3
$\begingroup$

For which sets of Gaussian integers $A\subseteq\mathbb{Z}[i]$ is the set of continued fractions with digits in $A$ $$ C(A)=\{a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cdots}}\mid a_i\in A\} $$ totally disconnected? Using dimension theory I get a partial result, but I like to know if there is a topological argument.

$\endgroup$
2
  • $\begingroup$ I corrected a few typos --- I hope "dimsion" was supposed to be "dimension", if not, I trust you'll change it to what it's supposed to be. $\endgroup$ Aug 23, 2013 at 0:19
  • $\begingroup$ Do you have the answer for the real case $A \subseteq \mathbb Z$ $\endgroup$ Dec 11, 2013 at 15:29

1 Answer 1

3
$\begingroup$

There exists a constant $c$, such that if $|\alpha| \ge c$ for all $\alpha \in A$, then $C(A)$ is totally disconnected.

To prove this, first let $a_0=0$, and consider $K= \{x+i y \mid x,y \in [-1/2,1/2)\}$. Let $T:K\to K$ be given by $$ Tz = \begin{cases} \frac{1}{z} - \left[ \frac{1}{z} \right] & z \neq 0\\ 0 & z = 0. \end{cases}, $$ where $[z]$ is defined as the point in $\mathbb{Z}[i]$ such that $z-[z] \in K$.

For each $z$, we can define a continued fraction expansion for it by $a_n(z) = [T^{n-1}z]$. Note that if we had picked $K$ differently, we may have obtained a different sequence of continued fraction digits.

The cylinder set $C[a_1]$ can be defined as the set of all $z \in K$ such that $a_1(z)= a_1$. We can define rank $n$ cylinder sets $C[a_1, a_2, \dots, a_n]$ as the set of points $z\in K$ such that $a_i(z) = a_i$ for $1 \le i \le n$. The cylinder sets of rank $n$ are necessarily disjoint from one another. Note that $T$ acts continuously on any cylinder set.

For all $\alpha$ with sufficiently large norm, the cylinder set $C[\alpha]$ is full - that is, $TC[\alpha]= K$. Let $\mathcal{F}$ denote the set of all $\alpha \in \mathbb{Z}[i]$ such that $C[\alpha]$ is full.

The fullness of cylinder sets implies that if you have a string $\{a_1, a_2, a_3, \dots \}$ with each $a_n \in \mathcal{F}$, then there exists a unique $z \in K$ such that $a_n(z) = a_n$ for all $n$. This implies that if $A \subset \mathcal{F}$, then for any $$ z = a_0+ \cfrac{1}{a_1+\dots} \in C(A) $$ we have $a_n(z-a_0)= a_n$, i.e., the continued fraction expansion given by $C(A)$ is the same as the expansion given by $T$.

Now, suppose that - in addition to $A \subset \mathcal{F}$ - all the $\alpha$ have norm sufficiently large so that $C[\alpha]$ is bounded away from the boundary of $K$. In particular, there exists an open set $U$ such that $C[\alpha] \subset U \subset K$, for all $\alpha \in A$.

To show that $C(A)$ is totally disconnected, it suffices to show that no two points $z, z' \in C(A)$ can exist in the same connected component. Without loss of generality we may assume $z, z' \in K \cap C(A)$. There exists a minimal $n$ such that $a_n(z) \neq a_n(z')$. We also have that $T^n z$ and $T^n z'$ are both in $U$. But then $$ z \in T|_{C[a_1(z)]}^{-1} T|_{C[a_2(z)]}^{-1} \dots T|_{C[a_n(z)]}^{-1} U \subset C[a_1(z), \dots, a_n(z)]$$ and $$ z' \in T|_{C[a_1(z')]}^{-1} T|_{C[a_2(z')]}^{-1} \dots T|_{C[a_n(z')]}^{-1} U \subset C[a_1(z'), \dots, a_n(z')]. $$ These preimages of $U$ are open, contain all points $C(A)$ that are in the corresponding rank $n$ cylinder set, and are necessarily disjoint since $a_n(z) \neq a_n(z')$, so $z, z'$ are in separate connected components.

I hope that helps.

$\endgroup$
2
  • $\begingroup$ Dear Joseph, many thanks for Your nice answer. I obtain the same result using a dimension estimate. What I still look for is a characterisation of sets $A$ in question. What is the case, if $A$ contains Gaussians of small modulus? $\endgroup$ Aug 25, 2013 at 0:48
  • $\begingroup$ Jörg, allowing elements of small modulus will be problematic for convergence. How do we define convergence if $a_n =0$, or if $a_n =1$ and $a_{n+1}=-1$? I would be intrigued to see if full solution of this problem is possible, but I suspect these convergence issues will need to be dealt with in some manner. $\endgroup$ Aug 25, 2013 at 2:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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