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Take a local gauge transformation $g(x)$ acting on the adjoint field by ${\Phi }^{\prime }=g\Phi g^{-1}$. The connection convention $D_{\mu }={\partial }_{\mu }+[A_{\mu },\cdot ]$ is compatible withIndeed, differentiating ${\Phi }^{\prime }$ gives the extra commutator $[({\partial }_{\mu }g)g^{-1},{\Phi }^{\prime }]$, exactly cancelled by the inhomogeneous term in the transformed gauge potential.

For a direct proof of the Yang-Mills gauge transformation of curvature, put ${\Omega }_{\mu }=({\partial }_{\mu }g)g^{-1}$ and $B_{\mu }=g{A}_{\mu }{g}^{-1}$. Differentiation givesSubstituting $A_{\mu }^{\prime }=B_{\mu }-{\Omega }_{\mu }$ into the Yang-Mills field strength cancels all terms involving $\Omega$ and leavesThis direct calculation also applies when the adjoint action has a kernel.

Expand $D_{\mu }{F}_{\nu \rho }+D_{\nu }{F}_{\rho \mu }+D_{\rho }{F}_{\mu \nu }$. The mixed second derivatives of $A$ cancel in pairs. The first-derivative terms from differentiating $[A_{\nu },A_{\rho }]$ cancel the terms from $[A_{\mu },{\partial }_{\nu }{A}_{\rho }-{\partial }_{\rho }{A}_{\nu }]$. The remaining sum isby the Jacobi identity. Thus the gauge-theory Bianchi identity is

For the variation of the Yang-Mills action, write $a_{\mu }=\delta A_{\mu }$. Differentiating the curvature gives $\delta F_{\mu \nu }=D_{\mu }{a}_{\nu }-D_{\nu }{a}_{\mu }$. Invariance of the Killing form impliesThis is invariant integration by parts for a gauge covariant derivative. Using symmetry of the Killing form, antisymmetry of $F$, and compactly supported variations,Since the Killing form of a semisimple Lie algebra is nondegenerate and the variations are arbitrary, the Yang-Mills equations are

For the constant Yang-Mills theta term, symmetry under exchanging the two antisymmetric index pairs givesup to the boundary variation of the Yang-Mills theta term. Contracting the gauge-theory Bianchi identity with ${ϵ}^{\mu \nu \rho \sigma }$ makes the last integrand vanish. Therefore a constant theta term does not alter the bulk Yang-Mills equations: $D_{\mu }{F}^{\mu \nu }=0$. Its variation is a boundary term, so this conclusion uses compact support or boundary conditions that remove that term.

Finally apply $D^{\rho }$ to the gauge-theory Bianchi identity, obtainingThe adjoint covariant derivative obeys $[D_{\alpha },D_{\beta }]X=[F_{\alpha \beta },X]$. Commute $D^{\rho }$ past the other derivative. The differentiated divergences vanish by the Yang-Mills equations, leavingIn the last line the second commutator needs its sign tracked carefully: directly,Thus the simplified sum gives the covariant wave equation for the Yang-Mills field strengthAll spacetime derivatives and index manipulations here use the fixed flat metric.

Absorb the gauge coupling into a matrix-valued connection 1-form, and usefor an adjoint-valued $p$-form $\alpha$. The matrix trace supplies the usual invariant nondegenerate pairing of the Lie algebra, and the spacetime metric defining the Hodge star is fixed during the gauge-field variation. Since $\delta F=D\delta A$ and the two-form Hodge pairing is symmetric,For variations fixed at the boundary, the Yang-Mills equations are thereforeThey accompany, rather than replace, the gauge-theory Bianchi identity $DF=0$, which follows directly by substituting $F=dA+A\wedge A$ and using $d^2=0$.

Fix the Yang-Mills gauge transformation conventionTaking $g=1+\eta +O({\eta }^2)$ givesThe curvature variation also follows from $D(D\eta )=[F,\eta ]$. If the group parameter is defined with the opposite sign, both displayed infinitesimal variations change sign together.

The Hodge star acts only on spacetime indices, so $\delta (\ast F)=[\ast F,\eta ]$. ConsequentlyThis proves exact off-shell gauge invariance, with no surviving boundary term, even for a gauge parameter nonzero at the boundary. It does not require the Yang-Mills equations.

One can check the surface terms explicitly in the integrated variation. The Lie-algebra-valued four-form $[F,\ast F]=F\wedge \ast F-\ast F\wedge F$ vanishes: writing $F=F^i{T}_i$, the coefficients $F^i\wedge \ast F^j$ are symmetric in $i,j$, and multiply the antisymmetric commutator $[T_i,T_j]$. Thus $D^2\ast F=[F,\ast F]=0$, and the gauge substitution in the variation givesThe last equality is Stokes theorem on the closed boundary; equivalently the cancellation is already pointwise in the preceding derivation. Keeping only the first surface term from an arbitrary variation would incorrectly produce a gauge boundary term.

For the additional Yang-Mills theta term, denote its coefficient-free integral by $P={\int }_M\mathrm{tr}(F\wedge F)$. Its gauge variation is again pointwise zero:Under an arbitrary connection variation, its boundary variation of the Yang-Mills theta term isThe bulk term vanishes by the gauge-theory Bianchi identity. Therefore adding a constant multiple of $P$ does not change the bulk Yang-Mills equations. With fixed boundary potential, it also leaves the usual variational problem unchanged. If boundary variations are allowed, it changes the boundary term: $I+\kappa P$ has surface variation $2{\int }_{\partial M}\mathrm{tr}(\delta A\wedge (\ast F+\kappa F))$, so compatible boundary conditions or a boundary action must account for that change.

Locally $\mathrm{tr}(F\wedge F)=d\mathrm{tr}(A\wedge dA+\frac{2}{3}A\wedge A\wedge A)$, the unnormalized Chern-Simons 3-form identity. That three-form need not be globally defined on a nontrivial bundle, so absence of a bulk variation does not force $P$ itself to vanish. Its exact gauge invariance is the curvature-conjugation calculation above, independently of the choice of local gauge potential.