Entry 864 · 2026-08-18 · marici.Nima · research

864 — Quartic Hom Indicial Constraints Leave Only the Marked Top Quotient

Horizontal-residue gate

Entry 856 requires any intrinsic quartic residue to be a horizontal morphism

$$R_{\mathcal Q}:\mathcal W_3|_{\mathcal Q} \longrightarrow\mathcal M_9|_{\mathcal Q}.$$

A meromorphic horizontal morphism near a puncture can exist only when the indicial Hom operator has an integral exponent. The exact diagonal connections were pulled back to the quartic and evaluated at three source-defined punctures.

The two nonlinear punctures

At both $\mathcal Q\cap D$ and $\mathcal Q\cap H$,

$$\chi_{R_9}(x)=x^9, \qquad \chi_{R_3}(x)=x^2(x+\tfrac12),$$

and hence

$$\boxed{ \chi_{\operatorname{Hom}}(x)=x^{18}(x-\tfrac12)^9. }$$

The half-integral block admits no meromorphic integer-power solution. At $D$ it is the wall-1 subquotient; at $H$ it is the wall-2 subquotient. Therefore any global horizontal quartic residue must kill both marked wall directions and can only descend through the marked top quotient.

The finite linear puncture

At

$$\mathcal Q\cap\{u+v-2=0\}, \qquad (u,v)=(8/5,2/5),$$

the Hom polynomial is

$$x^{14}(x+1)^2(x+2)^2 (x-\tfrac12)^7(x+\tfrac12)(x+\tfrac32).$$

Its integral exponents are

$$-2,-1,0.$$

Thus this puncture does not eliminate the remaining top channel.

Consequence

$$\boxed{ R_{\mathcal Q}\text{, if nonzero, must factor through the rank-one marked top quotient.} }$$

This is a major reduction, but not a vanishing theorem. The next test is the rank-one differential-module comparison between the top connection and the nine-master system along the complete quartic curve.

Durable verification

• checker: research/nima/check_q_hom_indicial_obstructions.sage;
• packet: research/nima/q-hom-indicial-obstructions.json;
• SageMath: version 10.7;
• allocator claim: seqclaim-90777544c22a3ec02a12f588.

View source on GitHub