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Let $n$ be a non-negative integer. $\;\;$ Let $\: f : \mathbb{R}^n \to \mathbb{R}^n \:$ and $\: g : \mathbb{R}^n \to \mathbb{R}^n \:$ be Lipschitz.

Define the relation $\stackrel{f}{\sim}$ on $\mathbb{R}^n$ by

$u \stackrel{f}{\sim} v \;\;$ if and only if $\;\; u$ and $v$ are in the same orbit of the dynamical system determined by $f$.

Define the relation $\stackrel{g}{\sim}$ similarly. $\:$ $\stackrel{f}{\sim}$ and $\stackrel{g}{\sim}$ are obviously equivalence relations.

Define $\:\langle X,\mathcal{T}_X\rangle\:$ and $\:\langle Y,\mathcal{T}_Y\rangle\:$ to be the quotient topological spaces of $\mathbb{R}^n$ by $\stackrel{f}{\sim}$ and $\stackrel{g}{\sim}$ respectively.

Suppose $X$ and $Y$ are homeomorphic. $\;\;$ Does it follow that there exists a homeomorphism

$h : \mathbb{R}^n \to \mathbb{R}^n \:$ such that for all members $u$ and $v$ of $\mathbb{R}^n$, $\: u \stackrel{f}{\sim} v \:$ if and only if $\: h(u) \stackrel{g}{\sim} h(v) \:\:$?

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up vote 4 down vote accepted

I assume that the dynamical systems in question are semigroups generated by $f$ and by $g$. Then the answer is negative: Take $n=2$, $f$, $g$ rotations of different orders. Then the quotient spaces are homeomorphic to ${\mathbb R}^2$ but the groups generated by $f$ and $g$ are not orbit-equivalent.

Or, maybe you mean that $f$ and $g$ are vector fields and you are looking at the flows they generate? Then the answer is still negative and the example I know is similar to the discrete example above. First, consider the circle action $\rho$ on the unit 3-sphere in ${\mathbb C}^2$; $\rho$ is given by $\rho(z)(u,v)=(z^pu, z^q v)$, where $|z|=1$. Think of this action as a flow generated by a smooth vector field. Then the quotient space of $S^3$ by this action is $S^2$. However, the actions are (typically) not orbit-equivalent, say, for $p=1,q=1$ and $p=2, q=3$. Regarding $S^3$ as the 1-point compactification of ${\mathbb R}^3$, one can modify the above example slightly so that the point at infinity is fixed (and the action is by ${\mathbb R}$, not $S^1$, although all but one orbits are circles). Then one gets an example on ${\mathbb R}^3$.

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(I meant the second choice.) $\:$ – Ricky Demer Apr 13 '12 at 20:46
OK, this explains the words ODE in the title... – Misha Apr 13 '12 at 20:54

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