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

Will Cook

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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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An isolated simple critical value

A disc containing $0$ and one simple critical value selects a component of degree two. Taking a square root removes its local branching and gives a coordinate in which the proposed connecting curve is a real segment. The ordinary argument below then bounds its length by area and capacity. Lean assumes this combined analytic construction as DiscSepBergmanArea; it does not derive it from the polynomial hypotheses. We therefore record the path criterion and its numerical consequences as unformalised remarks.

The simplicity assumption means that the chosen critical point is a simple zero of the derivative, so two inverse branches meet there. A multiple chosen critical point is excluded; the other critical points need not be simple. Also, separation of roots in the $z$ -plane is not enough: different critical points can have nearly equal values. An exact quartic example below makes this distinction explicit. For a positive example, take $f (z) = z^{3} - 3 a^{2} z$ with $0 < a < 1 / \sqrt{3}$ . Its roots $0, \pm \sqrt{3} a$ lie in the unit disc. At $c = a$ the critical value is $v = - 2 a^{3}$ , and the other normalised critical value is $- 1$ . Thus the disc centred at $1$ with radius $4 / 3$ contains $0$ and $1$ but excludes the other critical value. Radius $2$ is allowed as well: the other critical value then lies on the boundary. The proof must therefore allow a nonregular outer level.

Remark 7.1 (unformalised isolated-value criterion). Let $P$ be a polynomial of degree $n \geq 3$ whose leading coefficient has modulus one, with

P (0) = 1, P^{'} (0) = 0, P^{″} (0) \neq 0.

Fix $w_{0} \in [0, 1]$ and $S > max (w_{0}, 1 - w_{0})$ . Suppose every other critical point $d \neq 0$ satisfies

\begin{matrix} (4) & | P (d) - w_{0} | \geq S . \end{matrix}

Put $p = w_{0} (1 - w_{0})$ . The two local solutions of $P (Z (ξ)) = 1 - ξ^{2}$ , $Z (0) = 0$ , continue along the real segment to one injective curve $Γ$ joining two roots. Its endpoints are distinct roots, $Γ \subseteq {| P | \leq 1}$ , and

\begin{matrix} (5) & length (Γ)^{2} \leq 2 (\frac{S}{n - 1})^{2 / n} \log \frac{S^{2} + S + p}{S^{2} - S + p} . \end{matrix}

Consequently this curve has length less than $2$ whenever

\begin{matrix} (6) & (\frac{S}{n - 1})^{2 / n} \log \frac{S^{2} + S + p}{S^{2} - S + p} < 2. \end{matrix}

Proof. Let $Q = D (w_{0}, S)$ and let $U$ be the component of $P^{- 1} (Q)$ containing $0$ . Both $0$ and $1$ lie in $Q$ , and (4) says that $0$ is the only critical point in $U$ . A component of a polynomial sublevel set is simply connected: on any Jordan curve in the component, the maximum principle bounds the same polynomial throughout its interior, which must therefore remain in the component. Condition (4) permits critical values on $\partial Q$ , so we must not assume that $\partial U$ is regular. For $1 - w_{0} < S^{'} < S$ , let $U^{'}$ be the component of ${| P - w_{0} | < S^{'}}$ containing $0$ . This level is regular, and $0$ contributes its only unit of ramification. The component count in the proof of [8] therefore gives $\deg (P |_{U^{'}}) = 2$ .

These nested components exhaust $U$ . Indeed, a point of $U$ can be joined to $0$ by a compact path in $U$ ; the maximum of $| P - w_{0} |$ on that path is strictly below $S$ , so the path eventually lies in $U^{'}$ . For any fixed regular value in $Q$ , all its finitely many preimages in $U$ therefore lie in one such $U^{'}$ once $S^{'}$ is sufficiently large. The proper map $P : U \to Q$ consequently also has degree two. This establishes the count without a smoothness assumption at the outer level.

Set $a = 1 - w_{0}$ . On $U$ , the function $1 - P$ has only its double zero at $0$ , so it has a single-valued square root

ξ (z)^{2} = 1 - P (z) .

Taking this square root removes the double branching at $0$ . The map $ξ$ is proper into

\tilde{Q} = {ξ : | ξ^{2} - a | < S} .

Away from $0$ , the identity $2 ξ ξ^{'} = - P^{'}$ gives $ξ^{'} \neq 0$ ; at $0$ , $(ξ^{'} (0))^{2} = - P^{″} (0) / 2 \neq 0$ . The target is star-shaped about $0$ : because $S > a$ , the convex disc $D (a, S)$ contains $0$ , so $t^{2} ξ^{2} \in D (a, S)$ whenever $ξ \in \tilde{Q}$ and $0 \leq t \leq 1$ . It is therefore connected and simply connected. A proper local biholomorphism has open and closed image, hence is a covering of this target. Since $U$ is connected, it is a conformal bijection. Its inverse $Z$ continues both local inverse branches through the critical point. Since $[- 1, 1] \subset \tilde{Q}$ , the curve $Z ([- 1, 1])$ joins the two distinct points over $P = 0$ , and $P (Z (ξ)) = 1 - ξ^{2} \in [0, 1]$ gives its containment.

We next map the parameter domain to the unit disc so that the curve becomes a straight interval. This allows its length to be estimated by the area of $U$ , without replacing the domain by a larger disc. Squaring sends $\tilde{Q}$ onto $D (a, S)$ . To map this disc to $D$ while fixing $0$ , compose $w \mapsto (w - a) / S$ with the disc automorphism taking $- a / S$ to $0$ . The result is the Möbius map

w ⟼ \frac{S w}{S^{2} + a w - a^{2}} .

Taking its square root after substituting $w = ξ^{2}$ gives the conformal bijection

ζ (ξ) = ξ \sqrt{\frac{S}{S^{2} + a ξ^{2} - a^{2}}} : \tilde{Q} ⟶ D,

where the square-root factor is chosen positive at $0$ . It sends $[- 1, 1]$ to $[- q, q]$ , with

q^{2} = \frac{S}{S^{2} + p}, S^{2} - S + p = (S - w_{0}) (S - (1 - w_{0})) > 0.

The last inequality is exactly where $S > max (w_{0}, 1 - w_{0})$ ensures $0 < q < 1$ , as required by the length estimate below. For $Φ = Z \circ ζ^{- 1} : D \to U$ , the Bergman segment inequality, proved by the kernel estimate of Section 4 with the double integral taken over $[- q, q]^{2}$ , gives

\begin{aligned} length (Γ)^{2} & \leq \frac{2}{π} \log \frac{1 + q^{2}}{1 - q^{2}} Area (U) \\ (7) & = \frac{2}{π} \log \frac{S^{2} + S + p}{S^{2} - S + p} Area (U) . \end{aligned}

It remains to bound the area of this component. The factor $n - 1$ comes from the roots outside it, rather than from Pólya’s inequality alone. The auxiliary capacity estimate is as follows: for a monic degree- $n$ polynomial $F$ , a regular value $t > 0$ of $| F |$ , and a component $V$ of ${| F | < t}$ containing $k < n$ zeros counted with multiplicity,

\frac{cap (\overset{―}{V})^{n}}{t} < \frac{k}{2 n - k} .

Multiplying $F$ by a unimodular constant leaves the estimate unchanged. This is a separate component estimate, not part of the Riemann–Hurwitz statement cited above. We give the harmonic-measure argument behind it, following Theorem 2 and Corollary 3 of the component-capacity proof.

Let $m = n - k$ and let $ψ$ map $| ζ | > 1$ conformally onto the exterior of $\overset{―}{V}$ , with positive leading coefficient $cap (\overset{―}{V})$ . The $m$ excluded roots have preimages $ξ_{j}$ with $| ξ_{j} | > 1$ . Put $b_{j} = 1 / \overset{―}{ξ_{j}}$ and form

B (z) = \prod_{j = 1}^{m} \frac{z - b_{j}}{1 - \overset{―}{b_{j}} z} .

Reflection in the unit circle shows that $F (ψ (ζ)) / t$ equals $ζ^{n} \overset{―}{B (1 / \overset{―}{ζ})}$ up to a constant of modulus one: their zeros and pole at infinity agree, and both have modulus one on the boundary. Comparing leading coefficients gives

| B (0) | = \frac{cap (\overset{―}{V})^{n}}{t} .

Regularity of the level permits analytic continuation of $ψ$ across the boundary. On the unit circle,

\frac{d}{d θ} \arg F (ψ (e^{i θ})) = n - | B^{'} (e^{i θ}) | > 0.

Positivity follows because a regular polynomial level is mapped locally in the positive boundary direction. Thus $| B^{'} | < n$ there.

The boundary-fibre identity used next is Lemma 1 of the same component-capacity proof. Here is its harmonic-measure argument. For a continuous real function $h$ on the unit circle, let $H$ be its harmonic extension to the disc. Since $H \circ B$ is harmonic, the mean value property and the Poisson formula give

\frac{1}{2 π} \int_{0}^{2 π} h (B (e^{i θ})) d θ = H (B (0)) = \frac{1}{2 π} \int_{0}^{2 π} h (e^{i ϕ}) \frac{1 - | B (0) |^{2}}{| e^{i ϕ} - B (0) |^{2}} d ϕ .

Differentiating the factors gives the positive angular derivative

\frac{d}{d θ} \arg B (e^{i θ}) = | B^{'} (e^{i θ}) | = \sum_{j = 1}^{m} \frac{1 - | b_{j} |^{2}}{| e^{i θ} - b_{j} |^{2}} > 0.

Changing variables through the $m$ boundary inverse branches in the first integral and comparing the continuous densities therefore yields

\sum_{B (ζ) = w} \frac{1}{| B^{'} (ζ) |} = \frac{1 - | B (0) |^{2}}{| w - B (0) |^{2}} (| w | = 1) .

In particular each boundary fibre consists of $m$ distinct points. The Poisson kernel on the right is smallest at the boundary point opposite $B (0)$ . Choosing $w = - B (0) / | B (0) |$ therefore yields

\frac{m}{n} < \frac{1 - | B (0) |}{1 + | B (0) |}, | B (0) | < \frac{n - m}{n + m} = \frac{k}{2 n - k} .

Here $B (0) \neq 0$ because every excluded-root preimage is finite. This proves the component estimate and explains how the excluded roots enter it.

For the regular inner components $U^{'}$ above, the degree-two count gives exactly two zeros of $P - w_{0}$ , counted with multiplicity. Since $2 < n$ , the preceding estimate applies with $k = 2$ and gives

\frac{cap (\overset{―}{U^{'}})^{n}}{S^{'}} < \frac{k}{2 n - k} = \frac{1}{n - 1} .

Pólya’s area–capacity inequality $Area (K) \leq π cap (K)^{2}$ [2] then yields $Area (U^{'}) < π (S^{'} / (n - 1))^{2 / n}$ . The exhaustion proved at the start and continuity of area from below give, without assuming regularity of the limiting boundary,

Area (U) \leq π (\frac{S}{n - 1})^{2 / n} .

Substitution into (7) proves (5), and (6) makes its right side strictly less than $4$ . ◻

Remark 7.2 (unformalised numerical consequences). Inequality (6) holds for every $n \geq 3$ , every $w_{0} \in [0, 1]$ , and every $4 / 3 \leq S \leq 2$ . Thus, if $f$ is monic with roots in the open unit disc, $c$ is a simple critical point with $v = f (c) \neq 0$ and $| v | < 1$ , and

| \frac{f (d)}{v} - w_{0} | \geq \frac{4}{3}

for every other critical point $d$ , then two roots of $f$ are joined inside ${| f | < 1}$ by a curve of length strictly below $2$ .

Proof. For $S \geq 4 / 3$ and $p \geq 0$ ,

\frac{S^{2} + S + p}{S^{2} - S + p} \leq \frac{S + 1}{S - 1} \leq 7.

Also $S / (n - 1) \leq 1$ when $S \leq 2$ and $n \geq 3$ , while $\log 7 < 2$ . This proves (6). For the unnormalised polynomial, set $z = c + | v |^{1 / n} w$ . Then $P (w) = f (z) / v$ has leading coefficient $| v | / v$ , of modulus one. Scaling the connector multiplies its length by $| v |^{1 / n} < 1$ and takes ${| P | \leq 1}$ to ${| f | \leq | v |} \subset {| f | < 1}$ . The selected value $v$ need not attain the minimum defining $μ$ . ◻

Taking $w_{0} = 1$ and $S = 2$ gives the earlier criterion $| 1 - f (d) / v | \geq 2$ in every degree $n \geq 3$ . If $c$ attains the minimum modulus among all critical values and $v \neq 0$ , the Fekete–resultant bound gives $| v | \leq R^{n} < 1$ when all roots lie in a disc of radius $R < 1$ . Deleting zero critical values before taking the minimum would invalidate that inference. For example, $(z - R)^{2} (z + R)$ with $0 < R < 1$ has critical values $0$ and $32 R^{3} / 27$ , so its least nonzero critical-value modulus exceeds $R^{3}$ . The repeated root already gives a constant path, but it does not give this nonzero value the required bound. For a critical point chosen in another way, $| v | < 1$ must remain an explicit hypothesis. In degree three the choice $w_{0} = 1$ already works at $S = 6 / 5$ , since

(\frac{3}{5})^{2 / 3} \log 11 < 2.

Lean checks this sharper cubic constant.

Remark 7.1 imposes separation in the value plane. Separation of the roots or critical points alone does not imply its hypothesis. To distinguish these conditions, consider the family

g_{ε} (z) = z^{4} - ε z^{3} - 2 z^{2} + 3 ε z + \frac{1}{2}, 0 < ε \leq \frac{1}{32} .

Its derivative and two critical values are

g_{ε}^{'} (z) = (z^{2} - 1) (4 z - 3 ε), g_{ε} (\pm 1) = - \frac{1}{2} \pm 2 ε .

At the endpoints of each of the intervals $(- 3 / 2, - 5 / 4)$ , $(- 3 / 4, - 1 / 2)$ , $(1 / 2, 3 / 4)$ and $(5 / 4, 3 / 2)$ , the values of $g_{ε}$ have opposite signs throughout this parameter range. Each interval therefore contains exactly one root. Together with the critical points $- 1, 3 ε / 4, 1$ , these give seven points at pairwise distances greater than $1 / 4$ , uniformly in $ε$ . Nevertheless the two displayed critical values differ by $4 ε \to 0$ . The earlier numerical choice is the exact member $ε = 1 / 262144$ :

g (z) = z^{4} - \frac{z^{3}}{262144} - 2 z^{2} + \frac{3 z}{262144} + \frac{1}{2}, g^{'} (z) = 4 (z + 1) (z - 3 / 1048576) (z - 1),

for which that difference is $1 / 65536$ . The monic polynomial $2^{- 4} g_{ε} (2 z)$ puts every root inside $| z | < 3 / 4$ and retains a spatial separation greater than $1 / 8$ . For the critical points $\pm 1 / 2$ of the scaled polynomial, the ratio of the two values is unchanged and obeys

\frac{g_{ε} (- 1)}{g_{ε} (1)} = \frac{1 + 4 ε}{1 - 4 ε} \leq \frac{9}{7} < \frac{4}{3} .

The reciprocal ratio lies between $0$ and $1$ . Hence, for any centre $w_{0} \in [0, 1]$ , neither selected value satisfies the theorem’s radius- $4 / 3$ isolation condition: the other normalised value is less than $4 / 3$ away. This does not exclude a suitable isolated value at a different critical point.

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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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