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

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

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 \begin{equation} 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} \end{equation} 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?

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    $\begingroup$ $U(1)_R\times U(1)_L=U(1)_V\times U(1)_A$ $\endgroup$ Aug 24, 2021 at 13:10
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    $\begingroup$ 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. $\endgroup$
    – tomtom1-4
    Aug 24, 2021 at 13:29
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    $\begingroup$ 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 $V\times A$. $\endgroup$ Aug 24, 2021 at 13:37
  • $\begingroup$ @AccidentalFourierTransform This is my true doubt: why the two are the same? How the relabeling of $U(1)$ lead to different conserved currents forms? $\endgroup$
    – Pipe
    Aug 24, 2021 at 14:31
  • $\begingroup$ @CosmasZachos Why the $\gamma_{5}$ part survives? $\endgroup$
    – Pipe
    Aug 24, 2021 at 15:27

1 Answer 1


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.


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