I think the story where abelian, i.e. $U(1)$, gauge symmetry comes from is pretty straight-forward:

We describe massless spin 1 particles, which have only two physical degrees of freedom, with a spin 1 field, which is represented by a four-vector. This four-vector has 4 entries and therefore too many degrees of freedom. A description of a spin 1 particle in terms of a four-vector field is necessarily redundant and we call this redundancy "gauge symmetry". Formulated differently: particles are representations of the little groups of the Poincare group, whereas fields are representations of the complete Poincare group. This is what leads to the gauge redundancy. However, as far as I know this story only works for the familiar $U(1)$ symmetry.

(This point of view is emphasized, for example, in Weinbergs QFT book Vol. 1 section 5.9. Someone who currently likes to emphasize this perspective is Arkani-Hamed, for example, in section 2 of his latest paper: https://arxiv.org/abs/1709.04891 or here https://arxiv.org/abs/1612.02797. I actually asked him a month ago if he knows any idea for an analogous explanation for non-abelian gauge redundancies, but unfortunately he didn't had a good answer.)

Is there any similar idea where non-abelian gauge symmetries come from?

The big difference, I think, is that non-abelian gauge symmetries also in some sense help us to explain the particle spectrum. For example, we have doublets and triplets of elementary particles and this is a real physical consequence and can not be regarded as an accident, because we use the "wrong" objects to describe elementary particles.

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    $\begingroup$ I've been wondering the exact same thing for years, and as of today I haven't come up with a single convincing answer. If you ever do, please let me know. Weinberg's motivation for $U(1)$ is perfectly clear, but in Vol.II he offers no good motivation for the non-abelian case. The fact that he had no good answer for your question either makes me think there is no good answer at all, except for the pragmatic one: as of today, the recipe for non-abelian theories is the only known method of constructing a consistent theory. Any other attempt has failed. Oh well, kind of disappointing, isn't it? $\endgroup$ Oct 9, 2017 at 16:09
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    $\begingroup$ I was thinking just the other day about the possiblity of large scale spacetime symmetries as the source of such invariance. For example a closed cyclic FLRW universe has $S^3XS^1$ topology which is equivalent (via peter weyl theorem) to $SU(2)XU(1)$ symmetry and all fields on said manfold are then representable by harmonic expansions in terms of unitary representations of precisely $SU(2)XU(1)$ fields (which is a generalization of the fourier transform. $\endgroup$
    – R. Rankin
    Oct 6, 2018 at 2:15
  • $\begingroup$ Oh! you would get a constant (in space) scalar field as well, as the universe isn't unit diameter (in our units anyway) and thus the fields would be represented as a conformal transform of the above mentioned unitary ones. $\endgroup$
    – R. Rankin
    Oct 6, 2018 at 2:21

1 Answer 1


It is a known fact that the only physical deformation of the abelian u(1) gauge algebra is a Lie gauge algebra of a compact group, thus gluons and their symmetry are the necessary way in which a collection of (electric) chargeless photon-like fields can interact/self-couple.

Bizdadea, C., Cioroianu, E.M., Miaută, M.T., Negru, I. and Saliu, S.O. (2001), Lagrangian cohomological couplings among vector fields and matter fields. Ann. Phys., 10: 921–933. doi:10.1002/1521-3889(200111)10:11/12<921::AID-ANDP921>3.0.CO;2-I

  • $\begingroup$ Can you perhaps explain what you mean by a "physical deformation" in this context in a bit more detail? $\endgroup$
    – ACuriousMind
    Oct 9, 2017 at 11:06
  • $\begingroup$ "Physical" comes from respecting some basic rules for Lagrangian field theory (Hamiltonian bounded from below, real with respect to involution on the "target" space for the fields, number of space-time derivatives = 2, Grassmann parity =0). "Deformation" comes from the fact that the solution to the master (Zinn-Justin) equation is sought in a perturbative manner, i.e. a series expansion in the coupling constant. $\endgroup$
    – DanielC
    Oct 9, 2017 at 11:12

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