Amenability Reiter Nets and Folner Conditions — Examples

The metric formula in [F1] gives $d_2(x+y,x^{\mathrm{\prime }}+y^{\mathrm{\prime }})\le d_2(x,x^{\mathrm{\prime }})+d_2(y,y^{\mathrm{\prime }})$ and $d_2(-x,-x^{\mathrm{\prime }})=d_2(x,x^{\mathrm{\prime }})$, so addition and inversion are continuous. By [F1], ${\mathbb{R}}^n$ is a locally compact Hausdorff topological group.

By [A1], countable choice is available for the Lebesgue Radon and box-volume results [F2, F4]. The box formula gives ${\lambda }_n([-1,1{]}^n)=2^n>0$, so ${\lambda }_n$ is nonzero; [F2] makes it Radon and [F3] makes it left invariant. With step 1.1, these are the Haar conditions of [F11], so ${\lambda }_n$ is a left Haar measure on $G$.

If $Q=\mathrm{\varnothing }$, choose $R=0$. Otherwise [F6] gives $x_0\in {\mathbb{R}}^n$ and $r>0$ with $Q\subseteq B_2(x_0,r)$. Put $R:=r+d_2(x_0,0)>0$. For $x\in Q$ and each $j<n$, [F6] and the triangle inequality give $\mathrm{\mid }{x}_j\mathrm{\mid }\le d_2(x,0)\le d_2(x,x_0)+d_2(x_0,0)<R$, so $Q\subseteq [-R,R{]}^n$.

Fix $t>0$ and $x\in Q$, and put $A_x:=x+C_t$. By [F2, F3, F4] the measurable sets $A_x$ and $C_t$ have equal finite measure $(2t{)}^n$. Their finite additive decompositions over $A_x\cap C_t$ therefore give ${\lambda }_n(A_x\setminus C_t)={\lambda }_n(C_t\setminus A_x)$, hence ${\lambda }_n(A_x\mathrm{△}{C}_t)=2{\lambda }_n(A_x\setminus C_t)$. If $y\in A_x\setminus C_t$, write $y=x+z$ with $z\in C_t$; since $x\in [-R,R{]}^n$ and $z\in [-t,t{]}^n$, $y\in [-(t+R),t+R{]}^n\setminus C_t$. By [F4, F5], this outer box minus $C_t$ has measure $(2(t+R){)}^n-(2t{)}^n$, and monotonicity bounds ${\lambda }_n(A_x\setminus C_t)$ by that value. Dividing by $(2t{)}^n>0$ yields ${\lambda }_n(A_x\mathrm{△}{C}_t)\mathrm{/}{\lambda }_n(C_t)\le 2((1+R\mathrm{/}t{)}^n-1)$. The bound is uniform in $x\in Q$, and when $Q=\mathrm{\varnothing }$ the defect is zero by [F7], proving the displayed inequality for every $Q$ and $t>0$.

Every $C_t$ is a closed, hence Borel, box with $0<{\lambda }_n(C_t)=(2t{)}^n<\mathrm{\infty }$ by [F4], so [F8] indexes a left Følner net by $t\in (0,\mathrm{\infty })$. Fix compact $Q$ and $\epsilon >0$, and use [F6] to choose its bound $R$ as in step 2.2. If $R=0$, then every $x\in Q$ equals $0$ and the defect is zero. If $R>0$, put $S_n:={\sum }_{k=1}^n\iota \left(\genfrac{}{}{0px}{}{n}{k}\right)$ and $T:=1+R+2S_{n}R\mathrm{/}\epsilon$. For $t\ge T$, $u:=R\mathrm{/}t\in (0,1]$, so [F9] and step 3.1 give ${\mathrm{\Delta }}_Q(C_t)\le 2S_{n}R\mathrm{/}t\le \epsilon$. Thus the net is eventually Følner on every compact $Q$, and ${\mathbb{R}}^n$ satisfies the left Følner condition.

By [F10], the left Følner condition proved in step 4.1 implies amenability under the stated AC. Hence ${\mathbb{R}}^n$ has all the properties in the Statement.

By [A1, F1], choose left Haar $\nu$ and set $\mu :=\nu \mathrm{/}\nu (K)$; since [F1] gives $0<\nu (K)<\mathrm{\infty }$, positive scalar rescaling preserves left Haar properties and $\mu (K)=1$. Set $f={\mathbf{1}}_K$. It is Borel and nonnegative, and [F3] gives $\mathrm{\parallel }f{\mathrm{\parallel }}_1={\int }_K{\mathbf{1}}_{K}d\mu =\mu (K)=1$, so $f\in \mathcal{P}$. For every $x,y\in K$, $L_{x}f(y)=f(x^{-1}y)=1=f(y)$ because $x^{-1}y\in K$. Thus $L_{x}f=f$ as an $L^1$ class and $\mathrm{\parallel }{L}_{x}f-f{\mathrm{\parallel }}_1=0$ for every $x\in K$; in particular ${\mathrm{\Delta }}_Q(f)=0$ for every compact $Q$, including $Q=\mathrm{\varnothing }$.

By [F10], the singleton $I=\{\ast \}$ with its unique preorder is directed; set $f_{\ast }:=f$. For every compact $Q\subseteq K$ and $\epsilon >0$, step 1.1 gives ${\mathrm{\Delta }}_Q(f_{\ast })=0\le \epsilon$, so this constant net witnesses Reiter (P1) by [F7]. The Reiter equivalence [F8], under the stated AC, proves that $K$ is amenable.

Define $m([\phi ]):={\int }_K\phi d\mu$ on complex $L^{\mathrm{\infty }}(K)$. By [F5] every such class is integrable; [F6] makes the value independent of the representative and complex-linear. If $[\phi ]\ge 0$, its nonnegative representative has nonnegative integral, so $m$ is positive by [F3]; and $m(1_K)=\mu (K)=1$ by [F1] and [F3]. For $x\in K$, [F2] gives $m(L_x[\phi ])={\int }_K\phi (x^{-1}y)d\mu (y)={\int }_K\phi d\mu$ for every $[\phi ]\in L^{\mathrm{\infty }}(K)$. Thus $m$ is a left-invariant mean in the sense of [F9], explicitly realizing the compact-group amenability clause of [F11] under the library's complex $L^{\mathrm{\infty }}$ convention. The local calculation verifies the normalized-Haar mean on all $L^{\mathrm{\infty }}$ classes.

The multiplication is associative because both bracketings of $(a,b),(a^{\mathrm{\prime }},b^{\mathrm{\prime }}),(a^{\mathrm{\prime }\mathrm{\prime }},b^{\mathrm{\prime }\mathrm{\prime }})$ give $(a{a}^{\mathrm{\prime }}{a}^{\mathrm{\prime }\mathrm{\prime }},a{a}^{\mathrm{\prime }}{b}^{\mathrm{\prime }\mathrm{\prime }}+a{b}^{\mathrm{\prime }}+b)$; $(1,0)$ is the identity and $(a,b{)}^{-1}=(a^{-1},-b\mathrm{/}a)$. Each coordinate of multiplication is a sum or product of continuous coordinate maps, and each inverse coordinate is a quotient with denominator $a>0$, so the group operations are continuous by [F1]. The set $G$ is open in ${\mathbb{R}}^2$. At every point of this open set, the compact-closure neighbourhood base in the locally compact Hausdorff Euclidean space supplies a compact neighbourhood still contained in $G$; Hausdorffness is inherited. Thus $G$ is a locally compact Hausdorff topological group.

The coordinate formulas give the subgroup laws and commutativity of $N$ and $H$. The conjugation formula in [F3] shows $gN{g}^{-1}=N$ for every $g=(a,b)$, so $N$ is normal, since $t\mapsto at$ is onto $\mathbb{R}$ for $a>0$. Also $(1,b)(a,0)=(a,b)$ for every $(a,b)\in G$, so every group element is a product from $NH$.

For each $n\in N$, the maps $x\mapsto nx$ and ${\mathrm{id}}_X$ are continuous into the Hausdorff space $X$, so their agreement set is closed by [F4]. Their intersection is $X^N$, hence $X^N$ is closed; it is compact because $X$ is compact, and it is convex because each action map is affine.

By [A1] and [A2], the positive density $\omega =a^{-2}\mathrm{\mid }dadb\mathrm{\mid }$ on the open smooth manifold $G$ defines a compact-finite Radon Borel measure $\mu (E):={\int }_E{a}^{-2}d{\lambda }_2(a,b)$ for Borel $E\subseteq G$. This measure is nonzero: on $K=[1,2]\times [0,1]\subset G$ the density is at least $1\mathrm{/}4$ and ${\lambda }_2(K)=1$, so [F13] gives $\mu (K)\ge 1\mathrm{/}4$.

The action restricted to the abelian topological group $N$ is continuous and affine on the nonempty compact convex set $X$. By [F5], it has a fixed point, which belongs to $X^N$. Therefore $X^N$ is nonempty.

The set $X^N$ is invariant under $H$: if $x\in X^N$, $h\in H$, and $n\in N$, then $n(hx)=h((h^{-1}nh)x)=hx$, since $h^{-1}nh\in N$ by normality. The restricted $H$-action on $X^N$ is continuous and affine, because it is the restriction of the given continuous affine action.

Fix $g_0=(a_0,b_0)\in G$. Left translation $T_{g_0}(a,b)=(a_{0}a,a_{0}b+b_0)$ is a $C^1$ diffeomorphism of $G$ onto itself with Jacobian determinant $a_0^2$. For every Borel $E\subseteq G$, [F9] gives $\mu (g_{0}E)={\int }_E(a_{0}a{)}^{-2}{a}_0^{2}d{\lambda }_2(a,b)=\mu (E)$. Thus $\mu$ is a left Haar measure by [F10].

The abelian topological group $H$ acts continuously and affinely on the nonempty compact convex set $X^N$ by step 2.2. By [F5], there is $x\in X^N$ fixed by every element of $H$.

For $g_0=(a_0,b_0)$, right multiplication by $g_0^{-1}$ is the $C^1$ diffeomorphism $T_{g_0^{-1}}(a,b)=(a\mathrm{/}{a}_0,b-a{b}_0\mathrm{/}{a}_0)$, whose inverse is $(A,B)\mapsto (A{a}_0,B+A{b}_0)$ and has Jacobian determinant $a_0$. Applying [F9] to the nonnegative positive and negative parts of the real and imaginary parts of $f\in C_c(G)$ gives ${\int }_{G}f(x{g}_0^{-1})d\mu (x)={\int }_{G}f(A,B)(A{a}_0{)}^{-2}{a}_{0}d{\lambda }_2(A,B)=a_0^{-1}{\int }_{G}f(A,B)A^{-2}d{\lambda }_2(A,B)$. By [F11] and the left Haar conclusion of step 2.3, ${\mathrm{\Delta }}_G(g_0)=a_0^{-1}$.

This $x$ is fixed by both $N$ and $H$. Since $G=NH$ by step 1.2, it is fixed by every element of $G$. The action was arbitrary, so $G$ has the fixed point property; [F6] makes $G$ amenable in the sense of [F7].

Taking $g_0=(2,0)$ in step 3.2 gives ${\mathrm{\Delta }}_G(g_0)=1\mathrm{/}2\mathrm{\ne }1$; therefore $G$ is nonunimodular by [F12]. Step 4.1 proves it is amenable, completing both claims in the Statement.

Suppose a left-invariant mean $m$ exists. For each subset $E\subseteq F_2$ put $\nu (E):=m({\mathbf{1}}_E)$, which is defined by [F1]. Positivity gives $\nu (E)\ge 0$ and monotonicity under inclusion; complex linearity gives finite additivity on disjoint sets. Since $L_g{\mathbf{1}}_E={\mathbf{1}}_{gE}$, left invariance gives $\nu (gE)=\nu (E)$ for every $g\in F_2$, and $\nu (F_2)=m(1)=1$.

Let $A$ be the set of reduced words whose initial maximal block is $a^k$ for some nonzero integer $k$; the empty word is not in $A$. Every word outside $A$ is either empty or begins with a nonzero power of $b$. In the first case $a^{-1}\in A$; in the second case the reduced word $a^{-1}w$ begins with $a^{-1}$ and is in $A$. Thus $F_2=A\cup aA$. By [F2], $\nu (aA)=\nu (A)$; monotonicity and finite subadditivity from step 1.1 give $1=\nu (F_2)\le \nu (A)+\nu (aA)=2\nu (A)$, hence $\nu (A)\ge \frac{1}{2}$.

The sets $A$, $bA$, and $b^{2}A$ are pairwise disjoint: their reduced words have initial maximal blocks respectively a nonzero power of $a$, exactly one $b$ followed by a nonzero power of $a$, and exactly two $b$'s followed by a nonzero power of $a$; no cancellation occurs at these joins. Therefore monotonicity, finite additivity, and left invariance from step 1.1 yield $1=\nu (F_2)\ge \nu (A)+\nu (bA)+\nu (b^{2}A)=3\nu (A)\ge \frac{3}{2}$, a contradiction. Hence no invariant mean exists, and the amenability definition shows $F_2$ is not amenable. This proves the claim and its stated discrete counterpart.