# The remained global flavor symmetries of massless quarks after gauging electromagnetic $U(1)$

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 $$\frac{SU(N_f-1)_V \times SU(N_f)_A \times U(1)_V}{Z_{N_f}} [\text{This is WRONG!}]$$

• Thanks, I know you cannot write $SU(n)_V \times SU(m)_A$ as a group. So I am asking how to write the global symmetry group. – ann marie cœur Sep 12 '17 at 21:32
• u and d have different charges, so cannot be in a single rep of $U(1)_e$. – Cosmas Zachos Sep 12 '17 at 21:34
• The $U(1)$ is diagonal, but the element in $N_f=3$, you can consider a Lie algebra generator matrix proportional to $diag(2/3,-1/3,-1/3)$. It is still a $U(1)$, but not the complex $U(1)$ as $+e^{i \theta} \mathbb{I}$. I am talking about the different $U(1)$. The $SU(N)$ has no $e^{i \theta} \mathbb{I}$., only $U(N)$ has $+e^{i \theta} \mathbb{I}$ But it is already taken care carefully in my question. [What you said is not part of my question.] Thanks – ann marie cœur Sep 12 '17 at 21:40
• I am talking about the $U(1)_e$ diagonal for $N_f=3$, you can consider a Lie algebra generator matrix proportional to $diag(2/3,-1/3,-1/3)$. – ann marie cœur Sep 12 '17 at 21:42