# Deformational gauge fields

We have already considered one class of irregularities of the cellular lattice model — the Wilson gauge fields. They allows to describe the influence, which the region between the cells has on the movement of the cells. But in this picture, the cells remain at their regular places (more accurate, oscillate around these places), the lattice remains regular, is not modified at all.

Here we want to describe another type of irregularities of the lattice — irregularities where the lattice itself has to be modified. Imagine we cut the lattice into two parts, left and right, and then we move the lattice down, one lattice spacing. What we obtain is, again, a regular lattice. The lattice equation should be, at least in principle, the same as before.

But now let's think about the oscillatory modes which we have found during the consideration of the doubling effect. After this deformation some of the original oscillating modes — those who oscillate in the vertical direction — look very strange near the critical plane. They, obviously, will not be the correct oscillating solutions.

No wonder — the correct equation will be very different from the original one near the plane. The nearest neighbour of a cell on the left will be a different node than before. To describe this modified equation in terms of the original enumeration, we need some correction term.

What are the properties required for this correction term? It has to act on lattice functions and has to shift them. Moreover, it has to preserve the lattice equation.

Fortunately, these properties are already sufficient to define the related operators up to their signs. We obtain eight different shift operators.

### The elementary lattice shift operators

How do these operators act? Their most important property is that their action is identical for all the different components $$a^i_\mu, \pi^i_\mu$$. They, therefore, preserve the doublets and act on all doublets in exactly the same way. This is the characteristic property of weak gauge fields.

Explicit computation shows that, indeed, all operators we need to define weak gauge fields, especially the weak isospin operators $$I_k$$ and the operator $$\gamma^5$$, appear in this way, as associated with shift operators on the lattice.

In more detail, the eight operators associated with the nontrivial elementary shifts are:

• 1 for the trivial shift,
• $$2I_k \gamma^5$$ for the three basic lattice shifts,
• $$2I_k$$ for the three planar diagonal lattice shifts,
• $$\gamma^5$$ for the volume diagonal lattice shift.

### The general form of the correction term

Last but not least, we should think about the intermediate states of the deformation. There should be a continuous deformation between the original lattice equation and the final lattice equation, where we have to apply the elementary shift operator.

The solution is quite simple — we have to introduce coefficients which we have to multiply with the basic generators. These coefficients have to depend on the local geometry of the lattice, thus, they define fields on the lattice. And these lattice fields may be used to define the weak gauge fields.

## Restrictions following from Euclidean symmetry

The first, most important, restriction for the weak gauge field is that they do not depend on all the other charges — generation, color, and baryon number. They preserve the electroweak doublets and act on all the doublets in exactly the same way. Then, we have also found the generators of the gauge group: the operators $$I_k$$ and $$\gamma^5$$ .

Let's consider now the additional restrictions which follow from Euclidean symmetry. The consideration of rotations gives nothing new — we already know that generations should be preserved and that the gauge fields have to act on all generations in exactly the same way.

A nontrivial additional restriction follows, instead, from commutation with translations. As a consequence, there has to be a preserved direction — the direction of translation. As a consequence, we obtain the following additional restriction:

### There should be an invariant direction in each doublet

This is in agreement with observation — the weak gauge fields preserve the right-handed neutrino.

Moreover, we can compute the maximal gauge group which may be generated by the generators we have found and which leaves some generation invariant. The maximal group is $$U(2) \times U(1)$$. Moreover, the action of U(2) has to be chiral. Without restriction of generality, this group can be taken as $$U(2)_L \times U(1)_R$$. The charges are the weak charges $$I^L_k$$ of the weak gauge group $$SU(2)_L$$ of the standard model $I^L_k = \frac{1}{2}(1-\gamma^5) I_k$

additionally the charge $$I^L_0$$ of it's diagonal $$U(1)_L$$, which is simply $I^L_0 = \frac{1}{4}(1-\gamma^5),$

together with the charge $$I^R$$ of the right-handed gauge field $$U(1)_R$$ which leaves the right-handed neutrino invariant, which is $I^R = \frac{1}{2}(1+\gamma^5)(I_3-\frac{1}{2}).$

## Comparison with the standard model

The weak gauge group $$SU(2)_L$$ is a subgroup of the gauge group we have constructed here. Thus, for this part we have an ideal fit with observation.

The EM charge does not seem to fit into this subdivision into strong (Wilsonian) and weak (deformational) gauge fields. But there exists a decomposition of the EM charge into a strong and a weak part: $Q = 2I_B + (I_3 - \frac{1}{2}) = 2I_B + (I^R + I^L_3 - I^L_0).$

The operator $$I_B$$ — the baryon charge — is the generator of the diagonal of the maximal possible strong gauge group U(3), thus, allowed by all criteria for Wilson gauge fields. The other part fits into the conditions for weak charges.

As a consequence, all SM gauge fields are part of the maximal possible gauge group we have found, which is $$U(3)_c \times U(2)_L\times U(1)_R$$.

The question, why we observe the special linear combination defining the EM field, but not the field defined by the baryon charge, has a simple answer: For the EM field and all other SM fields the sum of the charges of all fermions is zero. This can be taken as an additional condition: With this condition, the maximal possible gauge group becomes $$S(U(3)_c \times U(2)_L\times U(1)_R)$$ and contains only one additional gauge field $$U(1)_U$$.

This field I have named upper axial gauge field. It is anomalous, thus, if we, additionally, forbid also anomalous gauge fields, we have obtained the whole SM gauge group exactly as the maximal possible group compatible with our construction.