The gauge principle and non-abelian algebras

Take a field ψ in some representation of a group G and rotate it by a constant element, ψ → Uψ with U = exp(iα^aTa) and α^a independent of position. Every derivative goes along for the ride, ∂_μ(Uψ) = U∂_μψ, so any Lagrangian built from ψ†ψ and ψ†∂_μψ is invariant for free; Noether hands back a conserved charge and the story ends. Now let α^a depend on x. The product rule bites: ∂_μ(U(x)ψ) = U(∂_μψ) + (∂_μU)ψ, and the second term has no partner to cancel it, so ψ†∂_μψ is no longer invariant. The reason is geometric rather than algebraic. A derivative subtracts field values at two neighbouring points, and those two points are now being rotated by different group elements, so the subtraction compares vectors in two different internal frames. Nothing can be differentiated until you supply an instruction for carrying an internal vector from one point to the next. That instruction is a connection, and the gauge field is it.

Two choices come before any field. The group supplies the generators T^a, Hermitian matrices obeying [T^a, T^b] = i f^abc T^c, and their number is what you will count: dim SU(N) = N² − 1, so SU(2) has three and SU(3) has eight, while U(1) has one. The representation says how large those matrices are and therefore what the field is. Quarks sit in the fundamental 3 of SU(3)_c, so T^a = λ^a/2 are 3 × 3; the gauge fields themselves sit in the adjoint, where (T^a)^bc = −i f^abc are 8 × 8. The structure constants are not free either. For SU(2) they are f^abc = ε^abc. For SU(3), with the standard normalisation tr(T^aTb) = ½δ^ab, the independent non-zero values are f¹²³ = 1, then f¹⁴⁷ = f²⁴⁶ = f²⁵⁷ = f³⁴⁵ = ½, f¹⁵⁶ = f³⁶⁷ = −½, and f⁴⁵⁸ = f⁶⁷⁸ = √3/2. Everything downstream — how many mediators there are, how they rotate, how strongly they couple to each other — is read off that table and nothing else.

Demand that D_μψ transform exactly as ψ does: D'_μ(Uψ) = U(D_μψ). Substitute D_μ = ∂_μ − igA_μ with A_μ ≡ A^a_μT^a and expand: (∂_μU)ψ + U∂_μψ − igA'_μUψ = U∂_μψ − igUA_μψ. The U∂_μψ cancels, leaving igA'_μU = ∂_μU + igUA_μ, and multiplying on the right by U⁻¹ gives A_μ → U A_μ U⁻¹ − (i/g)(∂_μU)U⁻¹. Nothing was invented; the law was forced by one requirement. Two readings matter. Infinitesimally, with U ≈ 1 + iα^aTa, it reads δA^a_μ = (1/g)∂_μα^a + f^abc A^b_μ α^c: the first term is the inhomogeneous piece that kills the derivative of the phase, and the second says the gauge field itself rotates in the adjoint representation, mixing colour directions among themselves. Set f^abc = 0 and only the first term survives, giving A_μ → A_μ + (1/g)∂_μα — the gauge freedom Maxwell's equations already had, recovered here rather than assumed.

Ask what the connection does around a closed path. Transport an internal vector a distance a along x¹ and then a along x², compare with the reverse order, and the mismatch is the commutator of the two covariant derivatives. That is the definition of the field strength: F_μν = (i/g)[D_μ, D_ν]. Working it out, the derivative pieces give the familiar ∂_μA_ν − ∂_νA_μ while the two −igA factors square to −ig[A_μ, A_ν], so in components F^a_μν = ∂_μA^a_ν − ∂_νA^a_μ + g f^abc A^b_μ A^c_ν. Two consequences follow. First, F transforms homogeneously, F_μν → U F_μν U⁻¹, which is why tr(F_μνF^μν) is invariant and can serve as a Lagrangian while A_μ alone cannot. Second, the extra term needs no derivative at all. Take the SU(2) configuration A¹₁ = A²₂ = 1.4 GeV with every other component zero and every derivative zero; at g = 1 the Maxwell-like part vanishes identically, yet F³₁₂ = g ε³¹² A¹₁A²₂ = 1.96 GeV². In electromagnetism a constant potential is no field. Here it is a field.

Put that curvature into L = −¼F^a_μνF^(aμν) and expand. The part quadratic in A is the ordinary kinetic term, three or eight copies of a massless vector. The cross term between ∂A and gfAA is cubic, of the form g f^abc (∂_μA^a_ν)A^(bμ)A^(cν); the square of the commutator term is quartic, g² f^abe f^cde A^a A^b A^c A^d. Those are the three-gluon and four-gluon vertices, and they are present in a pure gauge theory containing no matter fields whatsoever. The statement behind the algebra is that a non-abelian gauge field carries the charge it mediates: a gluon is coloured, so it can emit another gluon, while the photon is electrically neutral and cannot. That one difference propagates all the way to the running coupling. The gluon and ghost loops that exist only because f^abc ≠ 0 supply the +(11/3)C_A in β₀ = (11/3)C_A − (2/3)n_f, and with C_A = 3 and n_f = 5 that is β₀ = 23/3 > 0 — asymptotic freedom, traceable to this one term.

Be exact about what has been established. The gauge principle does not choose the group: SU(3) × SU(2) × U(1) is read off experiment, not derived, and nothing here forbids SU(5). It does not choose the representations either — anomaly cancellation and measurement do that — nor the couplings, which are inputs fitted to data. It also forbids something you may want: m²A^a_μA^(aμ) is not invariant under A_μ → UA_μU⁻¹ − (i/g)(∂_μU)U⁻¹, so every gauge boson built this way is massless and, at this classical level, infinite-ranged, and the 80.4 GeV W must come from breaking. Finally, all of the above is classical. Quantising it in the path integral means integrating over gauge orbits, which counts each physical configuration infinitely many times; the Faddeev–Popov procedure fixes a gauge and pays with ghost fields, anticommuting scalars in the adjoint that never appear as external states but must circulate in loops for the amplitudes to stay unitary. Ghosts are an artefact of covariant gauges, not of the physics: axial gauge has none and an uglier propagator instead.