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 EndSk(nV)

Let V be a vector space of dimension k over a field K of characteristic zero, with basis {va} indexed by a set A of size k. The symmetric group Sk is linearly represented on V by σ(va)=vσ(a). Thus for each n, nV is an Sk-module. As a vector space, EndSk(nV) is just the fixed points for the representation of Sk on End(nV) by conjugation. But this is a permutation representation for if XEnd(nV) has matrix Xa1a2anb1b2bn with respect to the basis {va1va2van} then σ(X) has matrix Xσ(a1)σ(a2)σ(an)σ(b1)σ(b2)σ(bn). Hence a basis for EndSk(nV) is given by {orbitXa1anb1bn} where each sum is taken over all {a1,a2,an,b1,b2,bn} in an orbit for the action of Sk on the 2n-fold Cartesian product A2n. Each orbit for this Sk action is given by a partition of {1,2,,2n} into subsets (obviously at most k of them). If denotes the equivalence relation defined by such a partition then the corresponding orbit on A2n consists of these 2n-tuples (a1,,an,,a2n) (setting bi=an+i) for which ai=aj iff ij. The number of equivalence relations with r classes is the Stirling number S(2n,r) so that the dimension of EndSk(nV) is i=1kS(2n,i). For 2nk this is the Bell number B(2n)=i=12nS(2n,i) (see [Sta1986] p.33). Compare the first few Bell numbers with the dimensions of the Bratteli diagram.

If is an equivalence relation on {1,,2n} we will call T the element of End(Vn) defined by the corresponding Sk-orbit. It clearly has the following matrix: (T)a1anan+1an= { 1 if(ai=ajij) 0 otherwise. Note that T is zero if the number of classes for is more than k. From our discussion {T|an equivalence relations} is a basis for EndSk(nV) with i=1kS(2n,i) elements (forgetting any T that may be zero).

In fact we will use another set which linearly spans EndSk(nV) and will also be a basis if k2n. For each as above define LEndSk(nV) by L= T. Where the sum is over all partitions coarser than . By Möbius inversion ([Sta1986] p.116) the T's can be expressed in terms of the L's so they also span EndSk(nV). Note though that the L's are always nonzero so they will not form a basis for k<2n.

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

Lemma 1.

1) If is defined by ii+n then L=id.
2) If is defined by ii+n for i>1 then L=ke1 with e1 as in the introduction.
3) If is defined by 12(n+1)(n+2) and ii+n for i>2 then L=e2 with e2 as in the introduction.

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