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](https://en.wikipedia.org/wiki/Coxeter_group)) for all $s\text{, }t\in S$ with $s\neq t$, and so on.



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> **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 $\left|w\right|\geq l$?
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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?



Vanishing of certain coefficients coming from Coxeter groups
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](https://en.wikipedia.org/wiki/Bruhat_order#Definition), 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](https://arxiv.org/abs/math/9801009).) 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.


