M Theory Lesson 57
Recall that the Bilson-Thompson diagrams for left and right handed electrons are formed with three strands, each with a full negative twist representing a one third charge. 
Ignoring the twists, the simplest braid composition of the form $e_L e_R e_L$ looks like this: 
This knot is called a $6_3$ knot, according to the amazing online knot atlas, which lists the polynomial invariants for this knot. For example, the Jones polynomial takes the form
$-q^3 + 2 q^2 -2q + 3 -2 q^{-1} + 2 q^{-2} - q^{-3}$
Since all negative twists add up on going around the knot, there is a total ninefold twist. There is now a button to the knot atlas on the sidebar.
Ignoring the twists, the simplest braid composition of the form $e_L e_R e_L$ looks like this: 
This knot is called a $6_3$ knot, according to the amazing online knot atlas, which lists the polynomial invariants for this knot. For example, the Jones polynomial takes the form$-q^3 + 2 q^2 -2q + 3 -2 q^{-1} + 2 q^{-2} - q^{-3}$
Since all negative twists add up on going around the knot, there is a total ninefold twist. There is now a button to the knot atlas on the sidebar.
posted by Kea | 1:38 PM
3 Comments:
Hi Kea,
RE: Bilson-Thompson diagrams for left and right handed electrons.
To satisfy QCD and make a more powerful presentation perhaps the stings need to be of different colors?
your reference and:
http://webphysics.davidson.edu/mjb/qcd.html
Kea
How could one represent e/5 charge using Bilson-Thompson-like diagrams?
How could one represent e/5 charge?
Wouldn't we need 5 strands (elements of B5) for that? Then a full twist on any strand would count for e/5. Hmm. The other possibility is Iteration: thinking of this 3 piece composition as giving charge 1 strands, then 5 finer strands internal to a single strand could count twists as e/5.
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