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I study about rotation maps on circle, and I have a question.

Let $Homeo(S^1)$ be the set of all circle homeomorphisms with sup-metric $d(f,g)= \sup \{ d(f(x),g(x)| x \in S^1 \}$, and rotation map $f_\alpha: S^1 \rightarrow S^1$ is given by $f_\alpha (x) = x + \alpha \ mod \ 1$.

For rational rotation map $f_\alpha$, we can construct a sequence of irrational rotation maps $(f_{\alpha_n})$ converging to $f_\alpha$.

Then, at least, all rotation maps is in $\overline{\{\text{rotation maps} \ f_\alpha| \alpha \in \mathbb{Q}^c \}}$.

Is it possible that $\overline{\{\text{rotation maps} \ f_\alpha| \alpha \in \mathbb{Q}^c \}} = Homeo(S^1)$?

If it is true, could you give me a hint for this question?

Thank you in advance!

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  • $\begingroup$ There are many homeomorphisms of $S^1$ that are not rotations. Such maps move the points of the circle varying distances along the circle, but preserve their relative ordering (you can take any rotation map and modify it by any function $m:[0,1]\rightarrow [0,1]$ that is continuous and monotone increasing to get a homeomorphism of $S^1$ that is not a rotation map). $\endgroup$ – Justin Benfield Apr 19 '16 at 9:13
  • $\begingroup$ @Justin Benfield: Thank you. Yes, there are many homomorphisms of $S^1$ that are not a limit of rotation maps. $\endgroup$ – user69833 Apr 21 '16 at 12:30
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Let $R=\{f_\alpha:\alpha\in\Bbb R\setminus Q\}$. $\langle\operatorname{Homeo}(S^1),d\rangle$ is a metric space, so $f\in\operatorname{cl}R$ if and only if there is a sequence $\langle\alpha_n:n\in\Bbb N\rangle$ in $\Bbb R\setminus\Bbb Q$ such that $\langle f_{\alpha_n}:n\in\Bbb N\rangle$ converges to $f$. Suppose that $f(0)=\alpha$. Then $\langle f_{\alpha_n}(0):n\in\Bbb N\rangle=\langle\alpha_n:n\in\Bbb N\rangle$ converges to $\alpha$ in $S^1$. But then $\langle f_{\alpha_n}:n\in\Bbb N\rangle\to f_\alpha$, so $f=f_\alpha$. Clearly, however, there are autohomeomorphisms of $S^1$ that are not rotations, so $\operatorname{cl}R\ne\operatorname{Homeo}(S^1)$.

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  • $\begingroup$ @user69833: You're welcome. $\endgroup$ – Brian M. Scott Apr 21 '16 at 16:00
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You seem to be thinking of the circle as the interval $[0,1]$ with endpoints identified, but it is more helpful to think of it as the unit circle in $\mathbb{C}$ or equivalently the real line $\mathbb{R}$ with a point at infinity added. In the latter model, the fractional linear transformations also known as Mobius transformations $x\mapsto \frac{ax+b}{cx+d}$ where $ad-bc\not=0$ give many examples of selfmaps of the circle that are not anywhere near the irrational rotations. The Mobius transformations are homeomorphisms of the circle.

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  • $\begingroup$ Thank you for your example. From Mobius transformations, I can get many examples. $\endgroup$ – user69833 Apr 21 '16 at 12:34
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To give an explicit counterexample, consider the map $m:[0,1]\rightarrow[0,1]$ given by $m(x)=x^2$, this map satisfies $m(0)=0$, and $m(1)=1$ and is monotone increasing (not strictly so). The map is continuous with continuous inverse, hence it is a homeomorphism between $[0,1]$ and itself (an autohomeomorphism). Applying this map to $S^1$ will move points along the circle in a non-uniform way, for instance $0=1$ is stationary, but $m(\frac{1}{2})=\frac{1}{4}$.

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