Timeline for How is the Julia set of $fg$ related to the Julia set of $gf$?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 4, 2012 at 15:32 | answer | added | Alexandre Eremenko | timeline score: 13 | |
| Mar 31, 2012 at 17:02 | answer | added | Luka Thaler | timeline score: 0 | |
| Jan 6, 2012 at 9:14 | vote | accept | Tom Leinster | ||
| Jan 4, 2012 at 5:32 | answer | added | Jim Belk | timeline score: 23 | |
| Jan 3, 2012 at 21:20 | comment | added | Tom Leinster | Thanks, Jim and Jacques. In answer to your question, Jim, I'm interested in more or less any properties that are known - e.g. results on their local resemblance, or examples of their global lack of resemblance. Or if you can just point me to a reference, that would be good too. | |
| Jan 3, 2012 at 18:23 | comment | added | Jacques Carette | I think that Jim Belk's is essentially the most general answer: they are branched covers of each other. If you have more information about the critical points of $f$ and $g$, whether they are in the Julia or Fatou set, will give quite different kinds of branched covers. In the case of polynomials, whether you can get a nice Riemann surface for either $f^{-1}$ or $g^{-1}$ (by say cutting along external rays) could give you a solid description of the covering relation. But I doubt there is much you can say 'in general', i.e. without reference to particular dynamical properties of $f$ and $g$. | |
| Jan 3, 2012 at 18:10 | comment | added | Jim Belk | 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. | |
| Jan 3, 2012 at 4:04 | comment | added | Will Sawin | 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 | |
| Jan 3, 2012 at 2:23 | comment | added | graveolensa | I have been generating Newton's method fractals of $H(z)=f(g(z))-g(f(z))$ and more often $S(z)=f(g(z))-g(f(z))-z$ on my blog, which (I'll link to one entry, I haven't been making these recently, but if you browse around that you will find more): goo.gl/7G6Za | |
| Jan 3, 2012 at 1:35 | comment | added | Tom Leinster | 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). | |
| Jan 3, 2012 at 0:36 | history | edited | Tom Leinster | CC BY-SA 3.0 |
added a few little details
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| Jan 3, 2012 at 0:33 | comment | added | Tom Leinster | 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 $\infty$. | |
| Jan 2, 2012 at 22:52 | comment | added | Will Sawin | 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? | |
| Jan 2, 2012 at 12:58 | history | asked | Tom Leinster | CC BY-SA 3.0 |