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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.

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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.

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  • $\begingroup$ PS I prefer the notation $I_2^{(m)}$, so that the subscript is always the rank. $\endgroup$ Commented Mar 21, 2017 at 13:52
  • $\begingroup$ thank you very much. $d_i =\sqrt{ \alpha_i^{\vee}(\alpha_i) }$ (the length of $\alpha_i$)?. $\endgroup$ Commented Mar 21, 2017 at 14:07
  • $\begingroup$ $\alpha_i^{\vee}(\alpha_i)$ is always $2$ under the conventions I'm used to. $d_i$ is the ratio between the proportional vectors $\alpha_i$ and $\alpha_i^{\vee}$ Probably best to read some standard book on Coxeter groups, e.g. Humphrey's, rather than me try to get it right. $\endgroup$ Commented Mar 21, 2017 at 15:58

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