Constructing Non-Projective Representation from Projective Ones - Derivation in Weinberg Vol I, Appendix 2B

Question

In Appendix 2B of Weinberg's QFT Vol I, he provided a proof for a theorem that the phases of the operators $U(T)$ for finite symmetry transformations $T$ may be chosen so that these operators form a representation of the symmetry group, rather than a projective representation.

Once he constructed a non-projective representation $U[\theta]$, he proceeds to prove any projective representation $\tilde U[\theta]$ of the same group with the same representation generators $t_a$ can only differ from $U[\theta]$ by a phase $$\tilde{U}[\theta]=e^{i \alpha(\theta)} U[\theta]$$ such that the phase $\phi$ in $$\tilde{U}\left[\theta^{\prime}\right] \tilde{U}[\theta]=e^{i \phi\left(\theta^{\prime}, \theta\right)} \tilde{U}\left[f\left(\theta^{\prime}, \theta\right)\right]$$ can be removed by a change of phase for $\tilde U[\theta]$

To prove this, he first considered the operator $$U[\theta]^{-1} U\left[\theta^{\prime}\right]^{-1} \tilde{U}\left[\theta^{\prime}\right] \tilde{U}[\theta]=U\left[f\left(\theta^{\prime}, \theta\right)\right]^{-1} \tilde{U}\left[f\left(\theta^{\prime}, \theta\right)\right] e^{i \phi\left(\theta^{\prime}, \theta\right)}\tag{1}$$ and he asserted that because $U[\theta]$ and $\tilde{U}[\theta]$ have the same generators, the derivative of the left-hand side with respect to $\theta^{\prime a}$ vanishes at $\theta^{\prime}=0$, and so $$0=\frac{\partial}{\partial \theta^b}\left\{U[\theta]^{-1} \tilde{U}[\theta]\right\}+i \phi_b(\theta) U[\theta]^{-1} \tilde{U}[\theta]\tag{2}$$ where $$\phi_b(\theta) \equiv h_b^a(\theta)\left[\frac{\partial}{\partial \theta^{\prime b}} \phi\left(\theta^{\prime}, \theta\right)\right]_{\theta^{\prime}=0}$$ and $h^a_b$ is defined implicitly as $$\left[h^{-1}\right]_b^a(\theta) \equiv\left[\frac{\partial f^a(\bar{\theta}, \theta)}{\partial \bar{\theta}^b}\right]_{\bar{\theta}=0} \tag{2.B.4}$$ Finally, by differentiating this result with respect to $\theta^c$ and antisymmetrizing in $b$ and $c$ we have $$0=\frac{\partial \phi_b(\theta)}{\partial \theta^c}-\frac{\partial \phi_c(\theta)}{\partial \theta^b}\tag{3}$$

Notice in $(1)$, the partial derivative is with respect to $\theta$, rather than the described $\theta'$, could this be a misprint? Also the the introduction of $h^a_b$ in the definition of $\phi_b(\theta)$ feels unmotivated, the indices in the definition of $\phi_b$ does not even match the summation convention. I wonder how one should go from $(1)$ to $(2)$ then to $(3)$, any help is appreciated.

Answer 1

OK, it seems I managed to work out the missing steps myself. First, bearing in mind the definition for $h^{-1}$ and by Taylor's theorem we have $$U[f(\theta',\theta)]^{-1}\tilde U[f(\theta',\theta)]\approx U[\theta^a+(h^{-1})^a_b\theta'^b]^{-1}\tilde U[\theta^a+(h^{-1})^a_b\theta'^b]$$ Let $\delta \theta^a=(h^{-1})^a_b\theta'^b$, and using again the Taylor's theorem we have $$U[f(\theta',\theta)]^{-1}\tilde U[f(\theta',\theta)]\approx U[\theta^a]^{-1}\tilde U[\theta^a]+(h^{-1})^a_b\theta'^b\frac{\partial}{\partial \theta^a}(U[\theta^a]\tilde U[\theta^a])$$ Thus, $$\begin{align} \frac{\partial}{\partial \theta'^b}\left(U[f(\theta',\theta)]^{-1}\tilde U[f(\theta',\theta)]e^{i\phi(\theta',\theta)}\right)=U[\theta^a]^{-1}\tilde U[\theta^a]\frac{\partial}{\partial \theta'^b}e^{i\phi(\theta',\theta)}+(h^{-1})^a_b\theta'^b\frac{\partial}{\partial \theta^a}(U[\theta^a]^{-1}\tilde U[\theta^a)e^{i\phi(\theta',\theta)}+(h^{-1})^a_b\theta'^b\frac{\partial}{\partial \theta^a}(U[\theta^a]^{-1}\tilde U[\theta^a])\frac{\partial}{\partial \theta'^b}e^{i\phi(\theta',\theta)} \end{align}$$ Now, take the limit as $\theta'\rightarrow0$, so the third term vanishes, simplify and cancelling the phases, then multiply through by $h^a_b$ yields the desired result. Notice there is a typo in Weinberg's book, in the definition of $\phi_b$, $\partial/\partial\theta'^b$ should have been $\partial/\partial\theta'^a$.