1
$\begingroup$

This paper "Local connectivity of Julia sets and bifuraction loci: three theorems of J.C.Yoccoz" (see http://pi.math.cornell.edu/~hubbard/Yoccoz.pdf) tells that it is quite easy to construct the topology of a puzzle to a given depth by hand (see Remark 5.3). I wonder that for the \frac{1}{3}-limb, we know that there are three external rays landing at the repelling fixed points $\alpha$, then it is easy to know the knowledge at depth-$0$, and the topology of depth-$1$ can be got by mapping each puzzle at depth-$0$ under $f^{-1}$. Now I am a little confused that how do we know the position of the preimage of $\alpha$, i.e. how do we know which piece does $f^{-1}(\alpha)$ belong to? Furthermore, how do we know that the topology of a puzzle at depth-$2$, or depth-$n$? Are there any more details can I get from here?

$\endgroup$
  • 1
    $\begingroup$ What is a puzzle in this context? $\endgroup$ – Vincent Sep 11 at 8:18
  • $\begingroup$ It is a theory which can be used in study the local connectivity of Julia sets, you can find the details in pi.math.cornell.edu/~hubbard/Yoccoz.pdf $\endgroup$ – Yee Neil Sep 11 at 10:29
1
$\begingroup$

The "rabbit" (1/3-limb) puzzle has three pieces, let's call them 0 (the piece containing the critical point $0$), 1 (the piece containing the critical value $c$), and 2 (the remaining piece). Combinatorially speaking, 1 is mapped 1-1 to 2, 2 is mapped 1-1 to 0. (More precisely, 1 is mapped to a piece bounded by the same rays as 2, but with the equipotential somewhat further out.) 0 is mapped over the whole puzzle, covering 1 twice, and the remaining two pieces once.

In particular, the non-periodic preimage of the $\alpha$ fixed point must necessarily be in the $0$ sector. The next level of the puzzle will therefore have one piece contained in 1, one piece contained in 2, but 3 pieces contained in the original piece 0, touching at the preimage of the $\alpha$ fixed point.

You can continue inductively to work out the structure of the level $n+1$ pieces from that of the level $n$ pieces, assuming that you know the position of the critical value with respect to the latter. Equivalently, you should know how the critical value maps through the original pieces under iteration. (I am being slightly vague here, but trust you can work out the details.)

In particular, the structure of the puzzle at higher levels does depend on where you are in the Mandelbrot set - it will not be the same for all points in the 1/3-limb. Indeed, for non-renormalisable maps, Yoccoz's theorem shows that the structure of how the puzzle develops determines the parameter $c$ uniquely.

$\endgroup$
  • $\begingroup$ Does it mean that we need to know how each preimage of those external rays and themselves maps to another one? If that is right, I think it must be very complicated to determine higher levels even we know that is determined by the initial state of the puzzle. $\endgroup$ – Yee Neil Sep 14 at 8:56
  • $\begingroup$ You need to know where your parameter is in the Mandelbrot set. One way to describe this is with an external angle. This is the angle of a parameter ray which, a posteriori, we know must land at the parameter. But it can also be described combinatorially, e.g. as an angle whose parameter ray is not separated from the given parameter by pairs of periodic or pre-periodic rays landing together. See Schleicher's paper on fibers of the Mandelbrot set, for example. $\endgroup$ – Lasse Rempe-Gillen 2 days ago
  • $\begingroup$ In terms of how complicated it is, it is not that complicated if you know where the parameter is. Of course, the number of rays involved in the puzzle does grow exponentially with the level of the pieces. $\endgroup$ – Lasse Rempe-Gillen 2 days ago

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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