If we consider the chiral Lagrangian after the spontaneous symmetry breaking, we have got fermion masses and Yukawa couplings to the physical Higgs boson. So it follows global symmetries in flavor space. Sometimes people refer to these symmetries as $U(3)_F$ symmetries and sometimes $SU(3)_F$ symmetries. Which one is correct?


2 Answers 2


In the absence of Yukawa couplings (only kinetic terms), the SM has the global flavor symmetry:

$$G_{y=0} = U(N_f)^5=U(3)^5$$

Because there are 5 distinct representations in the SM (3 for quarks: $u_R$, $d_R$, $Q_L$; and 2 for leptons: $e_R$, $L_L$).

However, $U(N) \sim SU(N)\times U(1)$, so the group can also be written as:

$$G_{y=0} = SU(3)^5 \times U(1)^5 = SU(3)_q^3 \times SU(3)_e^2 \times U(1)^5$$

The important part for flavor physics are the non-abelian factors (some of these $U(1)$ are anomalous and are broken by quantum effects).

For more details, check out a review on flavor physics (your question should be answered in the very first pages), e.g.,


I often see $\mathrm{SU}{(3)}_\text{flavor}$. However, I have seen $$\mathrm{U}(3)_\mathrm L \times \mathrm{U}(3)_\mathrm R = \mathrm{SU}(3)_\mathrm L \times \mathrm{SU}(3)_\mathrm R \times \mathrm{U}(1)_\text{vector} \times \mathrm{U}(1)_\text{axial}$$ where the last one is broken by the quantum anomaly.

See slide 14 in this lecture summary of Theoretical Hadron Physics which I attended last semester. From this I would conclude that both variants are sensible and have an interesting relationship with each other.


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