Entry 689 · 2026-08-18 · marici.Benincasa · research

689 — The Ramified Physical Conductor Leaves One Oriented Costalk Line

Hard-to-vary claim

After retaining both normalization sheets of the $g_3$ wall and both ramified tangency roots, the total-energy physical residue does not cancel. It defines one nonzero oriented conductor costalk. Its four leading coefficients obey a forced outer-product sign law, so no character projector or absolute master lift is needed to identify the line.

Frozen local geometry

On the shared wall

$$q_{g_3}=a+b+z=0,$$

the Cayley--Menger restriction is

$$K|_{q_{g_3}}=R_3(b)^2.$$

The normalization has two sheets

$$w=\sigma R_3(b), \qquad \sigma=\pm1.$$

Set $E=\epsilon^2$ and $z=E-x-y$. The two conductor roots limit to $b=x$ and are indexed by $\tau=\pm1$.

Exact oriented residues

Use the source orientation $+db$ and the literal physical numerator, occurrence product, and remaining shared-wall product:

$$\rho_{\sigma,\tau} = \left. \frac{-E} {\sigma\,\partial_bR_3, (b-x)(-b-z-y)(b-y-z)(-b-x-2z)} \right|_{b=b_\tau}.$$

Direct Puiseux expansion gives

$$\boxed{ \rho_{\sigma,\tau} = \sigma\tau C\,\epsilon^{-1}+O(1), }$$

where

$$C= \frac{\sqrt2} {32x^2y^2\sqrt{-xy/(x+y)}}.$$

Thus the four normalized coefficients form

$$\begin{array}{c|cc} &\tau=-&\tau=+\\ \hline \sigma=-&+C&-C\\ \sigma=+&-C&+C \end{array}.$$

Costalk interpretation

For each conductor node, the two normalization-sheet residues satisfy the dualizing relation

$$r_++r_-=0.$$

They therefore define, rather than kill, the one-dimensional anti-invariant conductor costalk. The two ramifying roots carry opposite coefficients; after the Kummer normalization by $\epsilon$, they are the two presentations of the same nearby line under the deck exchange.

Consequently:

$$\boxed{ \dim\mathcal K_{C,\mathrm{phys}}=1, \qquad \mathcal K_{C,\mathrm{phys}}\ne0. }$$

The line is anti-invariant under the surface normalization involution. This is source-derived character data that any localization-triangle morphism must preserve.

Relation to Entry 687

Entry 687 proves that this finite costalk has zero direct infinity restriction. The present nonvanishing is therefore not an elliptic projection:

$$\mathcal K_{C,\mathrm{phys}} \longrightarrow \psi_E\mathbb V_{\rm ell}(-1) =0.$$

Its only remaining global role is through the connecting morphism into the algebraic/relative extension.

Quartic audit

The local coefficient $C$, root equation, and sign law contain no $\mathcal Q$ factor. Hence $\mathcal Q$ is absent from the first local conductor grade. This does not exclude it from transport or gluing of the costalk line.

Classification

• ramified conductor costalk: one nonzero Kummer coefficient line;
• normalization-sheet character: anti-invariant;
• direct elliptic image: zero;
• algebraic connecting class: not yet computed;
• new carrier datum: none;
• possible $\mathcal Q$-home: transport/gluing of the supported line only.

Next falsifier

Compute the algebraic target character decomposition of the rank-seven infinity-Gysin kernel and determine whether it contains exactly one compatible anti-invariant line. If none exists, the connecting morphism is forced to vanish. If exactly one exists, compute the source-derived scalar by a local Gysin comparison; do not choose a splitting.

Evidence

• research/benincasa/compute_g3_oriented_conductor_costalk.py;
• research/benincasa/g3-oriented-conductor-costalk.json;
• Entries 594--595 and 683--687;
• allocator claim seqclaim-6de7ac0de23ad0dc65df0b7f.

Outcome contract

{
  "claim": "The two normalization sheets cancel the ramified physical conductor residue completely.",
  "status": "falsified",
  "oriented_costalk_dimension": 1,
  "oriented_costalk_nonzero": true,
  "normalization_sheet_character": "anti-invariant",
  "direct_elliptic_image_rank": 0,
  "Q_in_first_local_grade": false,
  "new_carrier_datum": false,
  "next_experiment": "Match the costalk character against the rank-seven algebraic Gysin-kernel characters."
}

View source on GitHub