In discrete dynamics, we say that two systems $f:X \to X$ and $g:Y \to Y$ are topologically conjugate if there exists a homeomorphism $p:X \to Y$ such that $p \circ f = g \circ p$.
My professor was discussing what the corresponding equivalence relation should be for flows. She said that a "naive" definition would be to say that the flows $f: X \times \mathbb{R} \to X$ and $g: X \times \mathbb{R} \to X$ are "equivalent" if there exists a homeomorphism $p:X \to Y$ such that for all $t$ we have $p \circ f_t = g_t \circ p$. This is called being "strongly topologically conjugate", but my professor said that this is too strong; if a flow is a parameterization of another one, then the two flows are not strongly topologically conjugate.
To see why this is true, I investigated an example. Consider the flows $f_t(x) = x + tv$ and $g_t(x) = f_{2t}(x) = x + 2tv$ where $x, v \in \mathbb{R}^2$. It seems to me that if we choose the homeomorphism $p(x) = 2x$, then the equation $(p \circ f_t) (x)= (g_t \circ p)(x)$ becomes
$$p(x+tv) = p(x) + 2tv \implies 2x + 2tv = 2x + 2tv,$$
so these systems are indeed strongly conjugate. Can someone please give me an example, or an intuitive explanation, of why reparameterizing does not necesarily give strongly conjugate systems?
Thank you!
EDIT: Throughout this I assumed that a reparameterization of $f_t(x)$ is a function of the form $f_{kt}(x)$ where $k$ is a real number. Is this the actual definition of a reparametrization?
Best Answer
$\mathbb{R}^2$ is very special because it has $\mathbb{R}^{\times}$ acting on it by dilation. Do the same calculation but for a vector field on $S^1$, which doesn't. Alternatively, pick a vector field which has a zero at some smallest finite time $t_0$; the value of $t_0$ doesn't change under strong conjugacy but it does change under reparameterization.