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I want to know why, when I look at the Julia sets of the quadratic family, I see only a finite number of repeating patterns, rather than a countable infinity of them.

My question is specifically about the interaction of these three theorems:

Theorem 1: Let $z_0\in\mathbb{C}$ be an repelling periodic point of the function $f_c:z\mapsto z^2+c$. Tan Lei proved in the 90s that the filled in Julia set $K_c$ is asymptotically $\lambda$-self-similar about $z_0$, where $\lambda$ denotes the multiplier of the orbit.

Theorem 2: (Iterated preimages are dense) Let $z\in J_c$, then the preimages of $z$ under the set $\cup_{n\in\mathbb{N}} ~ f^{-n}(z)$ is dense in $J_c$

Theorem 3: $J_c$ is the closure of repelling periodic points.

Let's expand on Theorem 1:
Technically it means that the sets $(\lambda^n \tau_{-z_0} K_c)\cap\mathbb{D}_r$ approach (in the Hausdorff metric of compact subsets of $\mathbb{C}$) a set $X \cap \mathbb{D_r}$ where the limit model $X \subset \mathbb{C}$ is such $\lambda$-self-similar: $X = \lambda X$.
Practically this means that, when one zooms into a computer generated $K_c$ about $z_0$, the image becomes, to all practical purposes, self-similar. No new information is gained by zooming again about $z_0$.

Lei also proved that $K_c$ is asymptotically $\lambda$-self-similar about the preimages of $z_0$, with the same limit model $X$, up to rotation and rescaling. This means that zooming in at each point in the repelling cycle of $z_0$, provides a basically the same spectacle, apart maybe rotated, that does zooming into $z_0$. Not only, but the preimages of $z_0$ are dense in $J_{c}$ (Theorem 2), meaning that this $X$ pattern can be seen throughout the Julia set.

Now, let consider a different repelling periodic point $z_1$. Lei tells us that $K_c$ will be asymptotically self-similar about $z_1$ and all its pre-images, with an a priori different limit set $Y$. Since the pre-images of $z_1$ are also dense in $J_c$ we may observe the limit model $Y$ all over $J_c$.

So, a priori to each repelling periodic orbit, there should be an associated limit model, and each of these limit models could be distinct. However, when I look at a computer generated Julia set, the parts of it that are asymptotically self-similar, seem to approach one of a finite set of limit models (up to rotation).

Why is it so? Maybe my eye cannot see the difference? Or the computer cannot generate all of the detail?

Or is it the case that the limit models are finite?

Simple Julia zoom In this image (read like a comic strip), I zoom into the neighbourhood of a point, four times, then purposely "miss the center", and zoom onto a detail for four more times. The patterns that emerge are very similar. Are they the same?
This is perhaps one of the simplest Julia set, but the experience is

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    $\begingroup$ The points on a "spine" where two "lobes" meet are preimages of the same repellent periodic point. (Think of them as analogues of rational numbers whose denominator is a power of 2.) The preimages of all those other repellent periodic points lie between them... $\endgroup$ – GNiklasch Aug 24 '16 at 12:54
  • $\begingroup$ What software did you use to generate those fractals? $\endgroup$ – Polygnome Aug 25 '16 at 8:36
  • $\begingroup$ @Polygnome it's a small iOS app called Fast Fractal, but I have seen similar things with JuliaTreck on MacOS $\endgroup$ – Andrea Aug 25 '16 at 10:52
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    $\begingroup$ @AndreaDiBiagio By the way, the result about asymptotic self-similarity at the repelling periodic points follows from the classical Koenigs theorem, 100 years before Tan Lei's work, which shows asymptotic self-similarity in the parameter plane (and that the picture is the same in dynamical and parameter plane). $\endgroup$ – Lasse Rempe-Gillen Aug 27 '16 at 19:03
  • $\begingroup$ @AndreaDiBiagio Do you feel that any questions from your post still remain open? If not, perhaps you might consider accepting an answer? $\endgroup$ – Lasse Rempe-Gillen Sep 9 '16 at 10:30
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Given that I have started making pictures, I thought it might be worthwhile adding another, shorter, direct answer to your questions, in addition to my longer, more detailed one.

Question 1. Are the limits in your pictures the same (up to a linear map)?

Answer. Yes. The only points in your "double basilica" picture at which two bounded Fatou components (interior regions of the filled Julia set) meet are preimages of the same periodic point. (This is the $\alpha$-fixed point of the first renormalisation.) Hence the Julia set near the two points is related by a conformal map, and the two scaling limits are the same, up to a linear transformation.

Question 2. Are there only finitely many scaling limits? Answer. No. But you must focus in on different periodic points to observe them. In other words, first fix your periodic point, then zoom in.

You did not give the precise parameter for your example, but here are scaling limits for the parameter $c=-1.3$. Full Julia set:

Double basilica Julia set

Scaling limits at the three real periodic points $x_1=-0.744989959798873$, $x_2=0.241619848709566$ and $x_3=1.131900530695346$ ($\alpha$ fixed point, $\alpha$ fixed point of the first renormalisation, and $\beta$ fixed point, respectively):

Finally, the scaling limit near the period 3 point $1.131900530695346 + 0.227896812185643i$. I give three successive zooms (each finer by a factor of 10), to emphasise the spiralling structure due to the non-real multiplier. The periodic point is at the centre of each picture.

You can clearly see that the scaling limits are different. You can pick more periodic points and obtain more scaling limits.

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Julia sets are all very closely related to self-similar sets - each one can be thought of as the invariant set of something like an iterated function system. Specifically, the Julia set of $f(z)=z^2 +c$ is the closure of the repelling periodic points of $f$. Thus, it makes sense that the Julia set itself should be attractive under an inverse of $f$ and there are two such inverses: $$f_{\pm}^{-1}(z) = \pm \sqrt{z-c}.$$

In fact, $$J = f_{+}^{-1}(J) \cup f_{-1}^{-1}(J)$$ so that it looks almost self-similar. Here's a gratuitous graphic illustrating the ides for $c=-1$:

enter image description here

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    $\begingroup$ That is a nice point - it is in some sense similar to the Dragon curve, en.wikipedia.org/wiki/Dragon_curve but non-linear. The similarities with en.wikipedia.org/wiki/Dragon_curve is quite evident... I wonder if there's a natural deformation interpolating these two fractals.... $\endgroup$ – Per Alexandersson Aug 24 '16 at 13:41
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    $\begingroup$ Iterated function systems were introduced with the intention of generalizing this property of quadratic Julia sets. See Example 14 in MR0799111 Barnsley, M. F.; Demko, S. Iterated function systems and the global construction of fractals. Proc. Roy. Soc. London Ser. A 399 (1985), no. 1817, 243–275. $\endgroup$ – Margaret Friedland Aug 24 '16 at 15:21
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    $\begingroup$ This is very clear, but does not answer my question, which is about the geometry in small and smaller neighbourhoods of specific points. Your answer addresses very clearly the global quasi-self-similairty. $\endgroup$ – Andrea Aug 25 '16 at 12:10
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    $\begingroup$ @MarkMcClure, you seem to have misunderstood my question. I am not surprised that I see the same patterns over and over again, this is a consequence of T1 and T2, and can be understood thanks to your answer too. I am surprised that there are so few of these patterns. T1,T2,T3 make me believe that since there are infinitely many repelling periodic points, each contributing to the self-similarity of $J$, I should be able to zoom in on arbitrarily many different shapes! $\endgroup$ – Andrea Aug 25 '16 at 13:20
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    $\begingroup$ @MarkMcClure You do get infinitely many different scaling limits in the same Julia set. Andrea is right to expect this, he just hasn't chosen the right places to zoom in (and moreover, unless the modulus of the multiplier is close to 1, you are unlikely to detect the different spiralling patterns by eye). $\endgroup$ – Lasse Rempe-Gillen Aug 27 '16 at 16:32
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To extend my comment and emphasize the self-similarity of Julia sets and the Dragon curve, here is an interpolation between the two.

enter image description here

Each frame is generated by two complex functions,

f1[z_, t_] := ((1.0 + I) z/2) t + (1 - t) (Sqrt[z + 0.9 I]);
f2[z_, t_] := (1 - (1.0 - I) z/2) t + (1 - t) (-Sqrt[ z + 0.9 I]);

where $t$ goes from $0$ to $1$ in the animation frames. For $t=0$, we have the classical Julia fractal for $c=-0.9i$, and at $t=1$, we have the two generators for the Dragon curve.

So what about the colors? Let $J$ be attracting set. Then $f_1(C)$ is the black set, and $f_2(C)$ is the blue set, and $C=f_1(C) \cup f_2(C)$. This puts emphasis of the self-similar nature.

So, note that in the Dragon curve case, since $f_1$ and $f_2$ are analytic and affine, they do not distort the picture at all, so you'll see exact copies at smaller levels. In the Julia case, we only have analytic maps, so there is some distortion caused by the square root, but the picture is more or less preserved (this is the nature of analytic maps).

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    $\begingroup$ I like this answer, upvoted it, and edited it a bit to improve the animation. Per my reputation, the edit won't be visible unless it gets approval through the review queue. You can view my version of the animation. I am a little puzzled why you chose a totally disconnected Julia set and a two dimensional self-similar tile, which can't be homeomorphic. $\endgroup$ – Mark McClure Aug 24 '16 at 18:21
  • $\begingroup$ @MarkMcClure: Yeah, maybe that wasn't the best example - but I think the self-similarity aspects are clear... perhaps there is a better Julia set one can interpolate with. I did not experiment that much with different parameters. $\endgroup$ – Per Alexandersson Aug 24 '16 at 18:48
  • $\begingroup$ For a direct connection between the dragon curve and Julia sets, see Milnor's "Pasting Together Julia Sets: A Worked Out Example of Mating" emis.de/journals/EM/expmath/volumes/13/13.1/Milnor.pdf $\endgroup$ – Lasse Rempe-Gillen Aug 24 '16 at 22:44
  • $\begingroup$ This is interesting, but does not answer my question. $\endgroup$ – Andrea Aug 25 '16 at 12:11
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As @GNiklasch points out, you seem to be zooming into two places which are both preimages of the same repelling periodic point. So the images of the Julia set are locally related by a conformal map, and hence indeed asymptotically the same.

If you zoom in at different repelling periodic points, then generally these will have different multipliers. For example, if you look at periodic points away from the real axis in your example, you would expect complex multipliers, and hence spiralling behaviours at small scales.

Look at this picture:quadratic Julia set

There is a periodic point within the "rabbit" parts, where there is lots of spiralling. There is also a fixed point connecting the big rabbit in the middle with the one to its left. (For those who know what this means, the latter is the $\alpha$ fixed point of the polynomial, while the former is the $\alpha$ fixed point of its renormalisation.) Finally, there is another fixed point to the very right of the picture.

The Julia set looks different near each of these.

EDIT. You can get an even clearer example by considering infinitely renormalisable quadratic polynomials. Consider the following procedure to select a parameter. Start at c=0, the centre of the main cardioid. Then move to the centre of the period 2 bulb at the left of the cardioid (c=-1, the "basilica"). This creates a periodic point at which two dynamic rays land, and which hence separates the Julia set into (exactly) two components.

Now move through a period 3 bifurcation from this component, creating a periodic point of period 6 having three rays landing at it. (This is the component containing the "dancing rabbits" shown above.) Continue, with a period 4 bifurcation, period 5, etc.

In the limit, you obtain a quadratic polynomial having infinitely many periodic points at which the Julia set is even topologically very different, in that they separate the Julia set into different components.

(For more details, on this kind of construction, refer to Milnor's Local connectivity of Julia sets: expository lectures, Section 3.)

I remark that, for a quadratic polynomial, the only points that can have more than two rays landing, and hence separate the Julia set into more than two pieces, are preimages of repelling periodic points. Each of these is associated to some small copy of the Mandelbrot set. Hence you can only get the above type of example by having an infinite number of renormalisations.

EDIT 2. As my original point does not seem to have come across to some, here are some pictures. For $$ c = 0.340095913765605+0.076587412582221i,$$ in the main cardioid, we obtain the following Julia set. Julia set of $z^2+c$ in the main cardioid

Here is the scaling limit near the $\beta$-fixedpoint, $$ z_0 = 0.618645316268697-0.322757842411465i:$$ Scaling limit near beta fixed point

Here is the scaling limit near a period 9 periodic point, $$ z_1 = 0.177144137748545 + 0.032520156063447i.$$ Scaling limit near period 9 point

You can see that the scaling limits are very different. (Images produced using the "Winfeed" fractal program by Richard Parris, 2012 version.)

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  • $\begingroup$ "As @GNiklasch points out, you seem to be zooming into two places which are both preimages of the same repelling periodic point". Indeed, this is very probable, since T2 says that the preimages of a specific point are dense on $J$. But T3 says that repelling periodic points are also dense on $J$, and each of these should have, a priori, a different multiplier, hence giving rise to different local behaviour. $\endgroup$ – Andrea Aug 25 '16 at 12:16
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    $\begingroup$ This is quite nice and shows that you can have multiple spiral patterns in the same Julia set but I wonder if it might miss the mark a bit. The specific spiral pattern that you see depends on where you zoom in on the Julia set but the set of all possible spiral patterns within that Julia set are dictated by the position of your parameter chosen from the point $c$ in the Mandelbrot set. As I understand it, there's still finitely many spiral patterns within one Julia set. $\endgroup$ – Mark McClure Aug 25 '16 at 13:43
  • $\begingroup$ @MarkMcClure No, there are infinitely many different periodic points in the Julia set, which you would expect to have different multipliers, and hence different scaling limits. $\endgroup$ – Lasse Rempe-Gillen Aug 27 '16 at 16:14
  • $\begingroup$ @AndreaDiBiagio As I understand your question, it is about the scaling limit at repelling periodic points. So you need to fix a periodic point and magnify near it, in order to see this scaling limit. It seems as though you have magnified near an "interesting" point, namely one that separates the Julia set - but in your case, there are only two periodic points that separate the Julia set, and any other cut point will be a preimage of one of these, and hence have the same scaling limit by conformality as I explain. $\endgroup$ – Lasse Rempe-Gillen Aug 27 '16 at 16:18
  • $\begingroup$ @Lasse Of course there are orbits of all periods but you can say the same thing for z^2 whose Julia set is just a circle. I don't think there are genuine spiral patterns evident in the Julia set of z^2+c for any c in the main cardioid. Spiral patterns become evident only after c passes through a bifurcation. For your example, there are two bifurcations as c passes from the main cardioid, into the period 2 bulb, and then into the period 3 bulb off of that. That is why we see the 2 specific types of spirals for that Julia set. $\endgroup$ – Mark McClure Aug 27 '16 at 17:50
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When you consider infinitely renormalizable polynomial, you can see infinitely many completely different picture. For example, there exists $c$ such that you can find periodic points $x_n$ such that $K_c\setminus \{x_n\}$ has exactly $n$ components.

Such an infinitely renormalizable $c$ is in a infinite nest of baby Mandelbrot sets. When you choose the baby Mandelbrot set of depth $n+1$ in the $1/n$-limb of the baby Mandelbrot set of depth $n$, then the so-called $\alpha$-fixed point $x_n$ of the $n$-th renormalization has rotation number $1/n$, so $K_c \setminus \{x_n\}$ has $n$ components.

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