Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

While studying a visual representation the Mandelbrot set, I have come across a very interesting property:

For any point inside the same primary bulb (a circular-like 'decoration' attached to the main body of the set), the periodicity of that point (i.e. the pattern of values that emerges when '$f(x) = z^2 + c$' is iterated with the '$c$' value that represents that point) is constant.

Does anyone know how to prove this property in a mathematical way? Is there more than one way in which this could be shown?

share|improve this question
    
I've edited the title to make it more descriptive. –  Rahul Jun 27 '12 at 19:10
    
Similarly, each bulb contains a root of $f^n(z)=0$. I seem to recall that the number of spokes was a factor of $n$, but it has been a while. –  Ross Millikan Jun 28 '12 at 0:22
    
@RossMillikan : Do you know how to prove any of these properties? –  Brian Jun 28 '12 at 0:40
    
@Brian: no, they were observations of mine long ago. The border of the bulb is where the derivative of the iterative loop is 1 in absolute value. The largest bulb off a root has twice the period of the main bulb, and the next two have three times the period, etc. Essentially the root of a bulb acts like the main bulb. I believe you can prove all this because the function is analytic, but don't have the details. –  Ross Millikan Jun 28 '12 at 3:58

2 Answers 2

For the first question: An application of the $\lambda$-lemma.

Theorem: Let $c_0$ and $c_1$ be in the same component $U$ of $C \backslash \partial M$, ($M$ is the Mandelbrot set) then $J_{c_0}$ and $J_{c_1}$ (the Julia sets) are homeomorphic (and even quasiconformal-homeomorphism) and the dynamics of the two polynomials $z^2+c_0$ and $z^2+c_1$ are conjugated on the Julia sets.

Proof: First notice that $M$ and $\mathbb C \backslash M$ are connected (Douady-Hubbard theorem). So $U$ is conformally equivalent to $\mathbb D$ (because simply connected).

Let $Q_k \subset U \times \mathbb C$ be set defined by the equation $P_c^k(z)=z$ where $P_c(z) = z^2+c$ and denote by $p_k : Q_k \longrightarrow U$ the projection onto the first factor. $Q_k$ is closed and $p_k$ is the restriction to $Q_k$ of the projection onto the first factor, so $p_k$ is a proper map. Moreover the two functions $(c,z)\mapsto P_c^k(z)-z$ and $(c,z)\mapsto (P_c^k(z))'-1$ vanish simultaneously at a discrete set $Z \subset U \times \mathbb C$, and the map

$p_k : Q_k \backslash p_k^{-1}(p_k(Z)) \longrightarrow U \backslash p_k(Z)$

is a finite sheeted convering map: it's proper and a local homeomorphism.

Let $c^\star$ be a point of $p_k(Z)$ and $U' \subset U$ a simply connected neighborhood of $c^\star$ containing no point of $p_k(Z)$. Denote by: $Q'_k = p_k^{-1}(U')$ and $Q_k^\star = Q'_k \backslash p_k^{-1}(c^\star)$. Denote by $Y_i$ the connected component of $Q^\star_k$; each of there is a finite cover of $U^\star = U' \backslash \{c^\star\}$. The closure of each $Y_i$ in $Q'_k$ is $Y_i \cup \{y_i\}$ for some $y_i \in \mathbb C$. If $(P^k_{c^\star})'(y_i)\neq 1$ the by the implicit function theorem $Q'_k$ is near $y_i$ the graph of an analytic function $\phi_i : U' \longrightarrow \mathbb C$.

Now let $Y_i$ be a component such that $(P^k_{c^\star})'(y_i)=1$. If $(c,z)\mapsto (P_c^k)'(z)$ is not constant on $Y_i$, the its image contains a neighborhood of $1$, in particular points of the unit circle, and the corresponding points of $Y_i$ are indifferent cycles that are not persistent. This cannot happen and $(c,z)\mapsto (P_c^k)'(z)$ is constant on every such component $Y_i$.

From the above it follows that if $R_k \subset Q_k$ is the subset of repelling cycles, then the projection $p_k : R_k \longrightarrow U$ is a covering map. Indeed, it is a local homeomorphism by the implicit function theorem, and proper since a sequence $(c_n,z_n)$ in $R_k$ converging in $Q_k$ cannot converge to a point $(c^\star,z^\star)$ where $P^k_{c^\star}(z^\star) = 1$. Hence the set of all repelling cycles of $P_c$ is a holomorphic motion. By the $\lambda$-lemma, this map extends to the closure of the set of repelling points, i.e. to the Julia set $J_c$, which also forms a holomorphic motion. $\square$

See also: Mane-Sad-Sullivan theorem.

I don't really understand your second question.

share|improve this answer
    
Is there really no more elegant, algebraic way of answering the first question? –  Brian Sep 3 '12 at 17:06
    
@Brian I'd be surprised if there were - I don't know of any elegant algebraic way of formulating the notion of being in the same bulb! It's possible that you could make some deformation argument with the 'pinch points' between bulbs taking on the role of critical points of the deformations, but that's unlikely to be any cleaner than this argument... –  Steven Stadnicki Sep 3 '12 at 17:44
    
We don't know any algebraic proof of this. –  Ilies Zidane Sep 4 '12 at 11:35
    
@Ilies Zidane: Is there any other way of proving this then, or is the way you showed the only way of doing it? Also, where could I learn more about this type of proof? –  Brian Sep 4 '12 at 12:16
    
I think this book has some proofs of things like this question. –  Old John Sep 4 '12 at 19:57

I am not sure on how to prove this. I myself have heard about other properties on periodicity. A particularly interisting one is as follows: Label a each primary bulb with a fraction $\frac pq$ where the bulb has period $q$ and the $p$th spoke the going counterclockwise from the main spoke is the biggest spoke of the antenna. Note, that $(p,q)=1$ for every primary bulb. Then the largest primary bulb between primary bulbs $\frac ab$ and $\frac cd$ is labeled with $\frac {a+c}{b+d}$ reduced to lowest terms.

share|improve this answer

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.