Amenable Groups and Folner Criteria — Examples

Let $M={\max }_{s\in S}\mathrm{\mid }s\mathrm{\mid }$. For every $s\in S$, the translate $s+F_n$ differs from $F_n$ only near the two ends of the interval, so $\mathrm{\mid }(s+F_n)\mathrm{△}{F}_n\mathrm{\mid }\le 2M$.

Since $\mathrm{\mid }{F}_n\mathrm{\mid }=2n+1$, the ratio $\mathrm{\mid }(s+F_n)\mathrm{△}{F}_n\mathrm{\mid }\mathrm{/}\mathrm{\mid }{F}_n\mathrm{\mid }$ is at most $2M\mathrm{/}(2n+1)$ and therefore tends to $0$ uniformly in $s\in S$. Hence $F_n$ is eventually $(S,\epsilon )$-Folner in the sense of [L1], illustrating the criterion [L2].

Let $M={\max }_{s\in S}\mathrm{\parallel }s{\mathrm{\parallel }}_{\mathrm{\infty }}$. For each $s\in S$, the translate $s+Q_n$ differs from $Q_n$ only inside the $M$-thick boundary layers of the cube, so $\mathrm{\mid }(s+Q_n)\mathrm{△}{Q}_n\mathrm{\mid }=O(n^{d-1})$.

Because $\mathrm{\mid }{Q}_n\mathrm{\mid }=(2n+1{)}^d$, the ratio $\mathrm{\mid }(s+Q_n)\mathrm{△}{Q}_n\mathrm{\mid }\mathrm{/}\mathrm{\mid }{Q}_n\mathrm{\mid }$ tends to $0$ as $n\to \mathrm{\infty }$ uniformly in $s\in S$. Hence the boxes are eventually $(S,\epsilon )$-Folner by [L1], illustrating [L2].

The finite case is exactly [L1].

Every finitely generated subgroup of $L$ is supported on finitely many coordinates and is therefore finite. Thus $L$ is locally finite, and [L2] makes it amenable.

The base group $B={⨁}_{n\in \mathbb{Z}}\mathbb{Z}\mathrm{/}2\mathbb{Z}$ is locally finite, because a finitely generated subgroup is supported on finitely many coordinates and is therefore finite. Thus [L2] makes $B$ amenable.

The quotient of $L$ by the normal base group $B$ is $\mathbb{Z}$, which is abelian and hence amenable by [L4]. Therefore [L3] applies to the semidirect product from [L1] and shows that the lamplighter group $L$ is amenable.

For every $n\ge 0$, one has $\mathrm{\mid }{B}_n\mathrm{\mid }=1+4{\sum }_{k=1}^n{3}^{k-1}=2\cdot 3^n-1$. Every element of $S{B}_n\setminus B_n$ has reduced-word length exactly $n+1$, and every reduced word of length $n+1$ is obtained by taking its length-$n$ prefix in $B_n$ and multiplying by its final letter in $S$. Hence $S{B}_n=B_{n+1}$ and $\mathrm{\mid }S{B}_n\setminus B_n\mathrm{\mid }=\mathrm{\mid }{B}_{n+1}\mathrm{\mid }-\mathrm{\mid }{B}_n\mathrm{\mid }=4\cdot 3^n\mathrm{.}$ Therefore $\frac{\mathrm{\mid }S{B}_n\setminus B_n\mathrm{\mid }}{\mathrm{\mid }{B}_n\mathrm{\mid }}=\frac{4\cdot 3^n}{2\cdot 3^n-1}⟶2.$

In particular the one-sided generator boundary ratio of the balls never approaches $0$. This explicit boundary expansion is the geometric obstruction behind the nonamenability recorded in [L1].

For $x\in \{a,a^{-1},b,b^{-1}\}$ let $W(x)$ be the set of nonempty reduced words beginning with $x$, and put $P=\{a^n:n\ge 0\}$ and $P_{+}=\{a^n:n\ge 1\}$. Define $A_1=W(a)\setminus P_{+}$, $A_2=W(a^{-1})\cup P$, $B_1=W(b)$, and $B_2=W(b^{-1})$.

The four sets in step 1.1 are pairwise disjoint and partition $F(a,b)$: the nonidentity reduced words have one of the four possible first letters, and $P_{+}$ has been moved from $W(a)$ into the piece containing the identity.

Reduction of the first letter gives $aW(a^{-1})=F(a,b)\setminus W(a)$ and $aP=P_{+}$, hence $F(a,b)=A_1\bigsqcup a{A}_2$. Similarly $bW(b^{-1})=F(a,b)\setminus W(b)$, hence $F(a,b)=B_1\bigsqcup b{B}_2$. Therefore the pieces in step 1.1 with translators $e,a,e,b$ satisfy [L1] and form a paradoxical decomposition.

Let $L=\left({⨁}_{\mathbb{Z}}\mathbb{Z}\mathrm{/}2\mathbb{Z}\right)⋊\mathbb{Z}$ with generators $t$ for the shift and $a$ for toggling the lamp at the origin. For every subset $E\subseteq \{0,\dots ,n-1\}$, the element obtained by walking from $0$ to $n$, toggling exactly the lamps in $E$ on the way, has word length at most $3n$; different subsets give different group elements.

Therefore the ball of radius $3n$ contains at least $2^n$ elements, so $L$ has exponential growth in the sense of [L1]. Together with [L2], this amenable group refutes the statement.