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For $F=dA$ on a five-dimensional oriented spacetime, the action $S=-\frac{1}{2}\int F\wedge \ast F+c\int A\wedge F\wedge F$ has bulk variation $\int \delta A\wedge (-d\ast F+3cF\wedge F)$. Two integrations by parts make the three variations of the Chern-Simons 5-form contribute equally. Under $A\mapsto A+d\lambda$, the extra action is $c\int d(\lambda F\wedge F)$, a boundary term by the Generalized Stokes theorem. Its bulk field equation is gauge invariant. The Chern-Simons term is metric-independent and has no bulk contribution to the stress-energy tensor when the differential-form field is held fixed in metric variation.

Hold the metric fixed when varying the gauge potential, put $F=dA$, and take variations with compact support or with appropriate fixed boundary data. The Hodge star operator then givesFor the Abelian Chern-Simons 5-form, the two occurrences of $F$ commute as even-degree differential forms. Since $dF=0$,The factor three comes from all three appearances of $A$, including those hidden in $F=dA$. The bulk variation is thereforeand the Abelian Chern-Simons five-form field equation is

Under an Abelian gauge transformation $A\mapsto A+d\lambda$, the gauge field strength stays fixed and the action changes byThus the local integrand is not gauge invariant, but its change is a boundary term by the Generalized Stokes theorem. With the stated boundary conditions the variational bulk equations are unchanged; they depend only on $F$ and the metric, so they are manifestly gauge invariant. On a spacetime with boundary, gauge transformations nonzero at the boundary require boundary conditions or boundary degrees of freedom; the bulk claim does not assert invariance of every possible boundary action.

For the metric variation, write the kinetic action in components,At fixed covariant components $A_{\mu }$ and $F_{\mu \nu }$,Substituting into the given definition of the stress-energy tensor yieldsThe Chern-Simons 5-form uses only wedge products of differential forms and orientation, with no metric or Hodge star operator. Its bulk metric derivative is zero, so the Chern-Simons term contributes no bulk stress-energy tensor. The displayed tensor is generally not traceless in five dimensions: $T^{\mu }{}_{\mu }=-F_{\rho \sigma }{F}^{\rho \sigma }/4$.

Use matrix multiplication together with the wedge product of differential forms, suppressing wedge symbols in this calculation. Set $F=dA+A^2$ and $a=\delta A$. The gauge field strength varies as $\delta F=da+Aa+aA$, the covariant exterior derivative of $a$. The matrix trace has graded cyclicity, $\mathrm{Tr}(\alpha \beta )=(-1{)}^{pq}\mathrm{Tr}(\beta \alpha )$ for forms of degrees $p,q$. Thus the Chern-Simons 5-form can be expanded asFor example, the two cross terms in $F^{2}A$ both become $\mathrm{Tr}(A^{3}dA)$; combining the remaining two terms gives the displayed coefficients. This expansion keeps track of the noncommuting products rather than treating $A$ as a scalar one-form.

Put $B=dA$. Vary the first term and use the graded exterior derivative to move derivatives off $a$:Here $dB=0$. For the next term use $d(A^3)=B{A}^2-ABA+A^{2}B$. Graded cyclicity then givesFinally all five contributions to $\delta \mathrm{Tr}(A^5)$ are equal under the matrix trace, giving $5\mathrm{Tr}(a{A}^4)$. Combining these results proves the variation of the Chern-Simons 5-form:Consequently, on an oriented five-manifold $N$,The boundary term vanishes for a closed manifold, for variations supported in its interior, or when the tangential boundary connection is held fixed. A different boundary variational problem would require boundary conditions or a boundary action as well.

Write $A=A^{\alpha }{T}_{\alpha }$ in a basis of the Lie algebra. Independent compactly supported variations of each component yield the bulk Euler-Lagrange equationsEquivalently, with $d_{\alpha \beta \gamma }=\frac{1}{2}\mathrm{Tr}[T_{\alpha }\{T_{\beta },T_{\gamma }\}]$, these are $d_{\alpha \beta \gamma }{F}^{\beta }\wedge F^{\gamma }=0$. The symmetry of the two curvature two-forms accounts for the anticommutator. Writing simply $F\wedge F=0$ as a full matrix equation requires an additional assumption: variations must span a matrix algebra paired nondegenerately by the matrix trace. For a smaller Lie algebra, only its trace-paired projection is constrained. If the trace is literally taken in the adjoint representation of a semisimple Lie algebra, its symmetric cubic invariant vanishes, and these bulk equations are identically zero. Indeed, the adjoint generators are skew with respect to the nondegenerate invariant Killing form; transposition in this pairing reverses the sign of the symmetrized trace of three generators. The usual nontrivial five-dimensional theory therefore needs an invariant cubic trace that does not vanish; saying that the gauge field transforms in the adjoint alone does not specify such a trace.

Under a Yang-Mills gauge transformation, use $A^g=g^{-1}Ag+g^{-1}dg$, so $F^g=g^{-1}Fg$. For any Lie-algebra element $T$,Since $gT{g}^{-1}$ is still in the Lie algebra, the right-hand side vanishes whenever the original field satisfies every equation. Every gauge transform of a bulk solution remains a bulk solution. This conclusion follows from covariance of the field equations; it does not require the action itself to be exactly invariant under all large gauge transformations or under transformations changing prescribed boundary data.