# Symmetries of two Massless Fermion

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?

The full symmetry group of the massless Lagrangian is actually: $$SU(2)_L\times U_L(1)\times SU(2)_R\times U_R(1)$$ which covers the additional symmetries you noticed.
It can be rearranged as: $$SU(2)_V\times U_V(1)\times SU(2)_A\times U_A(1)$$ where $$U_A(1)$$ (related to your $$q_1$$ portion) is usually broken by the quantum anomaly, and $$SU_A(2)$$ (related to your $$q_2$$ portion) might be dynamically violated by chiral symmetry breaking, depending on the circumstances (e.g. via QCD or NJL interactions).