Hi all, I am sure I am missing something really elementary, but I would really appreciate someone pointing it out to me. So, if we consider the situation in abelian gauge symmetry, say for fermion matter ψ, of charge q. The transformation law for ψ is, ψ→ψ' = e[- i q θ(x)] ψ. We then have to introduce a abelian gauge field, Aμ(x) to construct a covariant derivative, Dμ ψ = ∂μ ψ + i q Aμ ψ in order to make the kinetic term in the matter action gauge invariant: Dμ ψ → D'μ ψ' = e[- i q θ(x)] Dμ ψ. From this equation, the gauge transformation of the gauge field follows, Aμ → A'μ = Aμ + ∂μ θ. Thus the gauge transformation for the gauge field is independent of the charge of the matter, q. In principle I could have started from a fermion or scalar with a different charge, say q' and the gauge field transformation equation won't care about that. This is all nice! However, things change the moment I replicate the steps for a non-abelian gauge symmetry. The transformation law for the matter is now, ψ→ψ' = U ψ, U = e[- i g Ta θa(x)] with g playing the role of charge q. Then we introduce the gauge-covariant derivative Dμ ψ = ∂μ ψ + i g Aμ ψ, Dμ ψ → D'μ ψ' = U (Dμ ψ). This results in the gauge transformation law for the gauge field itself, Aμ → A'μ = U Aμ U-1 + (i/g)(∂μ U) U-1 . Infinitesimally, Aμ → A'μ = Aμ - g [Aμ, Ta] θa +Ta ∂μ θa . Now this looks not so nice because the color charge of the matter, g appears explicitly in the gauge field transformation law. Had we started from a matter field with a different color charge, g', then one would be led to a different gauge field transformation law! Does it mean unlike in the abelian case there can be only one common color charge g for all color-charged particles in the same representation? In electrodynamics electron and proton can in principle have different charge, at least classically. What is the problem here?