1
$\begingroup$

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?

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks so much, that's really helpful :) $\endgroup$
    – IGY
    Feb 24, 2023 at 11:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.