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Let $X$ be the space of all continuous maps from $S^{1}$ to $S^{1}$. $X$ is equipped with $C^{0}$ topology. Let $Y\subset X$ be the union of all finite order homeomorphisms i.e: all homeomorphisms $\phi$ with $\phi^{n}=Id$ for some natural number $n$. Assume that $g\in X$ belongs to $\bar{Y}-Y$. Some questions about such $g$:

1.Is $g$ necessarilly a homeomorphism?If yes, is it necessarilly order preserving?

  1. Is it true to say that $g$ satisfies at least one of the following conditions:

I: The periodic points of $g$ is dense

II: $g$ has a dense orbit

More generally, to what extent, the elements of $\bar{Y}-Y$ are known, dynamically?

What about if we replace $S^{1}$ with another manifold and study the elements of $\bar{Y}-Y$? What about if we consider $C^{k}$ topology for $k>0$?

Note: A possible operator theoretical version for question $1$ could be the following:Let $A$ be a $C^{*}$ algebra. Let $X$ be the space of all unital morphisms and $Y\subset X$ be the union of all finite order automorphisms. Assume that $T\in \bar{Y}-Y$. Is $T$ necessarilly an automorphisms.(Actually, is it surjective?)

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    $\begingroup$ If it is indeed the case that $g$ is a homeomorphism, then this should help you with $2)$. $\endgroup$
    – Hachino
    Jan 14, 2015 at 9:40
  • $\begingroup$ @Hachino thanks for your comment. Is the situation similar to $g=$ irrational rotation? Is there an example of a $g$ for which periodic orbits are dense, in contrast to irrational rotation? Could you please more explain on your comment? $\endgroup$ Jan 14, 2015 at 9:44
  • $\begingroup$ I don't know, that's just something your question reminded me of. I am no expert in DS, sorry. $\endgroup$
    – Hachino
    Jan 14, 2015 at 9:52
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    $\begingroup$ @AliTaghavi notice that if periodic orbits of $g$ are dense, then $g\in Y$. In fact, any periodic orientation-reversing homeo has period $2$, and hence there is no orientation-reversing map in $\overline{Y}\setminus Y$. On the other hand, every periodic orbit of an orientation-preserving homeo has the same period. So, any $g$ with dense periodic orbits must be periodic itself. $\endgroup$
    – Alejandro
    Jan 14, 2015 at 15:14
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    $\begingroup$ The elements that are not in the closure of the periodic elements, they're things like homeomorphisms that have some non-empty fixed-point sets, but the fixed point set is not the entire circle. So on the intervals between the fixed points they're shift operations. $\endgroup$ Jan 14, 2015 at 17:47

2 Answers 2

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The answer to both questions 1 and 2 is false, due to the following example. Consider $(n+1)$ consecutive intervals $I_0,...,I_n$ of lengths $1,2,4,...,2^{n-1}, 2^n$. Let the map $T_n$ cyclically permute them in an affine way. In fact, if you choose $I_j=[2^j,2^{j+1}]$, then $$ T_n(x)=\begin{cases} 2x, & x \not\in I_{{n}}\\ 2^{-n}x, & x\in I_{n}. \end{cases} $$

Now, make an affine change of the coordinates so that the union $I_0\cup\dots\cup I_n=[1,2^{n+1}]$ becomes (always the same) circle $[0,1]/(0\sim 1)$. Then, the constructed maps on this circle $C^0$-converge, as $n\to\infty$, to $$ T(x)=\begin{cases} 2x, & x \in [0,1/2]\\ 0, & x\in [1/2,1]. \end{cases} $$

This map is not a homeomorphism (it collapses the interval $[1/2,1]$ to a single point 0). Also, the orbit of any point falls to the fixed point 0 in a finite number of steps, so there is only one periodic point, and there is no dense orbit.

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  • $\begingroup$ thank you very much for your answer. Lets imagine these $I_{j}$ as arcs in the unit circle. So it seems that you fix $I_{0}$ at whole upper hemi circle, $I_{1}$ approachs to the whole lower hemi circle, and the remaining $I_{j}$'s are limited to the small remainig part. So it seems that the familly $T_{n}$'s is not an equicontinuous familly of maps. So this contradicts to uniforms convergence. So could you please more explain about your construction? $\endgroup$ Jan 15, 2015 at 13:16
  • $\begingroup$ In the other words, are you sure that $T_{n}$, $C^{0}-$ converges to $T$? $\endgroup$ Jan 15, 2015 at 13:24
  • $\begingroup$ In fact the last acr $I_{n}$ is very small for large n, and is mapped to $I_{0}$ which has a fixed size. so we loose equicontinuity.Am I mistaken? $\endgroup$ Jan 16, 2015 at 9:46
  • $\begingroup$ Sorry, but yes, you are: the length of $I_n$ is $2^n$, not $2^{-n}$, so this is almost half-circle arc, not very small one. On the other hand, $I_0$ has length 1 in a circle of length $2^{n+1}-1$, so it is a very small arc. $\endgroup$ Jan 17, 2015 at 0:13
  • $\begingroup$ I appologize for my misunderestanding. Now I underestand your counter example. $\endgroup$ Jan 17, 2015 at 7:54
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The answer to Question 2 is no. In fact, there is an orientation-preserving homeomorphism $g\colon \mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ such that $g\in\overline{Y}\setminus Y$ and such that it has exactly one fixed point (consequently, it has no dense orbit and it is not periodic):

For each $t\in\mathbb{R}$, let $\tilde g_t\colon\mathbb{R}\to\mathbb{R}$ be given by $\tilde g_t(x):= x + 1/8(1-\cos(2\pi x))+t$. Notice $\tilde g_t$ is the lift of a circle homeomorphism $g_t$. Let $g=g_0$ and observe $0\in\mathbb{R}/\mathbb{Z}$ is the only fixed point of $g$.

We want to show that $g\in\overline{Y}\setminus Y$. To do this, let $\rho_t=\rho(\tilde g_t)$ denote the rotation (also called translation) number of $\tilde g_t$. Observe that $\rho_t$ depends continuously on $t$ and $$ \rho_t > \rho_0=0, \quad\forall t>0.$$

In particular, this implies that there exists a strictly decreasing sequence $(t_n)$ with $t_n\to 0$ such that $\rho(t_n)$ is a Diophantine (irrational) number. Hence, by Herman-Yoccoz linearization theorem (see for instance https://eudml.org/doc/82144), for each $n$ there exists a $C^\infty$-diffeomorphism $h_n\colon\mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ such that $h_n\circ g_{t_n}\circ h_n^{-1}=R_{\rho_{t_n}}$, the rigid rotation of angle $\rho_{t_n}$. Then one can choose a sequence of sufficiently small positive real numbers $(\epsilon_n)$ such that $t_n+\epsilon_n\in\mathbb{Q}$ and $$h_n^{-1}\circ R_{t_n+\epsilon_n}\circ h_n\to g, \quad \text{in the } C^\infty\text{-topology}.$$

Finally, since $t_n+\epsilon_n\in\mathbb{Q}$, we have that $h_n^{-1}\circ R_{t_n+\epsilon_n}\circ h_n\in Y$ and consequently, $g\in\overline{Y}$.

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  • $\begingroup$ thank you very much for your interesting answer. What do you think about the first part? $\endgroup$ Jan 15, 2015 at 13:19
  • $\begingroup$ could you please more explain on $$h_n^{-1}\circ R_{t_n+\epsilon_n}\circ h_n\to g, \quad \text{in the } C^\infty\text{-topology}.$$ $\endgroup$ Jan 15, 2015 at 13:31
  • $\begingroup$ In the other words, is the "composition" a continuous map(not separately but strongly in its two variables)?(In C^0 or C^k topology?) $\endgroup$ Jan 15, 2015 at 13:49
  • $\begingroup$ Thanks again for your very interesting answer. I am sorry that I can not accept two answers:) meta.mathoverflow.net/questions/1491/… $\endgroup$ Jan 17, 2015 at 7:56
  • $\begingroup$ @AliTaghavi the above convergence is in the $C^\infty$-topology and that implies it also converges in the $C^0$-topology. $\endgroup$
    – Alejandro
    Jan 17, 2015 at 23:43

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