Timeline for Are the algebraic numbers dense everywhere on the boundary of the Mandelbrot set?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Oct 12, 2015 at 0:05 | comment | added | Per Alexandersson | @LasseRempe-Gillen: Yeah, I realized that halfway through my thought process, so a natural question is then - does the Julia set (of a polynomial, say) always contain algebraic points? | |
| Oct 11, 2015 at 22:29 | comment | added | Lasse Rempe | @PerAlexandersson If the coefficients of the function are algebraic, then clearly the periodic points are algebraic, and repelling periodic points are in the Julia set. On the other hand, if your function is not algebraic, it seems unclear why one should expect algebraic points in the Julia set in general. | |
| Oct 11, 2015 at 22:01 | comment | added | Per Alexandersson | Ok it is not entirely certain, but we have the following: Start with any point, and repeat $z \to \pm \sqrt{z-c}$ and pick a branch randomly each time. Then every point in the Julia set will be $\epsilon$-close to a point in this sequence, for every $\epsilon>0$. Now, if the starting point in the sequence is already IN the Julia set (and algebraic), all points in the sequence are also in the Julia set (and algebraic). Thus, it is enough to find ONE algebraic point in the Julia set, for algebraic points to be dense in it. | |
| Oct 11, 2015 at 21:51 | answer | added | Lasse Rempe | timeline score: 27 | |
| Oct 11, 2015 at 20:17 | comment | added | Vesselin Dimitrov | The Mandelbrot set has capacity $1$, with equilibrium measure supported on its boundary. (Baker, DeMarco: Preperiodic points and unlikely intersections). It follows that there is a sequence of algebraic integers whose Galois orbits accumulate to $\mathcal{B}$. What you ask, with the algebraic numbers lying exactly on $\mathcal{B}$, is entirely different of course. Still I thought I would make this remark. | |
| Oct 11, 2015 at 20:08 | comment | added | Vesselin Dimitrov | @PerAlexandersson: Could you please briefly explain why would that be, for a Julia set $J$? (with algebraic numbers lying exactly in $J$). Also, do you restrict your $z^2 + c$ to have $c \in \bar{\mathbb{Q}}$ | |
| Oct 11, 2015 at 20:00 | comment | added | Per Alexandersson | Hm, if you asked the same questions about some julia set of a quadratic map, the answer to 1. should be yes. | |
| Oct 11, 2015 at 19:45 | history | asked | Stefan Kohl♦ | CC BY-SA 3.0 |