# Can the topologist's sine curve be realized as a Julia set?

Does there exist a rational function $$f\in\Bbb{C}(z)$$ whose Julia set coincides with $$T:=\left\{\left(x,\sin\left(\frac{1}{x}\right)\right)\,\Big|\,x\in\left(0,\frac{1}{\pi}\right]\right\}\cup\big(\{0\}\times[-1,1]\big)\;?$$ What if we require the Julia set to only be homeomorphic with $$T$$?

• The complement of $T$ is simply connected in the Riemann sphere, so the Fatou set of any map $f$ with $J(f) = T$ has a single component $U$, and the classification of Fatou components tells us what $U$ could be like. $U$ is neither a Hermann ring (because it's simply connected) nor a Siegel disk (because it's clearly not the image of the unit disk under an analytic map). So either $U$ is parabolic or it contains a single attracting fixed point. Have you investigated what either of these possibilities would imply about $f$? Commented May 24 at 0:07
• Slightly subtle aspect of the simple connectedness issue (because this caused me to doubt myself after I posted that comment): $U$ isn't a Hermann ring because it's simply connected in the Riemann sphere $\mathbb{C} \cup \{ \infty \}$, and it's not a Siegel disk because it's not simply connected in the plane $\mathbb{C}$. Commented May 24 at 6:48
• @SophieM Thanks for your comments. I think the easiest way to argue that $f$ cannot admit a Siegel disk or Herman ring is to notice that $f$ should be injective on such a component. But there is only one Fatou component and a map of degree $d\geq 2$ restricts to a $d$-to-$1$ self-map on its Fatou set. Commented May 24 at 14:55

The answer is negative. Since every neighborhood of a point on the Julia set is mapped onto the whole Julia set by some iterate of the rational function, it follows that if a small piece of the Julia set is a simple curve, then the whole Julia set is either a simple curve or a Jordan curve.

Addressing the comment. Use the following Lemma. Let $$g$$ be a germ of an analytic function is a neighborhood of $$z_0$$, $$w_0=g(z_0)$$ and $$E$$ is an arbitrary set containing $$w_0$$. If the component of $$g^{-1}(E)$$ containing $$x_0$$ is a simple curve, then intersection of $$E$$ with a neighborhood of $$w_0$$ is either a simple curve, or a semi-closed simple curve with one end at $$w_0$$.

Proof. WLOG $$z_0=w_0=0$$, and $$g(z)=z^m$$. This makes the statement evident. In the first case, $$m=1$$, in the second case $$m=2$$.

It follows that the Julia set $$J$$ is a bordered closed 1-manifold, and thus $$J$$ is homeomorphic to a circle or to a segment.

Edit. There is also a reference on this result: Theorem A in this paper.

• I see why the Locally Eventually Onto property implies that the Julia set would have to be locally connected if it contains an arc with interior. But I cannot see why it would have to be a simple curve or Joran curve. Could you give a bit more explanation about this? Commented May 26 at 14:16
• @AlexandreEremenko Thanks for your answer. Indeed, once a small piece of the Julia set is a simple curve, the whole Julia set must be path connected by your argument because it is a surjective image of that piece. And we know that is not the case for the topologist's sine curve. Commented May 28 at 2:37
• KhashF: yes. But I sketched a proof and gave a reference to a stronger result. Commented May 28 at 12:50