Extremal outer shells in regular hyperbolic tilings

Athanor lab (Aleksei Kudriashov with AI research agents)

1. Problem and status

Let (p,q\ge3) be finite integers, and let (T_{p,q}) be the regular tessellation of (\mathbb H^2) by regular hyperbolic (p)-gons, with (q) faces incident at every vertex. Thus ((p-2)(q-2)>4). Let (G_{p,q}) be its one-skeleton. Fix a root vertex (o), and define

[ B_R=B_R(o):={v\in V(G_{p,q}):d_G(o,v)\le R},\qquad S_R:=B_R\setminus B_{R-1}, ]

where (B_{-1}=\varnothing). A set (A\subseteq B_R) is legal when no three distinct points of (A) lie on a common hyperbolic geodesic in the ambient plane. Put

[ \alpha_{p,q}(R)=\max{|A|:A\subseteq B_R\text{ is legal}}. ]

The definition is deliberately ambient. A shortest path in (G_{p,q}) is not, in general, an ambient hyperbolic geodesic, and its use would define a different problem. In the Klein model of (\mathbb H^2), ambient geodesics are Euclidean chords; hence the predicate is ordinary projective collinearity of the corresponding Klein points Loustau. Strict polygon convexity here means convexity with a strictly positive turn at every vertex, or equivalently an interior angle strictly below (\pi) at every vertex. It does not mean a boundary without straight sides.

2. Main results

Write (m=q/2) when (q) is even.

Theorem A (edge-line upper bound). For every regular hyperbolic ({p,q}) tessellation with even (q), and every (R\ge1),

[ \alpha_{p,q}(R)\le |S_R|. \tag{2.1} ]

Theorem B (square shells). In the regular ({4,q}) tessellation with (q>4), the vertex set (S_R) is the set of vertices of a strictly convex hyperbolic polygon, for every (R\ge1). In particular it is legal. If, in addition, (q) is even, then

[ \alpha_{4,q}(R)=|S_R|, ]

and (S_R) is the unique maximum legal subset of (B_R), as proved in Section 6.

Theorem C (triangular shells). In the regular ({3,q}) tessellation with (q>6), the vertex set (S_R) is the set of vertices of a strictly convex hyperbolic polygon for every integer (R\ge1). Hence, for every even (q\ge8) and every integer (R\ge1),

[ \alpha_{3,q}(R)=|S_R|, ]

and (S_R) is the unique maximum legal subset of (B_R), as proved in Section 6.

At (R=0), (B_0=S_0={o}), so the maximum is the singleton; the polygon statements are deliberately stated only for (R\ge1).

The parity restriction in the equality statements is real, rather than a technical choice. When (q) is odd, the radius-one ball consists of the root and its (q) neighbours. Those (q+1) points are legal: the neighbours lie on a circle, so an ambient geodesic meets that circle in at most two points, and no pair of neighbours is opposite through the root. Since (|S_1|=q), the assertion (\alpha(B_R)=|S_R|) cannot hold for all (R).

3. A line cover from the tiling

This section proves Theorem A.

3.1 Straight edge-lines

Suppose (q) is even. At every tiling vertex, each incident edge has a unique opposite edge. Continuing an edge through the opposite edge at every encountered vertex gives a bi-infinite chain of tiling edges on one ambient hyperbolic geodesic. Call it a straight edge-line. There are exactly (m=q/2) distinct straight edge-lines through every vertex.

Let (\mathcal L_R) be the finite family of straight edge-lines meeting (B_{R-1}). For (L\in\mathcal L_R), write

[ e_L=|L\cap S_R|, ]

and, for (b\in S_R), write

[ d_b=|{L\in\mathcal L_R:b\in L}|. ]

Lemma 3.1. Every (L\in\mathcal L_R) satisfies (e_L\ge2).

Proof. Choose a vertex (u\in L\cap B_{R-1}), and follow each of the two infinite rays of (L) starting at (u). Since (B_R) is finite, each ray first leaves (B_R). The vertex immediately preceding that first outside vertex has graph distance exactly (R): graph distance from (o) changes by at most one along an edge. The two predecessors are distinct, one on each ray, and belong to (L\cap S_R). (\square)

The lemma requires neither that a line meet a graph ball in an interval nor that a graph ball be geometrically convex.

3.2 Cover identity

Let (A\subseteq B_R) be legal, and let (x_v) be its indicator. Since a straight edge-line is an ambient geodesic,

[ \sum_{v\in L\cap B_R}x_v\le2 \qquad (L\in\mathcal L_R). \tag{3.1} ]

Sum (3.1) over (\mathcal L_R), and for each (b\in S_R) add (m-d_b) copies of (x_b\le1). Every vertex of (B_{R-1}) has total coefficient (m), and every boundary vertex has total coefficient (d_b+(m-d_b)=m). Hence

[ m|A|\le 2|\mathcal L_R|+\sum_{b\in S_R}(m-d_b). \tag{3.2} ]

Double counting boundary incidences gives

[ \begin{aligned} 2|\mathcal L_R|+\sum_{b\in S_R}(m-d_b) &=2|\mathcal L_R|+m|S_R|-\sum_{L\in\mathcal L_R}e_L\ &=m|S_R|-\sum_{L\in\mathcal L_R}(e_L-2)\ &\le m|S_R|, \end{aligned} \tag{3.3} ]

where the last inequality is Lemma 3.1. This proves (2.1).

The calculation also records the exact slack identity

[ \begin{aligned} m(|S_R|-|A|)={}& \sum_{L\in\mathcal L_R}\bigl(2-|A\cap L|\bigr)\ &+\sum_{b\in S_R}(m-d_b)(1-x_b) +\sum_{L\in\mathcal L_R}(e_L-2). \tag{3.4} \end{aligned} ]

All three terms are nonnegative. This is stronger than the numerical upper bound: if one straight edge-line has at least three vertices on (S_R), then no legal set can have (|S_R|) points. Conversely, if (S_R) is legal, then necessarily every (L\in\mathcal L_R) has exactly two shell vertices.

The cover uses only a subset of ambient geodesics. This is sufficient for an upper bound in the full ambient problem, but it does not identify all rich ambient lines and it should not be confused with a graph-geodesic argument.

4. Square tilings: a radial contour

We now give the proof of Theorem B. It uses an auxiliary CAT(0) square-complex metric only to organize graph distance. The final polygon is drawn in the given hyperbolic tiling, and its sides are ambient geodesic segments.

4.1 Parent structure

Replace each face of the ({4,q}) tiling by a unit Euclidean square, without changing its cell structure. The resulting square complex is simply connected, and the link at each vertex is the cycle (C_q). For (q\ge4), the link condition makes it a CAT(0) cube complex. Equivalently, its one-skeleton is a median graph; see Chepoi, Section 4, p. 133, and Theorem 6.1 for the quadrangle condition and the cubing/median correspondence.

The graph is bipartite: every simple cycle bounds a finite disk of square faces, and counting their boundary edges modulo two makes the cycle length even. Consequently an edge never joins vertices at the same root distance.

For a vertex (v\ne o), call a neighbour at distance (d_G(o,v)-1) a parent. If (u,w) are two parents of (v), the quadrangle condition for a median graph gives a vertex adjacent to both (u,w) at the preceding level. Every graph four-cycle in this CAT(0) square complex is the boundary of a square cell. Thus the two incoming edges at (v) bound a square face; equivalently they are adjacent in the link (C_q). The same condition excludes three parents, because three incident edges cannot be pairwise adjacent in the cycle (C_q). Hence every nonroot vertex has one or two parents, consecutive when there are two. All remaining incident edges go to children. Moreover, every square has levels, in cyclic order,

[ r,\ r+1,\ r+2,\ r+1. \tag{4.1} ]

Indeed, the alternative (r,r+1,r,r+1) would contradict uniqueness of the median of the root and the two high vertices. These facts can also be obtained from the rooted generation description for even-faced tessellations in Paul–Pippenger.

4.2 The level polygon

In every original hyperbolic square with levels as in (4.1), draw the hyperbolic diagonal joining its two equal-level vertices. It divides the square into two geodesic triangles. Assign each tiling vertex its graph distance from (o), and extend that value affinely in Klein coordinates over each triangle. Denote the resulting continuous, piecewise-affine function by (f).

Lemma 4.1. For every integer (R\ge1), the set (f^{-1}(R)) is a single embedded polygonal circle. Its vertices are exactly the members of (S_R).

Proof. Globally, the integral level (f^{-1}(R)) consists of the middle diagonals of the squares whose two middle vertices have level (R), together with their endpoints. A single face can also meet this level only at an extremal corner; that corner is already an endpoint of middle diagonals in its adjacent mixed-level faces. At a vertex of (S_R), the block of one or two parent edges and the complementary block of child edges have two transitions in the cyclic link. The two squares at these transitions supply exactly two incident level-diagonals. This also covers (R=1): each shell vertex has its unique parent (o), and the two faces adjoining that parent edge supply its two diagonals. The root itself is not on the level set because (R\ge1).

The parent block makes (K_R={f\le R}) locally a closed half-disk at each of its boundary vertices; away from them the same assertion is immediate in one of the affine triangles. Thus (K_R) is a compact planar surface with boundary, and its boundary is a finite disjoint union of embedded polygonal circles.

It is compact, because every triangle meeting it has a vertex in the finite graph ball (B_R). It is connected: in each triangle, descent along the affine function reaches a vertex of no larger level, and repeated parent steps reach (o).

It has no relatively compact complementary component in (\mathbb H^2). Such a component contains a tiling vertex: from any point in it, move within its triangle to a maximum-level vertex; the segment stays in the open superlevel set. Its closure meets finitely many vertices. Choose one of largest level. This vertex has a child, and the child edge lies in ({f>R}), so the child belongs to the same component and has larger level, a contradiction. Hence (K_R) is a connected compact planar surface with no relatively compact complementary component in (\mathbb H^2). Its boundary has one component, so it is one polygonal circle, proving the assertion. (\square)

The assertion that a vertex has a child in the last paragraph follows from the parent structure: a vertex with at most two parents in a (q)-regular tiling has at least (q-2\ge2) children.

4.3 Convexity

The angle of each face at a tiling vertex is (\theta=2\pi/q). The diagonal of a regular hyperbolic square bisects its corner angle. At a level-polygon vertex with one parent, its interior angle is (\theta); with two parents, it is (2\theta). Thus for (q>4), every interior angle is strictly less than (\pi).

Lemma 4.2. A simple polygon in the Klein disk whose sides are ambient geodesic segments and whose interior angles are all strictly less than (\pi) is strictly hyperbolically convex.

Proof. In Klein coordinates its sides are Euclidean segments and the sign of each turn is preserved. A simple Euclidean polygon all of whose turns have the same strictly positive sign is convex; strictness rules out a collinear run of vertices. Its Euclidean convexity is precisely hyperbolic geodesic convexity in the Klein model. (\square)

Lemmas 4.1 and 4.2 show that all vertices of (S_R) are extreme points of a strictly convex polygon. Hence no three of them are on one ambient hyperbolic geodesic. This proves the legality assertion of Theorem B. When (q) is even, Theorem A supplies the matching upper bound.

5. Triangular tilings

The proof for the triangular family is shorter because its radial structure is already recorded in the enumeration of regular tessellations by Paul and Pippenger Section 4. For ({3,q}), (q\ge6), every nonroot vertex has one or two parents and two same-generation neighbours, and the same-generation edges in each sphere form a single nested simple cycle.

Lemma 5.1 (radial-angle lemma). At a vertex of the level-(R) cycle having (k\in{1,2}) parents, the interior angle of the cycle is

[ (k+1)\frac{2\pi}{q}= \begin{cases}4\pi/q,&k=1,\6\pi/q,&k=2.\end{cases} \tag{5.1} ]

Justification. The inward sector from one same-level edge to the other contains exactly (k+1) triangular faces. Each contributes the face angle (2\pi/q). This counts the sectors described by the parent/sibling order in the cited radial structure.

For (q>6), both angles in (5.1) are below (\pi). The Klein-polygon argument of Lemma 4.2 therefore makes the cycle strictly convex and makes (S_R) legal. If (q) is even, Theorem A gives equality in Theorem C; Section 6 proves uniqueness.

The (q=6) equality case is the Euclidean triangular tiling, where a two-parent angle becomes (\pi); it is a useful boundary control, not part of the hyperbolic theorem.

6. Rigidity near the shell

Assume one of the two families in Theorems B and C, with even (q), and continue to write (m=q/2). Convexity and Lemma 3.1 imply

[ e_L=2\qquad (L\in\mathcal L_R). \tag{6.1} ]

Further, the inward angular sector at a shell vertex contains exactly its parent-edge directions. Each such edge reaches (B_{R-1}), which lies strictly inside the shell polygon. Child-edge rays point outside this sector; convexity prevents such a ray from leaving the vertex and later re-entering the polygon. In the triangular case, the two same-level directions lie along supporting sides and cannot meet the interior either. Hence the straight edge-lines through the vertex that meet (B_{R-1}) are exactly those containing a parent edge. Distinct parent edges here are not opposite and therefore give distinct lines. Thus

[ d_b=#{\text{parents of }b}\in{1,2}. \tag{6.2} ]

For a legal (A\subseteq B_R), put

[ t=|S_R|-|A|,\qquad r=|S_R\setminus A|,\qquad i=|A\cap B_{R-1}|.

]

Then (t=r-i). From (3.4), (6.1), and (6.2),

[ mt\ge (m-2)r, ]

and consequently

[ r\le\frac{m}{m-2}t,\qquad i=r-t\le\frac{2}{m-2}t. \tag{6.3} ]

These inequalities say that every near-optimal set is concentrated near the outer shell. At (t=0), they give (r=i=0); therefore the shell is the unique maximum.

There is a small extra consequence at deficit one.

Corollary 6.1 (initial deficit-one reduction). In either family with even (q\ge8), the legal sets of cardinality (|S_R|-1) are exactly the (|S_R|) sets obtained by omitting one shell vertex. For ({4,6}), the only additional possibility allowed by (6.3) has two selected interior vertices and three omitted shell vertices.

Proof. If a legal set contains one interior vertex (v), the (m) straight edge-lines through (v) have pairwise disjoint pairs of shell endpoints: two distinct geodesics through (v) cannot share a second point. Legality forces omission of at least one point from each pair, so (r\ge m) and (t=r-i\ge m-1) when there is only one interior point. At (t=1), (6.3) allows at most one interior point when (m\ge4), so this is impossible. When (m=3), it allows at most two interior points and the only non-shell numerical possibility is ((r,i)=(3,2)). (\square)

The remaining numerical possibility is excluded by the following local argument.

Proposition 6.2 (two-interior repair bound). In ({4,6}), a legal set containing two distinct vertices (v,w\in B_{R-1}) must omit at least five shell vertices. If (v,w) share a straight edge-line, it must omit at least six.

Proof. The curvature is normalized to (-1). Distinct straight edge-lines intersect only at tiling vertices, since they are unions of edges in an embedded tiling. A nondegenerate triangle whose sides lie on such lines would have three angles at least (\pi/3), contradicting the strict hyperbolic triangle angle-sum inequality. Also, a simple quadrilateral of such lines has area at most (2\pi-4\pi/3=2\pi/3). Its boundary is a simple Jordan cycle in the tiling's one-skeleton, so its interior is a nonempty finite union of whole tiles, each of area (2\pi/3); hence it is one square face. These use the standard angle/area formula Loustau, Chapter 14.

Each of (v,w) lies on three straight edge-lines, with two shell endpoints per line. If they share a line, both of its shell endpoints must be omitted. The other four lines have disjoint shell endpoint pairs: an additional common endpoint of the two pencils would form the excluded triangle with (v,w). Their endpoints also avoid the common line's endpoints. Thus four further deletions are needed.

Suppose instead that the two pencils share no line. If they had two common shell endpoints (b_1,b_2), the segments (v b_1,w b_1,w b_2,v b_2) would form a nondegenerate simple quadrilateral. Indeed, two common endpoints on a single pencil line would create an excluded triangle with the other pencil centre, and a crossing of opposite segments would likewise create such a triangle. The four sides are therefore distinct and do not cross. By the area argument this quadrilateral is an actual square face. Its alternating corners (b_1,b_2) have level (R); adjacency and interior membership force (v,w) to level (R-1). Both would then lie in all three graph intervals between pairs of (o,b_1,b_2), giving two medians. This contradicts median uniqueness, equivalently the square-level pattern (4.1).

Consequently the six distinct lines in the two pencils have at most one common shell endpoint between them. A shell deletion hits at most one line of each pencil. With only one possible shared endpoint, hitting all six lines takes at least five deletions. Other ambient geodesics can only add constraints. (\square)

Corollary 6.3 (complete deficit-one classification). In both families of Theorems B and C with even (q), for every (R\ge1), the legal sets of cardinality (|S_R|-1) are exactly

[ {S_R\setminus{b}:b\in S_R}. \tag{6.4} ]

Proof. Corollary 6.1 covers (q\ge8). In the remaining ({4,6}) case it leaves only two interior insertions with three shell deletions, which Proposition 6.2 excludes. Every single shell deletion is legal by convexity. (\square)

Thus the labelled deficit-one count is (|S_R|) at every radius. Under unrestricted atomic one-out/one-in swaps, these states form the complete graph (K_{|S_R|}). With single-point toggles, any two connect through the unique maximum by insertion then deletion. These are immediate consequences of (6.4), not claims about swaps along tiling edges.

7. Exact finite evidence for ({4,6})

The original exact integer-projective certificates cover (R\le4). An outside GPT-6.1-Sol replay rebuilt the balls and finite edge-line covers through (R=6), checked global supporting-hull relations, and counted all deficit-one states using the exhaustive stability/repair reduction. No floating-point collinearity is used.

| (R) | (|B_R|) | (|S_R|=\alpha_{4,6}(R)) | deficit-one sets | exceptions | boundary multipliers (3-d_b) | |---:|---:|---:|---:|---:|:---| | 1 | 7 | 6 | 6 | 0 | (2:6) | | 2 | 31 | 24 | 24 | 0 | (1:6,\ 2:18) | | 3 | 121 | 90 | 90 | 0 | (1:24,\ 2:66) | | 4 | 457 | 336 | 336 | 0 | (1:90,\ 2:246) | | 5 | 1,711 | 1,254 | 1,254 | 0 | (1:336,\ 2:918) | | 6 | 6,391 | 4,680 | 4,680 | 0 | (1:1254,\ 2:3426) |

Here (a:b) means that the value (a) occurs for (b) boundary vertices. Positivity of every listed multiplier proves uniqueness at these six radii directly from the finite cover, independently of the all-radius contour proof.

The source enumeration records every rich ambient line through radius four, not only tiling-edge lines. At radius four the all-pair audit found 651 rich non-edge ambient lines in addition to the edge-line constraints. The upper certificate needs only the edge-line subset; checking all shell triples is what certifies the displayed shell itself as legal.

The separately written outside replay reconstructed every ball with rational matrices, checked 32,220,702 point-versus-supporting-line relations, and independently excluded all 846,207 candidate interior pairs. The reduction checks every possible mixed triple: pairs of shell points give a blocker matching for each interior insertion, and shell points on the line through two insertions are forced deletions. The resulting two-matching vertex-cover calculation is an elementary repair formula, not a new general algorithm. This is a complete reduced census, not enumeration of all ball subsets.

A separate determinant-based direct enumeration, using neither stability nor repair pruning, checked all nonempty subsets at (R=1) and the following complete layers at (R=2):

| selected size | subsets visited | legal sets | |---:|---:|---:| | 23 | 7,888,725 | 24 | | 24 | 2,629,575 | 1 | | 25 | 736,281 | 0 |

The phrase “complete (R=2) enumeration” here refers to these cardinality layers, not all (2^{31}) subsets. Independent replay means separately written code within the outside GPT-6.1-Sol core, not two independent human authorship teams.

Finite evidence and archival scope. The self-contained supplement supplement/README.md supplies exact homogeneous integer coordinates, rational transformations, edge-line and slack certificates for every (1\le R\le6), and a standard-library checker. It also supplies a separate determinant enumerator for the exact small layers. The stated radius-three through radius-six calculation is a complete reduced stability/repair census, not a full enumeration of all ball subsets.

For the two-insertion reduction, after two interior points are fixed, every shell point on their common line is forced deleted and every remaining mixed triple gives a two-element shell obstruction. The further deletion cost is the minimum vertex-cover number of the union of these obstruction matchings. This elementary formula is stated here, so no companion manuscript is needed.

8. Growth and what is not new

The shell sizes are prior work. Bartholdi–Ceccherini-Silberstein record earlier growth work of Floyd--Plotnick and Cannon--Wagreich; Paul and Pippenger give, for the square family Equation (3.3) of Paul–Pippenger,

[ \sum_{R\ge0}|S_R|z^R= \frac{(1+z)^2}{1-(q-2)z+z^2}. \tag{8.1} ]

Thus

[ |S_1|=q,\qquad |S_2|=q(q-2),\qquad |S_R|=(q-2)|S_{R-1}|-|S_{R-2}|\quad(R\ge3). \tag{8.2} ]

For ({4,6}) this gives

[ 6,\ 24,\ 90,\ 336,\ 1254,\ 4680,\ldots ]

as the shell sequence. The proposed new content of this manuscript is not that sequence, but its role as the exact ambient no-three-in-line capacity, with uniqueness, stability, and the deficit-one classification. The latter counts reproduce the old shell sequence because of (6.4).

For ({4,6}), the dominant root of the recurrence is (\lambda=2+\sqrt3). By Theorem B, the density of a maximum legal set in the graph ball has the limit

[ \frac{\alpha_{4,6}(R)}{|B_R|} =\frac{|S_R|}{\sum_{j=0}^R|S_j|} \longrightarrow 1-\lambda^{-1}=\sqrt3-1. \tag{8.3} ]

This asymptotic combines a prior growth formula with the incidence theorem, proved above. It is not a newly discovered coordination sequence.

9. Relation to prior work and limits

Regular-tessellation generation counts and parent types were systematically studied by Paul and Pippenger Paul–Pippenger. Geodesic and growth structure in negatively curved planar tessellations has also been studied in graph language by Baues and Peyerimhoff Baues–Peyerimhoff and by Keller Keller. These sources are relevant tools and context, but do not, to the best of the present bounded reading, state the ambient-incidence extremal theorems proposed here.

The literature comparison here is deliberately limited. The sources cited below give close combinatorial, graph-convexity, and extremal context, but no claim of first occurrence is made for the ambient-incidence theorem.

Bandelt, Chepoi, and Eppstein Bandelt--Chepoi--Eppstein study squaregraphs, median structure, and a dual hyperbolic line-arrangement viewpoint; their dual lines are not the vertex edge-lines used here. Kisfaludi-Bak, Poon, and van Wordragen Kisfaludi-Bak--Poon--van Wordragen, and Ingervo and Kisfaludi-Bak Ingervo--Kisfaludi-Bak, study graph shortest paths, graph convexity, and substructure recognition in regular tilings. Those graph predicates do not imply ambient no-three collinearity. The published work of Malen, Roldán, and Toalá-Enríquez Malen--Roldán--Toalá-Enríquez concerns tile animals and their edge/vertex counts at fixed tile number, a different objective from selecting ambient-general-position subsets of (B_R).

Several nearby statements are different.

• In the Klein model, arbitrary finite hyperbolic point configurations are simply projective no-three-in-line configurations. The content here lies in the special point sets (B_R), not in a new notion of collinearity.
• Graph-general-position problems forbid triples on graph shortest paths; they are not the same as ambient hyperbolic collinearity.
• Finite projective-plane arcs have another incidence structure and should not be cited as proving this statement.
• Even (q) by itself gives the upper bound only. The legality and strict-convexity arguments presently cover (p=3,4), not all (p).
• The Euclidean tilings at (q=4) for squares and (q=6) for triangles have flat shell sides. They illustrate sharp angle thresholds but are not hyperbolic counterexamples.

10. Open questions

1. Classification for other faces. For hyperbolic ({p,q}) with even (q) and (p\notin{3,4}), Theorem A gives the upper bound, but shell legality has not been classified. The smallest cases ({5,4}) and ({6,4}) are natural tests.
2. Deficit two and beyond. Deficit one is now completely classified. At the next level, which interior configurations attain the stability bounds, and can their legal completions be counted uniformly in (R)?
3. Near maxima and maximal sets. Inequality (6.3) controls the support of a near maximum but does not count maximal legal sets or describe their reconfiguration landscape.
4. Novelty and alternative proofs. Is the line-cover identity already known in this setting, perhaps under a different terminology? Is there a Coxeter-theoretic proof of the convex-shell claim that extends beyond (p=3,4)?
5. Other balls. Metric balls, horoball truncations, and annuli are genuinely different finite point sets. The present proof should not be assumed to transfer to them.

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Contributions

Aleksei Kudriashov initiated and directed the research programme. athanor_lab developed the finite coordinate model and line-cover formulation. Astra (gpt-6-astra) audited the ambient-versus-graph distinction and the all-radius square-shell argument. The Guide (Claude Code) orchestrated the research record. An independent GPT-6.1-Sol challenge core supplied an outside proof audit, the two-interior deficit-one argument, and a separate finite replay. The finite-geometry research core (gpt-6.1-sol) implemented the original finite coordinate model and cover certificates. The literature core (gpt-6.1-sol) supplied source comparisons. The independent manuscript referee (gpt-6.1-sol) audited all proofs, replayed the finite certificates and identified the publication fixes incorporated here. The submitting account is athanor_lab; Aleksei Kudriashov is the responsible human author. Manuscript and data are licensed CC BY 4.0; lab code is MIT-licensed.