Rank-Three Obstruction and Associahedral Saturation
Record
Date: 2026-08-13
Status: the naïve rank-three closure hypothesis is falsified at ten points. The signed square/pentagon carriers constructed in entry 28 do not cover every facet of the three-dimensional associahedral faces generated by three physicalizing steps.
The failure is structured and repairable on the scalar presentation side. The missing carriers are actual marked facets of the same scalar associahedron, not formal cells adjoined from outside. Closing under those facets gives an integral, deck-equivariant rank-three saturation whose complete two-sphere boundaries vanish exactly.
This advances the source complex but does not yet construct its QTDS or twisted-worldsheet image.
Hypothesis tested
Entry 28 supplied a rank-two carrier for every pair of steps encountered directly in a marked Catalan transfer. The first higher question was:
Do those inherited square and pentagon carriers already form the complete boundary of every rank-three dependency block?
If yes, the pairwise homotopies would assemble without new two-cells. The exact audit answers:
[ \boxed{\text{No.}} ]
The first counterexamples occur at (n=10), the first multiplicity with three physicalizing steps.
The three universal rank-three faces
Let the three selected steps be distributed among dependency chains. There are three partitions of the rank:
[ 1+1+1, \qquad 2+1, \qquad 3. ]
They give the three possible three-dimensional associahedral faces.
Three independent chains
The carrier is
[ I\times I\times I, ]
a cube with
[ 8\ \text{vertices}, \qquad 6\ \text{square facets}. ]
All six square facets are inherited from entry 28.
Two consecutive steps and one independent step
The carrier is
[ K_2\times I, ]
a pentagonal prism with
[ 10\ \text{vertices}, \qquad 5\ \text{square facets}, \qquad 2\ \text{pentagon facets}. ]
The direct rank-two system contains
[ 3\ \text{squares}+2\ \text{pentagons} ]
and misses
[ \boxed{2\ \text{squares}.} ]
Three consecutive steps in one chain
The carrier is the three-dimensional associahedron of a hexagon,
[ K_3, ]
with
[ 14\ \text{vertices}, \qquad 3\ \text{square facets}, \qquad 6\ \text{pentagon facets}. ]
The direct rank-two system contains four pentagons and misses
[ \boxed{3\ \text{squares}+2\ \text{pentagons}.} ]
Thus the universal coverage table is:
| rank-three carrier | covered squares | covered pentagons | missing squares | missing pentagons |
|---|---|---|---|---|
| cube | 6 | 0 | 0 | 0 |
| pentagonal prism | 3 | 2 | 2 | 0 |
| hexagon associahedron | 0 | 4 | 3 | 2 |
The same table occurs at ten and twelve points and for both polarity sheets.
Topology of the uncovered part
The failure is not just a count mismatch. The covered and missing facets have characteristic topologies.
For the cube, the inherited carriers already form the complete boundary sphere:
[ (F,V,E,\chi,c,b)=(6,8,12,2,1,0), ]
where (c) is the number of connected components and (b) the number of boundary circles.
For the pentagonal prism, the five covered facets form a cylinder:
[ (5,10,15,0,1,2). ]
The two missing squares are two disconnected disks:
[ (2,8,8,2,2,2). ]
Both disks are independently required. A single extra global filler cannot replace them while preserving the cellular incidence structure.
For the hexagon associahedron, the four covered pentagons form one disk:
[ (4,14,17,1,1,1), ]
and the five missing facets form the complementary disk:
[ (5,14,18,1,1,1). ]
Here the missing cells constitute one connected complementary filling of the inherited boundary circle.
Core grading of the missing facets
Every facet is obtained by fixing one additional diagonal inside the rank-three face. Its parity determines whether it preserves or raises the physical-core degree.
For each pentagonal prism, the two missing squares are:
- one scalar facet, preserving core degree (p);
- one physical facet, raising core degree (p\mapsto p+1).
For each hexagon associahedron, the five missing facets are:
- two scalar squares;
- one physical square;
- one scalar pentagon;
- one physical pentagon.
At twelve points the same pattern appears over both (p=0) and (p=1). The grading is translated uniformly:
[ \text{scalar facet}:p\longmapsto p, \qquad \text{physical facet}:p\longmapsto p+1. ]
This is the key structural lesson. Higher coherence necessarily mixes:
- same-core scalar refinement;
- core-raising physical incidence.
A complex containing only the physical-core Hasse diagram cannot carry the required coherence.
Why the rank-two theorem remains correct
Entry 28 proved that every pair encountered directly along the dependency-chain transfer has a square or pentagon carrier with exact signed boundary. Nothing here contradicts that statement.
The rank-three failure is instead a failure of face completeness. A three-dimensional associahedral face has facets that are not themselves encountered as direct rank-two blocks from a zero-core source with the same marked contact. They occur along the complementary scalar refinement routes introduced by the lower homotopies.
This distinction is essential:
[ \text{pairwise carrier existence} \not\Rightarrow \text{higher face closure}. ]
Canonical scalar-side saturation
Let (\mathcal D_2^\epsilon) denote the marked rank-two carriers inherited directly from the Catalan transfer. For every rank-three dependency block, let (F_3) be the unique three-dimensional associahedral face determined by the common dissection of its direct states.
Define the rank-three saturation by adjoining every missing marked facet of (F_3):
[ \operatorname{Sat}_3(\mathcal D_2^\epsilon)
\mathcal D_2^\epsilon \cup \left{ (d,F_2): F_2\subset\partial F_3 \text{ for a direct rank-three block} \right}. ]
This is not a formal horn filler. Each (F_2) is an existing square or pentagon in the scalar associahedron. The marked diagonal (d) lies in the common dissection, so the contact weight
[ -X_d ]
is constant on the whole facet. At a retained partial core, the remaining denominator is also constant on the local face.
Give every polygonal facet its exact integral edge boundary and orient the facets coherently. Then every completed rank-three carrier satisfies
[ \boxed{ \partial \left( \sum_{F_2\subset\partial F_3} \epsilon(F_2,F_3),[F_2] \right) =0. } ]
Every edge occurs twice with opposite sign. No division by two, inverse Laplacian, random kinematics, or amplitude-level cancellation is used.
Deck covariance
One-step rotation exchanges the two alternating polarity sheets. The audit rotates:
- the zero-core source;
- the marked scalar diagonal;
- the rank-three common dissection;
- every rank-two facet;
- every signed flip-edge cycle.
The inherited coverage deficit is carried exactly from the plus sheet to the minus sheet, and so is the saturated two-sphere boundary. Hence
[ \rho\operatorname{Sat}_3(\mathcal D_2^+)
\operatorname{Sat}_3(\mathcal D_2^-)\rho ]
as a marked cellular statement, up to the single overall orientation sign of each three-dimensional face.
Exact finite certificate
There are no rank-three dependency blocks at six or eight points.
At ten points:
| carrier | marked occurrences | distinct unmarked faces |
|---|---|---|
| cube | 20 | 20 |
| pentagonal prism | 40 | 40 |
| hexagon associahedron | 10 | 10 |
The direct marked rank-two carrier set has (300) elements. Rank-three saturation adds (120) distinct marked facets:
[ 300\longrightarrow 420. ]
It completes (70) marked rank-three surfaces.
At twelve points:
| carrier | marked occurrences | distinct unmarked faces |
|---|---|---|
| cube | 480 | 400 |
| pentagonal prism | 720 | 600 |
| hexagon associahedron | 144 | 120 |
The direct marked rank-two carrier set has (3276) elements. Rank-three saturation adds (2028) distinct marked facets:
[ 3276\longrightarrow 5304. ]
It completes (1344) marked rank-three surfaces.
Run:
python -B research/nima/check_core_incidence_rank_three.pyThe script enumerates exact triangulations, exact face incidences, integral signed boundaries, covered/missing subcomplex topology, and full path-sheet rotation. It uses no stochastic or floating-point test.
Corrected source object
The direct Catalan transfer should not be regarded as a complete cellular object. It selects a monotone path system inside a larger canonical resolution.
The corrected candidate is its marked associahedral envelope:
[ \operatorname{AssEnv}(\Phi_\epsilon)
\text{the cellular system generated by the complete associahedral faces of dependency blocks}. ]
For a block taking (r_a) consecutive steps from dependency chain (a), the predicted local cell is
[ \prod_a K_{r_a}, \qquad \sum_a r_a=r, ]
with the convention (K_1=I). Entry 28 verifies the rank-two cases. This entry verifies the rank-three cases and shows that taking the complete face, rather than only its direct-path facets, is mandatory.
The formula at arbitrary rank is presently a strong structural conjecture. Its planar proof should identify a consecutive chain block with a contiguous polygonal region and distinct chain blocks with disjoint regions.
Relation to the earlier critiques
This saturation avoids the vacuity identified in the Stasheff and MacPherson audits:
- the cells are not freely adjoined formal cylinders;
- every cell is an actual face of the scalar presentation associahedron;
- the mark and scalar coefficient extend over it;
- all cellular boundaries are explicit and integral.
But one part of those critiques remains fully active:
A scalar presentation cell is not yet a loaded twisted cycle or a logarithmic Cousin class.
The construction still lacks an augmentation into the scalar twisted-chain complex with composition-stable acyclic kernel. It also lacks the chain-level image of the newly exposed scalar and physical facets on the QTDS/worldsheet side.
What is established
- the naïve inherited rank-two system fails to close at rank three;
- the first failure occurs at ten points;
- the failure has three universal local types determined by dependency-chain partitions;
- cubes are already complete;
- prisms require two independent missing squares;
- hexagon associahedra require a complementary five-facet disk;
- missing facets mix same-core scalar refinement and physical-core incidence;
- every missing cell is canonically present in the scalar associahedron;
- adjoining those cells gives exact integral rank-three boundary closure;
- the saturation and its deficit are deck-equivariant through twelve points.
What is not established
This entry does not claim:
- a QTDS image for the saturated cells;
- a cellular chain map from the scalar envelope;
- a filtered Pochhammer/Cousin comparison;
- higher-rank saturation beyond rank three;
- acyclicity or contractibility of the full saturated kernel;
- equality of chain representatives with ((\operatorname{Pf}’A)^2);
- resonance-safe inversion of a global KLT pairing.
Coefficient equality and fixed-core factorization remain established, but factorization naturality at the chain level remains conditional on constructing these maps.
Primary next test
There are two coupled tasks.
First, test the marked associahedral-envelope conjecture at rank four. The five dependency partitions predict the local cells
[ I^4, \qquad K_2\times I^2, \qquad K_2\times K_2, \qquad K_3\times I, \qquad K_4. ]
At twelve points these have respectively
[ 16,\ 20,\ 25,\ 28,\ 42 ]
vertices. Their complete rank-three facet boundaries should determine the next saturation layer.
Second, and more decisively, define a coefficient cosheaf on the saturated marked cells and a filtered comparison
[ \chi_{\rm cell}: C_^{\rm AssEnv}(\text{scalar grade}) \longrightarrow \operatorname{gr},C_^{\rm Poch/Cousin} ]
at finite nonresonant (\alpha’). The new scalar and physical missing facets must acquire loaded-current images whose boundaries agree with the already fixed edge transports.
Failure to lift either universal missing-facet type would be a genuine obstruction to the intrinsic half-object, not merely a presentation deficit.
Decision
Promote:
Direct Catalan paths and their pairwise homotopies are not the final scalar half-object. The correct source candidate is their marked associahedral envelope. Rank-three closure forces new same-core and core-raising facets, all canonically present in scalar geometry, and the resulting saturated boundaries are integral and deck-equivariant.
The immediate Nima frontier is now sharply typed: verify the envelope at rank four while constructing the coefficient-cosheaf/Pochhammer image of the two universal missing-facet mechanisms.