M Theory Lesson 213
As well as a $2 \times 2$ Kasteleyn matrix with $3 \times 3$ circulant elements, we can also look at a $3 \times 3$ operator with two dimensional elements, of the form 
This is associated to a three point Fourier transform with noncommutative coefficients described by matrices in the braid group $B_{2}$. On two strands it is only possible to create two kinds of knot: a multicrossing Hopf link, corresponding to even powers of the pseudo Pauli swap operator, or a single loop with $2n + 1$ crossings, such as the trefoil knot.
Allowing only Hopf braids, so that $B = D = Y = 0$, such a $6 \times 6$ element of $B_{6}$ looks like a diagram of three ribbon strands, because $A$, $C$ and $X$ just count powers of the $q$ factor defining the group generator. In this case, the matrix acts on a 3-vector $V$ via the ordinary Fourier transform
$V \mapsto (A \cdot 1 + C \cdot (231) + X \cdot (312)) V$
When $A = C = X = i = e^{\frac{2 \pi i}{4}}$ then $\sigma^{4} = 1$, so a full positive twist equals a full negative twist and the information of particle charge is lost. However, at a 12th root of unity a full positive twist is distinct from a full negative twist, which may be represented by the 10 crossing $\sigma^{10}$. By convention, let $\sigma^{6} = \sigma^{-6}$ determine the maximum number of crossings to be 6, that is three full twists.
This is associated to a three point Fourier transform with noncommutative coefficients described by matrices in the braid group $B_{2}$. On two strands it is only possible to create two kinds of knot: a multicrossing Hopf link, corresponding to even powers of the pseudo Pauli swap operator, or a single loop with $2n + 1$ crossings, such as the trefoil knot. Allowing only Hopf braids, so that $B = D = Y = 0$, such a $6 \times 6$ element of $B_{6}$ looks like a diagram of three ribbon strands, because $A$, $C$ and $X$ just count powers of the $q$ factor defining the group generator. In this case, the matrix acts on a 3-vector $V$ via the ordinary Fourier transform
$V \mapsto (A \cdot 1 + C \cdot (231) + X \cdot (312)) V$
When $A = C = X = i = e^{\frac{2 \pi i}{4}}$ then $\sigma^{4} = 1$, so a full positive twist equals a full negative twist and the information of particle charge is lost. However, at a 12th root of unity a full positive twist is distinct from a full negative twist, which may be represented by the 10 crossing $\sigma^{10}$. By convention, let $\sigma^{6} = \sigma^{-6}$ determine the maximum number of crossings to be 6, that is three full twists.
posted by Kea | 11:28 AM
3 Comments:
Good. Now you're working on the same stuff I am, except that I let those 2x2 matrices be a little more general, and I prefer to write them as products of primitive idempotents taken from a MUB or two.
Yes, I think the relation to Bilson-Thompson's ribbons is becoming clearer now. The generations end up permitting more crossings than in B-T's scheme, but there is still a fixed finite number permitted, and it's very natural from a knot theoretic point of view, although not from a spin foam point of view, where of course everything is far too arbitrary.
You got the 2x2 matrices by looking at the permutations on two elements, I suppose. Their multiplication rule is
(A, B) x (C, D) =
(AC+BD, BC+DA),
which is evidently commutative.
I think that this is the elliptic version of the complex numbers. That is, it's the (trivial) real Clifford algebra with one basis vector (call it s), with positive signature. The presentation for the algebra is simply:
ss = 1.
If you replace that presentation with
ss = -1,
then you get a trivial Clifford algebra that is equivalent to the complex numbers, which are the hyperbolic version of the algebra you're using for the twists on the braid group.
Post a Comment
<< Home