Let $\left(W\text{, }S\right)$ be a Coxeter system. For every $w\in W$ let us write $\left|w\right|$ for the length of $w$. Set $\lambda\left(e\right)=1$ where $e\in W$ denotes the neutral element of the group and define coefficients $\lambda\left(w\right)\in\mathbb{Z}$ recursively via \begin{eqnarray} \sum_{v\in W \text{: }\left|v^{-1}w\right|=\left|w\right|-\left|v\right|}\lambda\left(v\right)=0 \end{eqnarray} for any $w\in W$. For example this gives us $\lambda\left(e\right)=1$, $\lambda\left(s\right)=\left(-1\right)$ for all $s\in S$, $\lambda\left(st\right)=1$ if $m_{st}=2$ and $\lambda\left(st\right)=0$ if $m_{st}\neq2$ (for the notation see Wikipedia) for all $s\text{, }t\in S$ with $s\neq t$, and so on.

My question is: Does there always exist some $l\in\mathbb{N}$ such that $\lambda\left(w\right)=0$ for all $w\in W$ with $\left|w\right|\geq l$?

In the case that $\left(W\text{, }S\right)$ is right-angled (i.e. $m_{st}\in\left\{ 2\text{, }\infty\right\}$ for all $s\text{, }t\in S$ with $s\neq t$) this is true and we can take $l=\left|S\right|$. I'm wondering if in the general case this property also holds. If no, is it possible to characterize the property in an instructive way?


The answer is yes. I claim there are at most $2^n$ elements of $W$ for which $\lambda(w) \neq 0$. Specifically, let $\vee$ be the join in weak order, which is a semi-lattice. If $\{ s_1, s_2, \ldots, s_j \} \subseteq S$ and $s_1 \vee s_2 \vee \cdots \vee s_j$ is defined, then I claim that $\lambda(s_1 \vee s_2 \vee \cdots \vee s_j)=(-1)^j$, and otherwise I claim that $\lambda(w)=0$. Thus $N$ can be taken to be the greatest length of $s_1 \vee s_2 \vee \cdots \vee s_j$, restricting ourselves to cases where this join is defined.

The condition $|v^{-1} w| = |w| - |v|$ says that $v \leq w$ in weak order. Therefore, this is the recursion defining the Mobius function of weak order. The Mobius function of a finite lattice $L$ can be computed by Rota's crosscut theorem. (The first online reference I could find was Theorem 1.3 here.) Namely, let $A$ be the set of minimal elements of $L$ -- in this case, this is the set $S$. Then $$\mu(x) = \sum_{B \subseteq A,\ \bigvee B = x} (-1)^{|B|}.$$

So $\mu(w)=0$ if $w$ is not a join of elements of $S$. If $w$ is a join of elements of $S$, then it is so in only one way (this is a way that weak order is simpler than a general lattice) so we get the description from the first paragraph.

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  • $\begingroup$ Thanks for your reply. Great answer! $\endgroup$ – worldreporter14 Jun 24 '19 at 9:34

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