# Vanishing of certain coefficients coming from Coxeter groups

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.