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Let $G$ be a finitely generated discrete group with a finite symmetric generating set $S=S^{-1}\subset G$. For every group element $g$, define $\|g\|_S$ to be the length with respect to $S$, i.e. the minimal length of any word in $S$ that represents $g$. For a natural number $n$, define $B_n=${$ g\in G : \|g\|_S\leq n $}.

$G$ is said to have polynomial growth, if there exist constants $C,r$ such that $$|B_n|\leq C\cdot n^r $$ for all $n$. Note that this is really a property of the group itself, i.e. it does not depend on the choice of the generating set.

One particular implication of this is that by going from $B_n$ to $B_{2n}$, the ratio of the cardinalities $|\frac{B_{2n}}{B_n}|$ can be controlled uniformly by a constant.


Now I would like to strengthen this a little bit. For that, I mean the following sort of uniform covering property: (just a pragmatic notion for now)

The pair $(G,S)$ has the uniform covering property, if there exists a constant $C\in\mathbb{N}$ such that for all $n$, there exist $g_1,\dots,g_C\in G$ such that $B_{2n}\subset \bigcup_{i=1}^C g_i\cdot B_n$.

This means that there exists a constant that controls the number of copies (i.e. translates) of $B_n$ that one needs to cover $B_{2n}$. Presumably, this should not depend on the choice of $S$ either, although I have not bothered to check. So I state my question this way:

If $G$ has polynomial growth, is there a generating set $S$ such that the pair $(G,S)$ has the uniform covering property? If not, could there be an alternative description of the class of groups that have this property?

I should say that I have only limited knowledge/skills in questions concerning advanced group theoretic problems. But the questions seems elementary enough to hope that someone with more expertise in these areas could answer it.

My motivation for asking this question is that this uniform covering property came up in my research as a sufficient condition for a theorem that states something very nice about free topological dynamical systems $(X,\alpha,G)$, where $G$ acts continuously via $\alpha$ on a compact metric space $X$. On the one hand, a positive answer to this question would mean that the theorem says something nice about a very large class of amenable groups, not just standard examples like $G=\mathbb{Z}^m$. On the other hand, if the answer is no, there could still be a description of what groups I actually talk about. My intuition says that assuming this property is something very artificial/technical and should be replaced by something more natural, e.g. polynomial growth.

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Note that the uniform covering property is often called the doubling property, and is linked to Poincaré inequalities in metric-measure spaces. –  Benoît Kloeckner Feb 8 '13 at 11:53
    
Thank you for pointing this out –  Gabor Szabo Feb 8 '13 at 12:31

1 Answer 1

up vote 8 down vote accepted

The answer is yes. Moreover, given symmetric generating set $S$, any sets $B_n$ has uniform covering property for all large $n$.

Indeed, let $C_n$ be the minimal number of $g\cdot B_{ n}$ needed to cover $B_{2\cdot n}$. Let $d$ be the word metric for $S$. Then the sequence $(G,d/n)$ is precomact in the Gromov--Hausdorff topology. I particular, any partial limit of $(B_{2\cdot n},d/n)$ is compact. It follows that $C_n$ is bounded.

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Thank you very much for this quick answer! So I can find all of the details of this argument in the proof of Gromov's theorem? –  Gabor Szabo Feb 7 '13 at 21:28
    
Yes the precompactness is proved in the Gromov's paper (it is the easiest but also the game-changing part in the paper). –  Anton Petrunin Feb 8 '13 at 0:15

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