= Solution
The <Adjoint representation of a Lie algebra> acts on the <Lie algebra> itself by $\operatorname{ad}_X(Y)=[X,Y]$. The <Jacobi identity> gives
$$
[\operatorname{ad}_X,\operatorname{ad}_Y]Z=[X,[Y,Z]]-[Y,[X,Z]]=[[X,Y],Z]=\operatorname{ad}_{[X,Y]}Z.
$$
Thus it is a <Lie algebra representation>. In the basis $T_a$ with $[T_a,T_b]=c^c{}_{ab}T_c$, its matrices are
$$
\boxed{(T_a^{\rm ad})^c{}_b=c^c{}_{ab},\qquad[T_a^{\rm ad},T_b^{\rm ad}]=c^c{}_{ab}T_c^{\rm ad}.}
$$
The 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 $\kappa(X,Y)=\operatorname{tr}(\operatorname{ad}_X\operatorname{ad}_Y)$. Its invariance follows from the representation identity and trace cyclicity:
$$
\kappa([X,Y],Z)=\operatorname{tr}([\operatorname{ad}_X,\operatorname{ad}_Y]\operatorname{ad}_Z)=\kappa(X,[Y,Z]).
$$
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 $x\in L$,
$$
\kappa([u,v],x)=\kappa(v,[x,u])=0,
$$
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 gives
$$
L=\mathfrak a\oplus\mathfrak a^\perp.
$$
The 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 $\ell\ne0$, invariance gives
$$
\kappa([T_a,T_b],T_c)+\kappa(T_b,[T_a,T_c])=0,
$$
so $c_{abc}=-c_{acb}$ 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 $\Lambda=\lambda_at_a$, $A_\mu=A_{\mu a}t_a$, 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 gives
$$
\delta(D_\mu\phi)=(\partial_\mu\Lambda)\phi+\Lambda\partial_\mu\phi+(\delta A_\mu)\phi+A_\mu\Lambda\phi.
$$
For this to equal $\Lambda D_\mu\phi$, choose the <gauge transformation in the derivative-plus-connection convention>
$$
\boxed{\delta A_\mu=-\partial_\mu\Lambda+[\Lambda,A_\mu],\qquad\delta A_{\mu a}=-\partial_\mu\lambda_a+c_{abc}\lambda_bA_{\mu c}.}
$$
Then 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 strength>
$$
[D_\mu,D_\nu]\phi=F_{\mu\nu}\phi,\qquad F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu+[A_\mu,A_\nu].
$$
Thus its components are
$$
\boxed{F_{\mu\nu a}=\partial_\mu A_{\nu a}-\partial_\nu A_{\mu a}+c_{abc}A_{\mu b}A_{\nu c}.}
$$
The already obtained variation can be written $\delta D_\mu=[\Lambda,D_\mu]$. The operator <Jacobi identity> then yields
$$
\delta F_{\mu\nu}=[[\Lambda,D_\mu],D_\nu]+[D_\mu,[\Lambda,D_\nu]]=[\Lambda,[D_\mu,D_\nu]]=[\Lambda,F_{\mu\nu}],
$$
or
$$
\boxed{\delta F_{\mu\nu a}=c_{abc}\lambda_bF_{\mu\nu c}.}
$$
Finally, with fixed space-time metric,
$$
\delta\left(-\frac14F^{\mu\nu}_aF_{\mu\nu a}\right)=-\frac12c_{abc}\lambda_bF^{\mu\nu}_aF_{\mu\nu c}=0.
$$
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.
Back to article page
