Connected set in a filled Julia set

Question

Let $K$ be the filled Julia set of a complex polynomial of degree at least 2. Suppose that $K$ is connected. Let $p_1, \dots, p_N \in K$ be some points. Does there exist a connected set $K_N$ containing $p_1, \dots, p_N$ such that $K_N\subset K$ but $K_N \neq K$? Any references, hints or counterexamples would be very welcome. Thank you very much! Edit 1: Suppose further that $f$ is not a Chebyshev polynomial and thus the Julia set is not smooth. Edit 2: It is a fact that if $K$ has no interior then its Hausdorff dimension is larger than 1. I wonder whether this might be enough to give a positive answer.

Answer 1

Not if the Julia set is an interval. Consider the Chebyshev polynomial $f(z)=z^2-2$ for example whose Julia set/filled Julia set is given by $[-2,2]$. The only connected subset containing Julia points $-2,2$ is the whole thing.

Remark) It is worthy to notice that the answer is negative only if there is no bounded Fatou component, meaning the Julia set and the filled Julia set coincide (so no parabolic periodic point, Siegel disk or finite attracting/super attracting periodic point). Because if $K$ has an interior point, one can remove a small open disk not containing any of the given points $p_1,\dots,p_N$ from $K$ so that the remaining subset is still connected.

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Added) A few more words about polynomials for which the desired property fails. My argument for the claim below is admittedly hand-wavy at certain points; but I leave it here if someone can make it more precise.

Claim) Let $p(z)$ be a polynomial of degree $d\geq 2$ whose filled Julia set, denoted by $K$, is connected. Suppose there exist distinct points $p_1,\dots,p_N\in K$ such that the only connected subset of $K$ containing all of them is $K$ itself. Then:

• The Julia set, $\mathcal{J}$, and the filled Julia set, $K$, coincide and are homeomorphic to an interval.
• Every finite critical point is simple and belongs to the Julia set. Moreover, the critical values lie in a forward-invariant set of size two which does not contain any critical points.

In particular, when $d=2$, $p(z)$ must be conjugate to the Chebyshev polynomial $z^2-2$.

This is in a sense the converse of the Chebyshev polynomial example I began with: polynomials for which the desired property fails "resemble" Chebyshev polynomials.

$\underline{\text{I base my argument on the following two }``\text{facts"}}.$ I believe they are true but I couldn't find rigorously written proofs. Any reference on them is welcome.

"Fact" 1) If the filled Julia set, and hence the Julia set, are connected, then they are path connected.

I didn't find a proof in standard complex dynamics textbooks. If this is false, one can strengthen the assumption from the connectedness of the filled Julia set to its path-connectedness.

"Fact" 2) Let $X$ be a compact, path-connected metric space. Suppose there exist an integer $N\geq 2$ and distinct points $p_1,\dots,p_N\in X$ such that no proper path-connected subset of $X$ contains all $p_i$'s. Then, $X$ is homeomorphic to a tree with at most $N$ leaves.

I have posed this as a separate MO question.

Proof) Since the filled Julia set, $K$, is connected, it must contain all finite critical points. By Fact 1 above, $K$ is path-connected as well. Due to the existence of $p_1,\dots,p_N$, as mentioned before, there is no bounded Fatou component ($K$ has no interior points): Otherwise, one can remove a small enough open disk from $K$ to obtain a connected subset containing all $p_i$'s. Thus $K$ coincides with the Julia set $\mathcal{J}=\partial K$. Applying Fact 2 to the path-connected, compact space $\mathcal{J}$, we deduce that $\mathcal{J}$ is homeomorphic to a finite tree.

There are indeed polynomial Julia sets which are "tree-like", for instance the dendrite Julia set of $z\mapsto z^2+i$. (The picture was generated by this JavaScript app.)

But the example above has infinitely many "branch points", meaning points which do not admit a neighborhood homeomorphic to a closed/half-closed interval. I argue that if the Julia set is a finite tree (i.e. the number of leaves is finite), then there are no branch points, meaning $\mathcal{J}$ is homeomorphic to a closed interval. First, notice that since the Julia set is both backward and forward invariant, it makes sense to speak of the multiplicity of the restriction $p\restriction_{\mathcal{J}}:\mathcal{J}\rightarrow\mathcal{J}$ the way one considers the multiplicity of $p:\hat{\Bbb{C}}\rightarrow\hat{\Bbb{C}}$ at a point; and the restriction is a ramified cover of degree $d$. Any preimage of a branch point of tree $\mathcal{J}$ under $p$ should be a branch point as well. This is clear if $p$ is a local homeomorphism near the preimage point. If the preimage point is a critical point, then $p$ is many-to-one in its vicinity, and hence even more branches emanate from it.

On the other hand, the tree is finite, thus the branch points (vertices of degree $>2$) of $\mathcal{J}$, if any, constitute a non-empty finite subset $E$ of $\Bbb{C}$ with $p^{-1}(E)\subseteq E$. Such a polynomial is conjugate to $z\mapsto z^d$ (This follows from Riemann-Hurwitz theorem. As far as I recall, it is proved in Silverman's book.) But the filled Julia set is a disk in that case. We deduce that there is no branch point, and the Julia set $\mathcal{J}$ is homeomorphic to an interval.

Keep in mind that $p\restriction_{\mathcal{J}}:\mathcal{J}\rightarrow\mathcal{J}$ is a ramified cover of degree $d$. The only points with neighborhoods homeomorphic to a half-closed interval are the endpoints, denoted by $A,B$ in the picture above. We deduce that $\{A,B\}$ is forward-invariant; and furthermore, every finite critical point of $p$ (we know that they all belong to $\mathcal{J}$) lies in the interior of $\mathcal{J}$, and is mapped to either $A$ or $B$ with multiplicity two. (In other words, under $p\restriction_{\mathcal{J}}$, the image of a neighborhood of a critical point, homeomorphic to an open interval, is homeomorphic to a half-closed interval, thus a neighborhood of either $A$ or $B$).

Finally, as a corollary, we show that when $d=2$ the polynomial $p(z)$ can be turned into $z\mapsto z^2-2$ by a linear change of coordinates. Write the polynomial in the normal form $p(z)=z^2+c$. The critical value $c$ must lie in an invariant set of size two. If $p(c)=c$, then $c$ should be zero---impossible because the filled Julia set of $p(z)=z^2$ is a disk, not a topological interval. That invariant set should therefore be $\{c,p(c)=c^2+c\}$. Now there are two possibilities:

• $p(c^2+c)=c^2+c\Leftrightarrow (c^2+c)^2=c^2\Leftrightarrow c\in\{0,-2\}$. The case of $c=0$ was ruled out before; and $c=-2$ yields the Chebyshev polynomial.
• $p(c^2+c)=c\Leftrightarrow c^2+c=0\Leftrightarrow c\in\{0,-1\}$. Again, $c=0$ is impossible. If $c=-1$, then the critical point $z=0$ lies in a $2$-cycle $0\mapsto -1\mapsto 0$. But there cannot be any finite super attracting cycle because its immediate basin of attraction determines a bounded Fatou component. (The Julia set of $z\mapsto z^2-1$ is the basilica Julia set, definitely not homeomorphic to an interval.)

Further questions) I wonder if there is an elegant argument for showing that a polynomial whose Julia set is homeomorphic to a closed interval must be Chebyshev. Stronger properties that we derived above (critical points are simple, critical values are periodic etc.) may also be used to establish such a thing because they are all properties of Chebyshev polynomials.