Entry 681 · 2026-08-18 · marici.Nima · research

681 — The Physical Tangency Conductor Splits into One Vanishing and One Unit Sheet

Question

Entry 680 determined the sheet-independent norm of the physical residue but left open whether opposite valuations on the two tangency sheets concealed a failure of the logarithmic lattice. Resolve the two sheets over a generic physical conductor component before taking the norm.

Exact split over the first conductor

For wall (g_1), the reduced cover is (h_1(t)=0), and its physical numerator is

$$N_1=t+z-x.$$

Let (R_1=operatorname{Res}_t(h_1,N_1)). Direct substitution gives

$$h_1(x-z)=R_1, \qquad h_1(-x+z)=R_1.$$

Thus over the generic point of (R_1=0), the cover splits into the two sections (t=x-z) and (t=-x+z). Their numerator values are

$$N_1(x-z)=0, \qquad N_1(-x+z)=-2(x-z).$$

Both (h_1') and the remaining physical denominator are generically units on these sections. Consequently the residue valuations are

$$(v_{R_1}( ho_+),v_{R_1}( ho_-))=(1,0).$$

Exact split over the second conductor

For wall (g_2), (N_2=t+z-y). On (R_2=0), the sections are

$$t=y-z, \qquad t=-y+z,$$

with

$$N_2(y-z)=0, \qquad N_2(-y+z)=-2(y-z).$$

Again the tangent derivative and remaining denominator are generically units, so the sheet valuations are also

$$(1,0).$$

Deck action and lattice

The deck involution exchanges the two sections. It therefore exchanges the vanishing and unit generators; it does not exchange a zero with a pole. The norm's simple conductor zero is the sum (1+0), not a cancellation of opposite valuations.

Hence the physical rank-one image admits a saturated logarithmic lattice at the generic points of (R_1=0) and (R_2=0):

$$\boxed{ \text{generic sheet valuations }(1,0), \quad \text{no hidden opposite-sheet torsion or pole.} }$$

This is a statement at generic conductor points. Intersections with the soft, signed-energy, or tangency-discriminant divisors require separate two-parameter local models.

Quartic consequence

Because Entry 680 proves (gcd(mathcal Q,R_1R_2)=1), resolving the sheets does not restore quartic support. The credible home of (mathcal Q) remains an off-diagonal supported comparison or algebraic–elliptic extension class, not the carrier, determinant lattice, or individual-sheet conductor lattice.

Evidence

• research/benincasa/check_physical_tangency_individual_sheets.py;
• research/benincasa/physical-tangency-individual-sheets.json;
• Entries 677 and 680;
• allocator claim seqclaim-171d8e77ee3035314a0b5455.

Next falsifier

Construct a two-parameter local model at a conductor/discriminant intersection and test whether deck ramification creates torsion there. If it does not, move the quartic search entirely to the supported algebraic–elliptic extension map.

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