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First please see the question:

Let $X = S^1 \times [0, 1]$. Does there exist a continuous map $f: X \rightarrow X$ satisfying, for any point $x \in X$, $\{f^{(n)}(x)\}_{n=1}^\infty$ is dense in $X$? Here we denote $f^{(n)}=f \circ f \circ ... \circ f$ by the composition of $f$ for $n$ times. (EDIT: Here "for any point" means "for all points", "$\forall$")

Note that if $X = S^1 \times S^1$ is the torus, then the answer is "yes": just let $f(a, b)=(ae^{2\pi ip}, be^{2 \pi iq})$ where $p, q$ are rationally independent irrational numbers and we see that for any $x \in X$, $\{f^{(n)}(x)\}_{n=1}^\infty$ is dense in $X$, by the property of irrational numbers.

Note also that if $X=[0, 1] \times [0, 1]$, then the answer is "no". From the Brouwer's fixed point theorem we see that given any continuous $f : X \rightarrow X$ we can find a fixed point $t \in X$, and $f^{(n)}(t)=t$ for all $n$.

I am trying to show that given a continuous $f: S^1\times [0, 1] \rightarrow S^1 \times [0, 1]$ , there always exists some $a \in X$ such that the closure of $\{f^{(n)}(a)\}_{n=1}^\infty$ is a closed curve but not the whole of $X$. I don't know if this is true for all $f$, but it works for many examples. The simplest example is as follows: let $f(r, b)=(re^{2\pi ip}, b)$, where $p$ is an irrational number.

I have not found any other clues. Maybe this question requires some knowledge in dynamical systems or algebraic topology, which is not what I am familiar with.

Any hint would be greatly appreciated!

Alp Uzman
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user444628
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2 Answers2

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Consider the map: $$g: \Bbb{T}^2 \to \Bbb{T}^2, \quad (r, s) = \big(r e^{2\pi i p}, r e^{2\pi i q}\big)$$ With $p, q$ irrationally independent. The sequence $\{g^{(n)}\}$ is nothing but: $$\{ g^{(n)} \} = \{ \big(re^{2\pi in p}, r e^{2\pi i (np + q)} \big)\}$$ You can see that much like the other sequence this sequence is dense in $\Bbb{T}^2$, because $np + q$ and $p$ are never rationally dependent and the second entry never repeats.

Now a continuous map preserves limit points, so that if $D$ is dense in $X$, then $\phi(D)$ is dense in $\phi(X)$. Consider the map: $$f: X \to X, \quad (r, b)=(re^{ip}, \Re[re^{iq}])$$ This map induces the same sequence except the second entry is the cosine of the angle. But the map: $$\phi: \Bbb{T}^2 \to X, \quad \big(e^{i\theta_1}, e^{i\theta_2} \big) = \big(e^{i\theta_1}, \Re[e^{i\theta_2}] \big)$$ is clearly continuous and onto. Since every point of $\Bbb{T}^2$ is a limit point of the $g$ sequence, every point of $X$ is a limit point of the $f$ sequence. So $f$ is the required map, and it is valid for any initial point $x\in X$ as the argument does not depend on $r$.

Niki Di Giano
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    It looks to me that irrational translation on the torus does not induce a map on $X$ via the quotient: A point $(r, b)$ in $X$ generally comes from two points of the torus under the quotient map, and these points do not generally map to the same point of $X$ under irrational translation followed by the quotient map. <> Alternatively, it this trick worked, we could similarly quotient all the way down to the square; but we know there is no fixed-point free mapping of the square to itself. – Andrew D. Hwang May 08 '22 at 20:45
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    Note that elements in the set ${g^{(n)} (r, s)}$ can be written in the form $(e^{2 \pi i (m+p)}, e^{2 \pi i (m+q)})$. Take some $s \neq q-p$, and the element $(1, e^{2 \pi i s})$ is not in the closure of ${g^{(n)} (r, s)}$. – user444628 May 09 '22 at 03:38
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    I'm convinced the answer is not correct now. Thanks to both! – Niki Di Giano May 09 '22 at 07:58
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There is no minimal map on $S^1 \times [0,1]$. The only connected 2-manifolds (compact or not, with or without boundary) admitting minimal maps are the torus and the Klein bottle. See

https://zbmath.org/1072.37010

and

https://zbmath.org/1171.37008

Roman Hric
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