## 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.

## §1. Two bases for ${\text{End}}_{{S}_{k}}\left({\otimes }^{n}V\right)$

Let $V$ be a vector space of dimension $k$ over a field $K$ of characteristic zero, with basis $\left\{{v}_{a}\right\}$ indexed by a set $A$ of size $k\text{.}$ The symmetric group ${S}_{k}$ is linearly represented on $V$ by $\sigma \left({v}_{a}\right)={v}_{\sigma \left(a\right)}\text{.}$ Thus for each $n,$ ${\otimes }^{n}V$ is an ${S}_{k}\text{-module.}$ As a vector space, ${\text{End}}_{{S}_{k}}\left({\otimes }^{n}V\right)$ is just the fixed points for the representation of ${S}_{k}$ on $\text{End}\left({\otimes }^{n}V\right)$ by conjugation. But this is a permutation representation for if $X\in \text{End}\left({\otimes }^{n}V\right)$ has matrix ${X}_{{a}_{1}{a}_{2}\cdots {a}_{n}}^{{b}_{1}{b}_{2}\cdots {b}_{n}}$ with respect to the basis $\left\{{v}_{{a}_{1}}\otimes {v}_{{a}_{2}}\otimes \cdots \otimes {v}_{{a}_{n}}\right\}$ then $\sigma \left(X\right)$ has matrix ${X}_{\sigma \left({a}_{1}\right)\sigma \left({a}_{2}\right)\cdots \sigma \left({a}_{n}\right)}^{\sigma \left({b}_{1}\right)\sigma \left({b}_{2}\right)\cdots \sigma \left({b}_{n}\right)}\text{.}$ Hence a basis for ${\text{End}}_{{S}_{k}}\left({\otimes }^{n}V\right)$ is given by $\left\{\sum _{\text{orbit}}{X}_{{a}_{1}\cdots {a}_{n}}^{{b}_{1}\cdots {b}_{n}}\right\}$ where each sum is taken over all $\left\{{a}_{1},{a}_{2},\cdots {a}_{n},{b}_{1},{b}_{2},\cdots {b}_{n}\right\}$ in an orbit for the action of ${S}_{k}$ on the $2n\text{-fold}$ Cartesian product ${A}^{2n}\text{.}$ Each orbit for this ${S}_{k}$ action is given by a partition of $\left\{1,2,\dots ,2n\right\}$ into subsets (obviously at most $k$ of them). If $\sim$ denotes the equivalence relation defined by such a partition then the corresponding orbit on ${A}^{2n}$ consists of these $2n\text{-tuples}$ $\left({a}_{1},\cdots ,{a}_{n},\cdots ,{a}_{2n}\right)$ (setting ${b}_{i}={a}_{n+i}\text{)}$ for which ${a}_{i}={a}_{j}$ iff $i\sim j\text{.}$ The number of equivalence relations with $r$ classes is the Stirling number $S\left(2n,r\right)$ so that the dimension of ${\text{End}}_{{S}_{k}}\left({\otimes }^{n}V\right)$ is $\sum _{i=1}^{k}S\left(2n,i\right)\text{.}$ For $2n\le k$ this is the Bell number $B\left(2n\right)=\sum _{i=1}^{2n}S\left(2n,i\right)$ (see [Sta1986] p.33). Compare the first few Bell numbers with the dimensions of the Bratteli diagram.

If $\sim$ is an equivalence relation on $\left\{1,\dots ,2n\right\}$ we will call ${T}_{\sim }$ the element of $\text{End}\left({V}^{\otimes n}\right)$ defined by the corresponding ${S}_{k}\text{-orbit.}$ It clearly has the following matrix: $(T∼)a1⋯anan+1⋯an= { 1 if(ai=aj⇔i∼j) 0 otherwise.$ Note that ${T}_{\sim }$ is zero if the number of classes for $\sim$ is more than $k\text{.}$ From our discussion $\left\{{T}_{\sim } | \sim \text{an equivalence relations}\right\}$ is a basis for ${\text{End}}_{{S}_{k}}\left({\otimes }^{n}V\right)$ with $\sum _{i=1}^{k}S\left(2n,i\right)$ elements (forgetting any ${T}_{\sim }$ that may be zero).

In fact we will use another set which linearly spans ${\text{End}}_{{S}_{k}}\left({\otimes }^{n}V\right)$ and will also be a basis if $k\ge 2n\text{.}$ For each $\sim$ as above define ${L}_{\sim }\in {\text{End}}_{{S}_{k}}\left({\otimes }^{n}V\right)$ by $L∼=∑∼′≤∼ T∼.$ Where the sum is over all partitions $\sim \prime$ coarser than $\sim \text{.}$ By Möbius inversion ([Sta1986] p.116) the ${T}_{\sim }\text{'s}$ can be expressed in terms of the ${L}_{\sim }\text{'s}$ so they also span ${\text{End}}_{{S}_{k}}\left({\otimes }^{n}V\right)\text{.}$ Note though that the ${L}_{\sim }\text{'s}$ are always nonzero so they will not form a basis for $k<2n\text{.}$

We leave the verification of the following lemma to the reader as it should make clear what is going on.

Lemma 1.

 1) If $\sim$ is defined by $i\sim i+n$ then ${L}_{\sim }=\text{id.}$ 2) If $\sim$ is defined by $i\sim i+n$ for $i>1$ then ${L}_{\sim }=k{e}_{1}$ with ${e}_{1}$ as in the introduction. 3) If $\sim$ is defined by $1\sim 2\sim \left(n+1\right)\sim \left(n+2\right)$ and $i\sim i+n$ for $i>2$ then ${L}_{\sim }={e}_{2}$ with ${e}_{2}$ as in the introduction.