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Let $R_\alpha(s)=s+\alpha $ mod 1, likewise for $R_\beta$ be the circle rotations on the unit interval. Prove that $R_\alpha$ and $R_\beta$ are conjugate by a homeomorphism iff $\alpha = \pm \beta$ mod 1.

From Brin & Stuck: Introduction to dynamical systems 1.2.4

My effort:

$\impliedby$

Let by definition of modulus, $\alpha = \beta + k$ with $k\in\mathbb{Z}$. Then \begin{align} R_\alpha(s) &= s+ \alpha~(\text{mod 1})\\ &=s+ \alpha + k_1\\ &= s+(\beta + k_2) + k_1\\ &= s+\beta + k_1 +k_2\\ &= s+\beta~(\text{mod 1})\\ &= R_\beta(s) \end{align} Therefore, I define $h(s) = Id_{[0,1)}$, the identity on the unit interval, which is an homeomorphism. Thus $h \circ R_\alpha = R_\beta \circ h$ and h conjugates the circle rotations.

Now let $\alpha = -\beta + k$ with $k\in\mathbb{Z}$. \begin{align} R_\alpha(s) &= s+ \alpha~(\text{mod 1})\\ &=s+ \alpha + k_1\\ &= s+(-\beta + k_2) + k_1\\ &= s-\beta + k_1 +k_2\\ &= s-\beta~(\text{mod 1})\\ &= -(-s+\beta~(\text{mod 1}))\\ &= -R_\beta(-s)~\qquad\qquad\qquad (*) \end{align}

Then to be conjugations we must have an homeomorphism $h$ such that

\begin{align} h \circ R_\alpha &= R_\beta \circ h\\ h(R_\alpha(s)) &= R_\beta(h(s))~\text{if we plug in (*)}\\ h(-R_\beta(-s)) &= R_\beta(h(s))\\ h(-R_\beta(-s)) &=h(s)+ \beta~\text{mod 1} \end{align}

Now I am not sure how to continue as I am not able to construct an function $h$ from the last expression that would be an homeomorphism.

EDIT 1: WRONG

The function $h(s) = -Id_{[0,1)}$ works in this case and is a homeomorphism.

Now the other way.

$\implies$

Let $h$ be an homeomorphism s.t the unit rotations $R_\alpha$ and $R_\beta$ are conjugate. There exist lifts for the rotations denoted $r_\alpha$ and $r_\beta$, and thus for the homeomorphism h, denoted $H$ that conjugates the lift rotations. I assume that the lift is preserved under direction and increasing, therefore $H(s+1) = H(s) + 1$. Then \begin{align} r_\alpha(s) &= H^{-1}(r_\beta(H(s)))\\ &= H^{-1}(H(s) + \beta) \end{align}

Then follows \begin{align} H(r_\alpha(s)) &= H(s) + \beta\\ H(s+ \alpha) &= H(s) +\beta \\ \end{align}

Then, because we have the property of preservation of direction for $H$ that $\alpha=\beta$ for the lift and thus $\alpha=\beta \text{mod 1}$ for the rotations.

For the second case I assume that the lift is preserved under direction and decreasing, therefore $H(s+1) = H(s) - 1$.

Following the same argument, I get

\begin{align} H(r_\alpha(s)) &= H(s) + \beta\\ H(s+ \alpha) &= H(s) +\beta \\ \end{align}

and therefore $\alpha = -\beta$ for the lift and $\alpha = - \beta~\text{mod 1}$ for the rotations.

Combining results returns $\alpha = \pm \beta~\text{mod 1}$

EDIT 2:

Then, because we have the property of preservation of direction for $H$ that $\alpha=\beta$ for the lift and thus $\alpha=\beta \text{mod 1}$ for the rotations.

This is not a correct statement. I redo the prove.

$\implies$

Let $h$ be an homeomorphism s.t the unit rotations $R_\alpha$ and $R_\beta$ are conjugate. There exist lifts for the rotations denoted $r_\alpha$ and $r_\beta$, and thus for the homeomorphism h, denoted $H$ that conjugates the lift rotations. I assume that the lift is preserved under direction and increasing, therefore $H(s+1) = H(s) + 1$.

From the definition of $r_\alpha$ and the conjugacy we can express the conjugacy in terms of $r^n_\alpha$, \begin{equation} H \circ r^n_\alpha = r^n_\beta \circ H. \end{equation} Then \begin{align} H \circ r^n_\alpha &= r^n_\beta \circ H \\ H \circ r^n_\alpha &= H(s) +n \cdot \beta\\ \frac{H(r^n_\alpha(s)) - H(s)}{n} &= \beta \end{align}

Because the conjugacy holds for each $n\in \mathbb{N}$, we can take the limit of $n\to \infty$,

\begin{align} \beta &= \lim_{n\to \infty}\frac{H(r^n_\alpha(s)) - H(s)}{n}\\ &= \lim_{n\to \infty}\frac{H(r^n_\alpha(s))}{n} - \lim_{n\to \infty}\frac{H(s)}{n}\\ &= \lim_{n\to \infty}\frac{H(s +n \cdot \alpha)}{n} - \lim_{n\to \infty}\frac{H(s)}{n} \end{align}

The second quotient is independent of $n$, $H(s)$ is bounded, and therefore the quotient converges towards zero.

We use the squeeze thereom to determine the first quotient.

Because $\alpha$ is irrational, $\lfloor n\cdot \alpha \rfloor$ and $\lceil n\cdot \alpha \rceil$ are integers and $\lfloor n\cdot \alpha \rfloor \leq n\cdot \alpha \leq \lceil n\cdot \alpha \rceil$.

Therefore, using the preservation of direction,

\begin{align} H(s + \lfloor n\cdot \alpha \rfloor) &\leq& H(s +n \cdot \alpha) &\leq& H(s + \lceil n\cdot \alpha \rceil)\\ \lim_{n \to \infty}\frac{H(s + \lfloor n\cdot \alpha \rfloor)}{n} &\leq& \lim_{n \to \infty}\frac{H(s +n \cdot \alpha)}{n} &\leq& \lim_{n \to \infty}\frac{H(s + \lceil n\cdot \alpha \rceil)}{n}\\ \alpha\lim_{n \to \infty}\frac{H(s + \lfloor n\cdot \alpha \rfloor)}{n\cdot \alpha} &\leq& \lim_{n \to \infty}\frac{H(s +n \cdot \alpha)}{n} &\leq& \alpha\lim_{n \to \infty}\frac{H(s + \lceil n\cdot \alpha \rceil)}{n\cdot \alpha}\\ \alpha\lim_{n \to \infty}\frac{H(s) + \lfloor n\cdot \alpha \rfloor}{n\cdot \alpha} &\leq& \lim_{n \to \infty}\frac{H(s +n \cdot \alpha)}{n} &\leq& \alpha\lim_{n \to \infty}\frac{H(s) + \lceil n\cdot \alpha \rceil}{n\cdot \alpha} \end{align}

Once again, $H(s)$ is bounded and independent of $n$while $\frac{\lfloor n\cdot \alpha \rfloor}{n\cdot \alpha}\to 1$, thus follows \begin{equation} \lim_{n \to \infty}\frac{H(s) + \lfloor n\cdot \alpha \rfloor}{n\cdot \alpha} =\lim_{n \to \infty}\frac{H(s)}{n \cdot \alpha} + \lim_{n \to \infty}\frac{\lfloor n\cdot \alpha \rfloor}{n\cdot \alpha} = 0 + 1 = 1 \end{equation}

Because $\frac{\lceil n\cdot \alpha \rceil}{n \cdot \alpha} \to 1$ we can conclude \begin{align} \alpha\lim_{n \to \infty}\frac{H(s) + \lfloor n\cdot \alpha \rfloor}{n\cdot \alpha} &\leq& \lim_{n \to \infty}\frac{H(s +n \cdot \alpha)}{n} &\leq& \alpha\lim_{n \to \infty}\frac{H(s) + \lceil n\cdot \alpha \rceil}{n\cdot \alpha}\\ \alpha \cdot (0+1)&\leq& \lim_{n \to \infty}\frac{H(s +n \cdot \alpha)}{n} \leq \alpha \cdot (0+1) \end{align}

Therfore we can conclude that

\begin{equation} \beta = \lim_{n\to \infty}\frac{H(s +n \cdot \alpha)}{n} - \lim_{n\to \infty}\frac{H(s)}{n} = \alpha - 0 = \alpha \end{equation}

Thus, $\beta = \alpha$, and returning to the original rotations, we conclude $\beta = \alpha~\text{(mod(1))}$.

The result for $\beta = -\alpha~\text{(mod(1))}$ follows when we assume that the lift is preserved under direction and decreasing, therefore $H(s+1) = H(s) - 1$

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  • $\begingroup$ You can take $h=-\text{id}$, but you still need to take care of the other direction. $\endgroup$ – John B Sep 23 '18 at 16:21
  • $\begingroup$ Ouch, how did I not notice that $h = -$Id was the other solution.. I am aware that I need to do $\Rightarrow$ aswell. I Will edit that in. $\endgroup$ – Bo5man Sep 23 '18 at 16:23
  • $\begingroup$ @JohnB I edited the reverse direction, would you mind taking a look at it? $\endgroup$ – Bo5man Sep 23 '18 at 16:46
  • $\begingroup$ You should have a look at the sentence "Then, because we have the property of preservation of direction for $H$ that $\alpha=\beta$ for the lift and thus $\alpha=\beta \text{mod 1}$ for the rotations." It makes no sense since we only know that $H(t+1)=H(t)+1$. $\endgroup$ – John B Sep 23 '18 at 19:12
  • $\begingroup$ I am not sure how to proceed then, do you have a hint? $\endgroup$ – Bo5man Sep 23 '18 at 22:05

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