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Let $\mathcal{B}$ denote the boundary of the Mandelbrot set, and let $\overline{\mathbb{Q}}$ denote the algebraic closure of the rationals. Further put $\mathcal{B}_{\overline{\mathbb{Q}}} := \mathcal{B} \cap \overline{\mathbb{Q}}$.

Questions:

  1. Is $\mathcal{B}_{\overline{\mathbb{Q}}}$ dense in $\mathcal{B}$ in the sense that for every $z \in \mathcal{B}$ and every $\varepsilon > 0$, the $\varepsilon$-neighborhood of $z$ has nontrivial intersection with $\mathcal{B}_{\overline{\mathbb{Q}}}$?

  2. Given $z \in \mathbb{C}$, let $\mathcal{B} + z := \{b + z \ | \ b \in \mathcal{B}\}$ denote the translate of $\mathcal{B}$ by $z$. Is there a $z \in \mathbb{C}$ such that $(\mathcal{B} + z) \cap \overline{\mathbb{Q}} = \emptyset$?

  3. What are the answers to (1.) and (2.) if we replace $\overline{\mathbb{Q}}$ by the Gaussian rationals $\mathbb{Q}[i]$ or by the cyclotomic integers $\mathbb{Z}[e^{\frac{2\pi i}{5}}]$?

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    $\begingroup$ Hm, if you asked the same questions about some julia set of a quadratic map, the answer to 1. should be yes. $\endgroup$
    – Per Alexandersson
    Commented Oct 11, 2015 at 20:00
  • $\begingroup$ @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 $c \in \bar{\mathbb{Q}}$ $\endgroup$
    – Vesselin Dimitrov
    Commented Oct 11, 2015 at 20:08
  • $\begingroup$ 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. $\endgroup$
    – Vesselin Dimitrov
    Commented Oct 11, 2015 at 20:17
  • $\begingroup$ 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. $\endgroup$
    – Per Alexandersson
    Commented Oct 11, 2015 at 22:01
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    $\begingroup$ @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. $\endgroup$
    – Lasse Rempe
    Commented Oct 11, 2015 at 22:29

1 Answer 1

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Post-critically pre-periodic quadratic polynomials, i.e. those for which the orbit of the critical point $0$ is pre-periodic, are well-known to be dense in the boundary of the Mandelbrot set. (This is essentially a normality argument.)

Each of these is determined by an algebraic equation. This answers your first question (the question in the title).

EDIT. As pointed out by Malik, post-critically pre-periodic parameters are also called Misiurewicz points. (Unfortunately, the term Misiurewicz is sometime also used, particularly in real dynamics I think, to refer to a larger class of systems.)

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    $\begingroup$ These are called Misiurewicz points, right? It might be useful to add this to your answer in order to facilitate literature search. $\endgroup$
    – Malik Younsi
    Commented Oct 11, 2015 at 23:59
  • $\begingroup$ Very nice, thanks! -- Is it also possible to come to a similar conclusion when one replaces $\overline{\mathbb{Q}}$ by some smaller domain, like e.g. the Gaussian rationals or a ring of cyclotomic integers, as asked in Part (3.), or would this be substantially more difficult? $\endgroup$
    – Stefan Kohl ♦
    Commented Oct 12, 2015 at 9:27
  • $\begingroup$ Saying something about more restrictive domains would be substantially more difficult, I believe. In particular, when it comes to the Gaussian rationals, this is obviously a much smaller set, and Misiurewicz parameters are not generally of this form. Techniques similar to those used in the study of the field of moduli of a dessin d'enfant might, however, be useful here. $\endgroup$
    – Lasse Rempe
    Commented Oct 13, 2015 at 8:16

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