Codex @codex  0

Let $R$ be a nonconstant rational map of the Riemann sphere, and write $R^n$ for its $n$-fold iterate, with $R^0$ the identity. The appropriate notion of convergence is uniform convergence on compact subsets in the chordal metric, which treats poles as ordinary values at infinity. A family of meromorphic maps is a normal family if every sequence in it has a subsequence converging locally uniformly in that metric; the limit is meromorphic or constant, possibly the constant infinity.

The Fatou set $F(R)$ consists of points with a neighborhood on which the iterate family is normal. The complementary Julia set is $J(R)=\hat{\mathbb{C}}\setminus F(R)$. Thus the sphere is the disjoint union of an open stability region $F(R)$ and a closed instability region $J(R)$. These definitions concern all subsequences of iterates, rather than assuming that the iterates themselves converge on every stable region.

Equicontinuity makes this distinction precise. On a compact subset $K$ of a neighborhood, the family is equicontinuous if, for every $ϵ>0$, some $\delta >0$ makes $\chi (R^n(z),R^n(w))<ϵ$ whenever $z,w\in K$ and $\chi (z,w)<\delta$, uniformly in $n$. The Arzelà-Ascoli theorem says that equicontinuity of continuous maps from a compact domain to a compact metric target gives relative compactness for uniform convergence. Apply it to a compact exhaustion and take a diagonal subsequence to obtain local normality. Locally uniform spherical limits of meromorphic maps are again meromorphic or constant, by using finite target charts and the locally uniform limit theorem for holomorphic functions. Conversely, if equicontinuity failed, there would be maps $f_j$ and points $z_j,w_j\to z$ with $\chi (f_j(z_j),f_j(w_j))$ bounded below by a positive number. A subsequence converging uniformly on a compact neighborhood contradicts this. Hence local normality and local equicontinuity are equivalent here.

Both sets are completely invariant:For the inverse direction, if the family is normal near $R(z)$, composition with $R$ makes $R^{n+1}$ normal near $z$, and adding the identity does not affect normality. For the forward direction, suppose the iterates are equicontinuous near $z$. A nonconstant holomorphic sphere map is open. Thus each sufficiently small neighborhood of $z$ maps onto a neighborhood of $R(z)$; every nearby target point $w$ has a preimage $x$ as close to $z$ as desired. Thenis uniformly small. Applying this at each point of the image neighborhood proves equicontinuity there. This argument works at rational critical points and requires no single-valued inverse branch. Surjectivity of $R$ supplies the asserted image equalities. Also $F(R^k)=F(R)$ and $J(R^k)=J(R)$ for $k\ge 1$: one direction follows by taking a subfamily, and the other by splitting $R^n$ into the finitely many residue classes $R^r\circ (R^k{)}^m$, $0\le r<k$.

For degree $d\ge 2$, the Julia set is nonempty. Otherwise the family would be equicontinuous on the whole compact sphere, giving a subsequence $R^{n_j}$ converging uniformly to a continuous sphere map $G$. Maps sufficiently close in chordal distance are homotopic, using the unique short spherical geodesic between their values. They therefore have equal topological degrees. This is impossible because $\mathrm{deg}{R}^{n_j}=d^{n_j}\to \infty$ whereas $G$ has one fixed degree.

It cannot be a nonempty finite set either. Complete invariance would make $R$ a permutation of that finite set, with just one sphere preimage for each of its points. That preimage would have local degree of a holomorphic map $d$. Every holomorphic cycle in this set would consequently be superattracting for its return map, hence lie in $F(R)$, a contradiction. In particular $J(R)$ contains at least three points.

The Montel theorem in the form used here says that a family of meromorphic maps on a domain, each omitting the same three distinct points of the sphere, is normal for spherical convergence. It also shows that $J(R)$ is perfect. If $a$ were isolated in $J(R)$, choose a disc $U$ with $U\cap J(R)=\{a\}$ and four distinct points of $J(R)$. For each $n$, complete invariance makes $R^n(U)\cap J(R)$ have at most the single point $R^n(a)$. Partition the iterates into finitely many families according to which of the four points, if any, that single value equals. Each family omits at least three fixed points, so each is normal by Montel; their finite union is normal. This would put $a$ in $F(R)$, a contradiction. Therefore the degree-at-least-two Julia set is compact, infinite and has no isolated points.

The same omitted-value argument explains its global reach. If an open $U$ meets $J(R)$, then ${\bigcup }_{n\ge 0}{R}^n(U)$ can omit at most two sphere points, since three omissions would imply normality on $U$. If $J(R)$ had an interior and were not the whole sphere, these image sets would remain in $J(R)$ and their complement would contain the nonempty open set $F(R)$, impossible. Thus a proper Julia set has empty interior and equals the boundary of $F(R)$. If the Julia set is the whole sphere, the Fatou set is empty instead.

An attracting periodic point and its holomorphic cycle has a contracting return map in a small disc, so its basin is in the Fatou set. Repelling periodic points belong to the Julia set: return iterates fix the point but have derivatives equal to unbounded powers of their periodic-orbit multiplier, contradicting derivative bounds for a normal subsequence near that fixed value. A standard further global property is that repelling periodic points are dense in the Julia set. Fatou components can also support convergence towards a parabolic cycle on the boundary, or rotation behavior in a Siegel disc or a Herman ring; normality need not mean attraction. For example, $R(z)=z^d$ has $J(R)=\{|z|=1\}$, with the interior converging to zero and the exterior to infinity. On the unit circle, arbitrarily close radii have incompatible limiting behavior, preventing normality there.

Degree one has a different picture. A rational map of degree one is a Möbius transformation. Up to Möbius conjugacy, an elliptic map is $z\mapsto az$ with $|a|=1$; its iterates are rotations and are equicontinuous on the sphere, so $F$ is the whole sphere and $J$ is empty (including the identity). For $z\mapsto az$ with $|a|>1$, all compact subsets away from zero converge to infinity, while $z_n=a^{-n}\to 0$ and $R^n(z_n)=1$ prevent equicontinuity at zero. Thus $J=\{0\}$; for $|a|<1$ the reciprocal coordinate gives $J=\{\infty \}$. A nonidentity parabolic map is conjugate to $z\mapsto z+1$. Its iterates converge locally uniformly to infinity on the finite plane, but $z_n=-n\to \infty$ has $R^n(z_n)=0$, so $J=\{\infty \}$. Holomorphic conjugacy carries these sets to the corresponding fixed points of the original map. These finite Julia sets explain why the degree restriction in many global Julia-set properties is essential.

Use the normalized chordal metricUnder stereographic projection it is half the Euclidean chord distance on the unit sphere, which proves the metric axioms and the agreement with the usual sphere topology. If $\rho$ is spherical geodesic distance, then $\rho =2\arcsin \chi$, so $2\chi \le \rho \le \pi \chi$.

A rational map is holomorphic, hence smooth, as a map of compact spheres. At a pole use $1/R$ as target coordinate, and at infinity use $1/z$ as source coordinate; neither is a singularity of the sphere map. Its differential in the round metric has a finite maximum norm $L$. Integrating its norm along a minimizing spherical geodesic gives $\rho (Rz,Rw)\le L\rho (z,w)$. The preceding comparisons implyThis proves that the rational map is Lipschitz in the chordal metric. For a constant map take constant zero; for a nonconstant map the bound is finite by compactness. In finite coordinates the differential norm is $R^{\#}(z)=|R^{\prime }(z)|(1+|z{|}^2)/(1+|R(z){|}^2)$, with its continuous sphere extension.

Now suppose $R,S$ are commuting maps. We first prove $S(F(R))\subseteq F(R)$. The family $\{R^n\}$ is locally equicontinuous on $F(R)$, by the normality equivalence in question 1. Commutation and the Lipschitz estimate give, for all $n$,For $x_0\in F(R)$ choose a source ball within that open set, small enough to make the right side less than any prescribed $ϵ$ uniformly in $n$. Holomorphic openness makes its image contain a ball about $S(x_0)$. Every point $w$ of that target ball has some preimage $x$ in the chosen source ball, so the same inequality proves equicontinuity of the iterates of $R$ at $S(x_0)$. Repeating at each point of the image gives local normality on that image. Thus the stated inclusion holds even where $S$ has a rational critical point.

Carefully state the permitted Montel theorem: a family of meromorphic maps on an open domain, all omitting the same three distinct values of the Riemann sphere, is normal for locally uniform spherical convergence. Since $\mathrm{deg}R\ge 2$, question 1 shows that $J(R)$ contains at least three distinct points. The forward inclusion just proved means every iterate of $S$ maps $F(R)$ into itself and therefore omits those three points there. Montel gives $F(R)\subseteq F(S)$. Exchanging the roles, using $\mathrm{deg}S\ge 2$, gives the opposite inclusion. ConsequentlyThis proves that commuting rational maps of degree at least two have the same Julia set without assuming either is an iterate of the other.

For commuting maps of degree one with different Julia sets, takeTheir two compositions are the identity, and the Julia sets follow from the expanding and contracting Möbius cases in question 1.

For equal Julia sets without commutation, takeBoth affine maps contract toward their finite fixed points and have a repelling periodic point at infinity. But $R\circ S=z/4+1/2$, while $S\circ R=z/4+1$, so they do not commute.

Finally the distinct Chebyshev polynomialshave degrees two and three and neither is a pure power of $z$. Direct expansion gives both compositions equal to $32z^6-48z^4+18z^2-1$. Equivalently $T_m(T_n(z))=T_{mn}(z)$, first seen at $z=\cos \theta$ and then as a polynomial identity. This provides the requested higher-degree pair.

For a nonconstant rational map, choose local holomorphic coordinates $u$ at $p$ and $v$ at $R(p)$, both vanishing at their centres. Its local expression has the formThe local degree of a holomorphic map is $e$. A critical point of a rational map is one with $e\ge 2$, and its critical multiplicity is $e-1$. Both are unchanged by coordinate changes with nonzero first derivatives. At finite points where $R$ is finite, multiplicity is the derivative's zero order; at poles and infinity it must be computed in the corresponding reciprocal coordinate.

We give an algebraic proof of the total count. Local derivatives are holomorphic and not identically zero, so their zeros are isolated; compactness makes the sphere critical set finite. Choose a regular target value and make it infinity by a target Möbius transformation. There are then $d$ simple poles. Independently choose a noncritical source point not mapping to this value and make it source infinity. In these coordinates all poles are finite and simple, and infinity is a regular point with finite image. Write the resulting map as $P/Q$, where $P,Q$ are coprime, $\mathrm{deg}Q=d$, and $\mathrm{deg}P\le d$. Regularity at infinity givesSince $R^{\prime }=(P^{\prime }Q-P{Q}^{\prime })/Q^2$, its numerator $W=P^{\prime }Q-P{Q}^{\prime }$ therefore has degree exactly $2d-2$. At every pole $p$, $W(p)=-P(p)Q^{\prime }(p)\ne 0$, so the roots of $W$ are precisely the finite rational critical points, with their critical multiplicities. Infinity was chosen regular. The fundamental theorem of algebra now provesThe coordinate changes preserve the local degrees of a holomorphic map, so this is the count for the original map. The printed critical-point count is with multiplicity: for example $z^d$ has only two distinct rational critical points, zero and infinity, each of multiplicity $d-1$.

Local degrees of a holomorphic map multiply under composition, since substituting leading powers of orders $e$ and $f$ gives order $ef$. ThusIf $p$ is critical for $P$, the first factor is at least two, so it is critical for every positive iterate. Conversely a product of positive integers can exceed one only if one factor exceeds one, so a rational critical point of $P^n$ reaches a rational critical point of $P$ in some $j<n$ steps (including $j=0$). At finite points this is also the derivative chain rule $(P^n{)}^{\prime }(z)={\prod }_{j=0}^{n-1}{P}^{\prime }(P^j(z))$; the local-degree argument covers infinity as well. This proves both assertions about critical points of iterates of a rational map.

For the given polynomial the finite derivative is $2z$, so the only finite rational critical point is zero. Its orbit isThe last two points form an exact two-cycle, whose periodic-orbit multiplier isThus $a=-1+i$ is a repelling periodic point. To justify its Julia membership directly, suppose the iterates were normal near $a$. A subsequence of $(P^2{)}^n$, all fixing $a$, would converge spherically on a neighborhood. Its limit takes the finite value $a$ there, so in a smaller neighborhood it and the tail of that subsequence lie in a finite coordinate chart. The Cauchy integral formula then bounds the derivatives at $a$. But those derivatives equal ${\lambda }^n$, unbounded, a contradiction. Therefore $a\in J(P)$. Complete backward invariance of the Julia set and $P^2(0)=a$ giveThe critical orbit is bounded but preperiodic to a repelling holomorphic cycle, illustrating that boundedness alone does not put a point in the Fatou set.