I want to draw a picture that projects the E8 root system to its coxeter plane.
The coxeter plane is defined as follows: the coxeter element of the weyl group of E8 has a simple eigenvalue $e^{2\pi i/h}$ where $h=30$. The real and imag part of the eigenvector of this eigenvalue span a 2d plane $P$ in $\mathbb{R}^8$. To project the root system to $P$, just find a pair of orthogonal unit vectors $u,v$ in $P$, and for each root $x$, compute the inner product $(u,x)$ and $(v,x)$, this is the 2d point we want.
My problem is: how to find such a basis? Of course one can write down explicitly the matrices of the simple roots, multiply them in any order, this is the matrix of the coxeter element, then compute the eigenvectors, but this way is too tedious.
I learned from http://www.math.ubc.ca/~cass/research/pdf/Element.pdf that the 2D plane $P$ can also spanned by two eigenvectors $u',v'$ of the cartan matrix $C$. Here one computes the max and min eigenvaluesof $C$,and $u',v'$ are the eigenvectors of them.
But when I followed this way, the figure produced is symmetric but doesn't display the 8x30 ring pattern. I wonder whether there is something wrong with my work.
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The cartan matrix $C$:(without a factor 2)
$$\begin{pmatrix} 1. & -0.5 & 0. & 0. & 0. & 0. & 0. & 0. \\ -0.5 & 1. &-0.5& 0. & 0. & 0. & 0. & 0. \\ 0. & -0.5 & 1. & -0.5 & 0.& 0.& 0. & 0. \\ 0. & 0. &-0.5 &1. & -0.5 & 0. & 0. & 0. \\ 0. & 0. & 0. &-0.5 & 1. & -0.5 & 0. & -0.5\\ 0. & 0. & 0. & 0. & -0.5 & 1. & -0.5 & 0. \\ 0. & 0. & 0. & 0. & 0. & -0.5 &1. & 0. \\ 0. & 0. & 0. & 0. & -0.5 &0.& 0. & 1. \end{pmatrix}$$