Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 306 3 b Solution Created 2026-10-03 Updated 2026-10-05
The Jacobi identity for the Poisson bracket impliesFor first-class constraints with linearly independent differentials this gives , the Jacobi identity for the structure constants of the constraint algebra. Independence is an implicit assumption: for constraints obeying identities or vanishing identically, only the contracted identity follows, and arbitrary coefficients multiplying such constraints need not satisfy a Lie algebra identity.
With and , the canonical variables transform asTo fix the sign convention, vary the phase-space action directly:Thus action invariance requires . For , the Faddeev-Popov determinant is that ofUsing anticommuting Faddeev-Popov ghost fields, the invariant-convention result isThe original PDF has a sign error in the stated multiplier transformation; the TeX transcription has the consistent plus sign. Keeping the PDF's displayed minus sign mechanically would instead give . That expression exponentiates the determinant of the printed transformation, but that transformation does not preserve the stated action with the canonical convention above. It cannot be used as the invariant result without changing another convention consistently.
As for the particle, constant multiplier moduli and any residual zero mode in field theory must be handled separately; the constant gauge fixing is understood locally on the gauge orbit.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 304 4 d i Solution 2026-10-03
Without gauge fixing, every gauge orbit is integrated infinitely many times and the quadratic gauge-field operator has a zero mode in field theory for every pure-gauge direction. It therefore has no inverse and no propagator. A gauge condition selects one representative per orbit, up to global transformations and the possible nonperturbative Gribov ambiguity, and makes perturbative Gaussian integration well defined.
Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 304 4 c Solution 2026-09-28
The quadratic Yang-Mills operator has zero directions along gauge orbits, so it has no propagator until one chooses a gauge fixing. The Faddeev-Popov determinant generated by this choice is represented by anticommuting Faddeev-Popov ghost fields, which cancel unphysical gauge-field contributions in loop calculations. A Nakanishi-Lautrup field imposes the gauge condition algebraically and lets the gauge-fixing plus ghost action be written as a BRST-exact term with off-shell nilpotency.
Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 304 3 a Solution 2026-09-28
For the convention in the question, a finite Yang-Mills gauge transformation acts covariantly on the field strength:or with and exchanged if the opposite convention is used for . Cyclicity of the matrix trace givesso the Yang-Mills theory Lagrangian is gauge invariant.
The quadratic gauge-field operator has zero directions along each gauge orbit. It therefore has no inverse on the full field space. Gauge fixing removes this degeneracy and produces a propagator, while the Faddeev-Popov determinant accounts for the corresponding Jacobian.
Regular gauge slice 2026-10-14
A gauge condition is regular at a representative if its derivative along the gauge orbit is an invertible matrix. The inverse function theorem then makes the gauge-condition values valid local coordinates along that orbit. A unique zero without this regularity is insufficient: the translation action on the line with has a unique zero but a zero derivative there. Multiple regular intersections give a Gribov ambiguity rather than a globally unique slice.
Symmetry up to gauge transformation 2026-10-14
A spatial transformation is a symmetry of a gauge-field configuration when its pullback lies in the gauge orbit of the original configuration. It need not fix a particular representative. For a planar vortex, rotation changes its winding phase and a constant transformation compensates. For a hedgehog ansatz for a monopole, a spatial rotation is compensated by the corresponding internal adjoint rotation. Gauge-invariant densities remain invariant without choosing a special gauge.