# Why are symmetries labelled by groups and not representations?

Physicists will say that a certain system has $$G$$ symmetry, where $$G$$ is some group, such as $$SU(2)$$ or $$S_3$$ or whatever. To show that this is the case, they will conjure up an explicit representation $$\rho_G$$ of that group and show that the equations of motions—or the action, or whatever—are still the same. But a group is more general than a specific representation of that group, so conflating the two seems wrong.

So what does "the system has $$G$$ symmetry" mean?

1. I don't think this can mean "There exists a representation $$\rho_G$$ of $$G$$ that is a symmetry of the system," since this is trivially true for all $$G$$.
2. I suppose it could mean "For all representations $$\rho_G$$ of $$G$$ on the system's vector space $$V$$, $$\rho_G$$ is a symmetry." If this is so, I've never seen this much stronger statement actually shown, but maybe I'm missing something obvious.
3. Knowing my colleagues, it could just mean "There is a specific representations $$\rho_G$$ of $$G$$ that is a symmetry. For cultural and linguistic reasons, we will just forget the representation information, which you can figure out on your own."
4. Something else entirely?

I would say that it means I have a lagrangian $$\mathcal{L}$$ that depends on a bunch of fields. I can transform those fields under $$G$$. They may or may not transform under the same representation $$\rho_G$$ of some group $$G$$. The objects in any given representation are not invariant under the group (unless they are in a trivial representation). It's the system as a whole. So the system really does have the symmetry $$G$$, not $$\rho_G$$.

Examples:

• In QCD one has SU(3) gauge symmetry. The quarks transform under the fundamental. The gluons transform under the adjoint. The baryons are scalars under SU(3). They are all different representations but the symmetry of the system as a whole is the group symmetry $$G=SU(3)$$.
• Global Lorentz symmetry. In the Standard model one has scalars (don't transform), fermions (spin 1/2 rep) and vector bosons (spin 1). All of which are in different representations but the whole system has a Lorentz symmetry.
• Conformal symmetry. One has different conformal weight operators but everything transforms under the same conformal symmetry.

Said in a slightly different way, the system has a symmetry $$G$$, it has different components all being affected differently by a symmetry transformation in $$G$$, but at the end of the day, the system is invariant under the action of $$G$$ as a whole, not any specific representation.

Clarification:

Let's take the lagrangian $$\mathcal{L}$$ to represent our "system" (You can also take the action or the partition function, depends on how general you want to be, but let's stick with the Lagrangian for now). The Lagrangian depends on different fields $$\phi_{\rho_G^i}$$ that are in different representations $$\rho_G^i$$. We may state this in the following way:

$$\mathcal{L} = \mathcal{L}(\phi_{\rho_G^1},\phi_{\rho_G^2},...,\phi_{\rho_G^n}).$$

Now, the statement that the system has a certain symmetry means that the Lagrangian $$\mathcal{L}$$ doesn't change. Or in other words $$\mathcal{L}$$ is in the trivial representation of $$G$$.

The list of fields that the system (Lagrangian) depends on may transform in any myriad of ways under the action of the group $$G$$, so long as the Lagrangian transforms trivially, then we may say that the system has the symmetry group $$G$$.

• I don't believe this answers my question: I understand that each field transform under a representation of G. The collection of fields then transforms as a specific representation of G, a product of the individual reps. This product representation respects the Lagrangian. The examples given seem to imply that the representation chosen matters very much! If I chose just one set of fields, and switched to a trivial representation on it, the Lagrangian is no longer symmetric under the action of G! But as far as representation theory goes, the trivial rep is a perfectly good rep. – F. Bardamu Jul 10 at 8:22
• @F.Bardamu I don't understand your point. If you have fields that transform under a different rep but now the lagrangian isn't invariant, then you don't have that symmetry. The representations just tell you how different objects change under a transformation of the group. If the lagrangian isn't invariant then that transformation isn't a symmetry of the system. – Stratiev Jul 10 at 8:55
• This is precisely the issue: If a system is symmetric under one representation of G, but not under another representation of G, then why say "The system is symmetric under the group G" when the more precise statement is "The system is symmetric under this specific representation rho_G of the group G"? – F. Bardamu Jul 10 at 13:04
• @F.Bardamu I really don't see how you're getting this from my comment. Any separate object in the Lagrangian that does not transform trivially, (read "transforms in any other representation") is not symmetric under it, precisely because it changes under the group G. We say that the system as a whole has a symmetry $G$ if the system as a whole doesn't change, even though the distinct components do. Take GR for example. You have vectors and tensors and they all change in different ways (different representations), but the Hilbert action does not, hence the equations don't. – Stratiev Jul 10 at 13:12
• Hence the system is diffeomorphism invariant, even though the separate objects in their distinct representations do, in fact, change. – Stratiev Jul 10 at 13:12