# Does the field $\Psi$ have different representation for different generator $T_R^a$?

I'm learning the non-abelian gauge theories. Suppose we have a set of (general) fields $$\Psi^\alpha(x)$$ transforming in a given representation $$R$$ of the gauge group, with $$\alpha, \beta = 1,..., \dim(R)$$ and $$a = 1,...,\dim(\mathfrak{g})$$, where $$\mathfrak{g}$$ is the Lie algebra. Then we have the gauge transformation $$\Psi\rightarrow U_R\Psi$$, or in components: $$\begin{equation*} \Psi^\alpha(x)\rightarrow U_R(x)^\alpha_{{\ }\beta}(x)\Psi^\beta(x)\quad\text{where}\quad U_R(x) = \exp(ig\theta^a(x)T^a_R) \end{equation*}$$ However, sometimes I see the generator $$T_R^a$$ is written in its fundamental representation $$t^a$$. My question is in that case, do we have a different representation of the field $$\Psi$$ also? In some reference, we have $$U(x) = \exp(i\theta^a(x)t^a)$$ instead (the coupling constant $$g$$ is dropped), is that because $$t^a$$, rather than $$T^a_R$$, is used here?

Yes you do! Most generally a representation of a Lie group $$G$$ is a map $$R$$ from the group to the space of linear transformations $$GL(V)$$ acting on a vector space $$V$$. What it means for your field $$\Phi$$ to transform in a representation $$R$$ is exactly that the field is an object in $$V$$ such that under a group transformation $$g$$ it transforms as $$(R(g))(\Phi)$$

Now we know that to every Lie group you can associate a Lie algebra and that we normally look at the transformations in the context of the algebra. In you notation this is done through the exponential. I.e. we write

$$g = e^{i\theta^a t_a}$$ to relate the element in the algebra to those in the group. Now note that this $$g$$ can't act on $$\Psi$$ as they are generally in very different spaces! However we know how we can act on $$\Psi$$ with such an element: Apply the representation $$R$$ such that $$\Psi$$ transforms as

$$\Psi \rightarrow (R(e^{i \theta^a t_a}))(\Psi)$$.

Now the important fact: Every group representation $$R$$ induces a algebra representation $$r$$ suc that $$R(e^{i \theta^a t_a}) = e^{i r(theta^a t_a)}$$. This map is linear and as such

$$r(\theta^a t_a) = \theta^a r(t_a)$$ we now define the generators in a representation $$R$$ by exactly this map: $$T^a_R = r(t^a)$$. So the generators do change depending on the representation you are looking at!

As for the last part of your question when we use $$t_a$$ notationally what we mean is that the group acts on its natural space (as a matrix lie group) e.g. $$3x3$$ matrices act on $$\mathbb{C}^3$$ etc. as such $$R$$ is just the identity map. Thus this just implies that if $$t_a$$ is used in this way we are always looking at the fundamental rep.

• Thanks so much, that's really helpful :)
– IGY
Feb 24, 2023 at 11:12