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Let $X$ be the Cantor set, and let $g$ be a minimal homeomorphism of $X$. Let $h$ be a homeomorphism in the topological full group of $g$, that is, for every $x \in X$, there is a neighbourhood of $x$ such that $h$ restricts to a power of $g$ on that neighbourhood. Write $\langle h \rangle$ for the group generated by $h$.

Certainly, the action of $\langle h \rangle$ is not minimal in general, because it can preserve proper nonempty clopen subsets of $X$. But suppose we take the closure $Y$ of an $\langle h \rangle$-orbit. Does $\langle h \rangle$ act minimally on $Y$?

Equivalently, let $x \in X$, let $U$ be a neighbourhood of $x$ and let $T$ be the set of elements $h^n$ of $\langle h \rangle$ such that $h^nx \in U$. Is $T$ necessarily syndetic in $\langle h \rangle$, i.e. $\langle h \rangle = FT$ for some finite set $F$?

There is a lot known about topological full groups of minimal homeomorphisms on the Cantor set, but I couldn't find any reference to this property.

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    $\begingroup$ I'm not aware of such a result, but Keane established something very close to this in his 1975 paper on interval exchanges (Keane, Michael Interval exchange transformations. Math. Z. 141 (1975), 25-31): essentially an interval exchange $f$ yields a decomposition in multitintervals $I_0\sqcup \dots \sqcup I_n$ with each $I_j$ $f$-invariant, f has finite order on $I_0$ and is minimal on $I_j$ for all $j\ge 1$. It would be interesting to find a similar result in the context of topological-full groups. $\endgroup$ – YCor Mar 27 '18 at 7:49
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Having worked on this problem, I can now say the answer is yes, $h$ has minimal orbit closures. Actually YCor's comment is spot on: analogous to Keane's interval exchange transformations, the space decomposes into finitely many clopen $\langle h \rangle$-invariant pieces such that on each piece, the restriction of $h$ is either finite order or minimal. The key idea is something François Le Maître suggested to me, which is to think about infinite orbits of $h$ as moving either 'in the same direction' as the $g$-orbit or 'in the opposite direction'. The main arguments are in section 3 of the following article:

https://arxiv.org/abs/1812.00480

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    $\begingroup$ Great. Here's a follow-up question which I believe is true in the IET case but maybe also in this context. The issue with this decomposition is that when one passes to a power of $h$ one can have more minimality pieces. Question: is it true that some power $g=h^k$ of $h$ has such a decomposition, such that in addition, on every minimality piece, every power $g^n$, $n>0$, of $g$, is minimal? (or at the contrary could the number of minimality pieces of $h^n$ be unbounded?) $\endgroup$ – YCor Dec 4 '18 at 7:59
  • $\begingroup$ The number of minimality pieces of $h^n$ is unbounded in general, e.g. take $h$ to be an odometer, or indeed any equicontinuous minimal action of $\mathbb{Z}$. I don't know if this happens though if $h$ belongs to the topological full group of some action of $\mathbb{Z}$ such that every nontrivial subgroup of $\mathbb{Z}$ acts minimally. $\endgroup$ – Colin Reid Dec 4 '18 at 12:44
  • $\begingroup$ Sure, I should at least have chosen $\tau g$ with $g$ a subshift. $\endgroup$ – YCor Dec 4 '18 at 13:12

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