D4 grading, region poset and simple-morphism divisibility

For each of the seven hyperplanes, two chambers lie on opposite sides precisely when exactly one of them lies opposite $C$. Thus the hyperplanes separating $A$ and $B$ form $S(A)\mathrm{△}S(B)$, and [F2] gives $d(A,B)=\mathrm{\mid }S(A)\mathrm{△}S(B)\mathrm{\mid }$. In particular, $S(A)=S(B)$ implies $d(A,B)=0$ and hence $A=B$. Set inclusion is reflexive, antisymmetric and transitive, so the displayed relation is a partial order in the sense of Partial order and partially ordered set. Also $d(C,A)=\mathrm{\mid }S(A)\mathrm{\mid }$.

For a label $(x,I)$, [F2] supplies a standard expression for the morphism $x:\{2\}\to I$. Every such expression lifts from $(e,2)$ to a geodesic ending at $(x,I)$, with one crossing per factor. Hence every expression has exactly $d(C,(x,I))$ factors. This proves the existence and independence required for the notation in [F5] and gives ${\mathrm{\ell }}_{\mathcal{B}\mathcal{H}}(x)=d(C,(x,I))=\mathrm{\mid }S((x,I))\mathrm{\mid }$. At $x=e$ the empty expression has zero factors, agreeing with the distance.

By [F3], for each chamber $A$ there is exactly one simple arrow $u(C,A)$, the class of its minimal galleries. It left-divides $u(C,B)$ precisely when a minimal gallery from $C$ to $B$ can pass through $A$. Indeed all relations preserve path length, so a factorisation between these minimal arrows has additive lengths; conversely such a gallery gives a factorisation. By [F2], length additivity here is $\mathrm{\mid }S(A)\mathrm{\mid }+\mathrm{\mid }S(A)\mathrm{△}S(B)\mathrm{\mid }=\mathrm{\mid }S(B)\mathrm{\mid }$, which is equivalent to $S(A)\subseteq S(B)$. Thus the simple-arrow divisibility order is exactly the region order.

If $S(A)\subseteq S(B)$, a shortest gallery from $A$ to $B$ crosses exactly the hyperplanes in $S(B)\setminus S(A)$, so its intermediate separation sets increase one at a time. Thus a cover in the region order is exactly an adjacent pair whose ranks differ by one. Every adjacent pair differs in one sign, so it is such a cover in one direction. The empty separation set gives the minimum $C$, vertex 7. By [F4], negation preserves the arrangement and sends $C$ to a chamber $-C$ lying across every hyperplane from $C$. Thus $S(-C)$ is the full seven-element set, so $-C$ is the maximum and has rank seven. This proves the grading and the geometric description of both extrema; the graph calculation below identifies the label of $-C$.

Use the first table to define the bijection from Ko's 32 vertices to the 32 chamber vertices. For each row of the second table, substitution gives adjacency pairs from [F1]; conversely their numbers are $3,6,12,6,12,6,3$, totaling 48, with distinct endpoints and no repetitions. These exhaust [F1]'s 48 pairs. Thus both vertices and edges correspond. Every listed edge joins consecutive rows, and every vertex in a positive-numbered row has a neighbour in the preceding row. Consequently a vertex in row $r$ has a path of length $r$ from vertex 7, and no shorter path can reach its row. Its graph distance and rank are therefore $r$. Row 7 consists only of vertex 26, so step 3.1 identifies this vertex as $-C$, with label $(stuvtsutvtu,2)$ from [F1]. This is a calculation from the complete adjacency graph, requiring no imported vertex sign strings. Ko places row 0 at the top, whereas the article's region diagram places $C$ at the bottom; this is the asserted opposite vertical orientation. No reversal of the definition of set inclusion is needed.

The second label of every chamber in the first table gives exactly the stated transported colour by [F1]. Ko's printed numerals instead count reduced atomic expressions. Starting with 1 at row 0 and summing along incoming edges gives successive rows $1$; $1,1,1$; six $1$'s; six $2$'s; six $2$'s; six $4$'s; three $8$'s; $24$, agreeing with his numerical annotations. The comparison does not identify these numerals with chamber labels or orbit colours.

The coloured graph with its specified base determines the chamber labels uniquely. To make the uniqueness explicit, graph distance from the base fixes every row. Row 1 has three different colours, so its vertices are fixed. In row 3, the three red vertices have rank-2 neighbours of colour pairs $\{B,G\}$, $\{B,U\}$ and $\{G,U\}$, respectively, so they are individually distinguished. Each rank-2 vertex is then distinguished by its rank-1 neighbour and its red rank-3 neighbour. Each remaining rank-3 vertex is distinguished by its two rank-2 neighbours. The row-3 to row-4 edges are a perfect matching, fixing row 4; every row-5 vertex has a distinct pair of row-4 neighbours, and every row-6 vertex a distinct pair of row-5 neighbours. The final vertex is unique. Hence a colour-preserving automorphism fixing the base fixes every vertex. The completed labelled graph [F1] is one reconstruction, so it is the unique one on this coloured graph; its labelled crossings recover the displayed chamber labels.