I know that flavor symmetry is only approximate, broken since the u,d, and s quarks have different masses. Also that trying to extend $SU(3)$ symmetry to $SU(4)$, $SU(5)$, or $SU(6)$ symmetry is useless, because the masses of the c, b, and t quarks are very different from the lighter quarks.
I understand that isospin symmetry, which is an $SU(2)$ subgroup of $SU(3)$ flavor symmetry, is broken by the electroweak interaction. That is, if u and d quarks only interacted via the strong interaction, then isospin symmetry would be exact. In some sense, then, the different masses of the u and d quarks are due to the electroweak interaction.
What about $SU(3)$ flavor symmetry? If the u, d, and s quarks interacted only via the strong interaction, would it be exact? Is the electroweak interaction ultimately responsible for the symmetry breaking?
Or are interactions with the Higgs boson somehow involved? This thought occurred to me because the different masses of the three quarks are always cited as the reason for the symmetry breaking.
To check my understanding of Zachos' answer and this stackexchange answer, I will now attempt to express their content in my own words.
The Hamiltonian for quark interaction consists of several terms, among them:
- "Characteristic" strong interaction terms
- Higgs Yukawa terms, aka mass terms
- Electromagnetic and weak interaction terms
The characteristic strong terms commute with the flavor operators, so if they were the only terms, flavor symmetry would be exact. The flavor operators commute with neither the mass terms nor the electoweak terms, so both of these break the symmetry.
Unlike SU(3) flavor symmetry, SU(6) flavor symmetry is useless for certain kinds of (approximate) calculations, but it has other uses.