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

Will Cook

Long paper

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

Erdős #1041 By Will Cook 29 September 2026 Authorship, AI use and citation

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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.

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In this paper

Polynomials obtained from a cubic by a power substitution

For $f (z) = P ((z - h)^{q})$ , each nonzero root of $P$ gives a regular $q$ -gon of roots centred at $h$ . We exploit this symmetry by lifting one contained segment for the cubic into two segments through $h$ . A generic polynomial of degree $3 q$ has no such fibre structure. The first theorem supplies the segment to be lifted.

Theorem 9.6 (a contained radial segment for a cubic). If $r, s, v \in C$ have modulus below one, at least one $u \in {r, s, v}$ satisfies

| (t u - r) (t u - s) (t u - v) | \leq 1 (0 \leq t \leq 1) .

Theorem 9.7 (a cubic composed with a power map). Let $q \geq 2$ , $h \in C$ , $P$ be monic cubic, and

f (z) = P ((z - h)^{q}) .

If every zero of $f$ lies in the open unit disc and $f$ has at least two distinct zero values, then two zeros are joined through $h$ by a two-segment path of length below $2$ inside ${| f | < 1}$ . Equivalently, under the same hypotheses the conclusion holds for polynomials of the form

(z - h)^{3 q} + A (z - h)^{2 q} + B (z - h)^{q} + C

in every degree $3 q \geq 6$ .

Proof. One nonzero root of $P$ will supply both endpoints of the path through $h$ . Before selecting it, we check that every zero of $P$ lies in the open unit disc. Fix a $q$ th root $y$ of a zero of $P$ and a primitive $q$ th root of unity $ζ$ . The full fibre consists of $h + y ζ^{k}$ . Since every fibre point lies in the open unit disc,

\begin{matrix} (14) & \frac{1}{q} \sum_{k = 0}^{q - 1} | h + y ζ^{k} |^{2} = | h |^{2} + | y |^{2} < 1. \end{matrix}

Thus $| y | < 1$ , and the corresponding quotient root has modulus $| y |^{q} < 1$ . This verifies the open-disc hypothesis for all quotient roots before selecting a radial segment for the cubic. The identity retains more information than these separate bounds: $| h | < 1$ and every fibre radius is strictly below $\sqrt{1 - | h |^{2}}$ . Consequently, any two contained fibre segments constructed below have total length strictly less than $2 \sqrt{1 - | h |^{2}}$ . The position of the symmetry centre therefore improves the length bound; replacing it by $2$ discards this information.

Write $P (w) = (w - r) (w - s) (w - v)$ and associate to $r$ the real number $A_{r} = ℜ (r \overset{―}{s + v})$ , cyclically. The exact sum is

A_{r} + A_{s} + A_{v} = | r + s + v |^{2} - (| r |^{2} + | s |^{2} + | v |^{2}) > - 3,

so one of these numbers, say $A_{r}$ , exceeds $- 1$ . For $0 \leq t \leq 1$ ,

\frac{| t r - s |^{2} + | t r - v |^{2}}{2} = t^{2} | r |^{2} + \frac{| s |^{2} + | v |^{2}}{2} - t A_{r} < 1 + t + t^{2} .

AM–GM now gives $| t r - s | | t r - v | < 1 + t + t^{2}$ . Consequently, for $0 \leq t < 1$ ,

\begin{matrix} (15) & | P (t r) | < (1 - t) (1 + t + t^{2}) = 1 - t^{3} \leq 1. \end{matrix}

At $t = 1$ the polynomial vanishes. Hence the entire segment, including the origin, lies in the open unit lemniscate.

For a nonzero quotient root satisfying the segment bound from (15), two distinct fibre points satisfy

f (h + t y ζ^{k}) = P (t^{q} r),

and the two segments through $h$ have total length $2 | y | < 2 \sqrt{1 - | h |^{2}} \leq 2$ . If the selected quotient root is zero, choose a nonzero quotient root $s$ and write $P (w) = w (w - s) (w - v)$ . For $0 < t < 1$ ,

| P (t s) | = t (1 - t) | s |^{2} | t s - v | < 2 t (1 - t) \leq \frac{1}{2} .

The values at both endpoints are zero, so this nonzero root supplies two distinct fibre points. If no nonzero quotient root exists, $f$ has only one distinct zero value, contrary to the hypothesis. ◻

Scope.

The theorem gives length below $2$ . Lean does not check the sharper bound $2 \sqrt{1 - | h |^{2}}$ obtained in the proof.

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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