4
$\begingroup$

My question is the following: suppose we have two homeomorphisms $f,g:[0,1]\to[0,1]$ such that $f(0)=g(0)=0$, $f(1)=g(1)=1$ and that neither $f$ nor $g$ have a fixed point in $(0,1)$. Can we show that $f$ and $g$ are topologically conjugate?

I understand that one of the fixed points of $f$ is attracting and the other one is repelling (similarly for $g$). So the orbits exhibited by the homeomorphisms are easy to classify: there are two fixed points and the rest of the orbits asymptotically tend to the attracting fixed point.

In addition, the maps $f$ and $g$ could be considered as lifts of circle homeomorphisms (restricted to the unit interval); however, I'm not sure if this helps here.

Any help in constructing a homeomorphism $h$ such that $f\circ h = h\circ g$ would be much appreciated. Thanks in advance.

$\endgroup$
  • $\begingroup$ Do you want an exact explicit formula or construction method will be fine too? $\endgroup$ – Evgeny Nov 7 '13 at 4:04
  • $\begingroup$ A construction method seems the only plausible way forward to me. The only explicit information I can give is: h maps the attracting(repelling) point of g to the attracting(repelling) point of f. $\endgroup$ – Zephos Nov 7 '13 at 8:29
4
$\begingroup$

The standard recipe for construction is the following. Find half-interval $D = \lbrack a, b) \subset \lbrack 0, 1\rbrack$ such that for any point $x \in D$ it contains exactly one point from orbit $\mathcal{O}(x)$ under mapping $f$. Find analogous interval $D' = \lbrack a', b')$ for $g$. (Usually such intervals are called "fundamental domain" and their points represent orbits of map action) If you consider $\bigcup \limits_{n \in \mathbb{Z}} f^n(D)$, then you can show that it is equal to $( 0, 1 )$. Let's construct conjugacy $h$. So, you take any homeomorphism $\hat{h}$ that maps $D$ to $D'$. Then you take an arbitrary point $y$ (EDIT: except fixed points $0$ and $1$) and find $m \in \mathbb{Z}$ such that $f^m(y) \in D$ (there exists only one such number). After that you define $h(y)$ as $ g^{-m}(\hat{h}(f^m(y)))$. This is a well-defined homeomorphism from $(0, 1)$ to $(0, 1)$ and it can be extended to homeomorphism of $\lbrack 0, 1 \rbrack$ by setting $h(0) = 0$, $h(1) = 1$. Also, it conjugates $f$ and $g$.

The key fact is that if $f$ is homeomorphism of $[0, 1]$, $f(0) = 0$, $f(1) =1 $, then $f(x)$ has to be strictly increasing function of $x \in [0, 1]$. (EDIT: Here I'll assume that $0$ is repelling point and $1$ is attracting point; the other case can be treated similarly) So if we take point $x_0$ (EDIT: that doesn't coincide with fixed points) and consider half-interval $[x_0, f(x_0))$, then we'll see it doesn't contain more than one point from each orbit: by monotonicity for any $x' \in [x_0, f(x_0))$ follows $x_0 < f(x_0) < f(x')$. $\bigcup \limits_{n \in \mathbb{Z}} f^n(D) = (0, 1)$ follows from monotonicity and from that $0$ and $1$ are the only fixed points.

$\endgroup$
  • $\begingroup$ Hi, can you list any dynamical systems references for finding a "fundamental domain" -it's not a term I have come across before. $\endgroup$ – Zephos Nov 7 '13 at 17:13
  • $\begingroup$ It's a cross-branch term really: it appears in group theory (fundamental domain of group action), in topology (factor-topology for space of orbits). For dynamical systems usage you can take a look at Palis, de Melo "Geometric theory of dynamical systems: An Introduction". Also you may take a look at de Melo & van Strien "One-dimensional dynamics" book. $\endgroup$ – Evgeny Nov 7 '13 at 17:27
  • $\begingroup$ I've found the reference in de Melo & van Strien, thanks for that. Also, can I check that you let $D=[x_0, f(x_0))$ and $D'=[x_0, g(x_0))$ in your answer? $\endgroup$ – Zephos Nov 7 '13 at 22:08
  • $\begingroup$ Also, as a minor nitpick your half-intervals are empty if $f(x_0)<x_0$, which occurs if $0$ is the attracting fixed point of $f$. $\endgroup$ – Zephos Nov 7 '13 at 22:19
  • $\begingroup$ 1. Yes, you can. 2. That's my fault :) when I was writing, I've forgot to add that it must be any wandering point of the interval, so fixed points can't be taken as "generators" for fundamental domain $\endgroup$ – Evgeny Nov 8 '13 at 4:04

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.