In the derivation of the Lie algebra in the first volume of Quantum Theory of Fileds by Weinberg, it is assumed that the operator $U(T(\theta)))$ in equation (2.2.17) is unitary, and the rhs of the expansion \begin{equation} U(T(\theta)))=1+i\theta^a t_a +\frac{1}{2} \theta_b\theta_c t_{bc} + \dots \end{equation} requires $$t_{bc}=-\frac{1}{2}[t_b,t_c]_+.$$ If this is the case there is a redundancy somewhere. In fact, by symmetry $$ U(T(\theta))=1+i\theta_at_a+\frac{1}{2}\theta_a\theta_bt_{ab}+\dots\equiv 1+i\theta_at_a-\frac{1}{2}\theta_a\theta_bt_at_b+\dots $$ and it coincides with the second order expansion of $\exp\left(i\theta_at_a\right)$; the same argument would then hold at any order, obtaining $$U(T(\theta))=\exp\left(i\theta_at_a\right)$$ automatically.
However, according to eq. (2.2.26) of Weinberg's book, the expansion $$U(T(\theta))=\exp\left(i\theta_at_a\right)$$ holds only (if the group is just connected) for abelian groups.
This seems very sloppy and I think that Lie algebras relations could be obtained in a rigorous, self-consistent way only recurring to Differential Geometry methods.
