Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 305 / 4 / c / Solution

For $m_u=m_d=0$ and electromagnetism neglected, the classical flavour symmetry isup to finite quotients. The chiral anomaly breaks $U(1{)}_A$ quantum mechanically, while $U(1{)}_V=U(1{)}_B$ remains. The quark condensatespontaneously breaksSmall quark masses and electromagnetism explicitly break parts of this approximate symmetry.