# Circle rotations are conjugate by a homeomorphism iff $\alpha = \pm \beta$ mod 1

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 $$r^n_\alpha$$, $$$$H \circ r^n_\alpha = r^n_\beta \circ H.$$$$ 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 $$$$\lim_{n \to \infty}\frac{H(s) + \lfloor n\cdot \alpha \rfloor}{n\cdot \alpha} =\lim_{n \to \infty}\frac{H(s)}{n \cdot \alpha} + \lim_{n \to \infty}\frac{\lfloor n\cdot \alpha \rfloor}{n\cdot \alpha} = 0 + 1 = 1$$$$

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

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

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 $$H(s+1) = H(s) - 1$$

• You can take $h=-\text{id}$, but you still need to take care of the other direction. – John B Sep 23 '18 at 16:21
• 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. – Bo5man Sep 23 '18 at 16:23
• @JohnB I edited the reverse direction, would you mind taking a look at it? – Bo5man Sep 23 '18 at 16:46
• 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 $H(t+1)=H(t)+1$. – John B Sep 23 '18 at 19:12
• I am not sure how to proceed then, do you have a hint? – Bo5man Sep 23 '18 at 22:05