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This is related to my previous question here:

http://mathoverflow.net/questions/76422/antichains-and-measure-preserving-actions-on-boolean-algebras

This time I will ask something more precise.

Let $G$ be a group acting by homeomorphisms on the standard Cantor set $X$ (or if you prefer, $G$ is acting on the countable atomless Boolean algebra, which can be realised as the clopen subsets of $X$). Say a non-empty clopen subset $\alpha$ of $X$ is minorising (for $G$) if the following holds:

For every non-empty clopen subset $\beta$ of $X$, there is some $g \in G$ such that $g\beta$ contains $\alpha$.

Now suppose that there is a minorising subset $\alpha$, and that $G\alpha$ (that is the set of $g\alpha$ for $g \in G$) covers $X$. Let $n$ be the smallest size of a subcover of $G\alpha$, that is the smallest number of $G$-translates of $\alpha$ needed to cover $X$. It is clear that $n$ is finite (by compactness) and at least $2$, and that $n$ does not depend on the choice of $\alpha$ (beyond ensuring that $\alpha$ is minorising), so it is an invariant of the action.

The condition $n=2$ is equivalent to the condition that every proper non-empty clopen subset is minorising. (This occurs for instance if $G$ consists of every homeomorphism of $X$.)

Questions: How large can Can $n$ be ? greater than $2$? What if $G$ is simple, and/or every orbit of $G$ on $X$ is dense?

If $n=2$, can $G$ still have infinitely many orbits on the clopen subsets of $X$?

NB: The existence of a minorising subset means that $G$ destroys most of the extra structure I can think of putting on $X$, such as non-trivial metrics and measures. (Can $G$ act by quasi-isometries?) Effectively $\alpha$ is `as small as possible' up to the action of $G$.

Edit: Here is an example of such an action, to give a flavour of things:

Start with a rooted binary tree $T$. Make a finite cut of this tree, so it breaks into a finite tree $R$ containing the root, and finitely many other pieces which are all isomorphic to $T$. Now replace $R$ with another rooted subtree $R'$ of $T$ with the same number of leaves, and reattach the other pieces rigidly, one to each leaf of $R'$. This describes a quasi-isometry of the boundary of $T$; let $G$ be the group of all such quasi-isometries. (I think this is one of the Higman-Thompson simple groups.) The action satisfies the given hypotheses and $n=2$ (I think).

show/hide this revision's text 4 added 3 characters in body

This is related to my previous question here:

http://mathoverflow.net/questions/76422/antichains-and-measure-preserving-actions-on-boolean-algebras

This time I will ask something more precise.

Let $G$ be a group acting by homeomorphisms on the standard Cantor set $X$ (or if you prefer, $G$ is acting on the countable atomless Boolean algebra, which can be realised as the clopen subsets of $X$). Say a non-empty clopen subset $\alpha$ of $X$ is minorising (for $G$) if the following holds:

For every non-empty clopen subset $\beta$ of $X$, there is some $g \in G$ such that $g\beta$ contains $\alpha$.

Now suppose that there is a minorising subset $\alpha$, and that $G\alpha$ (that is the set of $g\alpha$ for $g \in G$) covers $X$. Let $n$ be the smallest size of a subcover of $G\alpha$, that is the smallest number of $G$-translates of $\alpha$ needed to cover $X$. It is clear that $n$ is finite (by compactness) and at least $2$, and that $n$ does not depend on the choice of $\alpha$ (beyond ensuring that $\alpha$ is minorising), so it is an invariant of the action.

The condition $n=2$ is equivalent to the condition that every proper non-empty clopen subset is minorising. (This occurs for instance if $G$ consists of every homeomorphism of $X$.)

Questions: Can How large can $n$ begreater than $2$? ? What if $G$ is simple, and/or every orbit of $G$ on $X$ is dense?

If $n=2$, can $G$ still have infinitely many orbits on the clopen subsets of $X$?

NB: The existence of a minorising subset means that $G$ destroys most of the extra structure I can think of putting on $X$, such as non-trivial metrics and measures. Effectively $\alpha$ is `as small as possible' up to the action of $G$.

Edit: Here is an example of such an action, to give a flavour of things:

Start with a rooted binary tree $T$. Make a finite cut of this tree, so it breaks into a finite tree $R$ containing the root, and finitely many other pieces which are all isomorphic to $T$. Now replace $R$ with another rooted subtree $R'$ of $T$ with the same number of leaves, and reattach the other pieces rigidly, one to each leaf of $R'$. This describes a quasi-isometry of the boundary of $T$; let $G$ be the group of all such quasi-isometries. (I think this is one of the Higman-Thompson simple groups.) The action satisfies the given hypotheses and $n=2$.n=2$ (I think).

show/hide this revision's text 3 added 605 characters in body

This is related to my previous question here:

http://mathoverflow.net/questions/76422/antichains-and-measure-preserving-actions-on-boolean-algebras

This time I will ask something more precise.

Let $G$ be a group acting by homeomorphisms on the standard Cantor set $X$ (or if you prefer, $G$ is acting on the countable atomless Boolean algebra, which can be realised as the clopen subsets of $X$). Say a non-empty clopen subset $\alpha$ of $X$ is minorising (for $G$) if the following holds:

For every non-empty clopen subset $\beta$ of $X$, there is some $g \in G$ such that $g\beta$ contains $\alpha$.

Now suppose that there is a minorising subset $\alpha$, and that $G\alpha$ (that is the set of $g\alpha$ for $g \in G$) covers $X$. Let $n$ be the smallest size of a subcover of $G\alpha$, that is the smallest number of $G$-translates of $\alpha$ needed to cover $X$. It is clear that $n$ is finite (by compactness) and at least $2$, and that $n$ does not depend on the choice of $\alpha$ (beyond ensuring that $\alpha$ is minorising), so it is an invariant of the action.

The condition $n=2$ is equivalent to the condition that every proper non-empty clopen subset is minorising. (This occurs for instance if $G$ consists of every homeomorphism of $X$.)

Questions: Can $n$ be greater than $2$? What if $G$ is simple, and/or every orbit of $G$ on $X$ is dense?

If $n=2$, can $G$ still have infinitely many orbits on the clopen subsets of $X$?

NB: The existence of a minorising subset means that $G$ destroys most of the extra structure I can think of putting on $X$, such as non-trivial metrics and measures. (Maybe $G$ can act by quasi-isometries?) Effectively $\alpha$ is `as small as possible' up to the action of $G$.

Edit: Here is an example of such an action, to give a flavour of things:

Start with a rooted binary tree $T$. Make a finite cut of this tree, so it breaks into a finite tree $R$ containing the root, and finitely many other pieces which are all isomorphic to $T$. Now replace $R$ with another rooted subtree $R'$ of $T$ with the same number of leaves, and reattach the other pieces rigidly, one to each leaf of $R'$. This describes a quasi-isometry of the boundary of $T$; let $G$ be the group of all such quasi-isometries. (I think this is one of the Higman-Thompson simple groups.) The action satisfies the given hypotheses and $n=2$.

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