Lemma 5

A vertex cannot belong to two distinct edges of order $\geq 4$.

Let $i$ be such a vertex. We would have $\sum_{j \neq i} q_{ij}^2 \geq (\frac{\sqrt{2}}{2})^2 + (\frac{\sqrt{2}}{2})^2 = 1$, which is impossible (Lemma 1).

Let $\{i, j\}$ be an edge of $X$. We are going to define a new Coxeter graph, which will be obtained from the graph of $M$ by identifying $i$ and $j$. The set $I'$ of its vertices is the quotient of $I$ obtained by identifying $i$ and $j$. Put $p = \{i, j\}$, which is an element of $I'$, and identify the elements of $I$ distinct from $i$ and $j$ with their canonical images in $I'$. Let $k, k'$ be two distinct elements of $I'$. Then, $\{k, k'\}$ is an edge of the new graph in the following cases:

1) $k$ and $k'$ are distinct from $p$ and $\{k, k'\}$ is an edge of $X$: in this case, the order of this edge is defined to be $m_{kk'}$:

2) $k = p$, and one of the sets $\{i, k'\}, \{j, k'\}$ is an edge of $X$; the order of $\{p, k'\}$ is defined to be $m_{ik'}$ if $\{i, k'\}$ is an edge of $X$, and $m_{jk'}$ if $\{j, k'\}$ is an edge of $X$ (these two possibilities are mutually exclusive because $X$ is a tree).

Let $M' = (m'_{ij})_{i,j \in I'}$ be the new Coxeter matrix thus defined, and put $q'_{ij} = -\cos \frac{\pi}{m'_{ij}}$. Then, for $k \neq p$, $t'_{pk} = q_{ik} + q_{jk}$. Thus, if $(\xi_i) \in \mathbf{R}^{I'}$,