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The factors $e^{i\xi\cdot A}$ and $e^{iu\cdot V}$ are chosen this way so there is a unique label $(u,\xi)$ for each group element $g=e^{i\xi\cdot A}e^{iu\cdot V}$. To see that $\xi$ is a label for $G/H$, it's enough to check that $g$ and $gh$ are associated with the same value $\xi$ when $h\in H$. This is easy to see: set $h=e^{iu_h\cdot V}$, so that ...
First of all, there are a few problems with your question: $J_{ab}^0 = \pi^a \epsilon^{ab} \Phi^b$ is not a valid expression, since there is a summation on the right hand side of the equation, but a and b are free indices on the left hand side. Your definition of $\epsilon$ is a bit weird, too. What you mean is $$J_{ab}^0 = \pi^i \epsilon_{ab}^{ij} ... 2 Let us consider the corresponding Hamiltonian theory, so that we have a notion of a commutator that we can use to form a Lie algebra bracket. Moreover, let us consider the classical theory for simplicity. Then the Poisson bracket$$\tag{1} \{\Phi^a({\bf x}),\Pi_b({\bf y})\}_{PB}~=~\delta^a_b~\delta^3({\bf x}-{\bf y}), \qquad \text{etc},$$plays the role ... 0 For convenience, I would first like to change the sign convention. That is, R_{z}(\alpha) = \begin{pmatrix}\cos\alpha&-\sin\alpha&0\\\sin\alpha&\cos\alpha&0\\0&0&1\end{pmatrix},\quad R_{y}(\beta) = \begin{pmatrix}\cos\beta&0&\sin\beta\\0&1&0\\-\sin\beta&0&\cos\beta\end{pmatrix},\quad ... 1 The following does not need the hint, but gives an easy way to retrieve the exact evolved state and prove the conservation of the average number of particles. The idea is to make use of the coherent state expression you had in a previous question. For t=0 let$$ |\alpha(0)\rangle = e^{\alpha_0 \, \hat a^\dagger - \alpha^*_0 \, \hat a} |0\rangle.  Use ...