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Auxiliary incidence bound at every scale

Proved
Esgk.maximally_circular_raw_auxiliary_incidence_bound

by mysticflounder · Oct 16, 2026 · Mathlib 0df444a (Lean v4.33.1)

combinatoricsgeometry

There is an absolute positive integer KKK with the following property.

Let GGG be an irreducible and absolutely irreducible real plane polynomial of degree ddd, with real zero set CCC. Assume that GGG is not a circle factor and is maximally circular: its local order at each of the two complex circular points at infinity is d/2d/2d/2. Let S⊂CS\subset CS⊂C be finite and put N=∣S∣N=|S|N=∣S∣.

Take any incidence configuration whose two carrier sets are both CCC, each represented by an irreducible real polynomial of degree at most ddd. Its finite point sets P1,P2P_1,P_2P1​,P2​ are subsets of CCC and form a disjoint split S=P1⊔P2S=P_1\sqcup P_2S=P1​⊔P2​. The carrier is neither a line nor a circle: in particular neither copy is a vertical line, the pair is neither parallel lines, perpendicular lines, nor concentric circles, and the conditional restrictions on line or circle carriers hold automatically. Assume

0<∣P1∣∣P2∣,∣P1∣≤2∣P2∣,∣P2∣≤2∣P1∣.0<|P_1||P_2|,\qquad |P_1|\le2|P_2|,\qquad |P_2|\le2|P_1|.0<∣P1​∣∣P2​∣,∣P1​∣≤2∣P2​∣,∣P2​∣≤2∣P1​∣.

Let III count ordered pairs of a label (a,b)∈P1×P1(a,b)\in P_1\times P_1(a,b)∈P1​×P1​ and a target (p,q)∈P2×P2(p,q)\in P_2\times P_2(p,q)∈P2​×P2​ satisfying ∥a−p∥2=∥b−q∥2\|a-p\|^2=\|b-q\|^2∥a−p∥2=∥b−q∥2. For every integer j≥24dj\ge24dj≥24d,

I≤K(dN3j+d2jN2+N2+d4j2+d3N2).I\le K\left(\frac{dN^3}{\sqrt j}+d^2jN^2+N^2+d^4j^2+d^3N^2\right).I≤K(j​dN3​+d2jN2+N2+d4j2+d3N2).

The constant KKK is independent of the polynomial, sample, incidence configuration, and scale. The balanced cell-cover construction supplies such a configuration for every sample with N≥2N\ge2N≥2.

Preamble
import Definitions.Def_Esgk_MaximallyCircularQuarticFromIncidence

open Esgk

Formal statement
theorem Esgk.maximally_circular_raw_auxiliary_incidence_bound :
    MaximallyCircularRawAuxiliaryIncidenceStatement := by sorry

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