garrett said:
Nature couples left-chiral fermions to the electro-weak gauge fields differently than it couples right-chiral fermions. It is hard to come up with a good geometric reason for why.
Yes, I did have trouble with this. The solution takes a bit of explaining.
First, you need to assume that the elementary fermions are the left-chiral and right-chiral ones. There is the problem that these are massless and you will later have to figure out a method of putting the masses in, but work is ongoing on that.
That done, the problem then becomes one of symmetry breaking. That is, the primitive idempotents of a Clifford algebra tend to be too danged symmetric.
I think the best way to break the symmetry is to modify the way that the tangent vectors (i.e. \partial_x, \partial_t, etc.) are connected up with the canonical basis vectors (i.e. \gamma_x, \gamma_y, etc.)
Use the standard Dirac matrices as an example. What we are going to do is to loosen the definition of the Dirac operator:
\nabla \psi = (\partial_x\gamma_x + \partial_y\gamma_y + ...) \psi
Instead of using the canonical basis VECTORS, \gamma_\mu, we will instead use arbitrary canonical basis ELEMENTS \Gamma_x, \Gamma_y, \Gamma_z, \Gamma_t provided only that the usual Dirac relations are satisfied. That is,
\Gamma_x^2 = \Gamma_y^2 = \Gamma_z^2 = -\Gamma_t^2 = 1
and the \Gamma_\mu anticommute.
Now in the usual spinor model of quantum mechanics, this modification leaves the Dirac equation unchanged. That is, \Gamma_\mu makes just as good a set of generators of a Clifford algebra as \gamma_\mu does. So the change doesn't alter any of the usual physics.
But in Trayling's model of the elementary particles, the change from \gamma_\mu to \Gamma_\mu amounts to a remapping of the elementary particles.
The remapping by the above naturally preserves all addition and multiplication in the Clifford algebra so all the remapped particles preserve their quantum numbers with respect to the remapped operators. But the remapping does not preserve the squared magnitude of the Clifford algebra. Since the squared magnitudes are what we associate with probabilities, this means that while the particle propagation is unchanged (same Dirac equation), the particle interactions are altered. This is exactly the kind of symmetry breaking you need.
I've been assuming just one small cyclic hidden dimension and a complexified Clifford algebra. This gives 8 primitive idempotents. Before breaking symmetry, these 8 idempotents all have squared magnitude 1/8. The symmetry breaking can only raise the squared magnitude of a primitive idempotent. But on the other hand, the primitive idempotents still sum to one so the squared magnitude of the sum of all the idempotents is still going to be 1.
Another way of saying this is to say that the primitive idempotents become unstable. Typically, four of them become slightly unstable (tending to form up in pairs), while the other four become extremely unstable (also tending to form up in pairs). A typical set of squared magnitudes is:
8|\iota_{---}|^2 = 1.31
8|\iota_{--+}|^2 = 1.31
8|\iota_{-+-}|^2 = 1.31
8|\iota_{-++}|^2 = 1.31
8|\iota_{---} + \iota_{--+}|^2 = 2.000
8|\iota_{-+-} + \iota_{-++}|^2 = 2.000
8|\iota_{+--} + \iota_{+-+}|^2 = 2.000
8|\iota_{++-} + \iota_{+++}|^2 = 2.000
8|\iota_{+--}|^2 = 231.7
8|\iota_{+-+}|^2 = 231.7
8|\iota_{++-}|^2 = 231.7
8|\iota_{+++}|^2 = 231.7
Without the symmetry breaking, all 8 of the primitive idempotents in the above would have squared magnitude 1/8, and the four pairs would have squared magnitude 1/4.
For example, you've now reduced your set of 8 primitive idempotents as potential elementary particles to 4 idempotents (the ones that have squared magnitude of 2.000). These idempotents can interact by making small changes, but the probabilities are very different between the two pairs. Hence, the symmetry is broken in probability / amplitude.
To match up with the weak hypercharge, weak isospin structure of the standard model, so far I've had to assume that the elementary particles correspond to certain combinations of four primitive idempotents each.
I presented a poster at the PANIC05 conference in Santa Fe last month on this subject that goes into more detail on this:
http://brannenworks.com/PPANIC05.pdf
The detailed calculations for the symmetry breaking are on pages 14-16 of this unfinished paper:
http://brannenworks.com/long_PANIC_Not_Complete_.pdf
An older paper with a less complete derivation of the particle symmetry breaking is here:
http://brannenworks.com/PHENO2005.pdf
Carl
