For the symmetric group on $n$ objects $S\_n$ the question of how to write its longest element $w\_0$ as a reduced decomposition is an important combinatorical problem. As example, in this [question](https://mathoverflow.net/questions/370333/number-of-reduced-decompositions-of-the-longest-element-of-the-weyl-group) the number of such decompositions is discussed.


Following [Stembridge](https://link.springer.com/article/10.1023/A:1022452717148), we define a relation $∼$ on the set of reduced expressions for $w\_0$. Let $\mathbf{w}$ and $\mathbf{w}'$ be two reduced expressions for $w\_0$ and define $\mathbf{w} \sim \mathbf{w}'$ if we can obtain $\mathbf{w}'$ from $\mathbf{w}$ by applying a single commutation. Now, define the equivalence relation $\simeq$ by taking the reflexive transitive
closure of $\sim$. Each equivalence class under $\simeq$ is called a commutation class.


The question of the number of commutation classes (of reduced expressions for $w\_0 \in S\_n$) is still an open problem. However, is there an formula for the number of equivalence classes containing just one element?



One element commutation classes of reduced decompositions of the longest element of the Weyl group
The reduced words that are in their own commutation classes are:


* $s = [123\cdots(n-2)(n-1)(n-2)\cdots321][23\cdots(n-3)(n-2)(n-3)\cdots32][3\cdots(n-4)(n-3)(n-4)\cdots3]\cdots$
* the reversal of $s$ (writing the word in reverse order)
* the complement of $s$ (mapping each letter $i$ to $n-i$)
* the reverse complement of $s$


This is just one word in $S\_2$ and two words in $S\_3$ (no big surprises there!). Otherwise it is four words. For example, in $S\_6$, they are


* 123454321234323
* 323432123454321
* 543212345432343
* 343234543212345


On a separate note, to add to the list of related topics, you might want to look at Fishel et al. (<https://doi.org/10.1016/j.ejc.2018.07.002>) about how commutation classes and braid classes interact.


