Let $\mathbb{G}$ be a discrete amenable semigroup, and $\left\{ F_{n}\right\} $ a Folner sequence.

For $S\subset \mathbb{G}$ define the upper density as $D^{\ast }(S)=\limsup_{n\rightarrow \infty }\frac{\left\vert S\cap F_{n}\right\vert }{% \left\vert F_{n}\right\vert }.$

Suppose the exists $m\geq 0$ such that for every $g\in \mathbb{G}$, we have that $gF_{m}\cap S\neq \varnothing .$

Is $D^{\ast }(S)>0?$

  • $\begingroup$ This isn't usually what "syndetic" means, is it? A set $S$ in a semigroup $G$ is syndetic if $G = g_1^{-1}S \cup \cdots \cup g_n^{-1} S$ for some elements $g_i \in G$. Such a set should have positive density with respect to any left Folner net. It is not enough to have the nonempty intersection condition you described, but what does that condition have to do with syndeticity? $\endgroup$ Feb 3 '20 at 23:32

This is true if $F_n$ is a right Følner sequence, i.e., $\frac{|F_n \Delta F_ng |}{|F_n|} \to 0$ for all $g \in G$. Indeed, if $F \subset G$ is finite, then the condition $g F \cap S \not= \emptyset$, for every $g \in G$, is equivalent to the condition $S F^{-1} = G$. Hence $D^*(S) = \frac{1}{|F|} \sum_{f \in F} D^* (Sf^{-1}) \geq \frac{1}{|F|} D^*(S F^{-1}) = \frac{1}{|F|}$.

This can fail however if we consider only left Følner sequences. A counter-example is given by letting $G$ be the infinite dihedral group $\mathbb Z \rtimes^\alpha (\mathbb Z / 2 \mathbb Z)$, where $\alpha$ implements the flip automorphism on $\mathbb Z$. If we set $S = \mathbb N \cup (-\mathbb N)\alpha \subset G$, and $F = \{ 0, \alpha \}$, then we have $g F \cap S \not= \emptyset$ for all $g \in G$. However, considering $F_n = [-n^2, n] \cup [- n, n^2]\alpha$ we have that $F_n$ is a left Følner sequence such that $D^*(S) = \limsup_{n \to \infty} \frac{2n + 2}{2n^2 + 2n + 2} = 0$.

  • $\begingroup$ I was thinking of left Følner sequences. Thanks! $\endgroup$
    – Rob
    Dec 21 '13 at 3:29
  • $\begingroup$ I'm not familiar with semigroup amenability but, if I understand correctly, a discrete countable amenable semigroup possesses, by definition, a (left and right)-Følner sequence. $\endgroup$
    – YCor
    Dec 24 '13 at 12:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.