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?


1 Answer 1


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.

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

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