Seven relative D4 hyperplanes and thirty-two chambers

Let $\rho$ be the sum of the twelve positive roots. Adding their coordinates gives $\rho =6{\alpha }_1+10{\alpha }_2+6{\alpha }_3+6{\alpha }_4$. The reflection formula gives $s_i\rho =\rho -2{\alpha }_i$ for each $i$. The orbit $W\rho$ is finite: it consists of sums of twelve elements of the finite root set. For any $f\in \mathrm{\Theta }$, choose $w$ maximizing $f(w\rho )$. Since $w{s}_i\rho =w\rho -2w{\alpha }_i$, maximality gives $f(w{\alpha }_i)\ge 0$ for every $i$. Thus $w^{-1}f$ belongs to the nonnegative fundamental chamber, so $f$ belongs to the Tits cone. Consequently the Tits cone is $\mathrm{\Theta }$; its intersection with ${\mathrm{\Theta }}_J$ and the interior of that intersection in ${\mathrm{\Theta }}_J$ are both ${\mathrm{\Theta }}_J=H_{{\alpha }_2}$.

Put $x=f({\alpha }_1)$, $y=f({\alpha }_3)$ and $z=f({\alpha }_4)$ for $f\in {\mathrm{\Theta }}_J$. These are independent coordinates on ${\mathrm{\Theta }}_J$. The twelve positive roots restrict, in the order listed by the positive-root supplier, to $x,0,y,z,x,y,z,x+y,x+z,y+z,x+y+z,x+y+z$. Hence only ${\alpha }_2$ has zero restriction, and the positive relative roots are exactly the other eleven. The seven distinct nonzero restricted forms are $x,y,z,x+y,x+z,y+z,x+y+z$; no two are nonzero scalar multiples. Their kernels are therefore seven distinct relative hyperplanes, represented by the seven roots in the Example. In particular both ${\alpha }_1$ and ${\alpha }_1+{\alpha }_2$ restrict to $x$.

The coordinates in step 1.2 identify ${\mathrm{\Theta }}_J$ with ${\mathbb{R}}^3$ with its Euclidean metric $d_2$. The inequalities $d_{\mathrm{\infty }}\le d_2\le \sqrt{3}{d}_{\mathrm{\infty }}$ follow by comparing the largest squared coordinate difference with the sum of the three squared differences; hence the two metrics induce the same topology, the product topology in [F4]. For each of the seven forms $\mathrm{\ell }$ in step 1.2, the triangle inequality gives $\mathrm{\mid }\mathrm{\ell }(q)-\mathrm{\ell }(p)\mathrm{\mid }\le 3d_2(p,q)$. If $\mathrm{\ell }(p)\mathrm{\ne }0$, the ball of radius $\mathrm{\mid }\mathrm{\ell }(p)\mathrm{\mid }\mathrm{/}6$ about $p$ therefore has the same sign of $\mathrm{\ell }$ as $p$. Thus $\{\mathrm{\ell }>0\}$ and $\{\mathrm{\ell }<0\}$ are open. If a subset $A$ of the complement met both half-spaces, their traces on $A$ would be disjoint nonempty relatively open sets covering $A$, a separation. Every nonempty connected subset of the complement consequently has constant signs for all seven forms.

Each nonempty sign region is convex: a convex combination of two points satisfying the same seven strict linear inequalities satisfies those inequalities. It is therefore connected by [F4] and the topology identification in step 2.1. For a point $p$ in that region, step 2.1 places every connected subset of the complement containing $p$ inside the same region; the region itself is such a connected subset. The union defining the connected component of $p$ in [F3] is thus exactly that region. Hence the nonempty sign regions are exactly the chambers.

In the two coordinate orthants where $x,y,z$ have the same sign, each orthant is one chamber. In each of the other six orthants, after an overall sign change and permutation, write $(x,y,z)=(a,b,-c)$ with $a,b,c>0$. The only additional equations are $c=a$, $c=b$ and $c=a+b$. They give five connected sign regions: $c<\min (a,b)$; $a<c<b$; $b<c<a$; $\max (a,b)<c<a+b$; and $c>a+b$. Each is nonempty and convex, as its displayed inequalities are linear. Therefore the total is $2+6\cdot 5=32$.