I seem to be missing something regarding why Yang-Mills theories are Lorentz invariant quantum mechanically.
Start by considering QED. If we just study the physics of a massless $U(1)$ gauge field then by the usual Wigner little group classification one-particle states form representations of the little group for $k^\mu=(k,0,0,k)$ which is just $ISO(2)$. As is standard, we assume that our states only transform non-trivially under the $SO(2)$ subgroup of $ISO(2)$ so that states are labeled by their momentum and their helicity, $h$. That is, under an $SO(2)$ rotation by angle $\theta$ we have $U(\theta)|k,h\rangle=e^{ih\theta}|k,h\rangle$. Correspondingly, the creation operator $a^\dagger_h(k)$ transforms as $U(\theta)a^\dagger_h(k)U(\theta)^\dagger=e^{ih\theta}a^\dagger_h(k)$. Because we assumed that the states transform trivially under the other $ISO(2)$ generators, the same equations will hold true if we replace $U(\theta)$ by $U(W)$ where $W$ is any member of the $ISO(2)$ group.
It's then well known that if we try to write down a field operator $A_\mu$ in the usual manner, we will find that the transformation properties of the creation and annihilation operators force $A_\mu$ to transform under $U(W)$ as $U(W)A_\mu U(W)^\dagger=W_\mu{}^\nu A_\nu+\partial_\nu\Omega$ for Lorentz transformation $W$ and some irrelevant function $\Omega$. This is the standard argument by which Lorentz invariance is found to demand gauge invariance for massless particles.
My question then is what happens when you consider instead a non-Abelian gauge theory? I'd assume the same story holds and all gauge bosons just acquire an internal index, say $a$. That is, I thought states would be labeled like $|k,h,a\rangle$ and field operators would look like $A_\mu^a$ and transform as $U(W)A_\mu^aU(W)^\dagger=W_\mu{}^\nu A_\nu^a+\partial_\nu\Omega^a$ for some function $\Omega^a$.
This seems to be wrong, though, since the YM lagrangian is not invariant under the above and it is instead only invariant under full non-Abelian gauge transformations, $A\to U^{-1}(A+d)U$.
The only way out I can see is that single particle non-Abelian states may change under $U(W)$ as $U(W)|k,h,a\rangle=e^{ih\theta}\sum_b D(W)_{ab}|k,h,b\rangle$, where $D(W)_{ab}$ is a unitary matrix which is rotating the internal index on the gauge states. Previously, I'd been assuming that $D(W)_{ab}$ is trivial. If this is the proper transformation, I could see the $A_\mu^a$ operator inheriting the more familiar non-Abelian gauge transformation behavior.
However, naively it seems wrong that a Lorentz rotation will rotate the gauge indices of states since this would seem to be a mix of spacetime and internal transformations and hence ruled out by Coleman-Mandula. An internal non-Abelian transformation generator $V(X)$ should act on our state as $V(X)|k,h,a\rangle=\sum_b D_{ab}|k,h,b\rangle$ for some unitary matrix $D_{ab}$ and hence $V(X)U(\theta)|k,h,a\rangle\neq U(\theta)V(X)|k,h,a\rangle$ so there would be symmetry generators which do not commute with Poincare.
In summary, my questions are:
1) How does the non-Abelian gauge operator change under Lorentz transformations?
2) Does a one-particle non-Abelian spin-1 state $|k,p,a\rangle$ indeed transform as $U(W)$ as $U(W)|k,h,a\rangle=e^{ih\theta}\sum_b D(W)_{ab}|k,h,b\rangle$ with $D(W)_{ab}$ a non-trivial element of the non-Abelian group? If so, why doesn't this violate Coleman-Mandula?
