Jordan mating is always possible for polynomials

Polynomial mating was an operation proposed by Douady and Hubbard to understand the dynamics of rational maps. Very roughly speaking, for two post-critically finite polynomials $P$ and $Q$ of degree $d\geq 2$ with both the Julia sets being connected, we may glue $f$ and $g$ along the Julia sets to get a topological map $F$. We say $f$ and $g$ are matable if $F$ is a branched covering map of the two sphere to itself, and moreover, $F$ is topologically conjugate to some rational map. Noting that the Julia set is the boundary of the immediate super-attracting basin of the infinity, the idea can be naturally extended to the situation of rational maps. Suppose $f$ and $g$ are two post-critically finite rational maps both of which have a simply connected immediate super-attracting basin of degree $d_{0}\geq 2$ such that there are no other critical orbits which intersect the immediate basins. Then one may construct a topological map by gluing $f$ and $g$ along the attracting basin boundaries and then copy this gluing for all the pre-images of the attracting basins. As in the case of polynomial mating, we say $f$ and $g$ are matable if $F$ is a branched covering of the two sphere to itself, and moreover, $F$ is topologically conjugate to some rational map $G$.

A particularly important case is that both the super-attracting basins are Jordan domains (Noting that all bounded immediate attracting basins of polynomials are Jordan domain [4]). In this case, no pinching happens when gluing $f$ and $g$ along the Jordan boundary and the topological map is always a branched covering of the two sphere to itself. Let us describe this topological construction as follows. Let $D_{f}$ and $D_{g}$ denote the two Jordan super-attracting basins and $D_{f}^{c},D_{g}^{c}$ be there complements respectively. Let $\phi:D_{f}\to\Delta$ and $\psi:D_{g}\to\Delta$ be the holomorphic isomorphism which conjugate $f$ and $g$ to $z\mapsto z^{d_{0}}$. Then for each $1\leq k\leq d_{0}-1$,

is a homeomorphism which reverses the orientation. We can extend it to a homeomorphism of the sphere so that it maps $\overline{D_{g}}$ to $D_{f}^{c}$ and maps $D_{g}^{c}$ to $\overline{D_{f}}$. Now we glue $D_{g}^{c}$ and $D_{f}^{c}$ by identifying the points $x$ and $\Phi(x)$. It is clear that $D_{g}^{c}\bigsqcup_{x\sim\Phi(x)}D_{f}^{c}$ is a topological two sphere. Define

by setting

Since by assumption no other critical orbits of $f$ and $g$ enter into $D_{f}$ and $D_{g}$ respectively, the way of extending $\Phi:\partial D_{g}\to\partial D_{f}$ dose not affect the combinatorially equivalent class of $F$. In the case that $F$ has no Thurston obstruction, we have a rational map $G$ which is combinatorially equivalent to $F$ (One can actually prove that $G$ is topologically conjugate to $F$). Unlike the usual mating, whose Julia sets is the disjoint union of $J_{f}$ and $J_{g}$ with those points in a ray equivalent class being identified, the Julia set of $G$ contains infinitely many copies of $J_{f}$ and $J_{g}$. To get $J_{G}$, one may start from $J_{g}\bigsqcup_{x\sim\Phi(x)}J_{f}$, and then iteratively fill the pre-images of $D_{f}$ and $D_{g}$ by copies of $J_{g}$ and $J_{f}$ respectively. To make a distinction with the usual mating, we call such mating a Jordan mating.

In contrast to the usual mating, for which there exist cubic polynomials which are topologically matable but not matable [6], Jordan mating is always possible for two polynomials. We will actually prove a stronger result.

Since the topological map $F$ in our case is always a branched covering of the sphere to itself, all we need to do is to show that $F$ has no Thurston obstructions. up to now there is no general way to check if a given topological map has Thurston obstructions or not, although many tools and ideas haven been developed[1][3][6][7][8]. The idea of our proof is to associate each non-peripheral curve a quantity which is monotonically increasing as we iterate the topological map. This property will lead us to get a Levy cycle from an irreducible Thurston obstruction. We then show that such a Levy cycle can be deformed into a Levy cycle of the rational map $g$, which is a contradiction. Our argument relies essentially on the assumption that one of the two rational maps is a polynomial.

In this section we give two examples of Jordan mating. Let $f$ be a cubic polynomial which has a degree two super-attracting fixed point at the origin so that the other finite critical point $c$ belongs to the boundary of the super-attracting basin, and moreover, $f^{2}(c)=f(c)$. Let $g$ be a cubic polynomial which has a a degree two super-attracting fixed point at the origin so that the other finite critical point $c$ belongs to the boundary of the super-attracting basin, and moreover, $g^{3}(c)=g^{2}(c)\neq g(c)$. Let $h$ be a post-critically finite cubic rational map so that it has a degree two super-attracting fixed point at $\infty$ and the Julia set is a Sierpinski carpet.

The reader may refer to [2] [5] for the details of the Thurston’s theory for characterization of post-critically finite rational maps. Throughout the paper we use $\widehat{\mathbb{C}}$, $\mathbb{C}$, $\mathbb{C}^{*}$, $\mathbb{T}$ and $\mathbb{D}$ to denote the Riemann sphere, the complex plane, the puncture complex plane, the unit circle and the unit disk respectively. Assume that $f,g\in\mathcal{R}_{d_{0}}$ with $f$ being a polynomial. Then there are $d_{0}-1$ ways to glue $f$ and $g$ along the boundary of the marked attracting basin, see (1.1). Let $F$ be one of such topological maps. All we need to do is to show that $F$ has no Thurston obstructions. We may identify the boundaries of the two marked immediate attracting basins with $\mathbb{T}$. We may assume that $F:\mathbb{T}\to\mathbb{T}$ is given by $z\mapsto z^{d_{0}}$, and up to combinatorial equivalence, $F=f$ outside $\mathbb{T}$ and $F=g$ inside $\mathbb{T}$. When a post-critical point $x\in P_{F}$ belongs to the forward orbit of some critical point of $f$, we write $x\in P_{f}$, and similarly, if it belongs to the forward orbit of some critical point of $g$, we write $x\in P_{g}$. In particular, we have

Suppose $\gamma$ is a non-peripheral curve of $F$. In the following we only concern those $\gamma$ so that $\gamma\cap\mathbb{T}\neq\emptyset$. By homotopy rel $P_{F}$ we may assume that $\gamma\cap\mathbb{T}$ is a finite set. Then $\gamma-\mathbb{T}$ consists of finitely many curve segments. Let $\sigma$ be any of such curve segments. We call $\sigma$ of type P (polynomial type) if it is outside $\mathbb{T}$, otherwise, we call it of type R (rational type). Let $I\subset\mathbb{T}$ be the arc so that $\partial I=\partial\sigma$ and $I$ is homotopic to $\sigma$ rel $\partial\sigma$ in $\mathbb{C}^{*}$. Let $D(\sigma)$ be the union of $I$ and the bounded domain bounded by $\sigma$ and $I$.

For $x\in P_{f}$ and $\gamma$ a non-peripheral curve in $\widehat{\mathbb{C}}-P_{F}$, let $\Sigma_{x}(\gamma)$ denote the set of all the type P curve segments $\sigma$ of $\gamma$ so that $x\in D(\sigma)$. Let $N(\Sigma_{x}(\gamma))$ denote the number of the elements in $\Sigma_{x}(\gamma)$. Let

where $\min$ is taken over all non-peripheral curves $\gamma^{\prime}$ which are homotopic to $\gamma$ in $\widehat{\mathbb{C}}-P_{F}$. We need more notations.

• •

  Let $\mathcal{O}$ denote the set of periodic points in $P_{f}$.

• •

  Let $\Sigma$ denote the class of non-peripheral curves $\gamma$ so that there is $x\in\mathcal{O}$ with $N(\Sigma_{x}([\gamma]))>0$ and let $\Pi$ denote the class of other non-peripheral curves.

• •

  Let $\Lambda$ denote the class of non-peripheral curves $\gamma$ so that $\Sigma_{x}([\gamma])=0$ holds for any $x\in P_{f}$.

Now suppose $F$ has an obstruction. Then by [5] $F$ has a canonical Thurston, say $\Gamma$, which consists of all homotopy classes of the non-peripheral curves whose length go to zero as we iterate the Thurston pull back induced by $F$. We claim that $\Gamma\cap\Lambda=\emptyset$. Let us prove the claim. Suppose $\Gamma\cap\Lambda\neq\emptyset$. Then $\Gamma\cap\Lambda$ must be $F$-stable by Lemma 3.1. Since the length of every curve in $\Gamma\cap\Lambda$ goes to zero as we iterate the Thurston pull back, the transformation matrix associated to $\Gamma\cap\Lambda$ must have an eigenvalue $\geq 1$. By the definition of $\Lambda$, one can deform the curves in $\Gamma\cap\Lambda$ so that it is a stable family of $g$ and with the same transformation matrix. This is a contradiction because $g$ has no obstruction. This, together with Lemma 3.2, implies that $\Gamma\cap\Pi=\emptyset$. We thus have

Now let us prove the main theorem. Suppose $F$ has an obstruction and let $\Gamma$ be an obstruction guaranteed by Lemma 3.3. We may assume that it is an irreducible one. That is, for any $\gamma,\eta\in\Gamma$, there is an $n\geq 1$ such that $\eta$ is homotopic to a component of $f^{-n}(\gamma)$. Now let us prove $\Gamma$ is a Levy cycle.

Claim: for each $\gamma\in\Gamma$, there exists exactly one non-peripheral component of $F^{-1}(\gamma)$. Suppose this were not true. Then we would have two non-peripheral components $\gamma_{1}\neq\eta_{1}$ of $F^{-1}(\gamma)$ and two sequence:

and

where $\gamma_{i+1}$ is homotopic to some component of $f^{-1}(\gamma_{i})$, $0\leq i\leq l+1$, and $\eta_{j+1}$ is homotopic to some component of $f^{-1}(\eta_{j})$, $0\leq j\leq k$. Since $\gamma\in\Sigma$, we have some $x\in\mathcal{O}$ such that $N(\Sigma_{x}([\gamma]))>0$. Let $y\in\mathcal{O}$ such that $x=F(y)$. Let $p$ be the period of the $x$. Repeating (3.1) $p$ times,

Apply Corollary 3.5 to the above sequence, we get

which implies that

Similarly, we may repeat (3.2) $p$ times and then apply Corollary 3.5, we get

which implies that

But by Lemma 3.4 and $\gamma_{1}\neq\eta_{1}$, we also have

This implies that $N(\Sigma_{x}([\gamma]))=0$. This contradicts the assumption that $\Sigma_{x}([\gamma])>0$. The Claim has been proved.

Now for any non-peripheral curve $\gamma$, let $K(\gamma)$ denote the number of the type P curve segments of $\gamma$ and let

where $\min$ is taken over all the non-peripheral curves which are homotopic to $\gamma$. From the claim above, it follows that $\Gamma$ must be a Levy cycle for $F$. Let $\Gamma=\{\gamma_{1},\cdots,\gamma_{n}\}$ so that

and for $1\leq i\leq n-1$, $\gamma_{i+1}$ is the unique non-peripheral component of $F^{-1}(\gamma_{i})$. Since the image of a type P curve segment contains at least one type P curve segment, we must have

So all type P curve segments of $\gamma_{i}$, $1\leq i\leq n$, are non-trivial in the sense that $D(\sigma)\cap P_{F}\neq\emptyset$. So for any type P curve segment $\sigma$ of $\gamma_{i+1}$, there is a type P curve segment $\sigma^{\prime}$ which is homotopic to $\sigma$ (Figure 8 illustrates the meaning that two type P curve segments are homotopic), such that $\sigma^{\prime}$ is mapped homeomorphically to some type P curve segment $\tau$ of $\gamma_{i}$. We thus get a cycle of type P curve segments $\{\sigma_{i}\}$, $1\leq i\leq m$ with $n|m$, such that for each $\sigma_{i}$, there is a type P curve segment $\mu_{i+1}$ which is homotopic to $\sigma_{i+1}$ so that $F:\mu_{i+1}\to\sigma_{i}$ is a homeomorphism. Since $f$ is post-critically finite which expands the orbifold metric, it follows that $D(\sigma_{i})$ contains exactly one point in $P_{F}$, say $x_{i}$, which lies in $I_{i}=D(\sigma_{i})\cap\mathbb{T}$, and moreover, $\{x_{i}\}$ is a periodic cycle. But one may then deform each $D(\sigma_{i})$ into a small neighborhood of $x_{i}$. In this way $\Gamma=\{\gamma_{i}\}$ becomes into a Levy cycle of $g$. This is impossible. The proof of the main theorem is completed.

$\mathbb{Acknowledgements.}$ The author would like to thank Fei Yang who provides all the computer generating pictures in the paper.