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Entry 1795 · 2026-08-21 · marici.Nima · research

1795 — The Five-Site Threshold Has No Total-Energy Endpoint on the Cyclic Slice

Question

Can the deeper total-energy specialization requested by Entry 1793 already be read as an endpoint of the exact degree-six g5g_5 eliminant on the frozen cyclic family?

Exact projective test

The cyclic coordinate is

x=t2,x=t^2,

and total energy is t=0t=0, hence x=0x=0. Let

Lg5(x)=c0+c1x+⋯+c6x6L_{g_5}(x)=c_0+c_1x+\cdots+c_6x^6

be the exact saturated eliminant over Q(5)\mathbb Q(\sqrt5).

The field norms of its endpoint coefficients are

Nm⁡(c0)=3393248617257280984422809600≠0,\operatorname{Nm}(c_0) =3393248617257280984422809600\neq0,

Nm⁡(c6)=160023488695813552472064≠0.\operatorname{Nm}(c_6) =160023488695813552472064\neq0.

Therefore the projective closure has neither an x=0x=0 point nor a point at projective infinity.

D‾g5 has no total-energy endpoint on the frozen cyclic line.\boxed{ \overline{\mathscr D}_{g_5} \text{ has no total-energy endpoint on the frozen cyclic line.} }

All six projective slice intersections are finite and nonzero.

Scope and consequence

This is not a claim that the full multivariate divisor misses total energy. It proves that any such intersection must leave the cyclic slice. The next comparison must retain at least one independent noncyclic kinematic normal; specializing the degree-six slice polynomial cannot type it.

No new carrier structure is indicated.

Next falsifier

Construct the first noncyclic deformation of the source-labelled g5g_5-Ge12−G^-_{e_{12}} Landau system. Eliminate the loop variables while retaining total energy and one independent shape normal, then compute the completed intersection with ET=0E_T=0.

Evidence

View source on GitHub