A3 relative arrangement and its six labelled chambers

Realize the root space as $V=\{(v_1,v_2,v_3,v_4)\in {\mathbb{R}}^4:\sum v_i=0\}$, with ${\alpha }_i={\epsilon }_i-{\epsilon }_{i+1}$, where ${\epsilon }_i$ are the standard coordinate vectors. The linear map from the simple-root basis to these differences is an isomorphism. The reflection formula in [F1] sends $s_i$ to the swap of coordinates $i,i+1$; these swaps generate all permutations, and [F5] makes the resulting identification $W\cong S_4$ faithful. Consequently the roots are precisely ${\epsilon }_i-{\epsilon }_j$ for $i\mathrm{\ne }j$, and the positive roots are ${\alpha }_1$, ${\alpha }_2$, ${\alpha }_3$, ${\alpha }_1+{\alpha }_2$, ${\alpha }_2+{\alpha }_3$ and ${\alpha }_1+{\alpha }_2+{\alpha }_3$.

A dual vector may be represented by a tuple $(t_1,t_2,t_3,t_4)$, unique up to adding a common constant, with $f({\alpha }_i)=t_i-t_{i+1}$. The closed fundamental chamber is $t_1\ge t_2\ge t_3\ge t_4$. Sorting any tuple shows that its permutation translates cover all of $\mathrm{\Theta }$; hence the Tits cone is all $\mathrm{\Theta }$. The condition $f({\alpha }_3)=0$ gives ${\mathrm{\Theta }}_J=\mathbb{R}{\alpha }_1^{\ast }\oplus \mathbb{R}{\alpha }_2^{\ast }$, so [F2] gives $\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{e}(J)={\mathrm{\Theta }}_J$ and its interior in ${\mathrm{\Theta }}_J$ is the same space.

In one-line permutation notation, multiplying the displayed words from right to left gives, in the stated order, $1234$, $2134$, $2341$, $3421$, $3412$ and $1342$. In each row the entries at positions $i,i+1$, where $I=\{i\}$ is the displayed second label, are $3,4$. Therefore $x{s}_i{x}^{-1}=s_3$ and $x{\alpha }_i={\alpha }_3$ is positive. Since the only positive root of the rank-one subsystem is ${\alpha }_i$, [F4] shows that $x$ is minimal in $x{W}_I$, and $W_{J}x=x{W}_I$. Thus each pair is a valid label in [F3].

Write $f=a{\alpha }_1^{\ast }+b{\alpha }_2^{\ast }$ on ${\mathrm{\Theta }}_J$. The positive roots in step 1.1 restrict respectively to $a,b,0,a+b,b,a+b$. Thus ${\alpha }_3$ gives the whole relative space and is excluded, while the distinct proper traces are $a=0$, $b=0$ and $a+b=0$. These are respectively $\mathbb{R}{\alpha }_2^{\ast }$, $\mathbb{R}{\alpha }_1^{\ast }$ and $\mathbb{R}({\alpha }_1^{\ast }-{\alpha }_2^{\ast })$. Their complement has six sectors.

Normalize the tuple by $t_3=t_4=0$; then $(t_1,t_2,t_3,t_4)=(a+b,b,0,0)$. By [F1], $f\in x{C}_{\{i\}}$ exactly when $t_{x(1)}\ge t_{x(2)}\ge t_{x(3)}\ge t_{x(4)}$, with equality exactly at positions $i,i+1$. The six permutations in step 2.2 therefore give, respectively, the strict conditions $(a>0,b>0)$, $(a<0,a+b>0)$, $(b>0,a+b<0)$, $(a<0,b<0)$, $(a>0,a+b<0)$ and $(b<0,a+b>0)$. These are the six sectors of step 3.1 in counterclockwise order from the positive quadrant. Applying the labelling bijection [F3] establishes every displayed label and completes the verification.