Representing SU(3) with 3 ropes in 3 dimensions

Question

The short question is: how exactly is SU(3) realized with ropes?

The long question: There is this idea that deformations of a configuration of three infinitely long, flexible ropes that cross each other can be mapped to the eight Gell-Matrices, the generators of SU(3).

SU(3) appears in the quantum harmonic oscillator and in qutrits. So a visualization of SU(3) with ropes is useful to quantum information theory.

The SU(3) idea is mentioned in http://arxiv.org/abs/0905.3905 on page 35. It seems that the idea started in this way: deformations of configurations of TWO ropes reproduce Dirac's string trick and behave like the Pauli matrices of SU(2). Deformations of configurations of ONE rope reproduce U(1).

THREE ropes apparently yield a relation between eight different versions of the third Reidemeister move and the eight Gell-Mann matrices. But the paper is too terse for me to see the relation in detail. A literature search does not bring up anything related to this idea. And I got no answer to my email.

Can anybody help to understand the details?

Added points:

Peter's answer below mentions a relation between the braid group $B_3$ and SU(3). A Google search does not yield anything about this topic. Can anybody provide a reference?

A graph similar to that of Joseph's answer below is also part of the paper. But I am not interested in QCD or unification: I'd like to understand how the deformations in that graph yield or correspond to the Gell-Mann matrices $\lambda_1$ to $\lambda_8$. I can see that the deformations of the graph correspond to $F_1$, $F_2$ and $F_3$, where $F_i=e^{i \pi \lambda_i / 2}$. This gives $F_1^4=F_2^4=F_3^4=1$, the unit matrix, as it should. Also the SU(2) subgroup is generated as it should. Next, the $\lambda_8$ deformation behaves as expected. But I cannot see (so far) that $\lambda_1 \lambda_4= \lambda_6/2 + i \lambda_7/2$. Can anybody provide a hint?

Answer 1

this strand model ("strand" = your "rope") of gauge interactions is the brain child of Christoph Schiller, who has written a 400+ page textbook on his theory.

Chapter 9 systematically goes through the various groups, U(1), SU(2), SU(3) --- I guess this is the source you want to study if you wish to pursue this idea.

For a discussion of the physics behind the whole approach, see for example here (and if you search for "schiller strand model" you'll find many more or less serious discussions).

Answer 2

This doesn't exactly answer the question, but it seems like it should be closely related. In general, if $G$ is a complex semi-simple Lie group, there is an associated braid group $B_{\mathfrak g}$ (edit: (the pure braid group) is the fundamental group of the complement in $\mathbb{C}[\mathfrak h]$ of the reflection hyperplanes of the Weyl group $W$ of $G$). (edit: The braid group is the fundamental group of the quotient of this space by $W$.) I have been told that there is a group homomorphism $B_{\mathfrak g} \to G$, although I unfortunately don't know a reference.

For $G = SL_n(\mathbb C)$, things can be made more explicit. The braid group $B_{sl_n}$ is the standard braid group (see https://en.wikipedia.org/wiki/Braid_group), and I believe the map $B_n \to SL_n(\mathbb C)$ is given by assigning the generator $\sigma_i \in B_n$ the identity matrix, with the $2\times 2$ submatrix in row $i,i+1$ and columns $i,i+1$ is replaced by

$\left[\begin{array}{cc} 0&-1\\1&0\end{array}\right]$.

(Sorry if the formatting is off - for some reason my computer isn't displaying things correctly at the moment.)

Answer 3

This is from Chapter 9 of the book which Carlo cited, Motion Mountain by Christoph Schiller, p.242. (Links to chapters here.) I wanted to see if there was some image of these three "strands." Not sure this is the most representative picture...