# What is the reason for $U(3)_{L} \times U(3)_{R} = U(1)_{V} \times U(1)_{A} \times SU(3)_{L} \times SU(3)_{R}$?

I am studying the QCD chiral symmetry, and by considering the $$u$$,$$d$$,$$s$$ quarks massless, the Lagrangian

$$$$\mathcal{L} = \sum_{i = u,d,s} \bar{q}_{k}i \gamma^{\mu}D_{\mu}q_{k}$$$$

where $$D_{\mu}$$ is the covariant derivativative containing the gluon gauge field, is invariant under $$U(3)_{L} \times U(3)_{R}$$ and as many textbooks says and also Wikipedia reported is possible to decompose $$$$U(3)_{L} \times U(3)_{R} \quad \mathrm{into} \quad U(1)_{V} \times U(1)_{A} \times SU(3)_{L} \times SU(3)_{R}$$$$ I am familiar with the relation $$U(N) \simeq SU(N) \times U(1)$$ but I am confused about the $$U(1)_{V} \times U(1)_{A}$$ product. What tells to the $$U(1)$$ to be vectorial or axial vectorial? Where the $$\gamma_{5}$$ of the conserved axial vector current comes from?

• $U(1)_R\times U(1)_L=U(1)_V\times U(1)_A$ Commented Aug 24, 2021 at 13:10
• Also the isometry is $U(N) \simeq SU(N) \times U(1) / \mathbb{Z}_N$. But we mostly ignore discrete transformations so that we get the group you are rightfully considering. Commented Aug 24, 2021 at 13:29
• Regarding factors of $\mathbb{Z}_n$ in the denominator (which the Wikipedia page seems to ignore): the paper arXiv:1807.07666 includes a relatively careful review of the symmetries of QCD, although I don't think it explicitly answers your question about the relation between $L\times R$ and $V\times A$. Commented Aug 24, 2021 at 13:37
• @AccidentalFourierTransform This is my true doubt: why the two are the same? How the relabeling of $U(1)$ lead to different conserved currents forms?
– Pipe
Commented Aug 24, 2021 at 14:31
• @CosmasZachos Why the $\gamma_{5}$ part survives?
– Pipe
Commented Aug 24, 2021 at 15:27

As stated in your question, the massless QCD Lagrangian $$\sum_{f= u,d,s} \bar{q}_{Lf} i \gamma^{\mu} D_{\mu} q_{Lf} + \bar{q}_{Rf} i \gamma^{\mu} D_{\mu} q_{Rf}$$ is invariant under $$U(1)_R \times U(1)_L$$. That just means that you can multiply the left-handed and right-handed quarks by independent (global) phases: $$U(1)_R \times U(1)_L: q_{Rf} \to e^{i\epsilon_R} q_{Rf}, \quad q_{Lf} \to e^{i\epsilon_L} q_{Lf},$$ where the $$\epsilon_L$$'s and $$\epsilon_R$$'s are the same for all f's. The corresponding Noether currents are $$L^{\mu} = \sum_{f= u,d,s} \bar{q}_{Lf} \gamma^{\mu} q_{Lf}, \quad R^{\mu} = \sum_{f= u,d,s} \bar{q}_{Rf} \gamma^{\mu} q_{Rf}.$$ The Noether currents of $$U(1)_V \times U(1)_A$$ are now just linear combinations of those: $$V^{\mu} = L^{\mu} + R^{\mu} = \sum_{f= u,d,s} \bar{q}_{f} \gamma^{\mu} q_{f},$$ $$A^{\mu} = R^{\mu} - L^{\mu} = \sum_{f= u,d,s} \bar{q}_{f} \gamma^{\mu} \gamma_{5} q_{f}.$$ You can check that these two transform as a vector and axial vector respectively. It is useful to remember how Dirac spinors and $$\gamma_5$$ look in the chiral basis $$q = \begin{pmatrix} q_L \\ q_R \end{pmatrix}, \quad \gamma_5 = \begin{pmatrix} -I_2 & 0 \\ 0 & I_2 \end{pmatrix} .$$ As we did for the currents, we can also take linear combinations of the generators $$\alpha=\frac{\epsilon_L + \epsilon_R}{2}, \quad \beta=\frac{\epsilon_R - \epsilon_L}{2}.$$ The $$U(1)_V \times U(1)_A$$ acts on Dirac spinors now as follows: $$U(1)_V \times U(1)_A: q_f \to e^{i \alpha} q_f, \quad q_f \to e^{i \beta \gamma_5} q_f,$$ which is equivalent to the transformation above.