For $N_f$ numebr of massless quarks, we know that there are global symmetries
$$
\frac{SU(N_f)_L \times SU(N_f)_R \times U(1)_V}{Z_{N_f}}
$$
here $U(1)_V$ is the same as $U(1)$-Baryon number conservation. The axial $U(1)_A$ is anomalous.

Question: What would be the remained global flavor symmetries of massless quarks after gauging electromagnetic $U(1)_e$? We can take $Nf=3$ where we have 3 quarks like $u,d,s$.

[WARNING]: Notice that $U(1)_e$ is part of the vector global flavor symmetry $SU(N_f)_V$, and gauging result in the remained $SU(N_f-1)_V$ global flavor symmetry. Namely, $U(1)_e$ is part of $SU(N_f)_L \times SU(N_f)_R$. However, we note that
$$SU(N_f)_L \times SU(N_f)_R \neq SU(N_f)_V \times SU(N_f)_A$$
because the left-right chiral flavor generators DO commute, but the vector and axial generators of $SU(N_f)_V$ and $SU(N_f)_A$ do NOT commute. So the answer won't be as simple as
$$
\frac{SU(N_f-1)_V \times SU(N_f)_A \times U(1)_V}{Z_{N_f}} [\text{This is WRONG!}]
$$

What is your answer?

This post imported from StackExchange Physics at 2020-10-29 11:42 (UTC), posted by SE-user annie marie heart