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I've been told that, given a knot projection, there is a way of associating a statistical system in such a way that the partition function of the system corresponds to the Jones polynomial of the original knot.

I have a rough understanding of how the associated graph is obtained (colour the regions checkerboard and put a vertex in each region, then edges are the crossings connecting regions and they get assigned a plus or minus depending on the crossing type), but I don't know how to get the associated partition function.

I'm looking for a detailed explanation of how to obtain the Jones polynomial of a knot via the partition function of the associated statistical system - an easy example would be particularly helpful.

EDIT: As suggested by Daniel Moskovich, I looked at Jones' explanation here, but when I actually do a computation I get something different from the usual Jones polynomial.

Consider the following picture of the unknot:

enter image description here

This corresponds to the graph:

enter image description here

where the edge is assigned a minus sign due to the type of crossing. Then fixing $Q \in \mathbb{N}$, the partition function of the Q-state Potts model is given by $$Z = \sum_{states} \, \prod_{edges} \omega_{\pm}(\sigma, \sigma ')$$ Jones claims that the choice $$\omega_{\pm}(\sigma, \sigma ') = \begin{cases} t^{\pm 1} &\mbox{if } \sigma = \sigma ' \\ -1 & \mbox{if } \sigma \not = \sigma ' \end{cases}$$ gives the Jones polynomial ("up to a simple normalisation") when $Q = 2 + t + t^{-1}$. For the given knot however, the partition function you get is $$Z = Qt^{-1} - Q(Q-1) = -t^{-1} - 3 - 3t - t^2 $$ On the other hand, the Jones polynomial of the Unknot should be 1, so clearly I need some normalisation factor. Unfortunately, I can't figure out the general form of the normalisation factor required, and haven't seen it written anywhere.

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  • $\begingroup$ If you think it's appropriate, you might like to add a reference-request tag. $\endgroup$ – Travis Dec 17 '14 at 0:32
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    $\begingroup$ As another example, the unknot with one crossing (with a positive sign this time) has $Z = Qt - Q(Q-1) = -t - 3 - 3t^{-1} - t^{-2}$, so it seems the normalisation factor should depend on the writhe of the knot. $\endgroup$ – Mark B Mar 19 '15 at 21:10
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Jones himself provided a nice and accessible explaination HERE.

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It is possible that the following paper would be useful

Kauffman

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