I've been using Unity3d (links to YouTube videos I've made are at the end of the post) and taking advantage of pixel shaders to explore (in real-time) the fixed-points of the complex quadratic map:


for values of $c$ given by a parametric equation $$c_r(t)=r(cos(t)+isin(t))=re^{it},t\in \Bbb{R^+},r\in [0,2]$$

The fixed point pair are given by $$z_0^*(c)=\frac{1\mp \sqrt{1-4c}}{2}, z_0^{*\alpha}(c)\equiv\frac{1-\sqrt{1-4c}}{2},z_0^{*\beta}(c)\equiv\frac{1+\sqrt{1-4c}}{2}$$ (I use $\mp$ becuase the alpha fixed point is the negative and the beta is the positive.)

I have noticed that the fixed points transform smoothly from their "alpha" fixed point to their equivalent "beta" fixed point (and vice-versa) after a full $2\pi$ rotation of $c$ about the origin, essentially permuting the roots. Generally, I keep $r=1$ and explore the unit circle. Therefore, I can use the Julia set known as the "Basilica" corresponding to $c=-1, (t=\pi)$ as a standard set or start/finish point when watching the fixed points transform through a full rotation. I choose this particular value/set because it's easiest (visually, I think) to see the relationship of the 2 fixed points and their preimages to the basins of attraction (at least, when $J_c$ is connected).

For example, the alpha fixed point (of $c=-1$) is the "pinch" (labeled $Z_0^\alpha$ in the picture) between the root basin and the period-2 basin to the left equal to $$z_0^{*\alpha}(-1)=\frac{1-\sqrt{1-4(-1)}}{2}=\frac{1-\sqrt{5}}{2}$$

and the beta fixed point is the far end of the set on the real axis (labeled $Z_0^{\beta}$ in the picture and is furthest right) $$z_0^{*\beta}(-1)=\frac{1+\sqrt{5}}{2}$$

and is basically a limit point. I consider it a limit-point in the sense that it falls at the end of a limit sequence of basins, whereas the alpha fixed point is the point between the 2 largest basins (For reasons of left-right symmetry, only looking at $\Re (z)<0$). The beta fixed point seems to be the equivalent of an infinitesimal for fixed points. All the other "pinch" points are preimages of the alpha fixed point (points labeled $Z_n^\alpha$ in image, plus more) and all of the other limit-points at the end of every limit-sequence of basins are preimages of the beta fixed point (points labled $Z_n^\beta$, plus more).

So when $t\to 2\pi n, n\in\Bbb{Z},n>0$, (one or more full rotations back to $c=r,t=0\mod 2\pi$), $$\lim_{t\to2\pi n}z_0^{*\alpha}(c(t))=z_0^{*\beta}(c(0)), \lim_{t\to2\pi n}z_0^{*\beta}(c(t))=z_0^{*\alpha}(c(0))$$

but there is a discontinuity in the equation at $t=2\pi n,n>0$ where the signs flip and $z^{*\alpha}$ and $z^{*\beta}$ go back to their original values. So if I'm watching the alpha fixed point and, right at the point of the sign change, I swap to watching the beta fixed point (and vice-versa after another $2\pi$), a continuous periodic curve (very, very close to but not exactly an ellipse) can be traced out with a period of $4\pi$ returning the "original" fixed points (when $t=0$) back to where they started.

(YouTube - Path Traced By Fixed Points)

Question 1) Is it possible to come up with a smooth parametric function for this continuous curve without having to swap signs of the fixed point due to the $\mp\sqrt{...}$? Could this idea lend itself to a Lie group of some kind? Is there any relation to spinors via a rotation of $2\pi$ transforming smoothly to an inverse (of sorts) while another $2\pi$ rotation transforms back to the original (a double covering)?

A "big" fixed-point (alpha-like, $Z_n^\alpha$) becomes a "small", or infinitesimal, fixed point (beta-like, limit-sequence, $Z_n^\beta$) and so on. The preimages of the fixed points behave similarily, alternating between alpha-like and beta-like every rotation.

Preimages of the fixed points are given by $z_{-(n+1)}=\pm\sqrt{z_{-n}-c}$ so the primary preimages are given by $$z_{-1}^*(t)=\pm\sqrt{z_0^*(t)-c(t)}=\pm\sqrt{\frac{1\mp\sqrt{1-4c(t)}}{2}-c(t)}$$

When this function is traced out, it creates the exact same shape as the fixed points themselves, only translated left, swapping between the first "pinch" point to the right of the main basin ($Z_1^\alpha$) and the far left limit point on the real axis ($Z_1^\beta$, far left).

(YouTube - Path Traced By Primary Preimages of Fixed Points)

Secondary preimages, given by $$z_{-2}^*(t)=\pm\sqrt{\pm\sqrt{\frac{1\mp\sqrt{1-4c(t)}}{2}-c(t)}-c(t)}$$ however, make a very different shape when traced out. There are four (unique) preimages for $n=-2$. There are two "big" or alpha-like preimages (labeled $Z_2^\alpha$) and two infinitesimal or beta-like preimages (labeled $Z_2^\beta$). These four fixed points also smoothly transform between each other over the course of a rotation of $2\pi$, except because there are four of them, they cycle between four positions every $2\pi$ and they require not $4\pi$ (or two rotations) to return to their original position (like the $0^{th}$ and $1^{st}$ preimages) but $8\pi$ (or four rotations) to return to their original position.

However, just like the other fixed points and preimages, they alternate every $2\pi$ between an alpha-like (or "big") fixed point and a beta-like (or infinitesimal) fixed point. A rotation of $2\pi$ essentially permutes all of the preimages of a given level of the fixed points while a rotation of $4\pi$ permutes the alpha-like fixed points separately from the beta-like fixed points. But the curve traced out by such a transformation is not a simple circle- or ellipse-like figure, but a slightly more complicated shape, shown in the diagram below:

(YouTube - Path Traced By Secondary Preimages of Fixed Points)

Tertiary preimages, of which there are 8 unique roots, permute between the four alpha preimages (labeled $Z_3^\alpha$) and the four beta preimages (labeled $Z_3^\beta$). That function, when traced out, leaves another strange curve:

(YouTube - Path Traced By Tertiary Preimages of Fixed Points)

(I have updated the pictures since the initial post, showing the actual path traced out instead of just a few points sampled along it, hopefully clarifying the issue a bit.)

The different colors of the sampled dots reflect the different piece-wise functions of the choices of $\pm\sqrt{...}$ for each level.

Question 2) Same as question 1, but for the preimages. Is it possible to describe those curves as continuous functions? Also, why does my spidey-sense tingle regarding this sort of permutation of roots, evoking thoughts of Galois Theory maybe?

(I'm also thinking about how to rephrase certain parts of this question based on answers given below...)

YouTube links

Path Traced By Fixed Points
Path Traced By Primary Preimages of Fixed Points
Path Traced By Secondary Preimages of Fixed Points
Path Traced By Tertiary Preimages of Fixed Points

  • $\begingroup$ can you show both parameter ( c) plane and dynamic ( z) plane at the same time? $\endgroup$
    – Adam
    Jun 2, 2018 at 4:12
  • $\begingroup$ On the parameter plane circle with radius 1 around origin is inside main cardioid. It croses polar coordinate system made by multiplier map : commons.wikimedia.org/wiki/… $\endgroup$
    – Adam
    Jun 2, 2018 at 4:21
  • $\begingroup$ Do you mean showing both planes side by side? Like the Mandelbrot on the left marking the relevant value of c and the corresponding Julia set on the right? $\endgroup$ Jun 2, 2018 at 21:36
  • $\begingroup$ Actually the unit circle doesn't intersect the main cardioid at all: upload.wikimedia.org/wikipedia/commons/5/56/Mandelset_hires.png $\endgroup$ Jun 2, 2018 at 21:38
  • $\begingroup$ yes. both sides like here : commons.wikimedia.org/wiki/… $\endgroup$
    – Adam
    Jun 3, 2018 at 5:44

1 Answer 1


Question 1) Is it possible to come up with a smooth parametric function for this continuous curve without having to swap signs of the fixed point due to the $\pm \sqrt{}$ ?

The answer is no, this is exactly what monodromy is about. No matter what parametrization or even what loop you choose, as long as the loop turns exactly once around $c=\frac{1}{4}$ (the parameter where the discriminant vanishes) the two fixed points will be switched.

If your loop doesn't turn around $c=\frac{1}{4}$, for instance if you pick a circle of radius less than $0.25$, then the roots will not be switched.

I'll leave it to you to generalize this to answer your second question.


OK, so probably covering space theory and algebraic topology is way too advanced for you at this point of your studies. Without going into the general theory, here is basically the reason of the phenomenon on a simple example.

Let $f: \mathbb C^* \to \mathbb C^*$ be the map $f(z)=z^2$. If you choose a path of the form $\gamma(t)= r(t) e^{it}$, where $r : [0,2\pi] \to \mathbb R_+^*$ and $r(0)=r(2\pi)=r_0>0$, you can ask if there exists a path $\tilde \gamma$ such that $f \circ \tilde \gamma = \gamma$. Such a path is called a lift of $\gamma$ under $f$. (To put it in a less rigorous way, you want to make sense of $\tilde \gamma(t)=\sqrt{\gamma(t)}$; think of your curves of fixed points as this $\tilde \gamma$).

First off, to find such a lift, you need to pick the starting point $\tilde \gamma(0)$; you have 2 choices, $\pm \sqrt{r_0}$. Let's say you pick $+\sqrt{r_0}$. Then you can check by hand that after this choice is made, $\tilde \gamma$ is completely determined, and must be equal to $$\tilde \gamma(t)=\sqrt{r(t)} e^{i t/2}.$$

In particular, after a full turn, you get $$\tilde \gamma(2\pi)=-\sqrt{r_0}.$$

So what's the point of all this? After $\gamma(t)$ makes a full turn around $z=0$, the two preimages of $\gamma(0)$ by $f$ are switched. You knew that after a turn, you would get again that $\tilde \gamma(2\pi)$ must satisfy $\tilde \gamma(2\pi)^2=\gamma(2\pi)=r_0$, but it wasn't obvious if you would come back to the starting point $+\sqrt{r_0}$, or go to the second preimage $-\sqrt{r_0}$. In other words, this operation of picking a starting point $\tilde \gamma(0)$ and then looking at the endpoint $\tilde \gamma(2\pi)$ gives you a permutation of the preimages of $\gamma(0)$, and this little discussion proves that this permutation is the non-trivial one (on a set of 2 elements). If you had picked $-\sqrt{r_0}$ for $\tilde \gamma(0)$, you would have found $\tilde \gamma(2\pi)=+\sqrt{r_0}$; and if your curve $\gamma$ turned twice around $z=0$, then your lifted curve $\tilde \gamma$ would come back to its starting point. (Prove it!)

In fancy terms, you would say that $f: \mathbb C^* \to \mathbb C^*$ is a covering map, and that its monodromy group (which describes the way that this operation of lifting the curve permutes preimages of $\gamma(0)$) is the $\mathbb Z/2\mathbb Z$, the group of permutations of a set of 2 elements. And here's something interesting : this whole discussion really doesn't depend on the function $r(t)$! the only important property about $\gamma$ is the number of times it turns around $0$ between $t=0$ and $t=2\pi$.

Here is how things generalize :

Theorem 1: If $f : X \to Y$ is a covering map and $X, Y$ are reasonnable spaces (never mind the technicalities), then every path $\gamma$ lifts to a unique path $\tilde \gamma$ (ie $f \circ \tilde \gamma = \gamma$), as long you chose the starting point of $\tilde \gamma$.

Theorem 2: Now suppose that $\gamma$ is a closed loop. Then the map $\tilde \gamma(0) \mapsto \tilde \gamma(2\pi)$ is a permutation of the preimages of $\gamma(0)$. Those permutations are called the monodromy group of $f$ (I'm not being very precise here); it doesn't depend on the parametrization of $\gamma$ (much better can be said, but let's leave it at that; key word is homotopy class).

Addendum 2

Now that you're an expert in algebraic topology ;) you can ask a very similar question to the one you ask: let's say I fix $f(z)=z^2-1$, and I choose a loop $\gamma(t)= \frac{1}{2} e^{it}$ around $0$. Let $\gamma_n$ be a lift of $\gamma$ by $f^n$, the $n$-th iterate of $f$. How does $\gamma_n$ permute the $n$-th preimages of $\gamma(0)$? What if I choose a loop around $z=1$ instead? (Why $0$ and $1$? because $f^n : \mathbb C \backslash f^{-n}(\{0,1\}) \to \mathbb C \backslash \{0,1\}$ is a covering map).

This is called the iterated monodromy group, and it is a subject of current research, with deep connections to complex dynamics, group theory, and automata theory.

So well done for asking this question, and keep wondering about things!

  • $\begingroup$ Thank you for pointing me in that direction! I had not heard of monodromy before this. It has given me a significant amount of material to read over, however it's also where the math starts getting a bit over my head personally. I'm trying to now learn about deck transformations, covering spaces and the fundamental group and it's a bit heavy in places. $\endgroup$ Jun 8, 2018 at 5:35
  • $\begingroup$ @GarrettMiller yes, it is not an easy topic to self-learn about. the classical reference is the book "Algebraic topology", by Hatcher, if you want to read about it in depth. If you do not, try to just work out the following: 1. $f(z)=z^2$ is a degree 2 covering map from the circle to the circle. 2. find its deck transformations, find its fundamental group. 3. work out by hand how the monodromy group. 4. is this cover Galois? $\endgroup$
    – Albert
    Jun 8, 2018 at 6:24
  • $\begingroup$ Sadly, Calc 2 (series, etc) and a little bit of linear algebra (in relation to video games/3d graphics) are the highest levels of math I've been able to take at a college level. I don't quite understand group theory enough to be able to answer those follow-up questions. I kinda understand the idea of a covering map from $S^1\to S^1$ (somewhat), but I don't feel like I would even know how to begin to answer the other questions. If you have the patience and time, I would love a (simple) walkthrough of how one would answer, but if not, no worries. :) $\endgroup$ Jun 9, 2018 at 8:54
  • $\begingroup$ Am I correct in figuring that the fundamental group of the circle is the integers (related to the winding number?) and therefore it is either $\Bbb{Z}$, or some sort of derivative of $\Bbb{Z}$? I'm still confused on two paths being combined and how the space that they are in determines a group. I now see "$p(z)=z^n$ is a regular cover with deck transformations given by multiplication by n-th roots of unity"? So, here, square roots of 1 are the deck transformations (at least for the fixed points)? And being regular (or Galois) indicates that $S^1$ is both connected and locally path-connected? $\endgroup$ Jun 9, 2018 at 9:49
  • $\begingroup$ duetosymmetry.com/tool/polynomial-roots-toy $\endgroup$
    – Adam
    Sep 11, 2021 at 16:28

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