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

Why the proposed spanning-tree estimate fails

The spanning-tree estimate in [10] fails already for a quadratic Cassini lemniscate. That manuscript, posted by shtuka on 24 March 2026, claims the unrestricted statement [10]. The defect was located publicly in the problem’s discussion thread: on 25 March 2026 Tao observed that the invocation of Lemma 8 there is unjustified and that the flow lines need not organise into connected trees, and on 26 March 2026 the manuscript’s author agreed that the statement of Proposition 12 itself is incorrect. We give the explicit numerical contradiction first. We then examine the saddle model and give an unformalised decomposition into inverse sheets, including the arcs that join them.

Recent work on polynomial lemniscates separates component counts from metric path questions. Ghosh and Ramachandran record that the open set ${| f | < 1}$ has $1 + # {j : | f (c_{j}) | \geq 1}$ components, where $c_{1}, \dots, c_{n - 1}$ are the critical points listed with multiplicity [5]; the underlying component-wise Riemann–Hurwitz count appears in the proof of [8]. For the binomial family $z^{n} - a$ , the condition $| a | < 1$ therefore puts the filled unit lemniscate in the connected regime. Connectedness alone gives no path-length bound. Bishop, Eremenko and Lazebnik describe the possible shapes: every rational lemniscate is a lemniscate graph whose vertices are the critical points on the level set, each of even degree at least four [20], and a lemniscate graph is realised by a polynomial lemniscate up to a homeomorphism of the plane exactly when it is the boundary of its unbounded face [20]. This topological description also gives no path-length bound.

The March manuscript’s Proposition 12 claims the following estimate. For $u = - \log | f |$ , a connected component $V$ of ${u > c}$ carrying $m \geq 2$ simple zeros, and the regularity and Morse hypotheses in that proposition, it constructs, for every $ε > 0$ , an embedded spanning tree $G_{ε} \subset V$ with

\begin{matrix} (5.1) & len (G_{ε}) \leq \frac{1}{2 π} \int_{2 α}^{\infty} P_{V} (t) d t + ε . \end{matrix}

Here $P_{V} (t) = H^{1} (V \cap {u = t})$ is the length of the level set in $V$ , and $α > 0$ is the truncation parameter in the proposed estimate. The final theorem depends on this bound.

The estimate itself is false. Take $f_{a} (z) = z^{2} - a^{2}$ with $a = 9 / 10$ and choose $2 α = - \log a$ . For $0 < c < α / 2$ , the component $V$ of ${- \log | f_{a} | > c}$ contains ${| f_{a} | \leq a}$ and both roots. The only critical point is a nondegenerate saddle at $0$ ; distinctness of critical values is vacuous.

Here the level-length integral can be bounded directly. Put $r = e^{- t}$ . On a regular level the inverse parametrisations $z = \pm \sqrt{a^{2} + r e^{i θ}}$ give

P_{V} (t) = r \int_{0}^{2 π} | a^{2} + r e^{i θ} |^{- 1 / 2} d θ .

The elementary identity

| a^{2} + r e^{i θ} |^{2} = (a^{2} + r)^{2} \cos^{2} (θ / 2) + (a^{2} - r)^{2} \sin^{2} (θ / 2)

implies

| a^{2} + r e^{i θ} |^{- 1 / 2} \leq (a^{2} + r)^{- 1 / 2} | \cos (θ / 2) |^{- 1 / 2} .

The angular factor has mean at most $2$ . Indeed, symmetry reduces its integral to $4 \int_{0}^{π / 2} (\cos s)^{- 1 / 2} d s$ , and $\cos s \geq 1 - 2 s / π$ bounds the latter integral by $4 π$ . Changing variables from $t$ to $r$ therefore gives

\frac{1}{2 π} \int_{2 α}^{\infty} P_{V} (t) d t \leq 2 \int_{0}^{a} \frac{d r}{\sqrt{a^{2} + r}} = 4 (\sqrt{a^{2} + a} - a) < \frac{41}{25} .

The isolated critical level is handled by the corresponding improper integral. Every connected set containing both roots has length at least $2 a = 9 / 5$ . Since $9 / 5 - 41 / 25 = 4 / 25$ , any $0 < ε < 4 / 25$ contradicts (5.1). Thus neither a different local saddle neighbourhood nor a perfect topological decomposition can recover the printed coefficient $1 / (2 π)$ . The exact numerical inequality and the resulting contradiction for a tree-length bound also have a Lean proof, in the repository’s research corpus, a library the checked build does not compile. The level-length majorant is an explicit analytic input.

At an interior index-one critical point $p$ , the proof invokes a Morse chart

u = u (p) + x^{2} - y^{2}

and replaces the saddle by a three-ended neighbourhood having one connected lower cross-section and two connected upper cross-sections. That local model is false as written. Because $p \in V$ and $V$ is open, a sufficiently small closed disc around $p$ lies entirely in $V$ . In that disc the full Morse chart has four sectors: two components of $u > u (p)$ and two components of $u < u (p)$ . The level set ${u = u (p)}$ is a lemniscate graph with a vertex of degree four at $p$ [20]. A global component argument cannot delete one local sector from a disc already contained in $V$ .

The four-sector calculation identifies an error in the proof, while the Cassini example above disproves the printed metric inequality itself. Possible alternative constructions include cutting an adjoining regular annulus along a separatrix or regular flow arc, retaining a four-pronged saddle neighbourhood, or using the inverse-sheet decomposition below. None can restore (5.1). A revised estimate would have to include the length needed to join the pieces, seek only one short pair instead of a tree spanning all roots, or use a different global metric inequality. The shorter descriptions of the same three-ended block do not repair the four-sector topology.

The following ordinary topological argument assumes that the critical values have pairwise distinct arguments and moduli and that the critical points are simple. Its ingredients are classical: the Newtonian graph of Shub, Tischler and Williams [7], the component-wise Riemann–Hurwitz count in the proof of [8], and the division of the Riemann surface of the inverse function into sheets by outward critical-value slits. Dubinin describes the sheet adjacency tree directly in [21] and uses a related network for capacity in [9]. The latter paper assumes bounded critical values and controls capacity; it does not supply a Euclidean tree-length bound. The remark below specifies the sheets, critical transpositions and descent arcs for the stated component.

The assumptions exclude a component containing a multiple critical point or two critical values on the same ray. For $z^{n} - b$ with $n > 2$ and $0 < | b | < 1$ , the component containing the origin fails the first condition. They hold, for example, for $f (z) = z^{3} - (3 / 25) z + 1 / 500$ : its critical points are $\pm 1 / 5$ and its critical values are $- 7 / 500$ and $9 / 500$ . On the unit circle the two lower terms have total modulus at most $61 / 500 < 1$ , so all three roots lie in the open unit disc by Rouché’s theorem. In the statement, “excellent” means that the critical levels of the Morse function are pairwise distinct. This regularity permits a decomposition into sheets; it does not bound the total length of the resulting tree.

Remark 13.1 (unformalised decomposition into inverse sheets). Let $f$ be monic, and let $U$ be a component of ${| f | < 1}$ containing $k \geq 2$ roots, counted with multiplicity. Suppose every critical point of $f$ in $U$ is simple, its critical value is nonzero, and these critical values have pairwise distinct arguments and pairwise distinct moduli. All preimages and sheets below are taken inside $U$ , and only critical points in $U$ determine the cuts. Then:

$- \log | f | : U \ f^{- 1} (0) \to (0, \infty)$ is a proper excellent Morse function, with exactly $k - 1$ nondegenerate saddles;
cutting $D ∖ {0}$ along the critical-value rays decomposes its preimage in $U$ into conformal strips;
cutting each ray only from its critical value to the unit circle gives $k$ conformal sheets, one per root, whose critical transpositions form a tree;
for each critical point $c \in U$ , the two inverse lifts of $[0, f (c)]$ join two roots through $c$ inside $U \cap {| f | \leq | f (c) |}$ , and the union of these arcs is an embedded geometric realisation of that tree.

Small neighbourhoods of the saddles can be chosen with diameter $O (\sqrt{δ})$ at value radius $δ$ .

Inverse-branch prerequisites.

For a nonconstant complex polynomial, each component of ${| f | < 1}$ contains a root. The restriction to that component is onto the unit disc and has finite fibres. Over values avoiding the critical values attained in that component, it is a covering map. Choose finitely many nonzero slit starts that include every critical value attained in the component, and remove the outward rays beginning at those starts from the value disc. On the remaining domain, each root in the component determines a unique continuous inverse branch, taking $0$ to that root. The archived Lean source states these prerequisites and supplies proof scripts; a kernel-check receipt for this module is not supplied here. The full conformal sheet decomposition, sheet count, monodromy and embedded-tree assertions remain ordinary arguments here; the prerequisites give no length estimate.

Proof of Remark 13.1. The map $f : U \to D$ is proper of degree $k$ , and $U$ is simply connected. To apply the count from the proof of [8] only at a regular level, join all roots and critical points in $U$ by finitely many compact paths in $U$ . Choose a regular $t < 1$ above the maximum of $| f |$ on their union. One component of ${| f | < t}$ then contains exactly the roots and critical points of $U$ ; Riemann–Hurwitz there gives $k - 1$ simple critical points. No regularity of the level-one boundary is needed. The holomorphic Morse lemma makes these points nondegenerate saddles of $- \log | f |$ , and their distinct moduli separate the critical levels. A compact range in $(0, \infty)$ stays away from both the level-one boundary and the deleted roots, proving properness of the Morse function.

Remove $0$ and the rays determined by these $k - 1$ values. Each remaining sector is simply connected and contains no branch value of $f : U \to D$ , even if it contains critical values coming from outside $U$ . Each component of its preimage in $U$ maps biholomorphically to the sector. If the sector has angular range $(θ_{1}, θ_{2})$ , the holomorphic coordinate $- \log f$ maps this component onto the semi-infinite strip

{x + i y : x > 0, - θ_{2} < y < - θ_{1}} .

Thus its real coordinate is $- \log | f |$ , while its imaginary coordinate is $- \arg f$ . The minus sign in both coordinates is needed for a holomorphic map; keeping $+ \arg f$ would reverse orientation.

For the sheet tree, remove only the outward slits ${t f (c) : 1 \leq t < 1 / | f (c) |}$ , for critical points $c \in U$ . The remaining value domain is star-shaped about $0$ , hence simply connected, and its preimage in $U$ splits into $k$ sheets labelled by the roots. Local monodromy at each slit endpoint is a transposition of two of these sheets. Deleting the finitely many fibres over the branch values from $U$ leaves a connected domain, so the monodromy action is transitive. The $k - 1$ transpositions therefore give a connected graph with $k$ vertices, which is a tree.

For $c \in U$ , the inward segment from $f (c)$ to $0$ contains no other branch value of $f : U \to D$ , by the distinct-argument hypothesis. Distinct moduli were needed only to separate the Morse levels. Its two inverse lifts converge to $c$ by the holomorphic Morse coordinate and start at the two root labels exchanged by its transposition. These labels are distinct, as are the roots, because a root in $U$ cannot also be a critical point under the nonzero-critical-value hypothesis. Lifts belonging to different critical points cannot cross away from roots, since an intersection would give one nonzero value on two critical rays. Their union is therefore an embedded tree. Each edge is rectifiable: at a simple critical endpoint, parametrising the value segment by its distance $s$ from the critical value gives speed $O (s^{- 1 / 2})$ , which is integrable. The root endpoint is regular, and the derivative is bounded on compact subsegments between the endpoints. This proves finiteness of each length, not a uniform bound for it. The same local Morse coordinate identifies a value disc of radius $δ$ with a spatial neighbourhood of diameter $O (\sqrt{δ})$ . The constant and the permitted range of $δ$ depend on $f$ and the chosen saddle. Indeed, in the coordinate $f (z) - f (c) = u^{2}$ , the inverse map has derivative of modulus $\sqrt{2 / | f^{″} (c) |}$ at $u = 0$ . No bound uniform over degenerating critical points is asserted. For a fixed polynomial there are finitely many saddles, so their individual constants have a finite maximum. ◻

The complete argument is also written out in the proof of the decomposition into inverse sheets. This remark supplies a topological construction with no useful bound on the sum of lengths: the canonical inverse-ray tree can already have the Cassini deficit discussed above.

The component restriction cannot be dropped. For example, take

f (z) = (z^{2} - 4)^{2} - \frac{1}{4} .

Its critical points are $0, \pm 2$ , with values $63 / 4, - 1 / 4, - 1 / 4$ . The positive roots $\sqrt{7 / 2}$ and $\sqrt{9 / 2}$ lie in one component $U$ : on the intervening real interval, $- 1 / 4 \leq f \leq 0$ . But $f (i y) = (y^{2} + 4)^{2} - 1 / 4 > 1$ for real $y$ , so this component cannot contain the negative roots or $- 2$ . Its only critical point is the simple point $2$ . Thus the theorem applies in $U$ , although the other component has the same critical value. This example concerns component topology; no root-unit-disc hypothesis is imposed here.

Corollary 11.2 also excludes finite saddle-to-saddle Newton connections inside $U$ . The theorem does not treat coincident arguments, simultaneous critical levels or multiple saddles in that component. Nor does it identify a slit domain with the quotient by all trajectories: the quadratic example following Problem 15.5 satisfies its hypotheses but has a non-Hausdorff orbit quotient. Boundary tangencies and quantitative estimates for joining the strips remain separate questions. The embedded tree supplies no length bound for the historical question.

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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