Least-weight lexicographic essential normal disks are vertex disks

Suppose $d$ were nonvertex and take the reduced diagram [F1]. By [F2], its two summands $X,Y$ are disks and its double curves are proper arcs. There is at least one: with none, each summand would survive as one full-resolution disk and have vector $d$. Each original disk meets the other, so its cut tree has at least two leaves. Every such original leaf patch has positive weight by [F5]. Let $m>0$ be the least weight among all these leaves. There are at least four leaves altogether, and their weights are part of the total $2w(D)$; hence $w(D)\ge 2m$. Neither resolved disk is untouched by all traces. An untouched full-resolution component would also be an unchanged connected original summand with vector $d$, contrary to nonproportionality.

For any minimum leaf $E$, let $\alpha$ be its sole original double arc. Its retained trace cuts off that same leaf on its resolved disk. If the other frontier trace lies on this disk, $E$ is an outer cap: it contains no other trace and therefore cannot contain the other frontier. Take the disjoint adjacent outer cap $B(E)$ from [F3]. If the other trace lies on the other resolved disk, take its unique cap adjacent to $E$, again by [F3]. In both cases $B(E)$ is disjoint from $E$ and $N(B(E))=N(E)$, so $w(B(E))=m$. Cut $B(E)$ along all traces in its interior, rooting the dual tree at its exterior trace side. Every non-root leaf is an original leaf disk patch and has weight at least $m$. If the tree has one vertex, $B(E)$ is itself a leaf. Otherwise nonnegativity and total weight $m$ force exactly one non-root leaf $L(E)$, of weight $m$, and all other regions have zero edge count. The tree is a path, since branching would yield two non-root leaves. Thus either $B(E)=L(E)$ or $R(E)=\overline{B(E)\setminus L(E)}$ is a nonempty zero-edge rectangle with two trace sides and two original-boundary sides. In either case $N(L(E))=N(E)$. These are actual cut patches on the original resolved disks; no three-dimensional cap parallelism is asserted.

The equality $B(E)=L(E)$ is impossible for any minimum leaf. In this case both adjacent opposite caps are original leaf disk patches at the same original arc. In a transverse rectangle label upper left and right rays $0,1$, and lower left and right rays $2,3$. Full regular pairs are $(0,2),(1,3)$ and original unswitched pairs are $(0,3),(1,2)$. Adjacency makes the leaf rays $0,1$, so they belong to different original families. Regularly switching just this arc swaps those two leaf disks between $X,Y$. Removing a boundary end disk from a disk leaves a disk; attaching the other end disk along the vacated interval again gives a disk. The swapped patches contain no other double curve, so every remaining curve still joins the unchanged parts of different families. The two new families are therefore embedded normal disks, and all their later partial resolutions are old partial resolutions with this arc already switched. The two swapped leaves have equal edge counts by [F3]. Thus each new disk has exactly its own original family's edge-count vector, and [F4] shows that its normal vector is unchanged. Both new vectors remain nonproportional to $d$. This gives an eligible diagram with one fewer double curve, contradicting reducedness.

First suppose $w(D)>2m$. The finite function $E\mapsto L(E)$ on the minimum leaves has no fixed point because $E$ is disjoint from $B(E)$. Iteration gives a cycle of distinct leaves $E_1,\dots ,E_r$, $r\ge 2$, with $E_{i+1}=L(E_i)$; write $B_i=B(E_i)$ and ${\alpha }_i$ for its exchange arc. The caps $B_i$ are pairwise disjoint. On different resolved disks this is automatic. On one disk nesting would put two distinct minimum leaves in one opposite cap, contrary to uniqueness in step 2.1; covering would give $w(D)\le 2m$, contrary to this case. These exhaust overlaps by [F6]. The ${\alpha }_i$ are distinct. At one original arc, two minimum leaf caps on the same frontier but opposite sides would cover its resolved disk and likewise contradict $w(D)>2m$. Minimum leaf caps on opposite frontiers must be adjacent: the nonadjacent alternative is the complement of the adjacent weight-$m$ cap, of weight $w(D)-m>m$. If two adjacent opposite caps at this arc were leaves, each would be the other's entire opposite cap, excluded by step 3.1. Thus no arc can occur twice in the cycle.

Each $R_i=\overline{B_i\setminus E_{i+1}}$ is a nonempty zero-edge rectangle by steps 2.1 and 3.1. Their interiors are pairwise disjoint by step 4.1. Leave precisely the cycle arcs ${\alpha }_i$ unswitched and retain the regular resolution elsewhere. Check the original diagonals with the four-ray labeling of step 3.1. At ${\alpha }_i$, $E_i$ occupies upper left ray $0$; the complementary lower left ray $2$ belongs to $R_{i-1}$. The adjacent cap $B_i$ occupies upper right ray $1$, belonging near that frontier to $R_i$; its complementary ray $3$ is outside all the $B_j$. Hence the original diagonals $(0,3),(1,2)$ join $E_i$ to the outside remainder and join $R_i$ to $R_{i-1}$. No irregular switch is used. At every other trace the regular continuation has either both incident parts among the $R_i$ or neither, since the only frontier traces of their union are the selected ones. Thus these nonempty rectangles with their short continuations form a union of whole normalization components, with any remaining boundary lying in $\mathrm{\partial }M$. This union misses all original edges. It contradicts the positive-weight property of actual partial normal resolutions [F5]. The case $w(D)>2m$ is impossible.

It remains that $w(D)=2m$. There are exactly four original leaf patches, each of weight $m$, and every other original patch has zero edge counts: each original disk has at least two leaves and their total weight is already $4m=2w(D)$. Each original disk's dual tree has exactly two leaves and is therefore a path. The resolved disks' cut patches are the same original patches, with the same trace-side counts. Each resolved disk has traces by step 1.1, so each has at least two leaves. Thus each has exactly two of these four leaf patches, all its other patches have zero edge counts, and its dual tree is also a path. A cap at any trace on such a resolved disk consists of its one end leaf and some zero-edge interior patches. Its count vector is precisely that end leaf's count vector.

There must be an exchange between the two resolved disks. Otherwise every original inverse continuation preserves its resolved-disk label. Each connected original component would then use patches from only one $D_i$. There are exactly two original components and both $D_i$ supply nonempty patches, so each original component uses all patches of one $D_i$. Its edge counts are $N(D)$, and [F4] makes its vector $d$, a contradiction. Now any same-copy exchange has disjoint outer caps by [F3]; on that resolved path these contain its two different end leaves. Their equal cap counts make those two leaf count vectors equal. A between-copy exchange equates an adjacent cap pair and its complementary pair by [F3], so propagates this equality to both leaves of the other resolved disk. Then all four leaves have one count vector $a$, and both original disks and both resolved disks have count vector $2a$ by step 6.1. Again [F4] contradicts original nonproportionality. Consequently every exchange in this remaining case joins different resolved disks.

Choose one such exchange. Label the two end leaves of each resolved path $A,B$ so the adjacent caps at this exchange contain the respective $A$ leaves. By [F3] their leaf counts agree, say $a$, and the complementary $B$ leaf counts also agree, say $b$. If $a=b$, all four counts agree and the preceding contradiction applies. Suppose $a\mathrm{\ne }b$. At every exchange, the cap on the $A$ side of its trace on $D_0$ must be adjacent to the $A$ cap on $D_1$: adjacency to its $B$ cap would force $a=b$. Orient the two finite resolved dual patch paths from their $A$ leaf to their $B$ leaf. Each interior patch then has one incoming frontier on its $A$ side and one outgoing frontier on its $B$ side. At an exchange, the original diagonal continuations pair an $A$-side patch germ from one resolved disk to a $B$-side germ from the other. These are exactly an outgoing and an incoming frontier; thus the directions paste on the ORIGINAL dual patch paths after all original continuations are restored. Each original path must start at an $A$ leaf and end at a $B$ leaf. This is only a direction on finite graphs, not an orientation assumption on the ambient manifold or its surfaces. Each original disk consequently has edge counts $a+b=N(D)$, since all its other patches have zero edge counts. By [F4] both its vectors equal $d$, the final contradiction.

Both possible values $w(D)>2m$ and $w(D)=2m$ have been excluded, so the assumed nonvertex decomposition cannot exist. Hence $d$ lies on an extreme ray of $C(T)$. Fundamentalness makes its coordinate gcd one: a larger common divisor would split it into two nonzero admissible integer vectors. Thus it is the primitive ray vector. For the consequence, normalize the supplied essential PL disk by Essential PL disks admit normal disk replacements. Choose a least normal disk weight in the resulting nonempty set, and then minimize the $7t$ coordinates successively; these are finitely many natural-number minima and require no choice principle. Apply the proved assertion to this disk. All switches, counts, matching coordinates and partial normal pieces belonged to the original triangulation throughout.