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.

§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 $\{v_{a_1}\otimes v_{a_2}\otimes \cdots \otimes v_{a_n}\}$ 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 $\{a_1,a_2,\cdots a_n,b_1,b_2,\cdots b_n\}$ 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 $\{1,2,\dots ,2n\}$ 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 $2n\text{-tuples}$ $(a_1,\cdots ,a_n,\cdots ,a_{2n})$ (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 $S(2n,r)$ so that the dimension of ${\text{End}}_{S_k}\left({\otimes }^{n}V\right)$ is $\sum _{i=1}^{k}S(2n,i)\text{.}$ For $2n\le k$ this is the Bell number $B\left(2n\right)=\sum _{i=1}^{2n}S(2n,i)$ (see [Sta1986] p.33). Compare the first few Bell numbers with the dimensions of the Bratteli diagram.

If $\sim$ is an equivalence relation on $\{1,\dots ,2n\}$ 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: $${\left(T_{\sim }\right)}_{a_1\cdots a_n}^{a_{n+1}\cdots a_n}=\{\begin{array}{cc}1& \text{if}(a_i=a_j\iff i\sim j)\\ 0& \text{otherwise.}\end{array}$$ Note that $T_{\sim }$ is zero if the number of classes for $\sim$ is more than $k\text{.}$ From our discussion $\left\{T_{\sim }\hspace{0.17em}\right|\hspace{0.17em}\sim \hspace{0.17em}\text{an equivalence relations}\}$ is a basis for ${\text{End}}_{S_k}\left({\otimes }^{n}V\right)$ with $\sum _{i=1}^{k}S(2n,i)$ 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 $$L_{\sim }=\sum _{\sim \prime \le \sim }{T}_{\sim }\text{.}$$ 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 $1\sim 2\sim (n+1)\sim (n+2)$ and $i\sim i+n$ for $i>2$ then $L_{\sim }=e_2$ with $e_2$ as in the introduction.

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