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C. Pugh and M. Shub showed in 1971 that, given an ergodic action of $G=\mathbb{R}^k$ on some separable finite measure space $(X,\mu)$, then all elements of $G$ , off a countable family of hyperplanes, are ergodic.

Is there an analogous statement in the topological setting, with ergodicity replaced by minimality (i.e. all orbits are dense) and $X$ assumed to be compact ?

edited Jun 23 2010 at 8:04

asked Jun 22 2010 at 8:39

coudy
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	Coudy, I guess you should add some condition on $X$, for example that it is compact. I will delit my answer then... – Dmitri Jun 23 2010 at 7:22
	@Dmitri. Yes, sorry, I forgot the compactness assumption. – coudy Jun 23 2010 at 8:03
	Just out of curiousity: Is the Pugh/Shub argument hard? And can one explain it and explain why it doesn't extend? – Helge Jun 23 2010 at 12:56
	@Helge. An invariant measure gives a unitary representation of G on $L^2$. Here is a proof of the Pugh-Shub result for $k=1$. Let f be a $g_{t_0}$ invariant function for some $t_0$. Then $F(x) = \int_{0}^{t_0}\ f(g_s(x))\ e^{-2\pi i s/ t_0}\ ds$ is an eigenvector for the flow $g_t$, associated to the eigenvalue $e^{2\pi i/t_0}$. Eigenvectors associated to different eigenvalues are orthogonal. The conclusion follows, assuming $L^2$ is separable. – coudy Jun 23 2010 at 18:17

coudy · Answer 1 · 2010-07-07 19:38:45Z UTC

A colleague pointed out the following counterexample. Let $h_t$ be the horocyclic flow on a negatively curved compact surface S. This R action is known to be minimal. Now Consider the $R^2$ action on $S\times S$ given by $(s,t)\rightarrow (h_s,h_t)$. This action is again minimal.

The action of the diagonal $\{(s,s), s\in R\}$ is not minimal since the orbit of any point (x,x) stays in the diagonal.

Let $\theta\in R$. The action of the line $\{(s,\theta s), s\in R\}$ is not minimal because it is conjugated to the diagonal action. This comes from the fact that the two actions $h_{\theta s}$ and $h_s$ are conjugated by the geodesic flow.

As a result, there are no elements in $R^2$ acting minimally, although $R^2$ itself acts minimally.

Minimal elements of minimal R^k actions

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