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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?

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    @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.2012-06-28

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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.

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    I don't know any algebraic proof. I'll be surprised otherwise. It is an analytic question. It is NOT difficult to define the objects in an algebraic way: the parameters in the same hyperbolic component have the same combinatorial properties. But it seems hard to prove that those components are conformal disks without using complex analysis.2013-10-01
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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.