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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.

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(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.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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