I have a very similar question to the one asked below:

Why are non-Abelian gauge theories Lorentz invariant quantum mechanically?

In particular, the setup to my question is essentially the same: In section 5.9 of Weinberg Volume I, it is stated that a massless helicty-1 field must transform like $A_{\mu} \rightarrow A_{\mu} + \partial_{\mu} \Omega$ under a Lorentz transformation. Because of this, when such a field is coupled to matter, it must be coupled to a conserved current, i.e. the action must take the form

$S = S_{photon} + S_{matter} + \int A_{\mu}J^{\mu}$,

in which $S_{photon} = \int (\partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu})^2$, $S_{matter}$ is the free part of action for the matter fields, and $J^{\mu}$ is some conserved current which comes about from applying the Noether procedure to $S_{matter}$. Because the current is conserved, when the action is varied under a Lorentz transformation one finds

$\delta_{Lorentz}S = \int(\partial_{\mu}\Omega) J^{\mu} = -\int\Omega(\partial_{\mu} J^{\mu}) = 0$.

This gives a concrete motivation for why, at least in the abelian case, we require the action to be gauge invariant--it's needed for the action to be Lorentz invariant.

I am confused as to how (or if) this argument can be generalized to the non-abelian case. I believe the setup should go as follows: First, suppose that we have $n$ species of massless, helicity-1 particles, $A^{a}_{\mu}, a = 1, 2, ..., n$. Momentarily ignore matter which will be included later. Because we require the action to be invariant under $\delta_{Lorentz}$, one can write

$S = S_{photon}$, where now $S_{photon} = \int(\partial_{\mu}A^a_{\nu} - \partial_{\nu}A^a_{\mu})^2$. Now turn on some collection of matter fields transforming under a representation of a group $G$, and include any self-interactions between the gauge fields themselves. Again, because one requires that the total action is invariant under $\delta_{Lorentz}$, the action must take the form

$S = S_{photon} + S_{matter} + \int A^a_{\mu}J^{a\mu}[A, \phi]$, in which I have emphasized the fact that $J^a_{\mu}$ may depend on the gauge fields and the matter fields. I believe $J$ should be interpreted as the Noether current corresponding to the global symmetry of the group G on the total action. With this information, one can show that under a global transformation $\delta_G$ the current transforms in the adjoint representation. Moreover, if one assumes the coupling term itself is invariant under $\delta_G$, one can show that the gauge fields also transform in the adjoint representation. At this point, however, I am stuck. I believe the next step should be to show that somehow, the statement $\delta_{Lorentz}S = 0$ is equivalent to $\delta^{local}_G S = 0$. By $\delta^{local}_G$, I mean a local gauge transformation under the group $G$. If this is true, then I could see how requiring non-abelian gauge invariance is necessary (and sufficient) for having Lorentz invariance. The main points to my question are as follows:

1) Is the statement that $\delta_{Lorentz}S = 0$ is equivalent to $\delta^{local}_G S = 0$ even true?

2) If it is true, is there a simple way to see it from the setup above, or does anyone know of a source which fleshes out the connection? I can find many sources which talk about the connection between Lorentz invariance, matter fields transforming under representations of abelian groups, and gauge invariance. However, I cannot find any sources which generalize the argument to the case in which matter fields transform under representations of non-abelian groups.


3) I have also not been able to show that the Yang-Mills Lagrangian without matter is even Lorentz invariant, i.e. invariant under the $A^a_{\mu} \rightarrow A^a_{\mu} + \partial_{\mu} \Omega^a.$ I thought maybe there was some $U(1)$ subgroup of $G$ so that the above transformation is a special case of the general gauge group transformation, but I haven't been able to make that work. I believe this to be my most significant point of confusion, because it seems as though the YM Lagrangian is not Lorentz Invariant. This is somewhat addressed in the answer to the question linked above, in which the author affirms that in the non-abelian case the operators do indeed acquire the extra divergence term under Lorentz transformations. However, there is no justification for why the YM Lagrangian is invariant under this transformation. If someone has any ideas for why this is true, I could be satisfied in seeing everything to do with the group $G$ as just extra structure.



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