Parameterization of an arbitrary element of $U(2)_L \times U(2)_R$ (Chiral symmetry with two quarks) 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.
 A: Let's see what relation can we find between $\alpha, \beta, \alpha^i, \beta^i$ and $\gamma, \delta, \gamma^i, \delta^i$
First using Baker Campbell Hausdorff lemma we deduce two things:
$$\alpha + \beta = \gamma \text{ and } \alpha - \beta = \delta$$
because $\mathbb{1}$ commutes with $\sigma$. And 
$$e^{i\vec{\alpha}\cdot \vec{\sigma}} = \mathbb{1}\text{cos }\alpha + i\sigma\cdot \hat{\alpha}\,\text{sin } \alpha$$
The latter gives us that
$$
e^{i\alpha\cdot \sigma}e^{\pm i\beta\cdot \sigma} =\left( \mathbb{1}\text{cos }a + i \sigma\cdot\hat{a}\,\text{sin } a\right)\left(\mathbb{1}\text{cos }b \pm i \sigma\cdot\hat{b}\,\text{sin } b\right) 
$$
$$
= \left(\text{cos } a \text{ cos b} \mp (\sigma\cdot \hat{a})(\sigma\cdot\hat{b})\text{ sin } a \text{ sin }b \right) + i \left(\sigma\cdot\hat{a}\text{ sin } a\text{ cos }b \pm \sigma \cdot\hat{b}\text{ sin } b\text{ cos }a \right)
$$
$$
= e^{i\gamma\cdot\sigma} \text{ or } e^{i\delta\cdot\sigma}
$$
Then you verify the ansatz suggested by user40085; $\vec{a} = (\vec{\gamma}+ \vec{\delta})/2$ and $\vec{b} = (\vec{\gamma}- \vec{\delta})/2$
