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 $2n$ bottom $\times \text{'s}$ to the $2n$ top $\times \text{'s}$ 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 $4n$ $\times \text{'s}$ into subsets of size 2. The multiplication is defined by a $2\text{-step}$ procedure.

**Step 1.** Stack the two rectangles on top of each other, lining up the $\times \text{'s.}$

**Step 2.** Remove the middle edges and middle $\times \text{'s.}$ One then has a new diagram possibly
containing some closed loops. If there are $p$ closed loops, the product is then the resulting diagram, with the closed loops removed, times
the scalar ${\delta}^{p}$ where $\delta $ is some fixed element of the field. One thus obtains a family
of algebras $K(2n,\delta )$ of dimension
$\frac{1}{2n+1}\left(\genfrac{}{}{0ex}{}{4n}{2n}\right)\text{.}$
We illustrate multiplication in $K(4,\delta )$ below.
$$\begin{array}{c}\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\alpha =\n\beta =\nso \alpha \beta =\delta =\n\n\n\n\end{array}$$

We now define special elements ${E}_{1},{E}_{2},\dots {E}_{2n-1}$ of $K(2n,\delta )\text{:}$ $$\begin{array}{c}\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\cdots \nE1\nE2\nEn\n\n\end{array}$$ One easily has $\{\begin{array}{c}{E}_{i}^{2}=\delta {E}_{i}\\ {E}_{i}{E}_{j}={E}_{j}{E}_{i},\phantom{\rule{2em}{0ex}}f|i-j|\ge 2\\ {E}_{i}{E}_{i\pm 1}{E}_{i}={E}_{i}\end{array}$ and it is easy to see that the ${E}_{i}\text{'s,}$ together with $1,$ generate $K(2n,\delta )\text{.}$ Counting reduced words on the ${E}_{i}\text{'s}$ one obtains the Catalan numbers (see [Jon1983]) so the above relations present $K(2n,\delta )\text{.}$

**Lemma 3.**
*
If $d$ is a diagram giving a basis element of $K(2n,\delta )$
as above, let ${\sim}_{d}$ be the equivalence relation it generates. Then the map
$$\phi \left({E}_{i}\right)=\{\begin{array}{cc}{L}_{\sim {E}_{i}}& i\phantom{\rule{1em}{0ex}}\text{odd}\\ k{L}_{\sim {E}_{i}}& i\phantom{\rule{1em}{0ex}}\text{even}\end{array}$$
extends to an algebra homomorphism from $K(2n,\frac{1}{k})$
to $\text{End}\left({\otimes}^{n}V\right)$ so that for a diagram
$d,$ $\phi \left(d\right)$ is a non zero multiple of
${L}_{\sim d}\text{.}$
*

*Proof.* Since the above relations present $K(2n,\frac{1}{k}),$
the existence of $\phi $ follows from the fact that the $\phi \left({E}_{i}\right)$
satisfy the same relations. We leave it to the reader to check that if $d$ and $d\prime $ are diagrams then
${L}_{\sim d}{L}_{\sim d\prime}$ is a
multiple of ${L}_{\sim dd\prime}\text{.}$

Q.E.D.

*Remark.* It seems suggestive that the mapping $\phi $ defined above is *injective* as soon as $k\ge 4$
whereas we have seen that for non-planar partitions there are linear relations between the ${L}_{\sim}$ as soon as
$n>k\text{.}$

The same thing happens with the $K(2n,\delta )$ as a subalgebra of the Brauer algebra. This linear independence is easily proved using the Markov trace technique of [Jon1983].

**Theorem.**
*
The commutant of ${S}_{k}$ on ${\otimes}^{n}V$
is generated by ${S}_{n},$ ${e}_{1}$ and
${e}_{2}$ where $V,$ ${e}_{1}$ and
${e}_{2}$ are as in Lemma 1 and ${S}_{n}$ acts by permuting the tensor factors.
*

*Proof.* It suffices to show that any ${L}_{\sim}$ is in the algebra generated by ${S}_{n}$
and ${e}_{1},{e}_{2}\text{.}$ We have seen that by multiplying
${L}_{\sim}$ on the left and right by elements of ${S}_{n},$ we can assume
that $\sim $ is planar. The elements $\phi \left({E}_{3}\right),\phi \left({E}_{4}\right),\dots \phi \left({E}_{n}\right)$
are just conjugates of $\phi \left({E}_{1}\right)$ and
$\phi \left({E}_{2}\right)$ under the action of ${S}_{n}$
so the theorem follows from Lemma 3.

Q.E.D.