Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 55 / 2 / Solution

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.