# How to get rid of the baryon charge field

The maximal possible gauge group for Wilson gauge fields is the group U(3). It contains, together with the group of strong interactions SU(3), which we need in the standard model, also its diagonal, a field $$U(1)_B$$ with the baryon charge $$I_B$$ as its charge.

On one hand, we need this additional field $$U(1)_B$$, to construct the EM field.

On the other hand, we have to get rid of it. There is no gauge field in the SM whose charge is the baryon charge. If we would like to include it, it's interaction constant should be many orders of magnitude smaller than that of the EM field: Indeed, the Earth, consisting of lots of baryons, has, in comparison with its electric charge, an extremely large baryon charge. If the corresponding force would be of similar strength, everything on Earth, as well heavily charged, and with the same charge, would fly away. Thus, even if the field $$U(1)_B$$ exists, it should be very heavily suppressed.

Fortunately, there is a good argument that such a field should indeed become suppressed. It would lead to a charged Dirac sea (ground state). This is something which is not allowed in condensed matter theory. (Here I rely on the research of others: Volovik writes: "An equilibrium homogeneous ground state of condensed matter has zero charge density, if charges interact via long range forces. For example, electroneutrality is the necessary property of bulk metals and superconductors; otherwise the vacuum energy of the system diverges faster than its volume. ... The same argument can be applied to "Planck condensed matter", and seems to work. The density of the electric charge of the Dirac sea is zero due to exact cancellation of electric charges of electrons, and quarks, ... in the fermionic vacuum".)

As a consequence of this restriction, the sum over the charges of all fermions has to be zero for all gauge field charges. Or, equivalently, the determinant of every element of the gauge group should be 1.

Without this restriction, the maximal possible gauge group compatible with our axioms would be $$U(3)_c\times U(2)_L \times U(1)_R$$, where the first part, $$U(3)_c$$ , describes the maximal possible Wilson gauge field, and the second part, $$U(2)_L \times U(1)_R$$, the maximal possible gauge-like field describing lattice deformations.

Now, the maximal possible group will be the subgroup of all elements of this group with determinant 1. Such a subgroup of a group G is usually denoted SG. Thus, the maximal possible gauge group compatible with our axioms is $G_{max} \cong S(U(3)_c\times U(2)_L \times U(1)_R)$

This is already almost exactly the gauge group of the standard model $G_{SM} \cong SU(3)_c\times SU(2)_L \times U(1)_Y.$

The remaining difference I have named upper axial gauge field.