Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $M$ be the Mandelbrot set and let $\lambda\in M, \mu\in \mathbb C$ be algebraic numbers. Let $t_{\lambda,\mu}$ be defined as $$ t_{\lambda,\mu} = \sup \lbrace t\in \mathbb R\colon \lambda +t\mu \in M\rbrace. $$

Question: Is $t_{\lambda,\mu}$ an algebraic number?

(If it's any easier - I'm also interested in the case when $\lambda$ and $\mu$ are both Gaussian rationals.)

share|improve this question
    
Most likely not, except in special cases (Alexandre notes one). Can you, for example, do this simple case: $\lambda = 0, \mu = 1+i$ ... ? –  Gerald Edgar Mar 11 '13 at 14:04
    
@Gerald, I don't know how to do this example. My initial intuition, however, is that there should be some relatively simple description of $t_{\la,\mu}$: if we first look at very large $t$, where the orbit of $0$ is unbounded, then we go to smaller and smaller $t$, suddenly the orbit becomes bounded. I thought this critical parameter should have some relatively easy relation to the polynomial which we iterate (e.g. being a critical point of some function, etc.) Then again, one could phrase a similarly sounding naive intuition about why should the whole Mandelbrot set be a "simple object"... –  Łukasz Grabowski Mar 11 '13 at 15:27

2 Answers 2

Clearly every piecewise real algebraic curve has this property. In general, I think you should expect that any curve with a simple, natural description, except for piecewise algebraic curves, does not have this property - it's simply too much of a conspiracy.

I will briefly review how to engineer such a conspiracy. Simply count the set of pairs of algebraic numbers, and step through them one by one. Start with any piecewise algebraic curve, and at each step, change it, either by increasing the degree or increasing the number of algebraic components, in a way that fixes the $t_{\lambda,\mu}$s of pairs $\lambda,\mu$ already counted. By the unbounded degrees of freedom available in such curves, this is clearly possible. One can also choose to do this while making only an incredibly small change to the overall shape of the curve, so that the sequence of curves converges uniformly to a limit, which cannot be algebraic as the complexity went to $\infty$, but which satisfies the requisite property.

But I see no reason to suspect the Mandelbrot set was created in this way. Note that the critical point is related not to one polynomial, but to an infinite sequence of polynomials, so even if it has a simple description, it is not so clear why that description should be an algebraic function. There are many other sorts of simple descriptions!

share|improve this answer
    
@Will, if you look at the Hofstadter's butterfly en.wikipedia.org/wiki/File:Gplot_by_Hofstadter.jpg and identify the vertical axis there with the interval [-i,i] and take $\lambda = iq$, where $q\in Q$ then $t_{\lambda,1}\in \bar{Q}$ (it's a trivial observation.) So it seems sensible to ask whether for the butterfly the function e.g. $x\mapsto t_{ix,1}$ maps $\bar{Q}$ to $\bar{Q}$. One definition of H's butterfly is as the parameters for which the orbit of 1 is bounded in a certain dyn. system defined by a linear recursion with non-const. coeff., so it's not too far away from $M$. –  Łukasz Grabowski Mar 11 '13 at 17:30
    
(the reason I asked my question for $M$ and not H's butterfly is I'm fairly sure this hasn't been studied for the H's butterfly, and I hoped perhaps it has been studied for $M$) –  Łukasz Grabowski Mar 11 '13 at 17:37

Of course, the answer depends on $\lambda$ and $\mu$. For example, if $\lambda=0$ and $\mu$ is $1$ or $-1$, then $t$ is rational.

On the other hand, even if $M$ is the unit disc, rather than Mandelbrot set, your number $t$ will be transcendental for some $\mu$ and $\lambda$.

So probably you want to rephrase your question somehow.

EDIT: Now I see that $\lambda$ and $\mu$ are algebraic numbers. (Sorry I did not see it when I read the question first time). Then the following results may be relevant.

  1. The boundary of the Mandelbrot set is not a semi-algebraic set (I believe this is in MR0974426 L. Blum, M. Shub, S. Smale, On a theory of computation and complexity over the real numbers: NP-completeness, recursive functions and universal machines. Bull. Amer. Math. Soc. (N.S.) 21 (1989), no. 1, 1–46.

  2. MR2466298 M. Braverman, M. Yampolsky, Computability of Julia sets. Algorithms and Computation in Mathematics, 23. Springer-Verlag, Berlin, 2009.

share|improve this answer
    
Thanks for you reply - however if $M$ is the unit disk, then certainly $t$ would be algebraic - the point in question would be in the intersection of two (real) curves $x^2+y^2 =1$ and a line $y=ax+b$, where $a$ and $b$ are algebraic numbers depending on $\mu$ and $\lambda$. So as it stands for the moment, I still think my question makes sense. –  Łukasz Grabowski Mar 11 '13 at 13:22
    
And I think your question still does not make sense. Take $M$ the unit disc, $\mu$=1 and $\lambda=\pi$. Then what is $t_{\mu,\lambda}$, according to your definition? –  Alexandre Eremenko Mar 12 '13 at 12:25
    
I clearly wrote that $\lambda$ and $\mu$ are assumed to be algebraic numbers –  Łukasz Grabowski Mar 12 '13 at 17:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.