Let's take the pendulum equation $\ddot{x} = -\sin x$. Here $x \in \mathbb{T}^{1}$. Now rewrite it as a coupled first order system $$\dot{y} = -\sin x, \quad \dot{x}=y.$$
Intuitively we know that $y$ corresponds to velocity, the norm of which (i.e. speed) can be as large or small as we want, thus $y \in \mathbb{R}$. Hence the phase space of the pendulum is the cylinder $\mathbb{T}^{1} \times \mathbb{R}$.
However $x(t) = x(t+t_{0})$ for some period $t_{0}$ and by the definition $ y =\dot{x}$ we also expect $y(t)=y(t+t_{0})$, i.e. we can say $y \in \mathbb{T}^{1}$.
Is this a contradiction? Why do we define $y$ to be in $\mathbb{R}$ and not in $\mathbb{T}$?
Best Answer
$x\in \mathbb{T}^1$ denotes the structure of the phase space itself, not the fact that the motion as a function of time is periodic. Any arbitrary motion of the pendulum can be represented in the phase space, not just the ones periodic (in time). We have $x\in \mathbb{T}^1$ because you can rotate the pendulum around the hinge for a full cycle and you end up with the same state. You cannot say the same for $y$.