Let $f$ and $g$ be complex rational functions (of degree $\geq 2$ if that helps). What can be said about the relationship between $J(fg)$ and $J(gf)$, the Julia sets of the composite functions $f \circ g$ and $g \circ f$?

If I'm not mistaken, $f$ restricts to a map $J(gf) \to J(fg)$, and $g$ restricts to a map $J(fg) \to J(gf)$. So $J(fg)$ and $J(gf)$ surject onto each other in a particular way (and, indeed, in a way that commutes with the actions of $fg$ on $J(fg)$ and $gf$ on $J(gf)$). Since Julia sets are completely invariant, the restricted map $f: J(gf) \to J(fg)$ is $deg(f)$-to-one, and similarly the other way round.

So there's some kind of relationship between the two sets.

If $f$ or $g$ has degree one then $J(fg)$ and $J(gf)$ are "isomorphic", in the sense that there's a Möbius transformation carrying one onto the other. Thus, the simplest nontrivial example would be to take $f$ and $g$ to be of degree 2. I don't know a way of computing, say, the example $f(z) = z^2$ and $g(z) = z^2 + 1$. That would mean computing the Julia sets of $gf(z) = z^4 + 1$ and $fg(z) = z^4 + 2z^2 + 1$.

My question isn't completely precise, I'm afraid. But here are some of the things that I would value in an answer: theorems on what $J(fg)$ and $J(gf)$ have in common, examples showing how different they can be, pictures of $J(fg)$ and $J(gf)$ for particular functions $f$ and $g$, and references to where I can find out more (especially those accessible to a non-specialist). Thanks.

  • $\begingroup$ This might not be true if the functions have poles. Suppose you had a point such that $f$ kept taking it very close to infinity and $g$ kept taking it back near $0$, then it would be in $J(gf)$ but $f$ of it would not be in $J(fg)$. Are they allowed to have poles? $\endgroup$
    – Will Sawin
    Jan 2, 2012 at 22:52
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    $\begingroup$ As to whether they can have poles: f and g are rational functions with complex coefficients, so they do in general have poles in the usual complex analysis sense. However, I think it's usually better to interpret them as holomorphic maps from the Riemann sphere to itself, in which case there's nothing special about the point $\infty$. $\endgroup$ Jan 3, 2012 at 0:33
  • $\begingroup$ Will, I don't get your argument. Can you produce a specific counterexample? When I try to turn what you wrote into a counterexample, I don't get one. And I thought I had a proof (which I'll supply if you want). $\endgroup$ Jan 3, 2012 at 1:35
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    $\begingroup$ I was thinking of the wrong definition of a Julia set, the one that only makes sense for polynomials because it looks at bounded orbits $\endgroup$
    – Will Sawin
    Jan 3, 2012 at 4:04
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    $\begingroup$ What kinds of properties are you interested in? As you point out, each Julia set is a branched cover of the other, so the local structures will be very similar, but the global structures may be very different. $\endgroup$
    – Jim Belk
    Jan 3, 2012 at 18:10

3 Answers 3


I'm not sure if this is helpful, but here is an example. The following picture shows the filled Julia set for $z^6 - 1$.

alt text

and the following picture shows the filled Julia set for $(z^2-1)^3$:

alt text

This is the case where $f(z) = z^2 - 1$ and $g(z) = z^3$. Note that the bottom image is a double cover of the top, while the top image is a triple cover of the bottom.

(These images were produced using Mathematica.)

  • $\begingroup$ That's very helpful! Thanks so much! $\endgroup$ Jan 4, 2012 at 8:59
  • $\begingroup$ Don't know what's going on here at Aug 2016 - the pictures are not displayed (although they exist at the given links) $\endgroup$ Aug 13, 2016 at 6:44

Answer. $J(fg)$ is the full $g$-preimage of $J(gf)$. (And vice versa with interchange of $f$ and $g$).

Proof. Let $A=fg$ and $B=gf$. Then we have a semi-conjugacy $gA=Bg$. Now it is a general fact, that whenever you have such a semi-conjugacy (of rational functions) the Julia set of $A$ is the $g$-preimage of the Julia set of $B$.

Proof. The semi-conjugacy can be iterated: $gA^n=B^ng$. Now, $z\in J(A)$, iff the family $gA^n$ is not normal, iff $B^ng$ is not normal, that is $z\in g^{-1}(J(B))$.

Added on 8.6.12. By the way, this demonstrates an amazing fact: for every $f$ and $g$, there exist sets $F$ and $G$ such that $G=f^{-1}(F)$ and $F=g^{-1}(G)$. This looks surprising to me. Finite sets $F$ and $G$ of cardinality greater than $2$ with such properties cannot exist, as a simple count shows. Are there other examples of such $F$ and $G$ ?

Added on 8.7.12. Let $f$ and $g$ be two rational functions. Let $F$ be a closed set, containing more than 2 points, and such that $(g^{-1}f^{-1}(F))=F$, then $F$ contains the Julia set of $fg$. And $J(fg)$ does have this property. (I am writing compositions $fg=f(g)$.) Trivial, but funny.

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    $\begingroup$ Thanks, Alexandre. Just to check: by "full g-preimage", you just mean "g-preimage" (i.e. preimage under g), right? So the word "full" is redundant? $\endgroup$ Aug 4, 2012 at 19:37
  • $\begingroup$ "Preimage" should be understood as "full preimage" in what I wrote. I think the general facts that a) the family $g(f_n)$ is normal on the set where $f_n$ is normal, and b) the family $f_n(g)$ is normal on the full $g$-preimage of the set where $f_n$ is normal are immediate consequences of definition of normality. $\endgroup$ Aug 4, 2012 at 19:54
  • $\begingroup$ Sorry, Alexandre, I'm still not certain that I understand you. I simply don't know what the term "full preimage" means (and my attempts to look it up have come to nothing). So: what is the definition of full preimage? $\endgroup$ Aug 4, 2012 at 21:41
  • $\begingroup$ Full $f$-preimage of a point $w$ is the set of all $z$ such that $f(z)=w$. Full preimage of a set is the union of full preimages of its points. $\endgroup$ Aug 5, 2012 at 8:27
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    $\begingroup$ OK, thanks. So "full preimage" is a synonym for "preimage". $\endgroup$ Aug 5, 2012 at 18:37

Hy I'm not shure that your restriction map $f: J(gf)\rightarrow J(fg)$ is right. Insted of looking at Julia set it is equivalent to look at Fatou set where is more natural work with mapings. Since you've put $f:\{(gf)^n\}_n \rightarrow \{(fg)^n\}_n$ it clearly does not work. Insted you have to use conjugation $f\circ-\circ f^{-1}$, since you have rational function of deg>1 the inverse is not properly defined. So in the neighbour of the periodic points of $gf$ which are not critical (you dont have inverse) for $f$ you get same dynamics.

As you know fixed points play important role in this theory so you can try to compare fixed points (or periodic) of $fg$ and $gf$.

I think there are some results if you have $g$ an $f$ that commutes and than you compare dimaics of $f$ and $g$ with dynamics of $fg$.

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    $\begingroup$ Luka, lease edit your message: I have difficulties reading it. $\endgroup$ Aug 4, 2012 at 17:09

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