I'm not exactly sure how to phrase my question, but I'm trying to ask the following: if I have an operator transforming in an irreducible transformation of some group, I get a corresponding symmetry transformation on my states, is this representation acting on my states also irreducible?
For example, suppose I had a lagrangian that was $L = \phi^\mu \phi_\mu$ then I can see that that it has $SO(n)$ symmetry in the following sense. Let $R(\omega)$ be a rotation (in the fundamental representation) then if I send $\phi_\mu \mapsto R(\omega)_\mu\ ^\nu \phi_\nu$ the lagrangian remains invariant. Corresponding to this I get a representation acting on the states by $R(\omega)_\mu\ ^\nu \phi_\nu = U(\omega)^{-1} \phi_\mu U(\omega)$
Now I know that the $R(\omega)$ is in the fundamental so that is necessarily an irreducible representation. However can I somehow conclude that the $U(\omega)$ representation is irreducible as well?
P.S. I know that in general states and operators dont even need to have the same symmetry group. I'm more interested in whether irreducibility of one implies irreducibility of the other
