Manuscript VII-F@1 · VII-F.tex · lemma · 2d81c03f

Lemma: Exact branch decomposition

Cellulation and the isolated particle event › Vertices and faces · lines 637–677 · proof included or linked

VII-F@1#lem:decomposition

The faces just listed form the graph of $F_{s}$ without overlap of face interiors or omitted regions.

Proof. For $s \leq 0$ , Lemma lem:active-region gives the asserted three core triangles and resting exterior.

For $s > 0$ , in the half-sector $ξ \geq η$ ,

h - k = \frac{2}{3} ξ + \frac{1}{15} η .

Hence the branch boundary is

10 ξ + η = 15 s .

It joins

B_{i} = (\frac{3}{2} s, 0) to C_{i} = (\frac{15}{11} s, \frac{15}{11} s) .

In the other half-sector the boundary is

ξ + 10 η = 15 s,

joining $C_{i}$ to $B_{i + 1}$ .

The region where $s + k$ dominates is precisely the union of the two triangles $[P, B_{i}, C_{i}]$ and $[P, C_{i}, B_{i + 1}]$ . The remainder of the sector triangle $ρ \leq 3$ is the stated pentagon $V_{i}$ . Since

ρ (C_{i}) = \frac{30 s}{11} \leq \frac{30}{11} < 3,

the inner boundary is strictly separated from the outer crest. The cap is active exactly in the prescribed exterior.

Displayed formulas in this passage

unlabeled · line 648
$h - k = \frac{2}{3} ξ + \frac{1}{15} η .$
unlabeled · line 652
$10 ξ + η = 15 s .$
unlabeled · line 656
$B_{i} = (\frac{3}{2} s, 0) to C_{i} = (\frac{15}{11} s, \frac{15}{11} s) .$
unlabeled · line 662
$ξ + 10 η = 15 s,$
unlabeled · line 672
$ρ (C_{i}) = \frac{30 s}{11} \leq \frac{30}{11} < 3,$

Nearby in this section

section: Vertices and faces