Chern-Simons 5-form ($Q_5(A)=\mathrm{Tr}(A{F}^2-\frac{1}{2}F{A}^3+\frac{1}{10}{A}^5)$)

For a matrix-valued gauge field $A$, let $F=dA+A\wedge A$ be its gauge field strength; products below combine matrix multiplication with the wedge product of differential forms. The five-form can be written $Q_5=\mathrm{Tr}(A(dA{)}^2+\frac{3}{2}{A}^{3}dA+\frac{3}{5}{A}^5)$. Applying the exterior derivative, using its graded product rule and the graded cyclic property of the matrix trace, gives $d{Q}_5=\mathrm{Tr}(F^3)$. The six-form is an invariant polynomial in the gauge curvature, while its local potential depends on the connection.

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.

The boundary four-form is $\Theta =\mathrm{Tr}[\delta A(FA+AF-\frac{1}{2}{A}^3)]$. To derive the identity, put $B=dA$ and $a=\delta A$. The variations of the three polynomial terms in $Q_5$ are respectively $3\mathrm{Tr}(a{B}^2)+d\mathrm{Tr}[a(BA+AB)]$, $3\mathrm{Tr}[a(A^{2}B+B{A}^2)]+\frac{3}{2}d\mathrm{Tr}(a{A}^3)$, and $3\mathrm{Tr}(a{A}^4)$. Their sum gives the identity because $F=B+A^2$. For variations in a Lie algebra with basis $T_{\alpha }$, the bulk field equations are $\mathrm{Tr}(T_{\alpha }{F}^2)=0$, not necessarily the unprojected matrix equation $F^2=0$. Conjugation of both the Lie algebra and the gauge curvature proves covariance of these equations under Yang-Mills gauge transformations.