Canonical Representation of Finitely Presented Coxeter Groups

Definition 1.  A Coxeter group is a group with presentation

\langle r_1,...,r_n:(r_ir_j)^{m_{ij}}=1\rangle

where m_{ii}=1 and m_{ij}\geq 2 for i\neq j.  Also m_{ij}=\infty says that (r_ir_j)^n\neq 1 for any n.

To be clear, the above is technically finitely presented.  Although we could have Coxeter groups with an infinite number of generators.

Proposition 2.  Let G be a Coxeter group with generators r_1,...,r_n.  Then we have

  1. r_i=r_i^{-1};
  2. If m_{ij}=2 for all distinct i,j, then G is abelian;
  3. m_{ij}=m_{ji}.

Proof.  (i)  Trivial since m_{ii}=1.  (ii)  We have (r_ir_j)^2=1.  We also of course have that r_i^2=r_j^2=1.  Thus


(iii)  (r_ir_j)^{m_{ij}}=1 so


Thus m_{ji}\leq m_{ij}.  The other inequality is dually shown.

Let M be the matrix defined by M_{ij}=m_{ij}.  This is called the Coxeter matrix of G.

Let G be a Coxeter group with n generators and define an inner product on \mathbb{R}^n

\displaystyle\langle e_i,e_j\rangle_G=-\cos\left(\frac{\pi}{m_{ij}}\right).

Let a\in\mathbb{R}^n  such that \langle a,a\rangle_G\neq 0.  Then define

s_a(x)=x-2\langle a,x\rangle_Ga.

We then define the representation \rho:G\to L(\mathbb{R}^n) of the Coxeter group on \mathbb{R}^n by


We also assume 1 goes to 1.  Let us now show that the relations carry over:  (s_{e_i}s_{e_j})^{m_{ij}}=1.  First suppose i=j, then

\begin{array}{lcl}(s_{e_i}s_{e_i})^{m_{ii}}(x)&=&s_{e_i}s_{e_i}(x)\\&=&s_{e_i}(x-2\langle e_i,x\rangle_Ge_i)\\&=&x-2\langle e_i,x\rangle_Ge_i-2\langle e_i,x-2\langle e_i,x\rangle_Ge_i\rangle_G e_i\\&=&x-4\langle e_i,x\rangle_Ge_i+4\langle e_i,x\rangle_G\langle e_i,e_i\rangle_Ge_i\\&=&x.\end{array}

For i\neq j, note that s_{e_i}s_{e_j} is acting on the subspace \mathbb{R}e_i\oplus\mathbb{R}e_j.  One can show that s_{e_i}s_{e_j} has order m_{ij} on this subspace.

[1]  Abramenko, Peter and Kenneth Brown.  Buildings.  Graduate Texts in Mathematics.  Vol. 248.  Springer Science and Business Media.  2008.


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