When you write down the Lagrangian for two quarks : \begin{equation} \mathcal{L}_\text{QCD}^0 = -\frac{1}{4} G_{\mu\nu}^a G^{a\mu\nu}+ \bar\Psi i \gamma^\mu D_\mu \Psi \end{equation} you find an $U(2)_L \times U(2)_R$ global symmetry because you can rewrite it : \begin{equation} \mathcal{L}_\text{QCD}^0 = -\frac{1}{4} G_{\mu\nu}^a G^{a\mu\nu}+ \mathcal{L}_\text{QCD}^L+\mathcal{L}_\text{QCD}^R \end{equation} with $\mathcal{L}_\text{QCD}^{L,R} = \bar\Psi_{L,R} i \gamma^\mu D_\mu \Psi_{L,R}$
An arbitrary element of $U(2)_L \times U(2)_R$ can be written : \begin{equation} (g_L, g_R) = \left(e^{ i \gamma +i \gamma^i \frac{\sigma_i}{2}},e^{ i \delta +i \delta^i \frac{\sigma_i}{2} }\right) \end{equation} where the $\sigma_i$s are the Pauli matrices. But you could, in principle, rewrite this element as : \begin{equation} (g_L, g_R) = \left(e^{ i \alpha} e^{ i \beta } e^ {i \alpha^i \frac{\sigma_i}{2}} e^{i \beta^i \frac{\sigma_i}{2}},e^{ i \alpha} e^{ - i \beta } e^{i \alpha^i \frac{\sigma_i}{2}} e^{-i \beta^i \frac{\sigma_i}{2}}\right) \end{equation} That expression shows that one can factor two $U(1)$s and obtain : \begin{equation} U(2)_L \times U(2)_R = SU(2)_L \times SU(2)_R \times U(1)_V \times U(1)_A \end{equation} What I don't understand is how to obtain explicitly the second expression of $(g_L, g_R)$ starting from the first one.
