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

Higher powers and vanishing complex power sums.

Lean checks only the quadratic case; the analytic argument reaches further. To estimate $G^{p}$ for $p > 0$ , we need an analytic function whose squared modulus is $G^{p}$ . Using the same logarithms as above, define

H_{p} = g^{p / 2} = 1 + \sum_{ν \geq 1} b_{ν} z^{ν} .

For interior centres, logarithmic differentiation and the Poisson majorant give

\begin{aligned} P & = 1 - \frac{4}{p} ℜ \frac{ζ H_{p}^{'}}{H_{p}}, \\ \sum_{j} w_{j} G (c_{j})^{p} & \leq \int | H_{p} |^{2} P d m = 1 - \sum_{ν \geq 1} (\frac{4 ν}{p} - 1) | b_{ν} |^{2} . \end{aligned}

The identity follows from the same two orthogonality calculations used for $g$ ; only the coefficient of the derivative term changes. For $0 < p \leq 4$ every displayed coefficient is nonnegative. Hence

\sum_{j} w_{j} G (c_{j})^{p} \leq 1 (0 < p \leq 4) .

Boundary centres follow by replacing $c_{j}$ by $r c_{j}$ : the degree- $ν$ coefficient becomes $r^{ν} b_{ν}$ , and the values on the left are $G_{r} (r c_{j}) = \prod_{k} | 1 - r^{2} \overset{―}{c_{k}} c_{j} |^{w_{k}}$ . First retain finitely many nonnegative terms, then let $r ↑ 1$ . Zero weights can be omitted throughout. This proves the closed-disc inequality without differentiating a boundary logarithm. The argument is the power-parameter Poisson identity; Lean does not check it.

In particular, equal weights and Lemma 10.3 give, with $m = n - 1$ ,

\frac{1}{m} \sum_{j} | f (c_{j}) |^{4 / m} \leq \frac{1}{m} \sum_{j} G (c_{j})^{4} \leq 1

on the unit disc. For roots in $\overset{―}{D} (h, R)$ this becomes

\sum_{j} | f (c_{j}) |^{4 / (n - 1)} \leq (n - 1) R^{4 n / (n - 1)} .

As before, $R > 0$ is handled by $R^{- n} f (h + R z)$ , and $R = 0$ by the vanishing of all critical values. The constant is attained by $(z - h)^{n} - λ$ with $| λ | = R^{n}$ . Thus the constant is sharp. Optimality of the exponent at each fixed degree is not asserted.

At $p = 4$ the coefficient of $| b_{1} |^{2}$ vanishes, so the previous quadratic equality proof cannot simply be reused. If equality holds, the finite-deficit bounds imply $H_{4} = 1 + b_{1} z$ . If $b_{1} \neq 0$ , the rational identity $H_{4}^{'} / H_{4} = 2 g^{'} / g$ shows, by comparing poles and their residues, that all nonzero centres coincide at $a = - \overset{―}{b_{1}}$ and have total weight $1 / 2$ . The other half of the weight is at zero. Their fourth-power average is

\frac{1}{2} (1 + (1 - | a |^{2})^{2}) = 1 - | a |^{2} + \frac{1}{2} | a |^{4} < 1 (0 < | a | \leq 1),

a contradiction. Thus $H_{4} = 1$ , and the pole argument already used for $g$ forces every centre of positive weight to be zero. Conversely that configuration gives equality. It follows that the fourth-power critical-value estimate has the same equality family as the quadratic one for a fixed containing disc.

For $p > 4$ , the coefficient $4 / p - 1$ is negative, so the nonnegative-deficit argument no longer applies. To extend its range we can force the first Taylor coefficients to vanish. Let $s \geq 1$ be an integer and suppose

\sum_{j} w_{j} c_{j}^{k} = 0 (1 \leq k < s) .

The expansion

\log H_{p} (z) = - \frac{p}{2} \sum_{k \geq 1} \frac{\overset{―}{\sum_{j} w_{j} c_{j}^{k}}}{k} z^{k}

then has no terms of degree below $s$ , so $b_{1} = \dots = b_{s - 1} = 0$ . All remaining coefficient weights are nonnegative for $0 < p \leq 4 s$ . The same proof therefore gives $\sum_{j} w_{j} G (c_{j})^{p} \leq 1$ throughout that range. For a monic polynomial with roots in $\overset{―}{D} (h, R)$ , $R > 0$ , the resulting endpoint is

\sum_{j} | f (c_{j}) |^{4 s / (n - 1)} \leq (n - 1) R^{4 s n / (n - 1)}, \sum_{j} (c_{j} - h)^{k} = 0 (1 \leq k < s) .

There is no cancellation hypothesis when $s = 1$ . For $s = 2$ , the condition says that the critical-point centroid is the disc centre. Comparing the coefficients of $f$ and $f^{'}$ shows

\frac{1}{n - 1} \sum_{j} c_{j} = \frac{1}{n} \sum_{i = 1}^{n} z_{i},

so this is also the root centroid. Recentring can increase the required radius: for $0 < r < 1$ , the roots of $(z - r)^{2} (z + r)$ lie in $\overset{―}{D} (0, r)$ , but their centroid is $r / 3$ , and a disc centred there needs radius $4 r / 3$ to contain the root $- r$ . Thus the exponent- $8 / (n - 1)$ statement cannot be applied at the old centre with the old radius. The hypotheses require exact cancellation of the moments; upper bounds for their moduli are insufficient. Lean does not check the higher-power conclusions in this paragraph; the Lean proof of Theorem 10.2 has exponent $2 / (n - 1)$ .

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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