Admissible normal vectors have rigid averages of tetrahedral edge counts

Fix one tetrahedron and put $\delta =\alpha u+\beta v-(\alpha +\beta )d$ on its seven coordinates. Write its triangle coordinates as $a,b,c,e$ at $A,B,C,D$, and its quadrilateral coordinates as $x,y,z$ at $q_1,q_2,q_3$. The hypothesis $B\delta =0$ and [F1] give the six equations $\begin{array}{rlrl}{\displaystyle a+b+y+z}& {\displaystyle =0,}& {\displaystyle c+e+y+z}& {\displaystyle =0,}\\ {\displaystyle a+c+x+z}& {\displaystyle =0,}& {\displaystyle b+e+x+z}& {\displaystyle =0,}\\ {\displaystyle a+e+x+y}& {\displaystyle =0,}& {\displaystyle b+c+x+y}& {\displaystyle =0.}\end{array}$ They are listed in opposite-edge pairs.

Subtracting the members of those pairs gives $a+b=c+e$, $a+c=b+e$ and $a+e=b+c$. The first two imply $b=c$ and $a=e$; substituting in the third gives $a=b$. Thus all four triangle differences equal one number $t$. The first equation in each pair now gives $y+z=x+z=x+y=-2t$, so $x=y=z=-t$. Hence the kernel of the six local edge equations is exactly the line generated by $(1,1,1,1,-1,-1,-1)$.

Put $\gamma =-t$. For each of the three quad types, $\alpha u_{q_i}+\beta v_{q_i}=(\alpha +\beta )d_{q_i}+\gamma \mathrm{.}$ If $\gamma <0$, choose a quad coordinate with $d_{q_i}=0$, which exists by [F2]. Its left side is nonnegative, whereas its right side is negative, a contradiction. If $\gamma >0$, all three left sides are strictly positive. Positivity of $\alpha ,\beta$ would then require the union of the positive quad supports of $u$ and $v$ to contain all three types. Each support has size at most one by [F2], so that union has size at most two. This is again a contradiction. Therefore $\gamma =0$.

It follows that $t=0$ and every coordinate of $\delta$ is zero. Apply the same argument independently in each tetrahedron to obtain $\alpha u+\beta v=(\alpha +\beta )d$ on the whole array. No matching equation was used.

For an actual normal surface, its local count on any tetrahedral edge is the number of its intersection points with that global edge: the normal disks in each incident tetrahedron meet exactly those points. Hence an equality of global edge counts gives the six local equalities in every tetrahedron, and step 4.1 proves the surface-vector assertion.

If $u\mathrm{\ne }d$, let $j$ be the first coordinate where they differ. The averaging identity gives $\beta (v-d)=-\alpha (u-d)$. Therefore $v$ also agrees with $d$ before $j$, and the differences at $j$ have opposite nonzero signs. One of $u,v$ is consequently lexicographically smaller than $d$, contradicting the assumed minimality. Thus $u=d$, and the same identity then gives $v=d$. Applying this to the set of essential normal disk vectors at the least edge weight proves the final assertion.