Repeating on smaller summands terminates in simple ideals. This proves the orthogonal ideal splitting for a nondegenerate Killing form and describes the required reduction; abelian one-dimensional factors do not occur in a semisimple algebra.

If $\kappa (T_a,T_b)=\ell {\delta }_{ab}$ with $\ell \ne 0$, invariance givesso $c_{abc}=-c_{acb}$ after lowering the output index with $\delta$. The bracket already gives $c_{abc}=-c_{bac}$. These two adjacent transpositions generate all permutations, proving that the structure constants of a Lie algebra are completely antisymmetric. This is the antisymmetry of Killing-lowered structure constants; the scalar $\ell$ cancels in the chosen normalization.

For the gauge calculation write $\Lambda ={\lambda }_a{t}_a$, $A_{\mu }=A_{\mu a}{t}_a$, and $D_{\mu }={\partial }_{\mu }+A_{\mu }$. The representation generators here obey $[t_a,t_b]=c_{abc}{t}_c$; factors of $i$ and the gauge coupling are absorbed in this connection convention. Expanding the varied derivative givesFor this to equal $\Lambda D_{\mu }\varphi$, choose the gauge transformation in the derivative-plus-connection conventionThen all extra terms cancel, proving covariance of the gauge covariant derivative for arbitrary space-dependent parameters.

Apply both derivatives to a test field. The second derivatives commute and the terms involving one derivative of the field cancel, leaving the Yang-Mills field strengthThus its components areThe already obtained variation can be written $\delta D_{\mu }=[\Lambda ,D_{\mu }]$. The operator Jacobi identity then yieldsorFinally, with fixed space-time metric,The contracted product is symmetric in $a,c$, while $c_{abc}$ is antisymmetric in those indices. Hence the Yang-Mills action density is gauge invariant, with the sign convention derived rather than assumed.