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I am looking for an example of a post-critically finite polynomial $P$ (i.e. all critical points have finite orbit), which has both the following:

  • a critical point on its Julia set (such as the polynomial $z^2+i$)

  • a super-attracting cycle, other than $\infty$, i.e. a periodic orbit $z_0\mapsto z_1\mapsto \cdots\mapsto z_{n-1} \mapsto z_0$ such that at least one $z_i$ is a critical point (such as the polynomial $z^2-1$ that has a $2$-cycle $0\mapsto -1\mapsto 0$)

Is there some known example from the multibrot families $z^q+c$ or any other example?

How about a post-critically finite polynomial with two critical points on the Julia set?

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    $\begingroup$ You can rule out the mulitbrot family, since they all have a single critical point, namely the origin. $\endgroup$
    – Mark McClure
    Commented Jul 20, 2015 at 21:30
  • $\begingroup$ @MarkMcClure Surely they have two critical points, $0$ and $\infty$? $\endgroup$
    – mdave16
    Commented Mar 14, 2017 at 22:53
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    $\begingroup$ @mdave16 Yes, definitely - when viewed as a dynamical system on the Riemann sphere, $\infty$ is a a critical point of any polynomial. Even if we restrict to just the complex plane, it is often useful to think of the point at $\infty$ as a super-attractive fixed point. It is also often useful, though, to distinguish the point at $\infty$ somewhat. In the question at hand, it is specifically stated that all critical points should have finite orbits. That immediately rules out the possibility that we consider $\infty$ to be a critical point. And thanks for the wake up call! :) $\endgroup$
    – Mark McClure
    Commented Mar 15, 2017 at 0:32
  • $\begingroup$ No problem about the wake up call, it's how we all get better. -- Why should we not consider $\infty$? as a critical point if all critical points should have finite orbit? Is it because $\infty \mapsto \infty$ has the most dull forward orbit possible? And so it's just a problem to forget it? $\endgroup$
    – mdave16
    Commented Mar 15, 2017 at 0:48

1 Answer 1

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The polynomial $2z^3-3z^2+\dfrac{1}{2}$ has the desired properties. The critical points are at 0 and 1, with $$ 0\mapsto \frac12 \mapsto 0 \qquad\text{and}\qquad 1\mapsto -\frac12 \mapsto -\frac12. $$ The point $1$ lies on the Julia set itself. The following picture shows the filled Julia set for this function, with the two critical orbits marked in yellow and red, respectively.

enter image description here

To find this example, I first decided to look for a cubic polynomial with one critical point of period two and one critical point that maps to a fixed point. I also decided (without loss of generality) that the critical points should be at $0$ and $1$, so the polynomial should have the form $f(z) = 2az^3-3az^2 + b$ for some constants $a$ and $b$. I then looked for values of $a$ and $b$ that satisfy the two equations $f(f(0))=0$ and $f(f(1)) = f(1)$.

As for having two critical points on the Julia set, the third Chebyshev polynomial $4z^3-3z$ has the line segment $[-1,1]$ on the real axis as its Julia set, and its two critical points are $\pm\dfrac12$.

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    $\begingroup$ Very nice! It might be worth pointing out that it's easy to show that $1$ is in the Julia set using the characterization that the Julia set is the closure of the repelling fixed points. First, $-1/2$ is in the Julia set since it's a repulsive fixed point (i.e. a point of period $1$). Then, $1$ is in the Julia set since it's the inverse image of $-1/2$ and the Julia set is backward invariant. $\endgroup$
    – Mark McClure
    Commented Jul 20, 2015 at 21:28
  • $\begingroup$ Very nice answer! What software are you using for the picture? $\endgroup$
    – Dimitris
    Commented Jul 20, 2015 at 23:47
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    $\begingroup$ @DimitriosNt It was made in Mathematica using code that I wrote myself. You can find my general Julia set code in the ColoredJuliaSets.nb file on my code web page. $\endgroup$
    – Jim Belk
    Commented Jul 21, 2015 at 0:18
  • $\begingroup$ @JimBelk Have you noticed this in V10: JuliaSetPlot[2 z^3 - 3 z^2 + 1/2, z]? $\endgroup$
    – Mark McClure
    Commented Jul 22, 2015 at 10:57
  • $\begingroup$ @MarkMcClure I don't have V10 yet, but that's nice to hear! $\endgroup$
    – Jim Belk
    Commented Jul 22, 2015 at 13:28

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