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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?

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3 Answers 3

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).

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Thank you, I did not know all this. –  Gina Martelli Sep 4 '13 at 5:43
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All the discussions seem about unification and quantum theory. The SU(3) issue is not explored. Thank you nevertheless! –  Gina Martelli Sep 4 '13 at 9:55

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.)

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Peter, that is very interesting! Could you find out about a reference or details on the case SU(3)? For SU(3), this would means that the braid group generated with ropes/strands is sufficient, as SU(3) is a subgroup of SL(3,C). (Is this correct?) Three Gell-Mann matrices are related to the substitution you mention: $e^{i \pi \lambda_2 /2}$, $e^{i \pi \lambda_5 /2}$ and $e^{i \pi \lambda_7 /2}$. The others are less clearly related - especially $\lambda_8$. What could be the way to proceed from SL(3,C) to SU(3)? –  Gina Martelli Sep 4 '13 at 5:26
    
The page http://en.wikipedia.org/wiki/List_of_simple_Lie_groups says that SU(3) is the compact dual of SL(3,C). Could this help here? –  Gina Martelli Sep 4 '13 at 5:40
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The fact that SU(3) is the compact dual of SL(3,C) is fairly involved, and unlikely to be of help here. What you want instead is the simpler fact that SU(3) is a maximal compact subgroup of SL(3,C). –  Tom Church Sep 4 '13 at 11:11
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Just a nitpick. I think the braid group you mention is the "pure" one. To get the full braid group you must quotient the complement of reflection hyperplanes by $W$ and take the fundamental group. –  BS. Sep 4 '13 at 11:57
    
The answer by Peter and the comment by Tom imply that there is a connection between the braid group $B_3$ and the Lie group SU(3). Where can I read more about this connection? –  Gina Martelli Sep 4 '13 at 16:10

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...
   p.242

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Thank you! A similar graph is also in the paper. –  Gina Martelli Sep 4 '13 at 5:44

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