# Universality of the Z interactions

In the SM the interaction of leptons and quarks with the Z boson are universal, and the way to see it is to start with the covariant derivatives which lead to terms of the form $$\delta_{ij}\bar{\psi}^i_{R/L}Z_\mu\gamma^\mu\psi^j_{R/L}$$ with $i,j$ the generation indices. By switching to the mass basis, we perform some unitary transformation $U$ , which from the above expression can be easily seen to cancel, since it operates on a different space than the gamma matrices, and then we simply have $U^\dagger U = Id$. My question is what kind of extension/mutation do we need to do to the SM to break this universaility?

I've read in my lecture notes that it's a feature that stems from the fact that all fermions of given chirality and given charge come from the same $SU(2)\times U(1)$ representation. I don't see how this has anything to do with it, because in any case even if we add some new irreps of $SU(2)$ which feature bare mass and Yukawa terms such that the mass basis doesn't diagonalize everything, the fact remains that switching to a mass basis boils down to rotating the flavor space by some $U$, and if we start with bare fields which are diagonalized in flavor space by $\delta_{ij}$, it doesn't matter what $U$ does - as long as it's unitary, it will cancel out.

Universality means that the interaction stays in the same flavor. Meaning, the interaction with the Z boson does not change the flavor. In terms of symmetries, each flavor has the same representation of $SU(2) \times U(1)$, which means that there isn't a copy of this symmetry group for each flavor.
For example, there are exotic models that proposes another boson that changes flavor, called Z-prime, which means it is introduced another symmetry to the fermions, namely another copy of $U(1)'$, or sometimes $SU(2)' \times U(1)'$ together with a W-prime boson.