The Potts model and the symmetric group

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

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.

§0. Introduction

In the Potts model ([Bax1982]) of classical equilibrium statistical mechanics, individual "atoms" can be in one of k possible "spin" states σ1,,σn. Two particles interact if they are neighbours and if their spins are the same. Thus the matrix w(σ,σ) of Boltzmann weights is invariant under the (diagonal) action of the symmetric group Sk. The usual presentation of the Potts model transfer matrices in the physics literature (eg. [Bax1982]) writes them as elements of an algebra generated by cyclic groups so only the cyclic symmetry is apparent. This is a little misleading and one should attempt to understand the full symmetry group. This seems especially important now that there are known to be solvable models whose Boltzmann weights have rather interesting symmetry groups (see [Jae1992]). The first question one may ask is to what extent the transfer matrices generate the commutant of the group of symmetries. The actual details of this question will depend on the kind of model (Spin, vertex, IRF, ...). Although we are only really concerned with the Potts model here, let us describe the situation for a certain vertex model (solvable by Temperley Lieb equivalence - see [Bax1982] chap 12) to illustrate clearly the connection between this problem and classical invariant theory.

A vertex model is specified by a k2×l2 matrix Rabxy(λ), a,b=1,k, x,y=1,l. In our special case we define eabxy=1kδaxδb,y (k=l), and R(λ)=f(λ)e+id for some conveniently chosen f(λ). The site-to-site transfer matrices are then obtained from the matrices e1,e2,,en-1 on nV, V being a vector space over a field of characteristic zero, where ei(w1wiwi+1wn)=w1e(wiwi+1)wn, where wiV for each i and e is the element of End(VV) having matrix eabxy with respect to the basis {vavx} of VV, {Va} being a basis of V. (Thus e(vavx)=b,yeabxyvbvy.)

The ei's are orthogonal projections satisfying ei2=ei*=ei eiei±1ei= 1k2ei eiej=ejei if|i-j| 2.

If V is considered to be a real Euclidean space for which {va} is an orthonorrnal basis, one recognizes e as being orthogonal projection onto the canonical subspace of VV, scalar multiples of avava, which is independent of the orthonormal basis. Thus this vertex model has as a natural symmetry group the orthogonal group O(k). The question of to what extent the site-to-site transfer matrices genrate the commutant of the symmetry group becomes the question of an explicit description of the commutant of the tensor powers of the vector representation of the orthogonal group. Thus question was answered in the 1930's by Brauer [Bra1937] who showed that this commutant is generated as an algebra by our ei's and the natural representation of the permutation group Sn on nV. He gave moreover an explicit abstract algebra with basis all possible partitions of a set of 2n elements onto subsets of size 2, represented by diagrams as below, n=5 and an obvious way of considering these diagrams as operators on nV. The element ei described above corresponds (up to a factor k) to the diagram 1 2 i i+1 n For k sufficiently large (and n fixed) the matrices corresponding to diagrams are a basis of the commutant of O(k), and they always span it linearly. Thus the commutant is always a quotient of the abstract Brauer algebra. Wenzl obtained the structure of the abstract Brauer algebra for generic values of the parameter k, using ideas from subfactors, knot theory and quantum groups, in [Wen1988-2]. (The interested reader might also want to consult [Jon1991].)

Thus we see that^ for this vertex model, the commutant of the symmetry group is generated by the site-to-site transfer matrices and the obvious action of the permutation group on nV. One may ask to what extent this is true in general. (For vertex models coming from quantum groups, the group symmetries would become quantum group symmetries and the group Sn would be replaced by the braid group - the result may well be true in general, through one would have to consider R(λ) for all values of λ.) In this paper we will show that the version of the result appropriate for spin models (i.e models of the same kind as the Potts model but with arbitrary Boltzmann weights w(σ,σ)) is true for the Potts model. The main differences are that nV will carry 2n matrices ei, that the parameter k2 in the ei relations becomes k, that the permutation group Sn has two orbits on the set of all ei's, and that the continuous group O(k) of the above vertex model is replaced by the finite group Sk.

Before getting on with the proof of the result let us point out that the algebra structure (and thus, if desired, the dimension) of the commutant of Sk is easily determined, at least inductively, from the representation theory of Sk. For the representation on V of Sk is a permutation representation, so tensoring by V corresponds to restriction to the stabilizer of a point, i.e. Sk-1 and then inducing back up to Sk. Since induction and restriction are particularly simply visible from the Young diagrams, one immediately obtains the Bratteli diagram for EndSk(nV): (illustrated for k=6): 1= 1= - dimension 2 2 3 1 1 - dimension 15 5 10 6 6 1 2 1 - dimension 203 etc.

In general for a given k the Bratteli diagram will stop growing wider after the point at which all irreducible representations of Sk appear. In this respect the Brauer algebra is different — the diagram will grow ever wider since the orthogonal group has infinitely many different irreducible representations. It is still true however that the number of simple components of the Brauer algebra is larger if k is larger (for large n).

I would like to thank Roger Howe for first pointing out to me the "e" matrix in the vertex model described above.

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