The Potts model and the symmetric group

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 18 April 2014

Notes and References

This is an excerpt of the paper The Potts model and the symmetric group by V.F.R. Jones. It appeared in: Subfactors: Proceedings of the Taniguchi Symposium on Operator Algebras (Kyuzeso, 1993), River Edge, NJ, World Sci. Publishing, 1994, pp. 259–267.

§2. Turning $L_{\sim }$ into a "planar" form

Consider a rectangle with $n$ marked points on the bottom and the same $n$ on the top as in the figure (where $n=5\text{)}$ $$\begin{array}{c} \end{array}$$ Surround each of the marked points on the top and bottom with two close neighbours (marked with a $\times$ in the figure). Now join the points marked with $\text{"}\times \text{"}$ to each other, within the rectangle, by any system of non-intersecting curves. The regions inside the rectangle can then be shaded black and white (with the regions touching the left and right sides of the rectangle shaded white. Any such diagram defines a partition of the original $2k$ marked points. We will call such a partition (or it's equivalence relation) "planar". Note also that a planar partition completely determines up to isotopy (rel the boundary) the system of curves joining the points marked $\text{"}\times \text{".}$ It is well known and easy to prove by induction that the number of planar partitions is the Catalan number $\frac{1}{2n+1}\left(\genfrac{}{}{0ex}{}{4n}{2n}\right)\text{.}$ The connection schemes of the points marked $\text{"}\times \text{"}$ were exploited by Kauffman in his "states model" for a knot polynomial (see [Kau1987]).

The result of this paper will follow from the following lemma which is almost obvious.

Lemma 2. If $\sim$ is any equivalence relation on the $2k$ marked points in a rectangle as above, one may choose $2$ permutations ${\alpha }_1$ and ${\alpha }_2$ of the top $k$ and bottom $k$ points respectively, so that $\sim$ becomes planar upon reordering the top and bottom points according to ${\alpha }_1$ and ${\alpha }_2\text{.}$

Proof. Begin with the bottom points and look at those equivalence classes for $\sim$ which contain only bottom points. Choose ${\alpha }_2$ so as to set all these points to the left. Do the same with the top and ${\alpha }_1\text{.}$ Now choose an equivalence class for $\sim$ involving both top and bottom points and extend ${\alpha }_1$ and ${\alpha }_2$ so that these points all occur, with no gaps, to the light of the points already decided upon. Continue until exhaustion. The resulting partition will have the form shown below and is obviously planar: $$\begin{array}{c} \end{array}$$

Q.E.D.

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