What is the Cartan matrix for a dihedral group? Dihedral groups are Coxeter groups of type $I_m$, $m \geq 3$. The Coxeter matrix of $I_m$ is
\begin{align}
\left( \begin{matrix} 1 & m \\ m & 1 \end{matrix} \right).
\end{align}
When $m=3,4,6$, $I_m$ are Coxeter groups of types $A_2,B_2,G_2$ respectively. They have the Cartan matrices 
\begin{align}
\left( \begin{matrix} 2 & -1 \\ -1 & 2 \end{matrix} \right), 
\left( \begin{matrix} 2 & -2 \\ -1 & 2 \end{matrix} \right), 
\left( \begin{matrix} 2 & -3 \\ -1 & 2 \end{matrix} \right),
\end{align}
respectively.
What are the Cartan matrices for $I_m$ in general? 
There is a definition of a root system for any Coxeter group in the book: $\Phi = \{w(\alpha_s) : w \in W, s \in S\}$. The definition of the action of Coxeter group on the root system is defined by: $s_j (\alpha_i)=\alpha_i− \beta_j^{\vee}(\alpha_i) \beta_j$. How to compute $\beta_j^{\vee}(\alpha_i)$ in the case of Dihedral group? 
Thank you very much.
 A: This is a little awkward to answer, because $I_m$ isn't crystallographic for $m \not \in \{2,3,4,6 \}$, which makes it unclear how to normalize the lengths of the roots. Let $\alpha_1$ and $\alpha_2$ be the simple roots and $\alpha_1^{\vee}$ and $\alpha_2^{\vee}$ be the corresponding co-roots. Let $d_i$ be the positive real scalar $\alpha_i^{\vee}/\alpha_i$. Then the matrix of inner products $\alpha_i^{\vee}(\alpha_j)$ is
$$\begin{bmatrix}
2 & - 2 d_1 d_2^{-1} \cos(\pi/m) \\
-2 d_1^{-1} d_2 \cos(\pi/m) & 2 \\ 
\end{bmatrix}$$
In the crystallographic cases, one makes the unique (up to switching $d_1$ and $d_2$, and up to rescaling both simultaneously) choice of $d_i$ which makes this matrix have integer entries; this corresponds to making the lattice $\mathrm{Span}_{\mathbb{Z}}(\alpha_1, \alpha_2)$ be invariant under the reflection action. These are the matrices you listed above.
For $m \not \in \{ 2,3,4,6 \}$, no choice of $d_i$ has this property. When $m$ is odd, it is natural to take $d_1=d_2$, because we would like to have $w_0 \alpha_i = -\alpha_{3-i}$ where $w_0$ is the longest element. When $m$ is even, I know of no natural normalization.
