Timeline for Contractibility of connected holomorphic dynamics?
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Nov 18, 2013 at 10:52 | vote | accept | Sebastien Palcoux | ||
| Nov 17, 2013 at 18:17 | answer | added | Lasse Rempe | timeline score: 4 | |
| Nov 17, 2013 at 10:24 | comment | added | Adam Epstein | @Sébastien Palcoux: I wonder if it is reasonable to expect to produce an informative picture. The reason I am skeptical is that a sequence of Siegel quadratic polynomials can converge to a Cremer quadratic polynomial, and in such a situation the filled-Julia sets will converge in the Hausdorff sense - so by definition, the Cremer picture will be visually indistinguishable from the Siegel pictures, many of which will be locally connected and contractible. As for the transcendental case, Alexandre and Lasse raise an important issue concerning closedness and connectedness. | |
| Nov 17, 2013 at 8:54 | comment | added | Sebastien Palcoux | @LasseRempe-Gillen : thank you for your comment. Meantime a proof, do you have a picture showing it's not true for these polynomials ? Next, I'm not sure to understand, isn't contractibility well defined for the entire functions ? | |
| Nov 17, 2013 at 7:01 | comment | added | Lasse Rempe | For entire functions, the question is even more weird, because the set is no longer closed (nor open). Distinctions such as whether the set itself is connected, or only connected when you add infinity, will also become important. | |
| Nov 17, 2013 at 6:59 | comment | added | Lasse Rempe | I believe that the answer is negative, probably for every quadratic Cremer polynomial. It shouldn't be too hard to prove, but I probably won't have a chance to think about it carefully enough for a couple of days. If no-one has written an answer by then, I will try. | |
| Nov 17, 2013 at 0:37 | comment | added | Sebastien Palcoux | @AdamEpstein : (after Alexandre's comment) I'm asking about all the functions holomorphic in $\mathbb{C}$, not only the polynomial ones. Do you know a counter-example of entire transcendental function, as Alexandre said ? | |
| Nov 17, 2013 at 0:13 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I've precised that it's holomorphic in C.
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| Nov 17, 2013 at 0:01 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I've removed "Filled Julia set" because it's only for polynomial function, as said in comment.
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| Nov 16, 2013 at 23:55 | comment | added | Alexandre Eremenko | Functions holomorpic in C are called entire. I am sure Adam can tell you a true (not numerical) counterexample. Just tell him that you are asking about entire transcendental functions. | |
| Nov 16, 2013 at 23:51 | comment | added | Sebastien Palcoux | @AlexandreEremenko : holomorphic on $\mathbb{C}$. Do you know a numerical counter-example? | |
| Nov 16, 2013 at 23:47 | comment | added | Alexandre Eremenko | Holomorphic WHERE? Entire? Adam already explained to you that this is unknown in the polynomial case. But in the entire case is is probably wrong. What is the exact definition of "contractible" for a set which is neither open nor closed ? | |
| Nov 16, 2013 at 23:46 | comment | added | Sebastien Palcoux | @AdamEpstein : do you know a "numerical" counter-example ("numerical" means a picture, on which we can "see" that the set is connected but not contractible) ? | |
| Nov 16, 2013 at 23:42 | comment | added | Sebastien Palcoux | @AlexandreEremenko : My question applies to all the holomorphic functions. What's the name of $K(f)$ in the general case ? Also, is it known to be true in the polynomial case ? | |
| Nov 16, 2013 at 23:31 | comment | added | Alexandre Eremenko | What do you really mean by "holomorphic function"? A polynomial? Or your question applies to transcendental entire functons (for which $K(f)$ is NOT called the filled in Julia set). | |
| Nov 16, 2013 at 22:57 | comment | added | Adam Epstein | Offhand I don't know. But any locally connected example will be a dendrite, hence contractible. | |
| Nov 16, 2013 at 22:55 | comment | added | Sebastien Palcoux | @AdamEpstein : thank you, for your comment. So it's an hard open problem... Do you know if "$K(f)$ path-connected $\Rightarrow$ contractible" is also open ? | |
| Nov 16, 2013 at 22:44 | comment | added | Adam Epstein | It is not even known if $K(f)$ must be path connected. | |
| Nov 16, 2013 at 22:15 | history | asked | Sebastien Palcoux | CC BY-SA 3.0 |