Plectis

Long paper

Paths in Polynomial Lemniscates: A Degree-Seven Counterexample and Radial Connections

Summary. The record gives the rational estimates and geometric proof for ani's degree-seven polynomial: every connected set in its strict unit lemniscate joining two roots has Hausdorff measure greater than two. It develops contained paths for trinomials, collinear roots and two sparse families. Ordinary analytic arguments for critical-value criteria retain their unformalised inputs beside the statements. Critical-value moments, counterexamples to prescribed paths and questions about flow-strip geometry complete the account. The one-polynomial formal statement and correspondence with the historical wording remain distinct.

This paper covers the exact geometry of ani's one-polynomial counterexample, proved positive families and critical-value arguments with their unformalised inputs identified; formal-statement status and historical correspondence remain owned by the claims registry.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or the current status of Erdős #1041, including the later one-polynomial Formal Conjectures path-image Hausdorff-length check and its unreviewed historical curve-length correspondence; those distinctions belong to docs/claims.json in the mathematics repository.

In this paper

The distance bound for the two nearest roots.

Corollary 10.6 (two nearest roots). If the roots lie in the open unit disc and c is a non-root critical point, the two nearest roots to c have total distance strictly below 2.

Let f have degree n≥2 and all roots in the closed unit disc, and let c be a critical point with f(c)≠0. List roots with multiplicity. Its two nearest roots have total distance at most 2. The root-disc hypothesis, rather than a bound on |f(c)|, is what gives this estimate.

Rotate so that c=t≥0, and order the distances as 0<d1≤d2≤⋯≤dn. Logarithmic differentiation gives ∑j(zj−c)−1=−f′(c)/f(c)=0. Dividing |zj−c|2+2tℜ(zj−c)≤1−t2 by dj2, then summing and using the reciprocal balance, gives

n≤(1−t2)∑jdj−2.

Thus t<1 and d1≤1−t2≤1. The same reciprocal balance, with the nearest term isolated, gives d2≤(n−1)d1.

Suppose d1+d2>2. Since d2≤1+t, we have d1>1−t, and hence

1−t2<d1(2−d1)<d1d2.

The first strict inequality uses that x(2−x) is increasing on [0,1]. It follows that

n≤(1−t2)(d1−2+(n−1)d2−2)<d2d1+(n−1)d1d2≤n.

For the last step, put x=d2/d1∈[1,n−1] and expand (x−1)(x−(n−1))≤0. This contradiction proves the closed-disc bound. For roots in the open unit disc, scaling by a containing radius 0<R<1 gives d1+d2≤2R<2. A research note gives the same argument. Lean checks the strict distance bound from the critical polynomial, including the root-disc reduction and the choice of the two nearest indices.

This argument selects the two nearest roots but does not put their segments inside a polynomial sublevel set. The next calculation supplies that separate containment estimate under a bound on |f(c)|.

About this paper

Authorship and AI use. Will Cook built and directed the research infrastructure and maintains the public release. He reviewed claims when he could. AI agents did most of the research and drafting. Cook did not independently verify every claim.

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