M Theory Lesson 117

A while back we looked at associating the number of strands in number theoretic braids with the size of the matrix operators in the Fourier transform, which is the same as the number of points on the circle (3 for mass). In the two strand case, the braids are easy to classify: m copies of the only generator, \sigma_{1}. In other words, an integer m labels all possible knots.

The homflypt polynomial for the torus knot \sigma_{1}^{2k + 1} is

p^{k} (1 + q^{2} (1 - p) \frac{(1 - q^{2k})}{(1 - q^{2})})

for two parameters q and p. Specialisations include p = q^{2} which results (effectively) in the Jones polynomial

q^{2k} (1 + q^{2} (1 - q^{2k}))

Since the endpoints of braid diagrams lie on a circle, there are two circles bounding a diagram. For the 3-strand case, there are two sets of three points defining the boundary, which thus looks like a 6 point torus.

1 Response so far »

  1. 1

    Matti Pitkanen said,

    Kea,

    I remember that you talked once about SL(3,C) representations of braid group for three strands. Some other blogger hand a nice article about this.

    They would be very interesting since in my model of DNA topological computation nucleotide-triplets could correspond to fundamental 3-braids defining fundamental sub-modules for topological quantum computation and quantum memory.


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