Lemma: Metric data and positivity

The finite lengths are those of Table tab:lengths. Every active finite edge has positive length. The compact face areas are

Every active compact face has positive area.

Proof. For oblique spatial displacement $(\Delta \xi ,\Delta \zeta )$ and time displacement $\Delta T$,

Substitution gives the first four rows and $P{B}_i$. For $P{C}_i$, the spatial displacement is $(\chi s,\chi s)$ and the time displacement is $b\chi s$. For $B_i{C}_i$, these are $((\chi -\beta )s,\chi s)$ and $b\chi s$; the reflected edge has the same length.

Every active spatial edge is nonzero and tangent to a spacelike face, so its induced length is positive. In particular the displayed radicals are positive. Also

The spatial area element is $S_{\alpha }d\xi d\zeta$. Its graph multiplier is $\sqrt{1-|g{|}^2}$: $1/3$ on $V_i$, $4/5$ on ${U_i}^{\pm }$, and $1$ on $Q_i$. The incoming projected triangle has area $S_{\alpha }{r}^2/2$. Each projected inner triangle has area $S_{\alpha }\beta \chi s^2/2$. Subtracting both inner triangles gives the outgoing core area. The collar is the difference of projected triangles of radial sizes $4$ and $r$.

Since $b<a$,

Therefore, for $0<s\le 1$,

Finally $S_{\alpha }>0$ and $r<4$, proving all area assertions.