Lie group symmetry in Weinberg's QFT book

Question

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 $C^a_{bc} = -f^a_{bc} + f^a_{cb}$).

Answer 1

You are correct, (2.B.10) is actually not true. But it's only used through (2.B.11), and (2.B.11) does hold, because the derivative $\frac{\partial^2 f^a(\theta, \theta_1)}{\partial\theta^b \partial\theta^c}$ is symmetric in $b$ and $c$.