Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 304 / 4

For a gauge orbit near a root ${\alpha }_0$ of $G(A^{\alpha })=0$, functional linearization givesThe multidimensional delta-function change-of-variables formula therefore givesComparison with the defining identity provesup to the usual field-independent normalization and a choice of determinant sign on the gauge patch.

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For axial gauge,so the Faddeev-Popov determinant is $\det (n^{\mu }{D}_{\mu })$. Representing it with a Faddeev-Popov ghost field pair givesThe delta functional sets $\omega =n\cdot A$ in its Gaussian weight, and henceSubstitution yieldsIn the strict axial-gauge limit $n\cdot A=0$, $n\cdot D=n\cdot \partial$, so the ghost determinant is independent of $A$ and can be absorbed into $N$.

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Write the gauge-fixing and ghost action as the BRST transformation of the gauge-fixing fermion:with harmless rescalings of $B$ accommodating the convention $s{\bar{c}}^a=\xi B^a$ used in the question. The gauge-invariant action obeys $sS[A]=0$, while nilpotence givesThus the entire gauge-fixed action is BRST invariant. Eliminating the auxiliary field $B^a$ by its algebraic field equation reproduces the axial gauge-fixing term and the ghost action found in part (b).

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Changing $n^{\mu }$ to $u^{\mu }$ changes the gauge-fixing fermion byThe corresponding change of the action is the BRST-exact term $s({\Psi }_u-{\Psi }_n)$. For a BRST-closed observable $\mathcal{O}$, invariance of the functional measure implies that the variation of its expectation value is the expectation of a total BRST variation and vanishes:Therefore physical states and observables, which are classes in BRST cohomology, do not depend on the fixed axial-gauge vector, provided there is no BRST anomaly or boundary contribution.

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