A3 doubled adjacency graph

Three distinct lines through the origin divide the plane into six sectors. Each sector shares a wall with precisely the preceding and following sectors in cyclic order. By [F1] these are $C_0,\dots ,C_5$, and by [F2] each shared wall supplies exactly the two indicated directed arrows. There are no other arrows.

Put $H=\{(t_1,t_2,t_3,t_4)\in {\mathbb{R}}^4:t_1+t_2+t_3+t_4=0\}$ and let ${\epsilon }_i$ be the coordinate vectors of ${\mathbb{R}}^4$. The map $T$ sending ${\alpha }_i$ to ${\epsilon }_i-{\epsilon }_{i+1}$ is an isomorphism onto $H$, since these three differences form a basis. In type $A_3$, [F5] gives $r_i({\alpha }_i)=-{\alpha }_i$, $r_i({\alpha }_j)={\alpha }_j+{\alpha }_i$ for $\mathrm{\mid }i-j\mathrm{\mid }=1$, and $r_i({\alpha }_j)={\alpha }_j$ for $\mathrm{\mid }i-j\mathrm{\mid }>1$. Hence $T{r}_i{T}^{-1}$ is the restriction of the coordinate transposition $(ii+1)$. The action of coordinate permutations on $H$ is faithful: fixing every difference ${\epsilon }_p-{\epsilon }_q$ forces every index to be fixed. Adjacent transpositions generate every permutation, because swapping an adjacent inverted pair reduces the number of inversions until the entries are sorted. Thus [F6] identifies $W$ with the symmetric group $S_4$, preserving all three simple generators. Products of permutations act from right to left. This derives the model locally without invoking internal proof steps of the arrangement supplier.

An inversion of a permutation $p$ is a pair $r<t$ with $p(r)>p(t)$. Right multiplication by an adjacent transposition swaps adjacent entries, changing the inversion count by exactly one. Starting at the identity gives a lower bound on every word length, and sorting adjacent inverted pairs gives a word of exactly that length. By [F7], Coxeter length therefore equals inversion count under step 1.2. For $i=1,2$, the parabolic generated by $s_i,s_{i+1}$ consists of all permutations of positions $i,i+1,i+2$, fixing the remaining position. It is finite; its unique permutation of greatest inversion count reverses those three positions, so $w_{\{i,i+1\}}=s_i{s}_{i+1}{s}_i=s_{i+1}{s}_i{s}_{i+1}$. On the root differences this reversal sends ${\alpha }_i$ to $-{\alpha }_{i+1}$ and ${\alpha }_{i+1}$ to $-{\alpha }_i$, so opposition interchanges the two vertices. The disconnected parabolic on $\{1,3\}$ consists of $e,s_1,s_3,s_1{s}_3$; its longest element is $s_1{s}_3$ and opposition fixes both vertices. A singleton parabolic has longest element its sole generator and fixes its vertex under opposition. Consequently [F3] gives crossing elements $s_1,s_2{s}_3,s_1{s}_2,s_3,s_2{s}_1,s_3{s}_2$ for $(a,I)=(1,3),(2,3),(1,2),(3,1),(2,1),(3,2)$ respectively. For connected pairs $D=I\cup \{a\}$; for the disconnected pair $w_I{w}_{I\cup \{a\}}=s_a=w_{D\setminus \{a\}}{w}_D$. Thus in all six cases the component formula [F3] agrees with the full-parabolic formula in [F4], whose finiteness hypothesis holds.

Represent the first components of $C_0,\dots ,C_5$ in one-line permutation notation as $1234,2134,2341,3421,3412,1342$. Right multiplication by the six crossing elements in step 2.1 takes each entry to the next, cyclically. The opposition rule in [F3] gives the corresponding second components $3,2,1,1,2,3$. Thus the six forward labels are exactly those displayed. All relevant parabolics are finite, so [F4] applies to each crossing.

For the connected pairs, opposition exchanges the two indices, so [F4] gives reverse labels ${\omega }_{3,2}$ after ${\omega }_{2,3}$, ${\omega }_{2,1}$ after ${\omega }_{1,2}$, ${\omega }_{1,2}$ after ${\omega }_{2,1}$, and ${\omega }_{2,3}$ after ${\omega }_{3,2}$. For the disconnected pair $\{1,3\}$, opposition fixes the crossing index, so the reverse labels remain ${\omega }_{1,3}$ and ${\omega }_{3,1}$ respectively. These are precisely the displayed reverse labels, completing the verification.