# Why is the standard model gauge group $SU(3) \times SU(2) \times U(1)$ and not $U(3) \times U(2) \times U(1)$?

The bilinears in the Lagrangian are invariant under the full group. Where does the restriction to unit determinant come from?

• Mar 17 '20 at 11:49
• Mar 17 '20 at 14:36
• Also related. Mar 17 '20 at 15:28

Indeed, the quark bilinears have an extra exact symmetry $$U(1)_B$$ (baryon number) and the lepton ones an extra symmetry $$U(1)_L$$ (lepton number). These two commute with each other and the 12 gauged ones of the SM. You are presumably asking why these are not gauged.
In the SM, nonperturbative ("topological") EW anomalies break the $$U(1)_{B+L}$$ combination, so gauging this one is not consistent. See links in comments.
The Higgs doublet of the SM couples to neither of them (B or L), and so cannot break them spontaneously. Other hypothetical ones at much, much higher energy scales could do that, and so $$U(B-L)$$ could be gauged and its gauge boson be unobservable, being Higgsed to ultra heavy, and break that symmetry to also unobservable levels. A logical possibility in the remote BSM territory.
But, if there were no such breaking, gauging $$U(B-L)$$ would require another massless gauge boson for it, distinctly not observed in our low-energy world.