Last update: 18 April 2014
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.
The diagrams joining the bottom to the top 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 into subsets of size 2. The multiplication is defined by a procedure.
Step 1. Stack the two rectangles on top of each other, lining up the
Step 2. Remove the middle edges and middle One then has a new diagram possibly containing some closed loops. If there are closed loops, the product is then the resulting diagram, with the closed loops removed, times the scalar where is some fixed element of the field. One thus obtains a family of algebras of dimension We illustrate multiplication in below.
We now define special elements of One easily has and it is easy to see that the together with generate Counting reduced words on the one obtains the Catalan numbers (see [Jon1983]) so the above relations present
Lemma 3. If is a diagram giving a basis element of as above, let be the equivalence relation it generates. Then the map extends to an algebra homomorphism from to so that for a diagram is a non zero multiple of
Proof. Since the above relations present the existence of follows from the fact that the satisfy the same relations. We leave it to the reader to check that if and are diagrams then is a multiple of
Remark. It seems suggestive that the mapping defined above is injective as soon as whereas we have seen that for non-planar partitions there are linear relations between the as soon as
The same thing happens with the as a subalgebra of the Brauer algebra. This linear independence is easily proved using the Markov trace technique of [Jon1983].
Theorem. The commutant of on is generated by and where and are as in Lemma 1 and acts by permuting the tensor factors.
Proof. It suffices to show that any is in the algebra generated by and We have seen that by multiplying on the left and right by elements of we can assume that is planar. The elements are just conjugates of and under the action of so the theorem follows from Lemma 3.