In Weinberg's QFT volume 1, section 2.2 and appendix 2.B discuss the Lie group symmetry in quantum mechanics and projective representation. In particular, it's shown in the appendix 2.B how a representation of the Lie algebra extends to a representation of the group in the neighborhood of the identity. Unfortunately I get stuck in one line of the derivation.
Loosely following the notation of the book, group elements are represented by the coordinates $\theta$ (identity element corresponds to $\theta=0$) and group multiplication is represented by the function $f$ \begin{align} g(\theta_1) g(\theta_2) = g(f(\theta_1,\theta_2)) \end{align} The function $f$ encodes the information of the group, and near the identity it expands to \begin{align} f(\theta_1,\theta_2)^a = \theta_1^a + \theta_2^a + f^a_{bc} \theta_1^b \theta_2^c + \ldots \end{align} Suppose we have the representation of the Lie algebra \begin{align} U(\theta) = 1 + i t_a \theta^a + \ldots \end{align} Representation of the Lie group can be obtained by exponentiation, namely by flow along a curve $\theta(s)$ via the differential equation (obtained by taking the $\Delta\theta\to0$ limit of $U(\Delta\theta)U(\theta) = U(f(\Delta\theta,\theta))$) \begin{align} \frac{d}{ds} U(\theta(s)) = i t_a U(\theta(s)) h^a_b(\theta(s)) \frac{d\theta^b(s)}{ds} \end{align} where ${h^{-1}}^a_b(\theta)=\frac{\partial f^a(\bar{\theta},\theta)}{\partial \bar{\theta}^b}|_{\bar{\theta}=0}$. It remains to prove the definition of $U$ doesn't depend on the path $\theta(s)$ chosen. That depends on the properties of the function $h^a_b$ (ultimately properties of $f$), in particular we need equation (2.B.10) \begin{equation} \partial_{\theta^c} h^a_b(\theta) = - f^a_{de} h^d_b(\theta) h^e_c(\theta) \end{equation} This is where I got stuck. Presumably this is derived from the group associativity condition \begin{align} f^a(f(\theta_3,\theta_2),\theta_1) = f^a(\theta_3,f(\theta_2,\theta_1)) \end{align} Differentiating with respect to $\theta_3^c$ we find \begin{align} \frac{\partial f^a(f(\theta_3,\theta_2),\theta_1)}{\partial f^d(\theta_3,\theta_2)} \frac{\partial f^d(\theta_3,\theta_2)}{\partial \theta_3^c} = \frac{\partial f^a(\theta_3,f(\theta_2,\theta_1))}{\partial \theta_3^c} \end{align} then differentiating with respect to $\theta_2^b$ \begin{align} \frac{\partial^2 f^a(f(\theta_3,\theta_2),\theta_1)}{\partial f^d(\theta_3,\theta_2)\partial f^e(\theta_3,\theta_2)} \frac{\partial f^d(\theta_3,\theta_2)}{\partial \theta_3^c}\frac{\partial f^e(\theta_3,\theta_2)}{\partial \theta_2^b}+ \frac{\partial f^a(f(\theta_3,\theta_2),\theta_1)}{\partial f^d(\theta_3,\theta_2)} \frac{\partial^2 f^d(\theta_3,\theta_2)}{\partial \theta_3^c \partial \theta_2^b} = \frac{\partial^2 f^a(\theta_3,f(\theta_2,\theta_1))}{\partial \theta_3^c \partial f^d(\theta_2,\theta_1)} \frac{\partial f^d(\theta_2,\theta_1)}{\partial \theta_2^b} \end{align} and setting $\theta_2=\theta_3=0$, we find \begin{align} \frac{\partial^2 f^a(\theta,\theta_1)}{\partial \theta^b \partial \theta^c}|_{\theta=0} + {h^{-1}}^a_d(\theta_1)f^d_{cb} = \partial_{\theta_1^d} {h^{-1}}^a_c(\theta_1) {h^{-1}}^d_b(\theta_1) \end{align} It seems I need to drop the first term $\frac{\partial^2 f^a(\theta,\theta_1)}{\partial \theta^b \partial \theta^c}|_{\theta=0}$ to get to the equation (2.B.10). I don't know why it should vanish. In addition, (2.B.10) enables us to compute the function $f$, it seems everything is encoded in the coefficient $f^a_{bc}$ (relating to the Lie algebra coefficient by $C^a_{bc} = -f^a_{bc} + f^a_{cb}$).
