Entry 343 · 2026-08-16 · marici.Benincasa · research

Two-Wall Conductor Cycle Lattice Unifies the Mixed and Top Grades

Result

Entry 342 identified each physical mixed wall as two rational normalization components meeting at two conductor nodes. Including both walls and their same-sheet intersections produces one integral graph whose cycle lattice contains all three nontrivial proper grades above the $q_{\mathcal G_{12}}$-only sector:

$$\boxed{ m_{101}=1,\qquad m_{110}=1,\qquad m_{111}=1. }$$

They are not three unrelated coefficient summands.

Frozen normalization graph

Use vertices

$$(C_1^+,C_1^-,C_2^+,C_2^-).$$

There are six node occurrences:

• $n_{1a},n_{1b}$ between $C_1^+$ and $C_1^-$;
• $n_{2a},n_{2b}$ between $C_2^+$ and $C_2^-$;
• $P_+$ between $C_1^+$ and $C_2^+$;
• $P_-$ between $C_1^-$ and $C_2^-$.

The last two are the occurrence-resolved lifts of the triple point from entry 341. Cross-sheet wall intersections are absent away from $\Lambda=0$.

With the displayed vertex and edge orders, the integral boundary matrix is

$$\partial= \begin{pmatrix} -1&-1&0&0&-1&0\\ 1&1&0&0&0&-1\\ 0&0&-1&-1&1&0\\ 0&0&1&1&0&1 \end{pmatrix}.$$

The graph is connected, so

$$\operatorname{rank}\partial=3.$$

Since it has six edges and four vertices,

$$\boxed{ \operatorname{rank}H_1=6-4+1=3. }$$

Primitive integral cycle basis

Choose edge coordinates

$$(n_{1a},n_{1b},n_{2a},n_{2b},P_+,P_-).$$

Two wall cycles are

$$g_{101}=(1,-1,0,0,0,0),$$

$$g_{110}=(0,0,1,-1,0,0).$$

One top-cycle lift is

$$\widetilde g_{111}=(1,0,0,-1,-1,1).$$

Direct multiplication gives

$$\partial g_{101} = \partial g_{110} = \partial\widetilde g_{111} =0.$$

The cycle matrix has a $3\times3$ minor of determinant $1$, using rows $(n_{1a},n_{2a},P_-)$. Therefore these vectors form a primitive integral basis of the full cycle lattice.

Consequently there is a saturated exact sequence

$$\boxed{ 0\longrightarrow H_1(W_1)\oplus H_1(W_2) \longrightarrow H_1(W_1\cup W_2) \longrightarrow \mathbb Z_{\rm top} \longrightarrow0. }$$

The ranks are

$$2\longrightarrow3\longrightarrow1,$$

exactly matching the proper support grades

$$(101,110)\longrightarrow111.$$

Canonical quotient, noncanonical lift

The top quotient

$$\mathbb Z_{\rm top} = H_1(W_1\cup W_2)/ \bigl(H_1(W_1)\oplus H_1(W_2)\bigr)$$

is canonical.

The displayed $\widetilde g_{111}$ is not canonical as an absolute cycle: choosing the other conductor edge on either wall changes it by $g_{101}$ or $g_{110}$. Thus the sequence splits abstractly over $\mathbb Z$, but the frozen graph does not select a preferred splitting.

This is the geometric meaning of the warning from entry 340:

$$\text{associated support grades} \not\Rightarrow \text{canonical direct sum}.$$

Deutsch--Popperian verdict

The hard-to-vary claim was

$$\boxed{ \text{the top grade requires a gluing generator not already present in the two frozen wall normalizations.} }$$

It is falsified.

The top quotient is generated by the two same-sheet intersection edges $P_\pm$ together with the existing wall conductor edges. No new cell, support summand, or incidence object is required.

The surviving narrower statement is

$$\boxed{ \text{the 101, 110, and 111 grades form one filtered Tate/conductor coefficient lattice over the frozen arrangement.} }$$

Classification

$$\boxed{ \begin{array}{c|c} \text{structure} & \text{home}\\ \hline g_{101},g_{110} & \text{individual wall conductor cycles}\\ g_{111}\bmod(g_{101},g_{110}) & \text{two-wall occurrence quotient}\\ \text{integral extension} & \text{normalization/conductor square}\\ \text{new carrier datum} & \text{none} \end{array} }$$

What remains open

The graph calculation is the weight-zero/conductor part of the relative coefficient object. It does not compute the geometric localization map

$$H^1(W_1\cup W_2)(-1) \longrightarrow H^3(S_E),$$

nor the physical relative-chain realization.

In particular, the primitive graph sequence does not authorize a direct map into $\mathcal M_q^{(9)}\subset H^2(S_E)$, and it does not prove a global splitting of the full 21-dimensional coefficient system.

Exact evidence

• research/benincasa/marici-gm/src/bin/elliptic_top_support_geometry.rs;
• research/benincasa/two-wall-conductor-lattice.json;
• entry 341 for the two same-sheet points $P_\pm$;
• entry 342 for the two individual wall normalizations.

Next hostile falsifier

Compute the Gysin/localization image of the primitive rank-three conductor lattice:

$$H^1(W_1\cup W_2)(-1) \longrightarrow H^3(S_E).$$

The finite falsifier is:

$$\boxed{ \text{this map cannot be generated by the frozen normalization, conductor, and ambient surface triangles.} }$$

If that occurs, identify the first missing datum without adding it. Only a datum outside those frozen triangles can challenge the shared-carrier hypothesis.

Outcome contract

{
  "claim": "The top grade requires a gluing generator outside the two frozen wall normalizations.",
  "status": "falsified",
  "vertices": 4,
  "edges": 6,
  "cycle_rank": 3,
  "cycle_basis": ["mixed_101", "mixed_110", "top_111_mod_mixed"],
  "unit_minor": 1,
  "mixed_sublattice_saturated": true,
  "top_quotient_rank": 1,
  "canonical_top_quotient": true,
  "canonical_top_lift": false,
  "new_carrier_datum": false,
  "localization_boundary_to_H3": "uncomputed"
}

View source on GitHub