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Let $W$ be a Coxeter group with associated graph $G$.

Define $$X(G) = \{w \in W : \text{any two simple reflections} \,S_{\alpha}\, \text{and}\, S_{\beta} \,\text{appearing in any of the reduced expression of $w$ commute} \} $$.

I have the following questions :

  1. Is this set $X(G)$ studied in the literature?

  2. What is the combinatorial significance of this set?

Thanks for your time and have a good day.

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If I see it correctly, there is not much going on in the set $X(G)$ from the viewpoint of Coxeter systems: Let $S$ be the simple system and let $G$ be the Coxeter graph of $(W,S)$ with vertex set $S$. Then $$X(G) = \big\{ s_1 \cdots s_k \in W \mid \{s_1,\ldots,s_k\} \subseteq S\text{ with } s_is_j = s_js_i \text{ for all } 1 \leq i<j \leq k\big\}.$$ Since different choices of $s_1,\ldots,s_k$ yield different elements, $X(G)$ is in bijection with totally disconnected subsets of $G$. This is, $\mathcal{D}(G) = \{ A \subseteq G \mid A \text{ finite and totally disconnected}\}$ and $\mathcal{D}(G) \rightarrow X(G)$ given by $A = \{ s_1, \ldots s_k\} \subseteq G$ of size $k$ is sent to $s_1\cdots s_k \in X(G) \subseteq W$ is a bijection. In particular, $X(G)$ is finite for finite $S$ even if $W$ is infinite.

Observe that I used that any two reduced words have the same simple generators involved. Thus, your desired property means that your element lives in a standard parabolic subgroup of type $\mathbb{A}_1^k$ for some $k$. This is exactly the describtion I gave.

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  • $\begingroup$ Since it took me hard time to understand your sentence "is in bijection..." let me rephrase it: the map from the set of [finite] totally disconnected subsets of $G$ to $W$, mapping $\{u_1,\dots,u_n\}$ (all distinct) to $u_1\dots u_n$, is a bijection into $X(G)$. Note that for a finite graph $G$, this is finite regardless of whether $W$ is finite. $\endgroup$
    – YCor
    Commented Mar 12, 2018 at 9:03
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    $\begingroup$ Maybe they want to actually count those independent subsets in G. Then there is "much going on" as the number of vertices grows. Like maximum independent set being NP-hard. $\endgroup$
    – AHusain
    Commented Mar 12, 2018 at 9:33
  • $\begingroup$ @YCor: edited for clarification. $\endgroup$
    – Christian Stump
    Commented Mar 12, 2018 at 10:13
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    $\begingroup$ @AHusain: edited that I meant "not much going on" from the viewpoint of Coxeter systems. $\endgroup$
    – Christian Stump
    Commented Mar 12, 2018 at 10:21

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