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Fix a (finite) generating set $S$ for $\Gamma$ (discrete) amenable. Given a Følner sequence (i.e. a sequence of finite sets $F_n$ whose boundary $\partial F_n$ in the Cayley graph of $S$ is such that $|\partial F_n|/|F_n| \to 0$) it's always possible to change the sequence or pass to a subsequence in order to assume some mild properties (e.g. the sets are connected, their complement has only infinite connected components, the sequence is "increasing", tempered, contains larger and larger balls around some point, ...)

$\mathbf{Question:}$ Is it always possible to find a Følner sequence so that there is a point $x_n \in F_n$ with the property that for any $y \in F_n$ there is a geodesic from $y$ to $x_n$ which lies in $F_n$?

It seems convexity (any pair of $x$,$y \in F_n$ has a geodesic which lies in $F_n$) would be too much (as some amenable groups are not almost-convex, since they do not have solvable word problem), though I cannot pretend to know any obstruction.

For groups of sub-exponential growth the natural Følner sequences are balls. In the elementary closure of this class of groups, one can take Følner sequences to be "poly-balls" (which, I believe, also have this property).

EDIT: (To complement the above) One could look at Reiter sequences (i.e. sequence of finitely supported functions $f_n$ with $\|f_n\|_{\ell^1} =1$ and $\|\lambda_s f_n - f_n \|_{\ell^1} \to_n 0$) instead of Følner sequence. Then ask the same question with occurences of $F_n$ replaced by $\mathrm{supp}\, f_n$. In this situation, a good candidate for $f_n$ is $\tfrac{1}{2}\mu^{*n} + \tfrac{1}{2}\mu^{*(n+1)}$ (where $\mu^{*n}$ is the $n^{\text{th}}$-step distribution of a random walk on $S$)). $f_n$ is not always Reiter (but this cover some other cases as the previously mentionned ones). But it's not clear (to me) whether the "slicing" which is used to get a Følner sequence out of a Reiter sequence preserves the "star-shapedness".

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  • $\begingroup$ By geodesic you mean a path in the Cayley graph that has minimum length among all paths connecting those two end points? $\endgroup$ – Joel Moreira Sep 27 '13 at 20:47
  • $\begingroup$ Indeed. There is no reason to expect uniqueness in this choice, hence the requirement that one geodesic lies in the Folner sequence. It might have been a better idea to ask, in a first time, whether all geodesics lie in the Folner set (as I also don't know a counterexample to this). $\endgroup$ – ARG Sep 29 '13 at 14:55

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