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.

§3. Proof of the theorem.

The diagrams joining the $2n$ bottom $\times \text{'s}$ to the $2n$ top $\times \text{'s}$ form the basis of an algebra sometimes called the Temperley Lieb algebra. It can be thought of as a subalgebra of the Brauer algebra defined in the introduction as each diagram gives a partition of the $4n$ $\times \text{'s}$ into subsets of size 2. The multiplication is defined by a $2\text{-step}$ procedure.

Step 1. Stack the two rectangles on top of each other, lining up the $\times \text{'s.}$

Step 2. Remove the middle edges and middle $\times \text{'s.}$ One then has a new diagram possibly containing some closed loops. If there are $p$ closed loops, the product is then the resulting diagram, with the closed loops removed, times the scalar ${\delta }^p$ where $\delta$ is some fixed element of the field. One thus obtains a family of algebras $K(2n,\delta )$ of dimension $\frac{1}{2n+1}\left(\genfrac{}{}{0ex}{}{4n}{2n}\right)\text{.}$ We illustrate multiplication in $K(4,\delta )$ below. $$\begin{array}{c} α= β= so αβ=δ= \end{array}$$

We now define special elements $E_1,E_2,\dots E_{2n-1}$ of $K(2n,\delta )\text{:}$ $$\begin{array}{c} ⋯ E1 E2 En \end{array}$$ One easily has $\{\begin{array}{c}{E}_i^2=\delta E_i\\ E_i{E}_j=E_j{E}_i,f|i-j|\ge 2\\ E_i{E}_{i\pm 1}{E}_i=E_i\end{array}$ and it is easy to see that the $E_i\text{'s,}$ together with $1,$ generate $K(2n,\delta )\text{.}$ Counting reduced words on the $E_i\text{'s}$ one obtains the Catalan numbers (see [Jon1983]) so the above relations present $K(2n,\delta )\text{.}$

Lemma 3. If $d$ is a diagram giving a basis element of $K(2n,\delta )$ as above, let ${\sim }_d$ be the equivalence relation it generates. Then the map $$\phi \left(E_i\right)=\{\begin{array}{cc}{L}_{\sim E_i}& i\text{odd}\\ k{L}_{\sim E_i}& i\text{even}\end{array}$$ extends to an algebra homomorphism from $K(2n,\frac{1}{k})$ to $\text{End}\left({\otimes }^{n}V\right)$ so that for a diagram $d,$ $\phi \left(d\right)$ is a non zero multiple of $L_{\sim d}\text{.}$

Proof. Since the above relations present $K(2n,\frac{1}{k}),$ the existence of $\phi$ follows from the fact that the $\phi \left(E_i\right)$ satisfy the same relations. We leave it to the reader to check that if $d$ and $d\prime$ are diagrams then $L_{\sim d}{L}_{\sim d\prime }$ is a multiple of $L_{\sim dd\prime }\text{.}$

Q.E.D.

Remark. It seems suggestive that the mapping $\phi$ defined above is injective as soon as $k\ge 4$ whereas we have seen that for non-planar partitions there are linear relations between the $L_{\sim }$ as soon as $n>k\text{.}$

The same thing happens with the $K(2n,\delta )$ as a subalgebra of the Brauer algebra. This linear independence is easily proved using the Markov trace technique of [Jon1983].

Theorem. The commutant of $S_k$ on ${\otimes }^{n}V$ is generated by $S_n,$ $e_1$ and $e_2$ where $V,$ $e_1$ and $e_2$ are as in Lemma 1 and $S_n$ acts by permuting the tensor factors.

Proof. It suffices to show that any $L_{\sim }$ is in the algebra generated by $S_n$ and $e_1,e_2\text{.}$ We have seen that by multiplying $L_{\sim }$ on the left and right by elements of $S_n,$ we can assume that $\sim$ is planar. The elements $\phi \left(E_3\right),\phi \left(E_4\right),\dots \phi \left(E_n\right)$ are just conjugates of $\phi \left(E_1\right)$ and $\phi \left(E_2\right)$ under the action of $S_n$ so the theorem follows from Lemma 3.

Q.E.D.

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