Density of Hyperbolicity for Rational Maps

Density of Hyperbolicity for Rational Maps

Definitions and mathematical statement

Represent a degree-$d$ rational map by $f(z)=P(z)/Q(z)$ with coprime polynomials, modulo common nonzero scalar multiplication, giving the parameter space $\operatorname{Rat}_d\subset{\mathbb{P}}^{2d+1}({\mathbb{C}})$. A critical point is a point of the sphere where the local mapping degree exceeds one. A periodic cycle is attracting if the modulus of its return-map multiplier is less than one. Call $f$ hyperbolic when every critical orbit converges to an attracting cycle. For each $d\ge2$, ask whether \[ \overline{\{f\in\operatorname{Rat}_d:f\text{ is hyperbolic}\}}=\operatorname{Rat}_d \] in the usual complex parameter topology. The target is density, not that every map is hyperbolic or structurally stable.

Formulation and concept references: [S1].

Short English statement

Can every rational dynamical system of a fixed degree be approximated arbitrarily closely by one whose critical points all settle into attracting cycles?

Research attempt

Current frontier.

Structural stability and hyperbolicity must not be identified: the density of stable parameters does not by itself exclude stable non-hyperbolic regions. Benini discusses rigidity, invariant line fields, and the special role of the flexible Lattès family [S1], Sections 3 and 6. A strong result directly relevant to simultaneous critical capture is Aspenberg's Theorem A: a rational Misiurewicz map which is not flexible Lattès can be approximated by hyperbolic maps; when its Julia set is the whole sphere, all critical points of the approximants can be made periodic [E1]. Neither hypothesis is imposed in the target. The searches for 2024--2026 work used for this entry did not identify a verified theorem covering all of $\operatorname{Rat}_d$.

Attack route.

An invariant-line-field route needs a rigidity theorem in precisely the rational-map setting, including the exceptional families; a renormalization route needs the necessary a priori bounds for every recurrent configuration. Here we pursue a more quantitative route: make the remaining critical points periodic by solving simultaneous critical-return equations. The missing lemma is not simply transversality. One needs transversality with control of the second derivatives on a sufficiently large parameter neighborhood. We prove a criterion, test it exactly in a small instance, and then show that the most tempting derivative-growth argument fails even at a familiar parabolic parameter.

Attempt: a simultaneous closing criterion.

Work in a holomorphic parameter slice $\lambda\mapsto f_\lambda$ inside $\operatorname{Rat}_d$, with $\lambda\in{\mathbb{C}}^m$ and center $0$. Assume that all critical points are simple and labeled holomorphically on the neighborhood under consideration. Designate $m$ critical points $c_1(\lambda),\ldots,c_m(\lambda)$ as the ones not yet captured. For each other critical point, assume that at $0$ a finite part of its orbit enters a strictly trapping neighborhood of an attracting cycle. After shrinking the parameter neighborhood, the same is true throughout it. This persistence follows from the strict inclusion defining the trap and continuity of the finitely many initial iterates.

Choose integers $n_i\ge1$ and compatible local coordinates near the initial and return points, and set \[ R_i(\lambda)=f_\lambda^{n_i}(c_i(\lambda))-c_i(\lambda), \qquad R=(R_1,\ldots,R_m). \] The displayed subtraction is in those coordinates; their domains must contain all points used on the closed parameter ball. Suppose $A=DR(0)$ is invertible. For a norm on ${\mathbb{C}}^m$ and its induced operator norms, write \[ \beta=\|A^{-1}R(0)\|,\qquad L=\sup_{\|\lambda\|\le2\beta}\|A^{-1}D^2R(\lambda)\|. \] Closing lemma. If $\beta>0$, the closed ball of radius $2\beta$ lies in the specified coordinate and trapping neighborhood, and \begin{equation}\label{wsref-0-65-71-31-35-35-2d-63-6c-6f-73-65} 2L\beta\le\tfrac12, \end{equation} then some $\lambda_*$ with $\|\lambda_*\|\le2\beta$ makes $f_{\lambda_*}$ hyperbolic. If $\beta=0$, the same conclusion holds at $0$ under the other assumptions.

Proof. For $T(\lambda)=\lambda-A^{-1}R(\lambda)$, the mean value formula along a segment gives \[ \|DT(\lambda)\| =\|A^{-1}(DR(0)-DR(\lambda))\| \le L\|\lambda\|\le\tfrac12. \] Thus $T$ is a contraction on that ball. Moreover $\|T(0)\|=\beta$, so $\|T(\lambda)\|\le\beta+\|\lambda\|/2\le2\beta$. The contraction theorem supplies a fixed point, and invertibility of $A$ gives $R(\lambda_*)=0$. Each marked critical point is periodic, with period dividing $n_i$, so its cycle is superattracting. All other critical points remain in the preserved attracting basins. Hence every critical orbit is attracted to an attracting cycle, exactly as required. $\square$

For example, in a Euclidean norm, a sufficient form of (eq155-close) is \[ 4H\|R(0)\|\le s^2, \qquad s=\sigma_{\min}(DR(0)),\qquad H\ge\sup_{\|\lambda\|\le2\beta}\|D^2R(\lambda)\|. \] Both the size of the coordinate neighborhood and the trapping radii remain separate conditions. The dimension of the slice, independence of its critical-return equations, and a lower bound on $s$ are not automatic.

Attempt: an exact arithmetic test.

For $f_c(z)=z^2+c$, infinity is already a superattracting fixed point, so only the finite critical point $0$ must be closed. The third-return polynomial is \[ R_3(c)=f_c^3(0)=c^4+2c^3+c^2+c. \] Take \[ c_0=-\frac{17549}{10000},\qquad r=\frac1{10000}. \] Exact rational calculation gives \[ R_3(c_0)=\frac{1261775714801}{10^{16}},\qquad R_3'(c_0)=-\frac{1412458905149}{250000000000}. \] On $|c-c_0|\le r$, one has $|c|<2$, and hence $|R_3''(c)|=|12c^2+12c+2|\le74$. With $A=R_3'(c_0)$, \[ \beta=|R_3(c_0)/A|<r/2,\qquad q=74r/|A|<1/2. \] The same contraction proof, now on the fixed radius-$r$ disk, produces a zero of $R_3$ in that disk. Since the disk excludes $c=0$, the critical point has exact period three: the only proper positive divisor of three is one, and $f_c(0)=0$ would require $c=0$. This is a finite certificate of a hyperbolic parameter, not a density argument. The attached checking script verifies the rational inequalities without floating-point decisions.

Attempt: a parabolic stress calculation.

A natural proposed general lemma would say that a sufficiently large parameter derivative makes the critical-return equation solvable nearby. The following calculation shows why that is not enough for the preceding criterion. Put $z_0(c)=0$ and $z_{n+1}(c)=z_n(c)^2+c$. At $c=1/4$, let \[ x_n=\tfrac12-z_n(1/4),\quad a_n=z_n'(1/4),\quad b_n=z_n''(1/4). \] Direct differentiation, with no interchange of infinite limits, gives \begin{align} x_{n+1}&=x_n-x_n^2, & x_0&=\tfrac12,\label{wsref-0-65-71-31-35-35-2d-78}\\ a_{n+1}&=(1-2x_n)a_n+1, & a_0&=0,\label{wsref-0-65-71-31-35-35-2d-61}\\ b_{n+1}&=(1-2x_n)b_n+2a_n^2, & b_0&=0.\label{wsref-0-65-71-31-35-35-2d-62} \end{align} The sequence $x_n$ decreases to zero, since a nonnegative limit in (eq155-x) must have square zero. Moreover \[ \frac1{x_{n+1}}-\frac1{x_n}=\frac1{1-x_n}\longrightarrow1, \] so the Cesàro theorem applied to the telescoping sum yields $nx_n\to1$.

For completeness, use the elementary recurrence fact \begin{equation}\label{wsref-0-65-71-31-35-35-2d-72-65-63-75-72-72-65-6e-63-65} y_{n+1}=\left(1-\frac{\alpha}{n}+o(n^{-1})\right)y_n +c n^k(1+o(1)) \quad\Longrightarrow\quad y_n\sim\frac{c}{\alpha+k+1}n^{k+1}, \end{equation} when $\alpha,c>0$, $k\ge0$, and $y_n\ge0$ eventually. Here is a proof rather than an appeal to a differential-equation approximation. Substitute $C n^{k+1}$ in the recurrence: its increment minus the right-hand increment is $((\alpha+k+1)C-c)n^k+o(n^k)$. Constants just above and below $c/(\alpha+k+1)$ therefore give eventual super- and subsolutions. An initial discrepancy is multiplied by products of eventually positive coefficients at most $1-\alpha/(2n)$; those products tend to zero, and their contribution divided by $n^{k+1}$ tends to zero. Comparison and then letting the two constants approach their common limit proves (eq155-recurrence).

Equations (eq155-a) and (eq155-b) now imply \[ a_n\sim\frac n3,\qquad b_n\sim\frac{2n^3}{45}. \] For the equation $R_n(c)=z_n(c)=0$, at the parabolic center this gives \begin{equation}\label{wsref-0-65-71-31-35-35-2d-66-61-69-6c-75-72-65} \left|\frac{R_n(1/4)}{R_n'(1/4)}\right|\sim\frac{3}{2n} \longrightarrow0, \qquad \frac{|R_n(1/4)R_n''(1/4)|}{|R_n'(1/4)|^2}\sim\frac n5 \longrightarrow\infty. \end{equation} Thus even the pointwise lower bound on the normalized Hessian already prevents (eq155-close) for large $n$. A small first Newton step conceals large distortion. This is a proved failure of this sufficient criterion at these centers, not a proof that the equations have no nearby roots.

Stress test.

The parabolic example is certainly not a counterexample to density. For $0<c<1/4$, the smaller fixed point $\alpha_c=(1-\sqrt{1-4c})/2$ has multiplier $2\alpha_c<1$. Starting at $0$, iteration increases monotonically and stays below $\alpha_c$, so it converges to that fixed point. Together with the attracting point at infinity, this proves hyperbolicity for these maps, which approach $c=1/4$. The failure in (eq155-failure) therefore identifies a weakness of the chosen proof device, not of the target.

One must also avoid requiring all critical points to be periodic in every neighborhood, even inside an attracting component: attracting orbits need not become periodic under arbitrary small perturbation. Our criterion leaves already attracted points alone. Multiple critical points can be avoided when first choosing a nearby generic parameter, but that does not establish the uniform return or distortion estimates needed at the next stage. Likewise, a closing result for one critical orbit does not control all the other critical orbits. Aspenberg's simultaneous control uses its nonrecurrence hypotheses [E1]; they cannot be removed by just applying the inverse function theorem repeatedly.

Outcome.

Outcome: Exploratory approach with unresolved gap.

The attempt yields a fully proved quantitative simultaneous-closing lemma, an exact rational period-three certificate, and a parabolic asymptotic obstruction to the naive derivative-growth implementation. These are lemmas and tests within the attempt, not a claim of a new density theorem or of priority for contraction-based closing methods.

A complete proof along this route would have to show that every parameter neighborhood contains a center and a collection of critical-return equations satisfying the radius, transversality, and distortion conditions, or replace this criterion with one that survives the demonstrated parabolic failure. No such uniform statement is proved, especially for recurrent critical orbits and the exceptional flexible families. The general density problem is not resolved here.

Sources

• S1 A. M. Benini, 'A survey on MLC, Rigidity and related topics', arXiv:1709.09869 (2017, revised 2018). https://arxiv.org/abs/1709.09869. Catalogue formulation source.
• E1 Magnus Aspenberg, Perturbations of rational Misiurewicz maps, arXiv:0804.1106v2, 22 June 2009, Definition 1.1, Theorem A, and Sections 3--5. https://arxiv.org/html/0804.1106v2. Consulted 27 September 2026. The exclusion of flexible Lattès maps and the Misiurewicz hypothesis are essential to the cited result.