- #71

CarlB

Science Advisor

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Recently, I've been looking at generalizations of the idempotency relation [tex]\rho^2 = \rho[/tex], for more general density matrices. Among the spinors, these define the available quantum numbers. An easy example is the Dirac algebra. A complete set of primitive idempotents are the diagonal matrices with only a single 1 on the diagonal and the rest zero. These are the density matrices made from the four orthogonal spinors which each has a single one and the rest zero entries.

The above definition relies on the choice of a representation. To make it rely less, one could write down the detailed equations for the matrix square: [tex]\rho = \rho^2.[/tex] This would be sixteen quadratic equations in sixteen unknowns, starting with:

[tex]\begin{array}{rcl}

\rho_{11} &=& \rho_{11}\rho_{11} + \rho_{12}\rho_{21} + \rho_{13}\rho_{31} + \rho_{14}\rho_{41},\\

\rho_{12} &=& \rho_{11}\rho_{12} + \rho_{12}\rho_{22} + \rho_{13}\rho_{32} + \rho_{14}\rho_{42},\\

&&...\\

\end{array}[/tex]

and going through the other 14 values. The above equations are fairly simple. There are sixteen variables that appear on the left. These variables appear in pairs on the right. A given pair either contributes in exactly one row or it does not contribute to any. With sixteen elements, there are 256 possible pairs but only sixteen of these contribute to a row.

We can associate an ugly finite group structure with these 16 idempotency equations. The structure is on a group with 17 elements. The elements are the above [tex]\rho_{mn}[/tex] plus zero, but since we're treating them as a group instead of complex numbers, we can call the group elements [tex]h_{mn}[/tex] and 0. The group multiplication rule is that [tex]h_{ij}h_{kl} = \delta_j^k h_{il}[/tex], and zero times anything is zero.

The whole thing can be reversed. Given a group, we can define an idempotency relation by examining each possible product, and putting a quadratic term into the row that is given by the group product.

Since I'm playing with a model where the elementary fermions are composed of three preons, the natural group to apply this to is the permutation group on three elements.

Before doing that, an easier problem is the even subgroup of the permutation group on three elements. This is an easy problem. Call the three elements [tex]I, J, K[/tex]. The group definitions and the resulting quadratic equations are:

[tex]\begin{array}{rcccl}

I&=& (123) &=& I^2 + 2JK,\\

J&=& (231) &=& K^2 + 2IJ,\\

K&=& (312) &=& J^2 + 2IK.\end{array}[/tex]

The above quadratic equations are what one gets when one enforces the idempotency relation on a circulant matrix:

[tex]\left(\begin{array}{ccc}

I&J&K\\K&I&J\\J&K&I\end{array}\right)^2 =

\left(\begin{array}{ccc}

I&J&K\\K&I&J\\J&K&I\end{array}\right)[/tex]

And as we've discussed before, the circulant matrices are a natural road to Koide's formula. This makes the permutation group of three elements worth suffering through. In the permutation group on three objects, there are six elements. You get six quadratic equations in six unknowns. The solution is arduous (at least for me), but the results are interesting.

The complex number that replaces the identity group element [tex]I[/tex] takes on the absolute values of weak hypercharge among the elementary fermions (and nothing else). That is, it takes on the values 0, 1/6, 1/3, 1/2, 2/3, 1. This seems natural because weak hypercharge is associated with a U(1) group.

The fact that it doesn't give the negative values doesn't bother me too much because the antiparticles take negative quantum numbers anyway. I suspect that if one changes the group slightly, one can get the negatives and positives. In particular, the permutation group on three elements does not include the Poincare symmetry. So the permutation group is a simplified problem.

There are three odd permutation elements. In the group, these are the elements that square to unity, which reminds one of the generators of SU(2). In solving the 6 quadratic equations, you will find that your task is eased considerably if you assume that the three permutation elements take the same complex values. In any case, the sum of those three complex values (i.e. triple one of them if they are all the same), turns out to be the weak isospin numbers that one associates with the weak hypercharge numbers of the fermions (i.e. consistent with the assignment of weak hypercharge to [tex]I[/tex]).

Having done this, there are still two group elements to deal with, [tex]J[/tex] and [tex]K[/tex]. These are the permutations that one thinks of as defining the twist of the corner of a cube as either a positive twist or a negative twist. Another way of saying this is that if we have a group consisting of the identity I, a left L, and a right R, we want to have IR = LL = R, IL = RR = L, LR = RL = I.

The values for these complex numbers can be assigned consistently to the elementary fermions so that the difference between the left and right handed particles of a pair always has the same value. That is, the difference between the neutrino left and right is the same as the difference between the down left and right, etc. This suggests that these two values can be interpreted as the quantum numbers of the Higgs.

I typed the algebra calculations into my paper

http://www.brannenworks.com/dmfound.pdf

which unfortunately includes a lot of denstiy matrix theory that no one wants to read before the algebra calculations above, which appears right now in section (5.1) page 33. To get anyone interested in reading it, I'll have to strip all that off.

Carl