Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 1 / 5 / v / Solution

For the usual diagonal Cartan subalgebra, write ${ϵ}_i(H)=H_{ii}$. The roots are ${ϵ}_i-{ϵ}_j$ for $i\ne j$. The three diagonal entries of the specified $H_0$ have orderingTherefore the positive roots selected by $\ell (\mu )=\mu (H_0)$ areTheir positive evaluations are respectively $2-\sqrt{2}$, $2\sqrt{2}-1$ and $1+\sqrt{2}$. For this nonstandard ordering the simple roots are ${\alpha }_1={ϵ}_2-{ϵ}_3$ and ${\alpha }_2={ϵ}_3-{ϵ}_1$; the third positive root is their sum. These simple roots differ from the usual choice used in Question 1.

Identify the real weight plane with triples $(u_1,u_2,u_3)$ of sum zero. Pairing with these simple coroots gives the differences $u_2-u_3$ and $u_3-u_1$. Consequently the fundamental chamber of a root system isIts closure replaces the inequalities by weak inequalities. The fundamental weights for this ordering are ${\omega }_1={ϵ}_2$ and ${\omega }_2=-{ϵ}_1$, represented by $(-1/3,2/3,-1/3)$ and $(-2/3,1/3,1/3)$. Their simple-coroot pairings are $(1,0)$ and $(0,1)$, so the two boundary rays areThe sketch uses ${ϵ}_1=(1,0)$, ${ϵ}_2=(-1/2,\sqrt{3}/2)$ and ${ϵ}_3=(-1/2,-\sqrt{3}/2)$ in a Euclidean plane. The shaded sector is the sixty-degree closed chamber; its interior is the open chamber. The ordering vector $H_0$, identified with its dual direction, lies inside it.