= Solution
For the usual diagonal <Cartan subalgebra>, write $\epsilon_i(H)=H_{ii}$. The roots are $\epsilon_i-\epsilon_j$ for $i\ne j$. The three diagonal entries of the specified $H_0$ have ordering
$$
1>\sqrt2-1>-\sqrt2.
$$
Therefore the <positive roots> selected by $\ell(\mu)=\mu(H_0)$ are
$$
\boxed{\Phi^+=\{\epsilon_2-\epsilon_3,\ \epsilon_3-\epsilon_1,\ \epsilon_2-\epsilon_1\}.}
$$
Their positive evaluations are respectively $2-\sqrt2$, $2\sqrt2-1$ and $1+\sqrt2$. For this nonstandard ordering the simple roots are $\alpha_1=\epsilon_2-\epsilon_3$ and $\alpha_2=\epsilon_3-\epsilon_1$; 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> is
$$
\boxed{C=\{u_1+u_2+u_3=0:\ u_2>u_3>u_1\}.}
$$
Its closure replaces the inequalities by weak inequalities. The fundamental weights for this ordering are $\omega_1=\epsilon_2$ and $\omega_2=-\epsilon_1$, 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 are
$$
\boxed{\overline C=\mathbb R_{\ge0}\epsilon_2+\mathbb R_{\ge0}(-\epsilon_1).}
$$
The sketch uses $\epsilon_1=(1,0)$, $\epsilon_2=(-1/2,\sqrt3/2)$ and $\epsilon_3=(-1/2,-\sqrt3/2)$ 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-1-weyl-chamber.png]
{title=Positive A2 roots and fundamental Weyl chamber for the specified ordering}
Back to article page
