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 the number of such decompositions is discussed.

Following Stembridge, 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?

  • $\begingroup$ Do you mean that you consider $S_n$ as a Coxeter group with generating set $s_1 = (12)$, $s_2 = (23)$, $s_3 = (34)$, etc.? What do you mean by a "single commutation"? Do you mean that you replace a single instance of $s_i s_j$ by $s_j s_i$ where $s_i$ and $s_j$ are commuting generators? $\endgroup$ Nov 24 at 13:59
  • $\begingroup$ @TomDeMedts: yes, and this set-up is very common, so I think the question is clear as it is. $\endgroup$ Nov 24 at 14:03
  • $\begingroup$ Not an answer to your question, but sort of related: there has been work done on computing the expected number of braid moves that one can apply to a random reduced expression for $w_0$: see Reiner (doi.org/10.1016/j.ejc.2004.06.010), Tenner (ajc.maths.uq.edu.au/pdf/62/ajc_v62_p147.pdf), Schilling et al. (doi.org/10.1016/j.ejc.2016.10.008) $\endgroup$ Nov 24 at 14:06
  • $\begingroup$ Another non-answer comment: we can create the graph of reduced expressions, with two kinds of edges - say red and blue - corresponding to the commutation and (long) braid moves. By the Tits-Matsumoto lemma, this graph is connected, so each vertex either has a red or blue edge coming out of it. The result of Reiner I cited above says that the total number of blue edges in this graph is one half the number of vertices. You are asking for an enumeration/classification of vertices that have only blue edges coming out of them. $\endgroup$ Nov 24 at 16:44

2 Answers 2


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.

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I believe that for $n \geq 4$ there will be exactly $4$ such reduced words. One such word, call it $R_n$ can be constructed by starting with $s_{n-1}s_{n-2} \cdots s_2s_1s_2 \cdots s_{n-2}s_{n-1}$ and appending $R_{n-1}$ with all indices shifted up by 1. For example, $$R_6=s_5s_4s_3s_2s_1s_2s_3s_4s_5s_4s_3s_2s_3s_4s_3.$$ Three more words can be obtained by replacing each $s_i$ with $s_{n-i}$, reversing the word, or doing both.

For $n=4,5,6$ these are the only possibilities, and this probably shouldn't be too hard to argue in general.


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