# Difference between limbs and bulbs in Mandelbrot Set

Taking a look to the picture of the Mandelbrot set, one immediately notice its biggest component which we call the main cardioid. This region is composed by the parameters $$c$$ for which $$p_c$$ is hyperbolic when its periodic point is a fixed point (that is whose period is $$n=1$$). It turns out that these parameters are all of the form $$c_{\mu}:=\frac {\mu}2\left(1-\frac {\mu}2\right)$$ when $$\mu$$ runs in the open unitary disk $$\Delta_1$$.

For every rational number $$\frac pq$$ ($$p,q$$ coprime), there is a circular shaped "bulb" tangent to the main cardioid at the point $$c_{\mu}$$ for $$\mu=e^{2\pi i\frac pq}$$, called the $$\frac pq$$-bulb (clearly $$0\le\frac pq\le1$$), consisting of all parameters $$c$$ whose polynomial $$p_c$$ admits $$q$$-periodic points.

The best current estimate known was proved by Yoccoz in the Hubbard paper Local connectivity of Julia sets and bifurcation loci: three theorems of J.C.Yoccoz, which states that the diameters of the $$\frac pq$$-limbs tends to zero like $$1/q$$. Nonetheless numerical experiments suggest the following conjeture:

The diameter of the limbs tends to zero like $$1/{q^2}$$.

Or maybe one can consider a broader problem:

Is the Yoccoz asymptotics $$1/q$$ sharp, or can it be improved?

Some questions for you:

-It is clear that we have different bulbs with the same denominator $$q$$; are they all the different circles tangents to the main cardioid, which have the same radius depending on $$q$$?

-In Wikipedia it seems that the same object is first called bulb, then called limb: are these only different names for the same object?

-it seems that these bulbs are exact circles; is this true for all of them?

I ask this because studying the decay rate of the diameter of the $$\frac pq$$-limbs, if we fix $$q$$ and we call $$p_1,\dots,p_n$$ all the integers coprime with $$q$$ (and $$\le q$$), it would be nice if all the $$\frac{p_1}q,\dots,\frac{p_n}q$$-limbs have exactly the same diameter. Then the question of ratio decay would be sharper defined (and I'm not sure if it is actually necessarily for the question to be well defined).

EDIT We can put concepts in order as follows:

There is a unique region $$\Omega_0$$ composed by the parameters $$c$$ for which $$p_c$$ is hyperbolic when its periodic point is a fixed point (that is whose period is $$n=1$$). It turns out that $$\Omega_0$$ is bounded by the main cardioid and the parameters composing $$\Omega_0$$ are all of the form $$c_{\mu}:=\frac {\mu}2\left(1-\frac {\mu}2\right)$$ when $$\mu$$ runs in the open unitary disk $$\Delta_1$$. For every rational number $$\frac pq$$ (such that $$0<\frac pq<1$$ and $$p,q$$ coprime), we define the $$\frac pq$$-limb $$M_{\frac pq}$$ as the connected component of $$M\setminus\overline{\Omega_0}$$ tangent to the main cardioid at the point $$c_{\mu}$$ for $$\mu=e^{2\pi i\frac pq}$$; the biggest circular shaped "bulb" in $$M_{\frac pq}$$, called the $$\frac pq$$-bulb and denoted as $$\Omega_{\frac pq}$$, consists of all parameters $$c$$ whose polynomial $$p_c$$ admits $$q$$-periodic points.

We note that even though the boundary of the bulb $$M_{\frac12}$$ %period $$2$$ hyperbolic component of the Mandelbrot set is a perfect circle of radius $$\frac14$$ centered at $$-1$$, not all the bulbs are circles; see the case of the bulbs of period $$3$$.

• There is an exact parametrization of the period-3 components presented in this paper and they are not perfect circles. I doubt that the $p/q$ bulbs for $q>4$ are equal sizes. If correct, that shouldn't be too hard to check. – Mark McClure Oct 23 '18 at 12:23
• Note that a related question was asked here. – Mark McClure Oct 23 '18 at 12:25
• Thanks. The related question, answers only partially to mine problems. Then main point is to have a precise definition of limbs and bulbs, to get the difference between them. – Joe Oct 23 '18 at 13:13
• pi.math.cornell.edu/~hubbard/hubbard.pdf is the paper mentioned, the asymptotics are Remark 4.3 therein. I'm not sure if the p/q limb of a primitive satellite component (aka baby Mandelbrot set cardioid) includes the decorations that extend beyond the filaments of the baby Mandelbrot set; what are the numerical experiments you refer to? – Claude Nov 6 '18 at 5:21
• "the $\frac{p}{q}$-bulb ... consists of all parameters $c$ whose polynomial $P_c$ admits $q$-periodic points." could be clarified by explicitly stating $c \in M_\frac{p}{q}$ – Claude Nov 6 '18 at 5:40

p/q-limb is a part of Mandelbrot set contained inside p/q-wake.

Strict definition is in paper : Periodic Orbits, Externals Rays and the Mandelbrot Set: An Expository Account by John W. Milnor

The size estimate that I use (though I forget where I found it, references would be appreciated) for bulbs bifurcating from cardioid-like shapes at rational internal angle $$p/q$$ (in lowest terms, measured in turns) is:

$$R_{p/q} = R_1 \sin\left(\pi\frac{p}{q}\right) \frac{1}{q^2}$$

For secondary bulbs bifurcating from circle-like shapes, omit the $$\sin$$ factor.

I think (though I'm not sure) the term "limb" might refer either to everything in the wake of the two external rays landing at the root of the bulb: eg $$1/3$$ limb is between $$.(001)$$ and $$.(010)$$ rays expressed in binary

Or that wake minus the wake of its principal hub (the largest $$3$$-way spiral above the antenna from the bulb's $$1/2 \to 1/2 \to \ldots$$ bifurcation sequence)

And I think the "diameter" of a limb might not be in Cartesian space, but in terms of the external angles. For example, the "width of a wake" is the difference between the two angles (eg the $$1/3$$ bulb's wake has width $$1/7$$).

• Thank you Claude! Very useful. I've edited my answer with the notion taken so far. Anyway: where did you found the notions about diameter? Thanks again – Joe Oct 23 '18 at 21:10
• @Joe arxiv.org/abs/math/9411238 "Internal addresses in the Mandelbrot set and Galois groups of polynomials" by Dierk Schleicher – Claude Oct 23 '18 at 21:38
• @Joe I found the paper, I think I was wrong about diameter (I now think it is in Cartesian space). – Claude Nov 6 '18 at 5:22