I'm reading Maggiore's book about QFT, and I'm having a trouble understanding the notation in the part about Lie algebras (Section 2.1):

The group generators are defined as $T^a_R=-i\frac{\partial D_R}{\partial \theta_a}|_{\theta=0}$, with $D_R\left(g\right)$ being a linear representation, depeding on parameters $\theta^a$. Later, while proving $e^{i\alpha_aT^a_R}e^{i\beta_aT^a_R}=e^{i\delta_aT^a_R}$, why is it said that $T^a_R$ is a matrix? if so, does the expression $\alpha_aT^a_R$ implies summation on repeating indices or not? I assume $\alpha_a$ to be a vector.


(I don't know Maggiore's book, but the question can be answered without, I guess.)

The notation suggests that $D_R\!\left(\theta_a\right)$ is a group element in some representation $R$, depending on a number of parameters $\theta_a$. (The index $a$ goes from $1$ to $d$, the dimension of the group, and the parameterization is chosen such that $\theta_a=0$ corresponds to the identity element.) Presumably, the representation is finite-dimensional, so you can think of $D_R$ as a matrix.

(Update: Note that $d$, the dimension of the group, i.e. the number of parameters you need to specify a group element (equivalently, the dimension of the Lie group as a manifold) is generically different from the dimension of the representation, $d_R$. A given group has infinitely many representations of various dimensions.)

Then clearly the generator $T_R^a$ is again a matrix of the same dimension, and there are $d$ independent such generators. Furthermore, the generators form a vector space (multiplication of group elements effectively turns into addition of generators), and indeed $\alpha_a T^a_R$ implies summation: This is a linear combination of generators with coefficients $\alpha_a$, forming a new generator.

  • $\begingroup$ If we have a $d$-dimensional space, $D_R$ is a $d \times d$ matrix, shouldn't we have $d^2$ of them to form a base for the space $d\times d$ matrices? $\endgroup$ – proton Jan 12 '18 at 14:20
  • 1
    $\begingroup$ @proton: I have included a clarification about dimensions. $D_R$ would be a $d_R\times d_R$ matrix, but the set of $D_R$'s or $T_R$'s does not in general form a basis for the set of such matrices. For example, the defining representation of $SO(2)$ is two-dimensional, but the group itself is only one-dimensional, so it has only one generator (which is $\big((0,1),(-1,0)\big)$), so that doesn't span the set of $2\times2$ matrices (and there's no reason it should). $\endgroup$ – Toffomat Jan 12 '18 at 14:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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