# Goldstone bosons in 2 and 3 quark flavor symmetries [closed]

In my (undergraduate) advanced elementary particles class last semester, we learnt that for a 2 quark (u/d) model the symmetry of the Lagrangian is (and breaks as)

$$U(2)_L \otimes U(2)_R = SU(2)_L \otimes SU(2)_R \otimes U(1)_L \otimes U(1)_R \rightarrow SU(2)_V \otimes U(1)_B$$

where $$SU(2)_V$$ is the isospin symmetry and $$U(1)_B$$ is the baryon number conserving symemtry. On a first glance, according to the Goldstone theorem, we should be getting 4 pseudo-Goldstone bosons, but that doesn't seem to be the case, as the only pseudo-Goldstone bosons that I can find seem to be the three pions that are related to the $$SU(2)$$ chiral symmetry breaking. Upon doing some research, I found that this is related to an anomaly on the so called $$U(1)_A$$ group.

Similarly, for a 3 quark (u/d/s) model we have a symmetry that breaks as: $$U(3)_L \otimes U(3)_R = SU(3)_L \otimes SU(3)_R \otimes U(1)_L \otimes U(1)_R \rightarrow SU(3)_V \otimes U(1)_B$$ According to the Goldstone theorem, we expect 9 pseudo-Goldstone bosons, and we can easily see that we have the three pions, the four Kaons, the eta meson and the eta prime meson. Except that the eta prime meson is much heavier than expected, and that's apparently due to (again) an anomaly in the breaking of the $$U(1)_A$$ group, which (apparently) also means that the eta prime is not a pseudo-Goldstone boson.

1. What is the $$U(1)_A$$ group and how is it related to $$U(1)_L \otimes U(1)_R$$?
2. What is this so called anomaly that's affecting $$U(1)_A$$?