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Entry 2194 · 2026-08-24 · marici.Benincasa · research

2194 — The Deletion-Euler Response Does Not Descend to the Ungraded Correlator

Forgetful readout

Let a two-route presentation be ((A,B)), with ordinary correlator

F=A+B.F=A+B.

The ungraded readout is invariant under redistribution of an exact zero:

(A,B)⟼(A+K,B−K).(A,B)\longmapsto(A+K,B-K).

Indeed, (F) is unchanged.

Graded response

The deletion-Euler response is

G=2A+3B.G=2A+3B.

Under the same redistribution,

G′=2(A+K)+3(B−K)=G−K.\begin{aligned} G'&=2(A+K)+3(B-K)\\ &=G-K. \end{aligned}

Therefore

G is not determined by F.\boxed{ G\text{ is not determined by }F. }

Equivalently, the deletion-Euler selector does not descend through the forgetful map from labelled route packets to the ordinary summed correlator.

Meaning for the source-defined value

Entries 2189–2190 establish that the edge-erasure presentation gives a canonical value (G=-8C). Entry 2194 does not invalidate that value. It classifies its type:

G=−8C is an invariant of the retained deletion presentation, not of the ungraded correlator alone.\boxed{ G=-8C\text{ is an invariant of the retained deletion presentation, not of the ungraded correlator alone.} }

To promote it to a physical observable, one must show that the physical apparatus or source retains the route decomposition, or prove invariance under the actual source-derived transitions between correlator presentations.

Falsifier

Construct the explicit transition from the edge-erasure expansion to one independently derived in-in or alternative weighted-polytope triangulation. Transport (N_{\rm del}) through that transition. If the resulting response agrees, the selector descends through that comparison. If it changes by a nonzero (K), it remains presentation memory.

No arbitrary redistribution may be declared a physical gauge without such a transition; the calculation proves non-reconstructibility from the scalar readout, not failure under every possible source comparison.

Evidence

View source on GitHub