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](https://books.google.co.il/books/about/Reflection_Groups_and_Coxeter_Groups.html?id=ODfjmOeNLMUC&redir_esc=y): $\Phi = \{w(\alpha\_s) : w \in W, s \in S\}$. 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.



What is the Cartan matrix for a dihedral group?
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.


