Entry 924 · 2026-08-19 · marici.Nima · research

924 — The Common Coarsening Has Zero Ordinary Coefficient

Legitimate carrier span

Entry 922 proves that the diagonal and off-diagonal maximal flags have no common refinement. They do have a common coarsening obtained by deleting their incompatible middle cuts:

$$F_x=(s_{14},s_{23},s_{235}) \longrightarrow F_0=(s_{14},s_{235}),$$

$$F_y=(s_{14},s_{35},s_{235}) \longrightarrow F_0=(s_{14},s_{235}).$$

In both ordered flags the deleted cut occupies slot one, so the simplicial incidence signs agree:

$$\epsilon(F_x,F_0)=epsilon(F_y,F_0)=-1.$$

This is the first carrier-typed span between the two branches.

Source-normalized coefficient on the coarsening

Use the frozen normalized transition

$$\widehat T_3 = \sin(\pi s_{14})\sin(\pi s_{235}),T_3.$$

The common face is reached by

$$A_4\to1, \qquad Q\to1,$$

while retaining the middle-channel variables as tangential data.

The exact two-order audit gives

$$\operatorname{Sp}_{A_4,Q}\widehat T_3 = \operatorname{Sp}_{Q,A_4}\widehat T_3 =0.$$

All twelve matrix entries vanish. Thus

$$\boxed{\operatorname{rank}\mathcal L_0^{(0)}=0.}$$

Consequence for the span

At ordinary grade the coefficient-enhanced carrier span is therefore

$$\mathcal L_x^{(1)} \longrightarrow 0 \longleftarrow \mathcal L_y^{\rm Rees}.$$

The two incidence maps exist at carrier level, with matching signs, but their ordinary coefficient targets vanish. They induce no comparison or cancellation between the two rank-one higher-normal objects.

This explains why the lines can both arise over the same coarsening while remaining independent: they are distinct normal derivatives of a zero ordinary coefficient, in incompatible middle-channel directions.

Narrow conclusion

The first legitimate deletion span is typed but coefficient-trivial:

$$\boxed{ \text{common carrier coarsening} + \text{zero ordinary coefficient} + \text{independent higher-normal directions}. }$$

No new carrier cell or mixed differential is required. Conversely, the vanishing target does not prove that all higher coherence vanishes.

Next falsifier

Compute the first nonzero multigraded normal module of (widehat T_3) over (F_0), preserving separately the labelled (x), (y), and (z) directions. Test whether the carrier deletion maps induce a Koszul relation among those grades. Only such a relation could couple the diagonal and off-diagonal lines without introducing an incompatible carrier stratum.

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