I'm considering the following free Lagrangian (density):
$$\mathcal{L}= \bar{\Psi} \left(i \displaystyle{\not}{\partial}\mathbb{1}_{2}\right)\Psi$$
Where $\Psi$ is a doublet of two fermion fields $\Psi = \begin{pmatrix}\psi_{1}\\\psi_{2}\end{pmatrix}$.
This Lagrangian is obviously invariant under $U(2)$ transformations:
$\Psi \rightarrow U \Psi$, such that $U^{\dagger} U=\mathbb{1}_{2}$,
Using the Pauli matrices as the generators of the $U(2)=SU(2)\times U(1)$ group we can write:
$$U=e^{-i \left(\theta^{0}\mathbb{1}_2+\theta^{1}\sigma_{1}+\theta^{2}\sigma_{2}+\theta^{3}\sigma_{3}\right)}$$
In all the book I have read they stops here with the symmetries of the lagrangian, the problem is that in my opinion there is an additional symmetry:
$$\Psi \rightarrow e^{-i \gamma^{5} (q_{1}\mathbb{1}_{5}+q_{2}\sigma_{1})} \Psi = \begin{pmatrix}e^{-i \gamma^{5} (q_{1}+q_{2})}\psi_{1}\\e^{-i \gamma^{5} (q_{1}-q_{2})}\psi_{2}\end{pmatrix}$$
That corresponds to the chiral (axial) symmetries for each $\psi_{k}$, and so the symmetry group should be $[U(2)]\otimes [U_A(1)]_{q_1+q_2}\otimes [U_A(1)]_{q_1-q_2}$.
This symmetry can't be obtained from the one above because the coefficients $\theta^{i}$ are not numbers anymore but rather 4x4 matrices $(\gamma^{5})$.
Is this right or am I missing something?
