Faithfullness of Weyl group action

Question

Let $I$ be a finite indexing set, and $A \in \operatorname{Mat}_I(\mathbb{Z})$ be a generalised Cartan matrix, i.e.

1. $a_{ii} = 2$,
2. $a_{ij} \leq 0$ for $i \neq j$, and
3. $a_{ij} = 0 \iff a_{ji} = 0$.

Then there is an abstract Weyl group $W_A$ associated to the matrix $A$, defined on the generators $\{s_i \mid i \in I\}$ subject only to the following relations:

1. $s_i^2 = 1$ for all $i \in I$, and
2. $(s_i s_j)^{m_{ij}} = 1$ for $i \neq j$, where $m_{ij} = 2, 3, 4, 6$ or $\infty$ depending on whether $a_{ij} a_{ij} = 0, 1, 2, 3$ or $\geq 4$ respectively. (If $m_{ij} = \infty$, we mean that $s_i, s_j$ generate an infinite dihedral group).

A based root datum of type $A$ is the data $R = (X, X^\vee, \langle - , - \rangle, (\alpha_i)_{i \in I}, (\alpha_i^\vee)_{i \in I})$ of:

1. Two $\mathbb{Z}$-modules $X, X^\vee$ of free finite rank in a perfect pairing $\langle -, - \rangle \colon X \times X^\vee \to \mathbb{Z}$, and
2. Choices of elements $\alpha_i \in X$ and $\alpha_i^\vee \in X^\vee$ such that $\langle \alpha_j, \alpha_i^\vee \rangle = a_{ij}$ for all $i, j \in I$.

Any time we have such data, we may define the simple reflections

$$ r_i \colon X \to X, \quad r_i(\lambda) = \lambda - \langle \lambda, \alpha_i^\vee \rangle \alpha_i, $$ $$ r_i^\vee \colon X^\vee \to X^\vee, \quad r_i^\vee(\mu) = \mu - \langle \alpha_i, \mu \rangle \alpha_i^\vee $$

and with a little work it is possible to show that the map $W_I \to \operatorname{GL}_{\mathbb{Z}}(X)$ defined by $s_i \mapsto r_i$ is in fact a group homomorphism, and similarly for $W_A \to \operatorname{GL}_{\mathbb{Z}}(X^\vee)$. Note that the perfect pairing is $W_A$-invariant, meaning that $\langle r_i \lambda, r_i^\vee \mu \rangle = \langle \lambda, \mu \rangle$, and so the $W_A$ action on $X$ is faithful iff the action on $X^\vee$ is faithful.

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Question: What are necessary and sufficient conditions on the based root datum $R$ such that the abstract Weyl group acts faithfully on $X$ and $X^\vee$?

A sufficient condition is that both the sets $\{\alpha_i\} \subseteq X$ and $\{\alpha_i^\vee\} \subseteq X^\vee$ are linearly independent. Then one may do some chamber geometry on $X \otimes \mathbb{R}$ to show that $\langle r_i \mid i \in I \rangle$ is an abstract Coxeter group, as is done in Kac' book Infinite dimensional Lie algebras. However, this condition is not necessary, as one can see by taking the affine $\mathfrak{sl}_2$ Cartan matrix $$\begin{pmatrix} 2 & -2 \\ -2 & 2 \end{pmatrix}$$ and taking a root datum where $X \cong \mathbb{Z}^2 \cong X^\vee$ and let $\alpha_1^\vee, \alpha_2^\vee$ be a basis of $X^\vee$, forcing $\alpha_1 + \alpha_2 = 0$ to be linearly dependent roots. It is not hard to see that $r_1$ and $r_2$ then have no relation, so the action here is faithful.

On the other hand, we could also take a rank-one realisation of that same Cartan matrix by choosing any $\alpha_1 \in \mathbb{Z}$ and $\alpha_1^\vee \in \mathbb{Z}$ satisfying $\langle \alpha_1, \alpha_1^\vee \rangle = 2$ and then setting $\alpha_2 = -\alpha_1$ and $\alpha_2^\vee = - \alpha_1^\vee$. In this case, both $r_1$ and $r_2$ are just multiplication by $-1$, and so we have the relation $r_1 r_2 = 1$.

Answer 1

The progress I made on this shortly afterwards is that a sufficient condition is for either the simple roots or coroots to be linearly independent. Assuming the simple roots are linearly independent, one finds during the proof of (rirj)mij=1 that mij is the smallest integer for which this holds, meaning that the Coxeter matrix coincides with the matrix [ord(rirj)]ij. Together with the fact that the ri satisfy the exchange condition (3.11 in Kac), the ri define the same Coxeter group as the si. – Joppy