How do I get the lift of a map in the unit circle?

Question

I'm working with the "An Introduction to Chaotic Dynamical Systems" book, and I'm having issues understanding the lifts of maps in the unit circle. I think that most of it is due to my lack of base knowledge in maths, so let's see.

I have a diffeomorphism $\tau_\omega : \mathbb{S}^1 \to \mathbb{S}^1$, where $\mathbb{S}^1 = \{e^{i\theta}, 0 < \theta \leq 2\pi\}$.

For example: $\tau_\omega(\theta) = \theta + 2\pi\omega$. This basically translates any angle by $2\pi\omega$, where $\omega$ is a parameter of the map.

The book says that a lift of a map is a map $T: \mathbb{R} \to \mathbb{R}$ where $\pi \circ T = \tau \circ \pi$, where $\pi$ is a map $\pi: \mathbb{R} \to \mathbb{S}^1$:

$\pi(x) = e^{2\pi i x} = \cos{(2\pi x)} + i\sin{(2\pi x)}$

So, the issue I'm having is that the book says that $T_{\omega, k} = x + \omega + k$ is a map of $\tau_\omega$ for all $k \in \mathbb{Z}$. The problem is that I'm unable to demonstrate this, or to create a lift of my own. If I calculate $\pi \circ T$ I get $e^{2\pi i (x + \omega + k)}$, which I'm unable to compare with $\tau_\omega\circ\pi = e^{2\pi i x} + 2\pi\omega$. Even using Wolfram Alpha, I won't get any equality demonstration.

The book has another example, where it says that the lift of $f_\epsilon(\theta) = \theta + \epsilon\sin{(\theta)}$ is $F_{\epsilon,k} = x + \frac{\epsilon}{2\pi}\sin{(2\pi x)}$.

The only pattern I'm seeing is that $\theta$ gets converted to $x$, and the rest of the elements in the function get divided by $2\pi$, while if there is a trigonometric function, $\theta$ gets converted to $2\pi x$. But I don't understand why.

What am I missing?

Answer 1

You made an algebra error which perhaps led you astray: $e^{2 \pi i(x+\omega)} = e^{2 \pi i x + 2 \pi i \omega}$, not $e^{2 \pi i x} + 2 \pi \omega$.

So your goal should be, instead, to compare $e^{2 \pi i(x + \omega + k)}$ to $e^{2 \pi i(x + \omega)}$. And to do this, you have to do more algebra: \begin{align*} e^{2 \pi i(x + \omega + k)} &= e^{2 \pi i(x+\omega) + 2 \pi i k} = e^{2 \pi i(x + \omega)} \cdot e^{2 \pi i k} = e^{2 \pi i(x + \omega)} \cdot (e^{2 \pi i})^k = e^{2 \pi i(x + \omega)} \cdot 1^k \\ &= e^{2 \pi i(x + \omega)} \end{align*} The key step here is the identity $$e^{2 \pi i}=1 $$ You can think of this as the complex valued version of two trig identities which I'm sure you know: $$\cos(2\pi)=1, \quad \sin(2 \pi)=0 $$ The link between these is Euler's identity $e^{\theta i} = \cos(\theta) + i \sin(\theta)$ with $\theta=2\pi$.

Answer 2

I think there are a few typos in the maps you wrote down (or, since they came directly out of the book, there are some implicit assumptions that should be made explicit). As written, the example you gave for $\tau_{\omega}$ is actually a map $\text{S}^1\rightarrow\mathbb{R}$. To make it a circle map, write it as: $$ \tau_{\omega}: \text{S}^1\rightarrow \text{S}^1, \quad\theta \mapsto (\theta + 2 \pi \omega) \; \text{mod }2\pi\,. $$ Then, the projection $\pi:\mathbb{R} \rightarrow \text{S}^1$ should be written as: $$ \pi: \mathbb{R} \rightarrow \text{S}^1, \quad x \mapsto \arg\left[ {e^{i 2 \pi x}}\right]\,, $$ i.e., you want the argument (angle) of the complex number that lies on the unit circle, not the complex number itself.

With these adjustments you can indeed see that $$ \pi \circ T_{\omega, k} = \tau_{\omega} \circ \pi\,, $$ where $T_{\omega, k}: \mathbb{R} \rightarrow \mathbb{R}, \, x \mapsto x + \omega + k$, for any $k \in \mathbb{Z}$. Or, as a diagram:

$$\require{AMScd} \begin{CD} \mathbb{R} @>{T_{\omega, k}}>> \mathbb{R}\\ @V{\pi}VV @V{\pi}VV \\ \text{S}^1 @>{\tau_{\omega}}>> \text{S}^1 \end{CD}$$