What is the Coxeter plane? I'm having trouble in understanding what the Coxeter plane is and how to get graph like this one. Suppose I have a Coxeter group with presentation
Suppose I have The Coxeter group  $$H_{4}=\left\langle R_{1},R_{2},R_{3},R_{4};\left(R_{i}R_{j}\right)^{m_{ij}}=1\right\rangle, $$ where the $m_{ij}$ are given by the Coxeter matrix $$\left(\begin{array}{cccc}
1 & 3 & 2 & 2\\
3 & 1 & 3 & 2\\
2 & 3 & 1 & 5\\
2 & 2 & 5 & 1
\end{array}\right)$$
Then how do I get the Coxeter plane? Should I consider the invariant plane with respect to the element $R_{1}R_{2}R_{3}R_{4}$? Why should always be the same?
 A: You need to first have an explicit description of a faithful representation of the Coxeter group before you can find the invariant plane which will be a subspace of this representation.
Start with a vector space $V=\mathbb{R}^S$ where $S$ is your set of Coxeter generators, and define a symmetric bilinear form on $V$ given by $$B(e_i,e_j)=-\cos\frac{\pi}{m_{ij}}.$$ In the example $H_4$ is finite, so this will be positive definite. Define a representation $\sigma\colon W\to O(V,B)\colon w\mapsto \sigma_w$ of your Coxeter group $W$ into the group of transformations of $V$ which are orthogonal with respect to $B$ by setting $$\sigma_{R_i}\colon V\to V\colon v\mapsto v-2B(v,e_i)e_i.$$
In the case of $H_4$ this yeilds
$$ \sigma_{R_1}=\begin{pmatrix}-1&1&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix},\;
\sigma_{R_2}=\begin{pmatrix}1&0&0&0\\1&-1&1&0\\0&0&1&0\\0&0&0&1\end{pmatrix},$$ $$\sigma_{R_3}=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&1&-1&(1+\sqrt{5})/2\\0&0&0&1\end{pmatrix},\;\textrm{and}\;
\sigma_{R_4}=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&(1+\sqrt{5})/2&-1\end{pmatrix}.$$
Hence the Coxeter element acts by the matrix $$\sigma_{R_1R_2R_3R_4}=\sigma_{R_1}\sigma_{R_2}\sigma_{R_3}\sigma_{R_4}=\begin{pmatrix}-1&1&0&0\\-1&0&1&0\\-1&0&0&(1+\sqrt{5})/2\\-(1+\sqrt{5})/2&0&0&(1+\sqrt{5})/2\end{pmatrix}.$$
Now compute the eigenvectors/values of this matrix in the normal way, and fine the plane which is fixed and on which it acts with eigenvalues $e^{2\pi i/h}$ and $e^{-2\pi i/h}$ where $h$ is the Coxeter number, which for $H_4$ is 30. Remember that if you are computing orthogonal complements of vectors, this orthogonality needs to be with respect to $B$.
Where did you read that the Coxeter plane is independent of the choice of Coxeter element? I don't this this is true.
