C08.2 · Derivation

Why the trace weights must agree

Direct Zorn multiplication gives \[ [s,u_i]=\sqrt2\,u_i,\qquad [u_i,v_i]=\sqrt2\,s. \] Therefore \[ B([s,u_i],v_i)=\sqrt2\,a,\qquad B(s,[u_i,v_i])=\sqrt2\,b. \] Compatibility forces \(a=b\). This is a necessary condition obtained from an actual bracket, not from inserting the desired readout.

Sufficiency must also be checked. The defect \[ \Delta(x,y,z)=B([x,y],z)-B(x,[y,z]) \] is trilinear, so checking it on every ordered basis triple is sufficient over the whole space. The exact run evaluates all \(343\) triples and finds \(12\) nonzero symbolic defects, taking only the two forms \[ +\sqrt2(a-b),\qquad-\sqrt2(a-b). \] The generated polynomial ideal is exactly \((a-b)\), as printed in `Frobenius.constraint_ideal`. This establishes both necessity and sufficiency in the **declared reflected two-weight class**.

A control with \(a=1,b=2\) fails the displayed witness by \(-\sqrt2\). The result is therefore not a test that silently assumed equal weights.

The normalized readout, and what remains free

The total weighted trace is \(6a+b\), while one triplet contributes \(3a\). Hence \[ r=\frac{6a+b}{3a}=2+\frac{b}{3a}. \] For \(a=b\ne0\), \[ r=\frac73. \] The ratio is computed by `finite_trace.normalized_ratio`. The zero form \(a=b=0\) also satisfies the homogeneous compatibility equations, but its ratio is undefined. The closure calculation fixes the **relative** weights, not a nonzero absolute scale.

This matters for the word “unique”: it is the normalized ratio that is unique in the stated nonzero class. It is not a proof of a unique physical normalization across different carriers, reflections or readout prescriptions.

The imaginary metric has a different signed count

The code computes the norm of each imaginary basis vector from the split product's norm: \[ \eta=\operatorname{diag}(1,1,1,-1,-1,-1,-1). \] Thus the imaginary norm has inertia \((3,4,0)\). The unweighted dimension is \(7\), but the full unit-channel signed contraction is \[ 3-4=-1. \] They are different contractions, not two approximate estimates of one observable.

For any subset of the registered orthonormal channels, write \(p\) and \(q\) for its positive and negative counts. Then \[ R(S)=p-q,\qquad 0\le p\le3,\quad 0\le q\le4, \] so \[ -4\le R(S)\le3. \] The script exhausts all \(128\) subsets and prints the histogram, endpoints and full-space value. The combinatorial argument establishes the bound directly; the exhaustive run checks every member of the finite class. Since \(7\) is outside that range, it cannot be this unit-weight signed residue count.

Appendix I §8, Lemma 8.3, sealed Rev32.7 PDF p.212:

> This closes the specific residue route on Im Os ; it does not assert that no other construction could realize 7/3 physically.

Two negative controls prevent a false universal no-go

First, change to a non-orthonormal basis by multiplying the first positive basis vector by \(2\). The raw trace of the metric matrix changes to \(2\), while its inertia remains \((3,4,0)\). The invariant content is the signature and the specified orthonormal unit-channel contraction, **not the coordinate trace of an arbitrary congruent metric matrix**.

Second, leave the unit-weight class. A single positive channel with weight \(7\) has contraction \(7\). This is an explicit outside-class control, not a recovery inside the registered class. It shows why the unit-weight hypothesis must stay attached to the bound. The source's “signed residue” terminology is quoted as a designation of its tested route; this finite sum does not independently establish physical unitarity.

Why this calculation cannot be called LSZ

A finite trace is an algebraic functional. A physical LSZ residue requires a specified physical field, its two-point function, a pole and the associated normalization in a field theory. None of those objects is supplied by a weighted trace or by an orthonormal channel inventory. No spacetime limit, mass shell, amputation or interacting two-point function is computed in this article.

Replacing a signed contraction by a positive sum of squares changes the functional. It may recover the dimension numerically, but that does not establish the missing physical map. The finite normalization remains valid on its own terms; the physical export remains separate.