For any polynomial map $f$ we can define the filled Julia $K$ to be closure of the complement of $ \Omega$ in $\mathbb{C}$ of the basin of infinity $$\Omega = \{z \in \mathbb{C}; f^{\circ n}(z)\rightarrow \infty \}.$$

Thus the boundary of $K$ is what is often called the Julia set. Now there are examples like $f=z^2$ for which the filled Julia set is connected and locally connected. My question is roughly speaking: How many such polynomial maps do exist? Or more precisely. Are there infinitely many non-equivalent such polynomial maps in any degree? By equivalence I mean the usual one. Two endomorphisms $f,g$ of the projective line are equivalent if there exists a Möbius transformation $\mu$ such that $\mu\circ f\circ \mu^{-1} =g$. If the second question has a positive answer I would be very interested in whether the proof is constructive. Any comment or reference would be greatly appreciated.


1 Answer 1


There infinitely many non-equivalent such polynomial maps of even degree:

Let $f$ be a polynomial of the form $f(z)=z^d +c$ with $d$ an even integer and $c$ real. Then the Julia set of $f$ is either totally disconnected or locally connected.

Genadi Levin and Sebastian Van Strien. "Local Connectivity of the Julia Set of Real Polynomials", Annals of Mathematics, 147, no. 3 (1998): 471—541. doi:10.2307/120958

  • $\begingroup$ same question; but what about $z \mapsto z^2+c$, for which values of c is the Julia set known to be locally connected? Are there valiues for which it is known that the Julia connected but not locally connected? $\endgroup$
    – Sheldon L
    Dec 5, 2017 at 22:31
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    $\begingroup$ @SheldonL, because of the result in my answer, the $c$ for which the Julia set is totally disconnected are exactly those outside the Mandelbrot set. Therefore, in the quadratic case, there are no Julia sets that are connected but not locally connected. $\endgroup$
    – lhf
    Dec 5, 2017 at 22:42
  • $\begingroup$ I know if $z \mapsto z^2+c$ starting at z=0 escapes, then the Julia is a disconnected cantor set - I don't know if I've seen the proof in Milner's book or Devaney's book or Carleson/Gamelin's book. But if iterating z remains bounded, then there is the more subtle question of the Julia set being locally connected vs just connected ... just asking. $\endgroup$
    – Sheldon L
    Dec 5, 2017 at 23:19
  • $\begingroup$ Thanks Ihf! The fact that local connectivity implies connectivity was not obvious to me. They seem to have proven that the Julia set is locally connected if and only if the only critical point 0 does not escape to infinity which implies that the Julia set is connected and locally connected in those cases. As an aside. From their result follows that for $f=z^2+c$ and $c$ in the whole interval $[-2,1/4]$ the Julia set is connected and locally connected. I am surprised that there are even uncountably many examples. $\endgroup$
    – Gari
    Dec 6, 2017 at 19:24

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