Let us say this as simply as possible. Here we use the ordinary quantum Fourier transform, not the twisted version (which sends one neutrino to the photon).
In passing the looking glass, the neutral particles are special. Recall that the set of charged leptons and quarks can be sent through a mirror set in two ways. One can (i) flip charges on all ribbons, by flipping the $B_2$ crossings, or (ii) flip the two crossings in the $B_3$ braid. In case (i) the neutrino sector is unaltered, but the $Z$ color triplets cross the mirror. In case (ii) the $Z$ triplets are unaltered, because they are bosons, but the neutrino sector is swapped.
In both cases, the photon is fixed at the mirror. The fun Fourier transform sends neutrino operators to either photons or $Z$ bosons, and vice versa. Thus (i) and (ii) are just two ways of looking through the glass. So we expect the relation between $Z$ bosons and photons to resemble the relation between neutrinos, mirror neutrinos and photons.
Altogether there are $12$ neutrino states, including generations, and $12$ $Z$ bosons, including color. This defines a $24$ dimensional space with Fourier supersymmetry. The $24$th roots of unity ($\pm \pi /12$) denote this arithmetic dimension. They can only appear as Koide phases for the neutrinos, because $Z$ bosons are diagonal. The left right pairing of braids to create rest mass is the color kinematic duality of Bern et al.
6 years ago