E(3) is an object everybody knows from everyday life, even if he does not know what is a "group" in mathematics and has never seen the denotation E(3). It is simply the group generated by translations and rotations of our three-dimensional Euclidean space.
An Euclidean transformation is, therefore, defined by a rotation - a 3×3 matrix (ωij) - and a translation - a three-dimensional vector (ti), with 1 ≤ i,j ≤ 3. It maps a point x with coordinates (xi) into a point y with coordinates (yi) by the following formula:
(1) xi → yi = ωijxj + ti
or, in more detail:
|x1 →||y1 =||ω11x1+||ω12x2+||ω13x3+||t1|
|x2 →||y2 =||ω21x1+||ω22x2+||ω23x3+||t2|
|x3 →||y3 =||ω31x1+||ω32x2+||ω33x3+||t3|
Another way to define it is that is the subgroup of the affine group Aff(3) which preserves all Euclidean distances.
Because an Euclidean transformation is, by this definition, also an affine transformation, we can compose Euclidean transformations with affine transformations. The composition rule can be easily computed given the action (1).
Nonetheless, because of the correspondence between SM doublets and affine transformations, the results are important for our approach. Therefore we give here the explicit results.
We apply an affine transformation (aiμ) to x and, then, apply a rotation (ωij) to the result y. Then we compare with the original formula to see which affine transformation (a'iμ) gives the same result in a single step. This gives:
(2) aiμ → a'iμ = ωijajμ
Thus, the "color index" μ remains unchanged. The rotation "rotates" only the upper index, the "generation index".
The consideration of the doubling effect does not give additional effects - all doublers are transformed in exactly the same way.
Now, all gauge fields of the SM preserve generations, and they act on all generations in exactly the same way. Once the rotations act only on the generation index, and the gauge actions do not depend on the generation index at all, all SM gauge fields commute with Euclidean rotations.
This consideration works in the other direction too: Assume, as a postulate, that Euclidean symmetry is preserved by all gauge actions. What follows?
We apply an affine transformation (aiμ) to x and, then, apply a translation (ti) to the result y. Then we compare with the original formula to see which affine transformation (a'iμ) gives the same result in a single step. The result is much simpler, only the coefficients ai0 are transformed at all:
(3) ai0 → a'i0 = ai0 + ti
Thus, the "quark sector" μ>0 remains completely unchanged. The translation acts only on the leptonic part μ=0.
The consideration of the doubling effect gives another interesting information: The shift vector (ti) is a constant function on the lattice. As a consequence, only one of the doublers — the non-oscillating doubler — is shiftet. The oscillating doublers remain unchanged as well.
We have to identify now the non-oscillating doubler with a linear combination of the right-handed neutrino and its antiparticle. After this identification, no gauge field acts on the non-oscillating doubler.
As a consequence, all SM gauge fields commute with Euclidean translations.
In the other direction, it follows from translational symmetry, that there should exist some common direction which has zero charge for all gauge fields. Indeed, the commutation between the gauge action φ→gφ and translation φ→φ+t gives g(φ+t) = gφ + t, thus, gt=g. That means, the direction of translation should be preserved by all gauge fields.