## The Potts model and the symmetric group

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 $×\text{'s}$ to the $2n$ top $×\text{'s}$ 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 $4n$ $×\text{'s}$ 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 $×\text{'s.}$

Step 2. Remove the middle edges and middle $×\text{'s.}$ 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 $K\left(2n,\delta \right)$ of dimension $\frac{1}{2n+1}\left(\genfrac{}{}{0}{}{4n}{2n}\right)\text{.}$ We illustrate multiplication in $K\left(4,\delta \right)$ below. $\alpha = \beta = so \alpha \beta =\delta =$

We now define special elements ${E}_{1},{E}_{2},\dots {E}_{2n-1}$ of $K\left(2n,\delta \right)\text{:}$ $\cdots {E}_{1} {E}_{2} {E}_{n}$ One easily has $\left\{\begin{array}{c}{E}_{i}^{2}=\delta {E}_{i}\\ {E}_{i}{E}_{j}={E}_{j}{E}_{i},\phantom{\rule{2em}{0ex}}f|i-j|\ge 2\\ {E}_{i}{E}_{i±1}{E}_{i}={E}_{i}\end{array}$ and it is easy to see that the ${E}_{i}\text{'s,}$ together with $1,$ generate $K\left(2n,\delta \right)\text{.}$ Counting reduced words on the ${E}_{i}\text{'s}$ one obtains the Catalan numbers (see [Jon1983]) so the above relations present $K\left(2n,\delta \right)\text{.}$

Lemma 3. If $d$ is a diagram giving a basis element of $K\left(2n,\delta \right)$ as above, let ${\sim }_{d}$ be the equivalence relation it generates. Then the map $φ(Ei)= { L∼Ei iodd kL∼Ei ieven$ extends to an algebra homomorphism from $K\left(2n,\frac{1}{k}\right)$ 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 $K\left(2n,\frac{1}{k}\right),$ 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 $K\left(2n,\delta \right)$ 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.