If we start with two charged quark hexagons, the third hexagon in the lattice could become a $(b,c,s)$ triplet.
Recall that the phase for the $(b,c,s)$ triplet is $18/27$, three times the charged lepton phase. The phase for a charged triplet is $2/27$ or $4/27$. Three hexagon tiles form the core of a honeycomb dual to a four qutrit extended tetractys. Note that the outer polygons here are rather asymmetrical. An alternative uses three charged triplets, including one non standard configuration:
This choice maintains (charge) triality on the trivalent subdiagrams.
6 years ago