Proposition: Lengths, areas, and nondegeneracy

The table gives the exact induced lengths. Physical lengths are $\ell$ times the displayed values. Normalized transverse face areas are

\operatorname{Area}(V_i)
 &=
 \begin{cases}
 S_\vartheta r^2/6,&s\le0,\\
 (S_\vartheta/3)(r^2/2-\beta\chi s^2),&s>0,
 \end{cases}\nonumber\\
 \operatorname{Area}(U_i^\pm)
 &=\frac25S_\vartheta\beta\chi s^2,\qquad s>0,\nonumber\\
 \operatorname{Area}(Q_i)
 &=\frac12S_\vartheta(16-r^2).
 \label{eq:areas}

Every active finite edge and compact face has positive size. Physical transverse areas are ${\ell }^2$ times these values. Extrusion through a normalized axial interval of length $L_z$ multiplies the area by $\ell L_z$.

Proof. For displacement $(\Delta \xi ,\Delta \eta ,\Delta \tau )$,

Substitution gives every table entry. Each nonzero edge is tangent to a spacelike face, so its radical is positive. The persistent radial factor satisfies $r-\beta s=\beta (2-s)>0$.

The spatial area element is $S_{\vartheta }\mathrm{d}\xi \mathrm{d}\eta$. The graph area multipliers are respectively $1/3$, $4/5$, and $1$ for $V,U,Q$, by the gradient computation. Each projected inner triangle has area $S_{\vartheta }\beta \chi s^2/2$. Subtracting both from the projected core triangle proves (eq:areas).

Since $r=2\beta$ and $\chi <\beta$,

The collar is positive because $r<4$. The axial extrusion is orthogonal to all transverse tangent vectors and has unchanged unit spatial metric.