How to prove a set of matrices form a representation of Lie algebra? When reading Paul Langacker's The Standard Model and Beyond, I am quite confused on equation 3.29, which says with a set of fields $\Phi _a$, where $a$ goes from 1 to $n$, are chosen to be transformed to it self by Lie algebra generators $T^i$. Therefore we make the following assumption
$$
\left[ T^i , \Phi _a \right] \equiv - L_{a b}^i \Phi _b
$$
It is then said $L^i$ can be easily proven to form a representation of Lie algebra $U_G = e^{- \mathrm{i} \beta ^i T^i}$.
Well, I think if one can prove that $L^i$ and $T^i$ satisfy the same commutation relation, this conclusion is then true. But how do we do that?
Thanks in advance!
 A: I have the answer now! It is explained in equation 3.36 and 3.37 of the same book.
First, take the adjoint of $\left[ T^i , \Phi _a \right] = - L_{a b}^i \Phi _b$, remember $T^i$ is hermitian, assume $L^i$ is real, which would lead to $\left[ T^i , \Phi _a^{\dagger} \right] = \left( L_{a b}^i \right)^T \Phi _b^{\dagger}$.
Now for a ground state $\left| 0 \right>$, assume $\Phi _a^{\dagger} \left| 0 \right> = \left| a \right>$. Also, the ground state is invariant for now, $T^i \left| 0 \right> = 0$.
Then we would have
$$
T^i \left| a \right> = T^i \Phi _a^{\dagger} \left| 0 \right> = \Phi _a^{\dagger} T^i \left| 0 \right> + \left( L_{a b}^i \right)^T \Phi _b^{\dagger} \left| 0 \right> = 0 + \left( L_{a b}^i \right)^T \left| b \right> = \left| b \right> L_{b a}^i
$$
However, it is true that $T^i \left| a \right> = \left| b \right> \left< b \left| T^i \right| a \right>$, which tells us
$$
L_{b a}^i = \left< b \left| T^i \right| a \right>
$$
Hence $L_{b a}^i$ is a representation of $T^i$ in the particle space.
