# A point set of power series with coefficients in {-1, 1}. Connected or not?

Let $z$ be a fixed complex number with $|z|<1$ and consider the set $$X_z := \Big\{\sum\limits_{i=1}^{\infty} a_i z^i \ \Big|\ a_i\in \{-1,1\} \forall i\Big\}.$$

What can be said about the set $M$ of those $|z|\lt 1$ such that $X_z \subset \mathbb{C}$ connected?

• @SebastianGoette Thx for your comment, but I think I had wronged the idea... For a fixed z it's not a power series. And the point is determined by the signature of $\{a_n\}$. Would you please give me more information if you know some details about this question? – Kirby Lee Nov 20 '15 at 8:52
• I took the liberty of rephrasing your question, let me know if this is not what you wanted to ask. I think the question is nice and deserves an attention it would not have gotten in your phrasing. – Benoît Kloeckner Nov 20 '15 at 8:57
• @BenoîtKloeckner Thanks a lot. It's exactly the question. I'm a little ashamed for my poor description... – Kirby Lee Nov 20 '15 at 9:03
• Take $|z|=1/3$. If $a_1=1$, then the corresponding element of $X_z$ is at distance of $\leq 1/6$ from $z$, otherwise it is at distance of $\leq 1/6$ from $-z$. Thus $X_z$ is disconnected. A more interesting question arises when $|z|>1/2$ (and now it may depend on the argument of $z$ as well)... – Ilya Bogdanov Nov 20 '15 at 9:17
• @IlyaBogdanov: you are right, but even the $|z|=1/2$ case is interesting. If $z=\pm1/2$ then $X_z$ is a segment, but I guess it is a Cantor set in every other case. Certainly, when $|z|>1/2$ there are more possibilities. – Benoît Kloeckner Nov 20 '15 at 9:21

There is a related iterated function system with two functions,

$f_0(x) = 1+zx$

$f_1(x) = -1+zx.$

$X_z$ is the unique nonempty compact fixed set of this iterated function system. It is sometimes called a generalized dragon set with parameter $z$, and particular values of $z$ can produce some well-known fractals called dragons.

A relevant result on iterated function systems is that the fixed set $X$ is connected iff it is arcwise connected iff the family of subsets $\{f_i(X)\}$ is connected, which in this case means $f_0(X_z) \cap f_1(X_z)$ is nonempty. (This paper refers to Kigami, Analysis on Fractals chapter 1 for the result.) So, the set is connected iff we can write

$$\begin{eqnarray} 1 + \sum_{i=1}^{\infty} a_i z^i &=& -1 + \sum_{i=1}^\infty a_i' z^i \newline 1 + \sum_{i=1}^{\infty} \frac{a_i-a_i'}{2} z^i &=& 0 \newline 1 + \sum_{i=1}^{\infty} b_i z^i &=& 0 \end{eqnarray}$$

where $b_i = (a_i-a_i')/2 \in \{0,1,-1\}$.

In particular, $X_z$ is connected when $z$ is real with $1/2 \le |z| \lt 1$ and when $z$ is a root of a polynomial with coefficients in $\{-1,0,1\}$. The intersection of the closure of those roots with the interior of the disk is $M$.

This image shows the nonzero roots of polynomials of degrees up to $9$ with coefficients in $\{-1,0,1\}$ with the circles of radii $\frac{1}{\sqrt{2}}$ and $1$.

The closures of roots of polynomials with restricted coefficients have been studied, and they are quite interesting. In some areas, there seems to be a Julia-Mandelbrot correspondence, where the set of roots of small degree near a point resembles the fixed set of the iterated function at that point. This preprint of Thierry Bousch proves some connectivity properties of the closure, and that the annulus $\frac{1}{\sqrt{2}} \lt |z| \lt 1$ is in $M$. So, some of the apparent holes in the picture above close up as the degree increases, including those between the two circles such as near some roots of unity. The paper of Calegari et al mentioned by Nikita Sidorov proves that there are many actual holes in $M$, among other results.

• Thanks for your professional answer to the question. It's hard for me to look for the relative results for knowing little about dynamical system... Your webpage links are also helpful and attractive. And it seems that I need a lot of time to go deeper and really understand it. – Kirby Lee Nov 20 '15 at 12:28

First, if $z=0$, then $X_z=\{0\}$ is obviously connected.

Let me prove that the answer is negative whenever $0<|z|<1/2$. In this case, we will show that $X_z$ is a Cantor set, thus totally disconnected, by showing that it is a bi-Lipschitz embedding of $\{-1,1\}^{\mathbb{N}}$ endowed with the ultrametric $d((a_i),(b_i))=\theta^{\inf\{i,a_i\neq b_i\}}$ where $\theta=|z|$ (this is a metric as soon as $z\neq 0$).

We thus consider the map defined by $$F((a_i))=\sum_{i=1}^\infty a_i z^i;$$ given two sequences $(a_i)$, $(b_i)$ of $\pm1$, let $i_0=\inf\{i,a_i\neq b_i\}$ and observe that \begin{align*} |F((a_i))-F((b_i))| &= \Big|\sum_{i=i_0}^\infty (a_i-b_i)z^i \Big| \\ &\le \sum_{i=i_0}^\infty |a_i-b_i| \theta^i \\ &\le 2\theta^{i_0}/(1-\theta) \end{align*} which precisely says that $F$ is a Lipschitz map. In the other direction, we have \begin{align*} |F((a_i))-F((b_i))| &\ge 2\theta^{i_0}-\Big|\sum_{i=i_0+1}^\infty (a_i-b_i)z^i \Big| \\ &\ge 2\theta^{i_0} - 2\theta^{i_0+1}/(1-\theta) \end{align*} and we are done as soon as $1-\theta/(1-\theta)>0$, i.e. $|z|<1/2$.

When $\theta\ge 1/2$, things seems a bit more complicated. One observes that $X_{\frac12}=[-1,1]$, thus is connected. (Indeed one inclusion is clear, the other is obtained by taking $(a_i)$ the base-$2$ development of $r\in [0,1]$, then $F((a_i))=2r-1$). In general, $X_z$ is a self-similar set with contraction ratio $|z|$, and you can probably find much information by the literature on these sets. Your question might be hard in general.

• You're right. And Zare has some results and references which you might be interested in. Also thanks for answering. – Kirby Lee Nov 20 '15 at 12:39

What you are referring to is called the connectedness locus. The most recent important paper on the subject is probably this paper by Calegari, Koch and Walker.

In particular, they show that the connectedness locus contains infinitely many holes. The paper also contains an extensive list of references.

• Note: they deal (as many previous authors) with the subset $X'_z$ of $\mathbf{C}$ defined in the same way but with $a_i\in\{0,1\}$ instead of $a_i\in\{-1,1\}$. If $a_i\in\{-1,1\}$ and we write $q_z=\sum (-1)^nz^n$ and $b_i=(a_i+1)/2\in\{0,1\}$, then $\sum a_i z^i=q_z+2\sum b_iz^i$, so $X_z=q_z+2X'_z$, so $X_z$ and $X'_z$ are homothetic. – YCor Nov 20 '15 at 20:06
• Yeah, it doesn't really matter which "digits" we use as long as there are two of them. All these sets are affine copies of each other. – Nikita Sidorov Nov 20 '15 at 20:44
• An easier introduction in this post by D. Calegari: lamington.wordpress.com/2014/12/21/… – YCor Nov 20 '15 at 20:55
• My classmate (from whom I get this question) says he finds what he needs from the paper you linked here. Thanks. – Kirby Lee Nov 21 '15 at 7:01