Bounded geometry version of property A

Abstract.

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For uniformly dicrete metric spaces without bounded geometry we suggest a modified version of property A based on metrics of bounded geometry greater than the given metric. We show that this version still implies coarse embeddability in Hilbert spaces, and that some examples of non-property A spaces of unbounded geometry satisfy this version. We also relate this version of property A to our version of uniform Roe algebras for spaces without bounded geometry and introduce an appropriate equivalence relation.

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Introduction

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Bounded geometry for discrete metric spaces is a useful property in coarse geometry. For Rips complexes of metric spaces it means their finite dimension. But many natural examples of metric spaces do not have bounded geometry, e.g. Cayley graphs of infinitely generated groups or some box spaces. Making in such spaces the distances between points much greater, we may obtain metrics of bounded geometry, and we suggest to use these metrics of bounded geometry to get information about the original metric. For Rips complexes this approach reduces to study of finitedimensional approximations instead of the complex itself. In [3] we have used this approach to define a bounded geometry version for the uniform Roe algebras. Here we use it to expose what we think should be a more suitable version of property A of Guoliang Yu for spaces without bounded geometry, which we denote by $\mathcal{B}\mathcal{G}$-A and call the bounded geometry version of property A.

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P. Nowak in [4] has shown that the unbounded geometry space $\sqcup_{i\in\mathbb{N}}(\mathbb{Z}/2\mathbb{Z})^{i}$ does not have property A, although should be viewed as ‘amenable’ in any reasonable sense. In particular, it admits a coarse embedding in a Hilbert space. Property $\mathcal{B}\mathcal{G}$-A coincides with property A on bounded geometry spaces and includes more spaces without bounded geometry, in particular, the Nowak’s example. Also property $\mathcal{B}\mathcal{G}$-A implies coarse embeddability in a Hilbert space by the standard argument with negative type kernels, thus justifying the approach based on bounded geometry metrics.

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Unlike spaces with bounded geometry, general metric spaces may be coarsely equivalent, but still very different. For example, an infinite space of finite diameter is coarsely equivalent to a single point, but the uniform Roe algebra for such space is even not an exact $C^{*}$-algebra. Considering on a metric space $(X,d)$ all metrics of bounded geometry greater than $d$ gives a more subtle structure on $(X,d)$ and allows to distinguish such spaces using $\mathcal{B}\mathcal{G}$-coarse equivalence relation. In particular, we show that if $(X,d_{X})$ and $(Y,d_{Y})$ are $\mathcal{B}\mathcal{G}$-coarsely equivalent and one of them has property $\mathcal{B}\mathcal{G}$-A then the other one has this property too.

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We also establish relations between this structure and our version of the uniform Roe algebras for uniformly discrete metric spaces.

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1. Notation and property A

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Let $X=(X,d)$ be a discrete metric space. We write $B^{d}_{S}(x)=\{y\in X:d(x,y)\leq S\}$ for the ball of radius $R$ with respect to the metric $d$ centered at $x$ (as we will consider several metrics on $X$, we specify the metric as a superscript), and, for $A\subset X$, we denote by $|A|$ the number of points in $A$. Recall that $(X,d)$ has bounded geometry if for any $R>0$ there exists $N\in\mathbb{N}$ such that $|B^{d}_{R}(x)|\leq N$ for any $x\in X$. A metric space is called uniformly discrete if there is $\alpha>0$ such that $d(x,y)\geq\alpha$ for any $x\neq y\in X$.

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Let $l_{2}(X)$ denote the Hilbert space of square-summable complex-valued functions on $X$. A map $\xi:X\to l_{2}(X)$, $x\mapsto\xi_{x}$, is said to have $(R,\varepsilon)$ variation if $d(x,y)\leq R$ implies $\|\xi_{x}-\xi_{y}\|<\varepsilon$.

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It is known that for metric spaces of bounded geometry the following conditions are equivalent (cf. Theorem 1.2.4 of [8]):

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• $(X,d)$ has property A;

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• for all $R>0$, $\varepsilon>0$, there exists $S=S(d,R,\varepsilon)>0$ and a map $\xi:X\to l_{2}(X)$ such that

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  1. $\xi$ has $(R,\varepsilon)$ variation, and $\|\xi_{x}\|=1$ for any $x\in X$;

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  2. $\operatorname{supp}\xi_{x}\subset B_{S}(x)$ for any $x\in X$.

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For general discrete metric spaces, not necessarily of bounded geometry, property A is defined as follows: $(X,d)$ has property A if for any $R>0$ and any $\varepsilon>0$ there exists $S>0$ and a family of finite subsets $\{A_{x}\}_{x\in X}$ in $X\times\mathbb{N}$ such that

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1. $d(x,y)<R$ implies $\frac{|A_{x}\Delta A_{y}|}{|A_{x}\cap A_{y}|}<\varepsilon$;

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2. $A_{x}\subset B_{S}(x)\times\mathbb{N}$ for any $x\in X$.

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Note that it was shown in [9] that, for bounded geometry spaces, one can get rid of the seemingly redundant factor $\mathbb{N}$ and to use subsets $A_{x}\subset X$, $x\in X$, in the above definition.

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2. Metrics of bounded geometry

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Given two metrics, $d_{1}$ and $d_{2}$, on a set $X$, we say that $d_{2}$ dominates $d_{1}$ and write $d_{1}\preceq d_{2}$ if $d_{1}(x,y)\geq d_{2}(x,y)$ for any $x,y\in X$. We say that $d_{2}$ coarsely dominates $d_{1}$ and write $d_{1}\preceq_{c}d_{2}$ if there exists a homeomorphism $\psi$ of $(0,\infty]$ such that $\psi(d_{1}(x,y))\geq d_{2}(x,y)$ for any $x,y\in X$. Both $\preceq$ and $\preceq_{c}$ give a partial order on the set of all metrics on $X$.

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Given a metric space $X=(X,d)$ with a uniformly discrete metric $d$, consider the set $\mathcal{B}\mathcal{G}(X,d)$ of all metrics $\rho$ on $X$ such that

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• $\rho\preceq d$;

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• $\rho$ is of bounded geometry.

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The following result was shown in [3]. As the set $\mathcal{B}\mathcal{G}(X,d)$ was defined in [3] a bit differently, we provide a proof here.

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Lemma 2.1.

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The set $\mathcal{B}\mathcal{G}(X,d)$ is directed, i.e. for any $\rho_{1},\rho_{2}\in\mathcal{B}\mathcal{G}(X,d)$ there exists $\rho\in\mathcal{B}\mathcal{G}(X,d)$ such that $\rho_{i}\preceq\rho$, $i=1,2$.

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Proof.

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Let $\rho_{1},\rho_{2}\in\mathcal{B}\mathcal{G}(X,d)$. As $(X,d)$ is uniformly discrete, we may assume without loss of generality that $d(x,y)\geq 1$ for any $x\neq y\in X$. Turn $X$ into a complete graph $G$ by connecting each two points, $x,y\in X$, by an edge, and assign length $l(x,y)=\min(\rho_{1}(x,y),\rho_{2}(x,y))$ to the edge that connects $x$ with $y$. As $\rho_{1},\rho_{2}\preceq d$, $l(x,y)\geq 1$ for any $x\neq y\in X$. If $\gamma$ is a path in $G$ that connects two vertices, define its length $l(\gamma)$ as the sum of lengths of the edges contained in $\gamma$. Set $\rho(x,y)=\inf_{\gamma}l(\gamma)$, where the infimum is taken over all paths that connect $x$ with $y$. Clearly, $\rho$ is a metric. Given $x,y\in X$ and $\varepsilon>0$, there is a path $\gamma$ connecting $x$ with $y$ such that $\rho(x,y)>l(\gamma)-\varepsilon$. Let $x_{0}=x,x_{1},x_{2},\ldots,x_{n}=y$ be the vertices on $\gamma$, i.e. $\gamma=(x=x_{0},x_{1},\ldots,x_{n}=y)$. Then $l(\gamma)=\sum_{i=1}^{n}l(x_{i-1},x_{i})$. Therefore,

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$$d(x,y)\leq\sum_{i=1}^{n}d(x_{i-1},x_{i})\leq\sum_{i=1}^{n}l(x_{i-1},x_{i})<\rho(x,y)-\varepsilon.$$

As $\varepsilon$ is arbitrary, we conclude that $d(x,y)\leq\rho(x,y)$ for any $x,y$.

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Note that if we assign length $l_{1}(x,y)=\rho_{1}(x,y)$ to each edge of $G$ and define a metric in the same way as the infimum of lengths of paths connecting two vertices then this metric would coincide with $\rho_{1}$. Therefore, $\rho(x,y)\leq\rho_{1}(x,y)$ for any $x,y\in X$. Similarly, $\rho(x,y)\leq\rho_{2}(x,y)$ for any $x,y\in X$.

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It remains to show that $\rho$ has bounded geometry. Fix $x\in X$, $R\in\mathbb{N}$, and let $\rho(x,y)<R$. Let $\gamma$ be a path connecting $x$ with $y$ with length $l(\gamma)<R$. As $\rho(u,v)\geq 1$ for any $u,v\in X$, the path $\gamma$ contains not more than $R$ edges, and the length of each edge does not exceed $R$. Let $\gamma=(x=x_{0},x_{1},\ldots,x_{n}=y)$, $n\leq R$. As $\rho(x,x_{1})<R$, either $\rho_{1}(x,x_{1})$ or $\rho_{2}(x,x_{1})$ is less than $R$, hence $x_{1}\in B^{\rho_{1}}_{R}(x)\cup B^{\rho_{2}}_{R}(x)$, so there are not more than $N(r)=\sup_{x\in X}(|B^{\rho_{1}}_{R}(x)|+|B^{\rho_{2}}_{R}(x)|)$ points $x_{1}$ with $\rho(x,x_{1})<R$. Similarly, there are not more than $N(R)$ points $x_{2}$ with $\rho(x_{1},x_{2})<R$, etc. Then $|B^{\rho}_{R}(x)|\leq N(R)^{R}$ for any $x\in X$. ∎

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Lemma 2.1 allows to consider $d$ as the limit metric for $\mathcal{B}\mathcal{G}(X,d)$.

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In the following two lemmas we construct bounded geometry metrics on $X$ that agree with $d$ on certain subsets.

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Lemma 2.2.

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Let $X=(X,d)$ be a countable metric space. There exists a sequence $\{\sigma_{m}\}_{m\in\mathbb{N}}$ of metrics on $X$ such that

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• $\sigma_{m}\in\mathcal{B}\mathcal{G}(X,d)$ for any $m\in\mathbb{N}$;

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• $\sigma_{1}\preceq\ldots\preceq\sigma_{m}\preceq\sigma_{m+1}\preceq\ldots\preceq d$;

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• for any $x,y\in X$ there exists $N\in\mathbb{N}$ such that $\sigma_{m}(x,y)=d(x,y)$ for any $m\geq N$.

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Proof.

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Let $X=\{x_{1},x_{2},\ldots\}$, $X_{m}=\{x_{1},\ldots,x_{m}\}$. For $x,y\in X_{n}$ set $\sigma_{m}(x,y)=d(x,y)$. For $x_{m+1}$ and $x\in X_{m}$ set

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$$\sigma_{m}(x_{m+1},x)=\max_{i=1,\ldots,m}\{d(x_{m+1},x_{i}),\operatorname{diam}(X_{m})\}+1.$$

Then, inductively, for $x_{m+k+1}$ and $y\in X_{m+k}$ set

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$$\sigma_{m}(x_{m+k+1},y)=\max_{i=1,\ldots,m+k}\{d(x_{m+1},x_{i}),\operatorname{diam}(X_{m+k})\}+k.$$

The triangle inequality trivially holds, so each $\sigma_{m}$ is a metric. Other claimed properties of $\sigma_{m}$ can be easily checked. ∎

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Lemma 2.3.

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Let $X$ be uniformly discrete, and let $h:X\to X$ be a map. Suppose that there exists $N\in\mathbb{N}$ such that $|h^{-1}(x)|\leq N$. Then there exists $\rho\in\mathcal{B}\mathcal{G}(X,d)$ such that $\rho(x,h(x))=d(x,h(x))$ for any $x\in X$.

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Proof.

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Construct a weighted graph with $X$ as the set of vertices, and two vertices $x,y\in X$ are connected by an edge of length $d(x,y)$ if $y=h(x)$. Let $A_{i}$, $i\in\mathbb{N}$, be the connected components of this graph. Fix a vertex $x_{i}$ in each $A_{i}$, and add the edges connecting $x_{i}$ with $x_{i+1}$ of length $d(x_{i},x_{i+1})$. Then the resulting graph is connected. Let $\rho$ be the wighted graph metric, i.e. the length of a path is the sum of lengths of its edges, and $\rho(x,y)$ is defined as the infimum of path lengths over all paths that connect $x$ with $y$ in the graph. If $\gamma=(x=y_{0},y_{1},y_{2},\ldots,y_{n}=y)$ is a path such that $\rho(x,y)\geq\sum_{j=1}^{n}d(y_{i-1},y_{i})-\varepsilon$ then, by the triangle inequality, $d(x,y)\leq\sum_{j=1}^{n}d(y_{i-1},y_{i})\leq\rho(x,y)+\varepsilon$, and as $\varepsilon>0$ is arbitrary, we conclude that $\rho\preceq d$. In particular, this implies that $\rho(x,h(x))=d(x,h(x))$, as these two points are connected by a path consisting of a single edge. Also, as $X$ is uniformly discrete, lengths of all edges are separated from zero. and each vertex has no more than $N+2$ adjacent vertices, therefore, $\rho$ has bounded geometry. ∎

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3. Property $\mathcal{B}\mathcal{G}$-A for discrete metric spaces

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Definition 3.1.

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We say that $(X,d)$ has the bounded geometry version of property A (property $\mathcal{B}\mathcal{G}$-A) if for any $\sigma\in\mathcal{B}\mathcal{G}(X,d)$ there exists $\rho\in\mathcal{B}\mathcal{G}(X,d)$ such that

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• $\sigma\preceq_{c}\rho$;

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• $(X,\rho)$ satisfies property A.

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Recall that a function $k:X\times X\to\mathbb{R}$ is a negative type kernel if $k(y,x)=k(x,y)$ for any $x,y\in X$ and if $\sum_{i,j=1}^{n}\alpha_{i}k(x_{i},x_{j})\alpha_{j}\leq 0$ for any $n\in\mathbb{N}$, $x_{1},\ldots,x_{n}\in X$ and $\alpha_{1},\ldots,\alpha_{n}\in\mathbb{R}$ such that $\sum_{i=1}^{n}\alpha_{i}=0$.

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Theorem 3.2.

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Let $X=(X,d)$ be a countable metric space. If $(X,d)$ has property $\mathcal{B}\mathcal{G}$- A then there exists a negative type kernel $k$ on $X$ and homeomorphisms $\phi_{1},\phi_{2}$ of $(0,\infty]$ such that

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$$\phi_{1}(d(x,y)\leq k(x,y)\leq\phi_{2}(d(x,y))$$

for any $x,y\in X$.

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Proof.

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Let $\varepsilon=\frac{1}{2^{n}}$, $R=n$. Let $F_{m}=\{x_{1},\ldots,x_{m}\}\subset X$, and let $\{\sigma_{m}\}_{m\in\mathbb{N}}$ be a sequence of metrics on $X$ such that $\sigma_{m}|_{F_{m}}=d$, and let $\rho_{m}$ is a metric that satisfies the properties from Definition 3.1, i.e. $\sigma_{m}\preceq\rho_{m}$, and there exists a map $\xi^{m,n}:X\to l_{2}(X)$ such that $\xi^{m,n}$ has $(n,\frac{1}{2^{n}})$ variation with respect to the metric $\rho_{m}$, $\|\xi^{m,n}_{x}\|=1$ for any $x\in X$, and $\operatorname{supp}\xi^{m,n}_{x}\subset B^{\rho_{m}}_{S_{n}}(x)$ for any $x\in X$. Set

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$$k^{m}_{n}(x,y)=\langle\xi^{m,n}_{x},\xi^{m,n}_{y}\rangle,$$

$$k^{m}(x,y)=\sum_{n\in\mathbb{N}}(1-k^{m}_{n}(x,y)).$$ (1)

For each $m$ the series (1) is convergent by Theorem 3.2.8 in [8] to a negative type kernel, and $|k^{m}(x,y)|\leq 2\rho_{m}(x,y)+1$ for any $x,y\in X$. By assumption, there exists a homeomorphism $\psi$ of $(0,\infty]$ such that for any fixed points $x,y\in X$, we have

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$$\rho_{m}(x,y)\leq\psi(\sigma_{m}(x,y))=\psi(d(x,y))$$

for sufficiently great $m$, therefore, the sequence $k^{m}(x,y)$, $m\in\mathbb{N}$, is bounded, hence contains a convergent subsequence. Let $(x_{i},y_{i})_{i\in\mathbb{N}}$ be a sequence of all pairs of points of $X$. Choose a subsequence $m_{1}(j)_{j\in\mathbb{N}}$ such that $k^{m_{1}(j)}(x_{1},y_{1})$ is convergent. Then, within this subsequence, choose a subsequence $m_{2}(j)_{j\in\mathbb{N}}$ such that $k^{m_{1}(j)}(x_{2},y_{2})$ is convergent, etc., and inductively we obtain, for each $i\in\mathbb{N}$, a subsequence $m_{i}(j)_{j\in\mathbb{N}}$ such that $k^{m_{i}(j)}(x_{i},y_{i})$, $j\in\mathbb{N}$, is convergent for each $i\in\mathbb{N}$, and each next subsequence is a subsequence of the preceeeding subsequence. Passing to the diagonal sequence $(k^{m_{j}(j)})_{j\in\mathbb{N}}$, we obtain a convergent sequence for each $x,y\in X$. Set $k(x,y)=\lim_{j\to\infty}k^{m_{j}(j)}(x,y)$. Being the pointwise limit of negative type kernels, $k$ is a negative type kernel too. Clearly, $|k(x,y)|\leq 2\psi(d(x,y))+1$ for any $x,y\in X$, and we may set $\phi_{2}(t)=2\psi(t)+1$.

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Let $S^{m}_{n}=S(\rho_{m},n,\frac{1}{2^{n}})$ be a constant from Definition 3.1. Without loss of generality we may assume that $S^{m}_{n}>m+n$. For each $n$ there exists $m(n)$ such that $S^{m(n)}_{n}>n$, and $\lim_{n\to\infty}m(n)=\infty$. Passing to the inverse function, for each $m$ there exists $n(m)$ such that $S^{m}_{n(m)}>n(m)$, and $\lim_{m\to\infty}n(m)=\infty$. Let $\phi$ be a homeomorphism of $(0,\infty]$ such that $\phi(2S^{m(n)}_{n})>n$ (equivalently, $\phi(2S^{m}_{n(m)})>n(m)$), $m\in\mathbb{N}$. Let $2S^{m}_{n(m)}<\rho^{m}(x,y)\leq 2S^{m}_{n(m)+1}$. Then $\langle\xi^{m,n}_{x},\xi^{m,n}_{y}\rangle=0$ for $n=1,\ldots,n(m)$, hence $k^{m}_{n}(x,y)=0$ for $n=1,\ldots,n(m)$, thus the first $n(m)$ terms in (1) equal 1, therefore,

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$$k^{m}(x,y)\geq n(m)\geq\phi(\rho_{m}(x,y))-1\geq\phi(d(x,y))-1$$

for any $x,y\in X$ and for any $m\in\mathbb{N}$, hence $k(x,y)\geq\phi(d(x,y))-1$, and we can set $\phi_{1}(t)=\phi(t)-1$. ∎

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Corollary 3.3.

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Let $X$ be a discrete metric space with property $\mathcal{B}\mathcal{G}$-A. Then $X$ is coarsely embeddable in a Hilbert space.

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4. Relation to Roe $C^{*}$-algebras

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Let $T$ be an operator on $l_{2}(X)$ with matrix entries $T_{xy}=\langle\delta_{x},T\delta_{y}\rangle$, where $x,y\in X$, and $\delta_{x}(y)=\left\{\begin{array}[]{cl}1,&\mbox{\ if\ }y=x;\\ 0,&\mbox{\ if\ }y\neq x.\end{array}\right.$ Recall that $T$ has propagation $\leq L$ if $d(x,y)\geq L$ implies $T_{xy}=0$. The norm closure of the set of operators of finite propagation is called the uniform Roe algebra of $X$ and is usually denoted by $C^{*}_{u}(X)$. We shall include dependence on a metric in this notation and write $C^{*}_{u}(X,d)$.

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Lemma 4.1.

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Let $d_{1},d_{2}$ be two metrics on $X$. If $d_{1}\preceq_{c}d_{2}$ then $C^{*}_{u}(X,d_{2})\subset C^{*}_{u}(X,d_{1})$.

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Proof.

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Obvious. ∎

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The following remarkable characterization was proved in [6].

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Theorem 4.2 ([6]).

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Let $X$ be a metric space of bounded geometry. Then the following are equivalent:

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1. $X$ has property A;

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2. $C^{*}_{u}(X)$ is nuclear;

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3. $C^{*}_{u}(X)$ is exact.

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Using this result we can modify the definition of property $\mathcal{B}\mathcal{G}$-A.

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Lemma 4.3.

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Let $\rho,\sigma\in\mathcal{B}\mathcal{G}(X,d)$, $\sigma\preceq_{c}\rho$. If $(X,\rho)$ satisfies property A then $(X,\sigma)$ satisfies property A too.

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Proof.

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By Lemma 4.1, $C^{*}_{u}(X,\sigma)\subset C^{*}_{u}(X,\rho)$ is a $C^{*}$-subalgebra. Property A for $(X,\rho)$ iplies that $C^{*}_{u}(X,\rho)$ is exact. As $C^{*}$-subalgebras of exact $C^{*}$-algebras are exact, $C^{*}_{u}(X,\sigma)$ is exact, therefore, $(X,\sigma)$ has property A. ∎

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Remark 4.4.

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It is challenging to find a direct geometric proof of this geometric result.

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Corollary 4.5.

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Let $(X,d)$ be a uniformly discrete space. It has property $\mathcal{B}\mathcal{G}$-A iff $(X,\sigma)$ has property A for any $\sigma\in\mathcal{B}\mathcal{G}(X,d)$.

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For discrete spaces without bounded geometry (but uniformly discrete) Theorem 4.2 does not hold. In particular, if $X$ is an infinite space of finite diameter then $C^{*}_{u}(X)$ equals $\mathbb{B}(l_{2}(X))$, which is far from being nuclear, while property A holds by trivial argument. This lead us to suggest in [3] a modification of the uniform Roe algebra for spaces without bounded geometry. Namely, as the set $\mathcal{B}\mathcal{G}(X,d)$ is directed when $(X,d)$ is uniformly discrete, we can set $C^{*}_{\mathcal{B}\mathcal{G}}(X,d)=\operatorname{dir}\lim_{\rho\in\mathcal{B}\mathcal{G}(X,d)}C^{*}_{u}(X,\rho)$. Clearly, if $(X,d)$ is already of bounded geometry then $C^{*}_{\mathcal{B}\mathcal{G}}(X,d)=C^{*}_{u}(X,d)$. For this modification it is easy to prove an analog of Theorem 4.2.

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Theorem 4.6.

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Let $(X,d)$ be a uniformly discrete space. Then the following are equivalent:

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1. $(X,d)$ has property $\mathcal{B}\mathcal{G}$-A;

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2. $C^{*}_{\mathcal{B}\mathcal{G}}(X,d)$ is nuclear;

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3. $C^{*}_{\mathcal{B}\mathcal{G}}(X,d)$ is exact.

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Proof.

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If $(X,d)$ has property $\mathcal{B}\mathcal{G}$-A then $(X,\rho)$ has property A for any $\rho\in\mathcal{B}\mathcal{G}(X,d)$, hence $C^{*}_{u}(X,\rho)$ is nuclear. The direct limit of nuclear $C^{*}$-algebras is locally nuclear, hence nuclear, hence (1) implies (2). (2) obviously implies (3). To show that (3) implies (1), note that if $C^{*}_{\mathcal{B}\mathcal{G}}(X,d)$ is exact then any its $C^{*}$-subalgebra is exact, in particular, $C^{*}_{u}(X,\rho)$ for any $\rho\in\mathcal{B}\mathcal{G}(X,d)$, hence $(X,\rho)$ has property A. ∎

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5. Example 1

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In this section we show that some spaces without bounded geometry related to abelian groups, including thr Nowak’s example, have property $\mathcal{B}\mathcal{G}$-A.

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Lemma 5.1.

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Let $(X,d)$ be a uniformly discrete metric space, and let a group $G$ acts transitively on $X$. Let $\sigma\in\mathcal{B}\mathcal{G}(X,d)$. Then there exists a $G$-invariant metric $\rho\in\mathcal{B}\mathcal{G}(X,d)$ such that $\sigma\preceq\rho$.

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Proof.

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Set

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$$\rho(x,y)=\inf_{n\in\mathbb{N};g_{1},\ldots,g_{n}\in G}\sum_{i=1}^{n}\sigma(g_{i}x_{i-1},g_{i}x_{i}),$$

where $x_{0}=x$, $x_{n}=y$, and all $x_{i}$, $i=0,\ldots,n$ are distinct. Then $\rho$ satisfies the triangle inequality. As $d$ is uniformly discrete and $\sigma\preceq d$, $\sigma$ is also uniformly discrete. Without loss of generality we may assume that $d(x,y)\geq 1$ for any two distinct points $x,y\in X$. Clearly, $\rho$ is invariant, i.e. $\rho(gx,gy)=\rho(x,y)$ for any $x,y\in X$ and any $g\in G$. Setting $n=1$ and $g=e$, we see that $\rho(x,y)\leq\sigma(x,y)$, hence $\sigma\preceq\rho$.

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It remains to show that $\rho$ is of bounded geometry. By invariance and by transitivity, it suffices to show that the balls centered at some fixed point $z\in X$. Fix $R>0$, and let $|B^{\sigma}_{R}(x)|\leq C_{R}$ for any $x\in X$.

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Let $\rho(z,x)\leq R$. Then there exist $n\in\mathbb{N}$ and $g_{1},\ldots,g_{n}\in G$ such that $\sum_{i=1}^{n}\sigma(g_{i}x_{i-1},g_{i}x_{i})<R+1$, where $x_{0}=z$, $x_{n}=x$. As each term satisfies

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$$\sigma(g_{i}x_{i-1},g_{i}x_{i})\geq d(g_{i}x_{i-1},g_{i}x_{i})\geq 1,$$

we have $n<R+1$. It also follows that $\sigma(g_{i}x_{i-1},g_{i}x_{i})<R+1$ for each $i=1,\ldots,n$.

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Let $F_{1}=\{x_{1}\in X:\sigma(g_{1}z,g_{1}x_{1})<R+1\}$. Then $g_{1}F_{1}=B^{\sigma}_{R+1}(g_{1}z)$, hence $|F_{1}|\leq C_{R+1}$. Similarly, for each $x_{1}\in F_{1}$, there are not more than $C_{R+1}$ points $x_{2}$ such that $\sigma(g_{2}x_{1},g_{2}x_{2})<R+1$. Proceeding, we see that there are not more than $C_{R+1}^{R+1}$ points $x_{n}$ such that $\sigma(g_{i}x_{i-1},g_{i}x_{i})<R+1$ for each $i=1,\ldots,n$. Thus, $|B^{\rho}_{R}(z)|\leq C_{R+1}^{R+1}$. ∎

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Let $G$ be a finite group with a metric $d_{G}$, and let $X=G^{\infty}$. Then any $x\in X$ can be written as a sequence $x=(x_{i})_{i\in\mathbb{N}}$ with $x_{i}\in G$, and $x_{i}=0$ for sufficiently great $i$. Let $\mathcal{C}$ denote the set of all sequences $c=\{c_{i}\}_{i\in\mathbb{N}}$ such that

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• $c_{i}\geq 1$ for any $i\in\mathbb{N}$;

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• $c_{i}\leq c_{i+1}$ for any $i\in\mathbb{N}$;

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• $\lim_{i\to\infty}c_{i}=\infty$.

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Then a sequence $c\in\mathcal{C}$ determines a weighted length metric $\rho_{c}$ on $X$ by $\rho_{c}(x,y)=\sum_{i\in\mathbb{N}}c_{i}d_{G}(x_{i},y_{i})$. Under the above assumptions on $c$, the metrics $\rho_{c}$ are of bounded geometry. We also consider the metric $d$ on $G^{\infty}$ given by $d(x,y)=\sum_{i\in\mathbb{N}}d_{G}(x_{i},y_{i})$ (note that $(1,1,\ldots)\notin\mathcal{C}$). Clearly, $d\preceq\rho_{c}$ for any $c$.

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Similarly, let $Y=\sqcup_{i\in\mathbb{N}}G^{i}$ be the coarse union of $G^{i}$, $i\in\mathbb{N}$, with the (generalized) metric $d$ such that $d|_{G^{i}}=d_{G}$, and $d(x,y)=\infty$ if $x\in G^{i}$, $y\in G^{j}$, $i\neq j$ (or we may set the latter distances to be finite but sufficiently greater than diameters of $G^{i}$’s).

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Lemma 5.2.

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Let $\sigma\in\mathcal{B}\mathcal{G}(X,d)$ (resp., $\sigma\in\mathcal{B}\mathcal{G}(Y,d)$) be invariant under translations by $X$. Then there exists $c\in\mathcal{C}$ such that $\rho_{c}(x,y)\leq(\sigma(x,y)+1)^{2}$ for any $x,y\in X$ (resp., $\in Y$).

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Proof.

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The argument is the same for $Y$ and for $X$, so we give it only for $X$. Due to invariance, we may assume that $y=e_{X}=(e,e\ldots)$, where $e\in G$ is the neutral element. For shortness’ sake we write $d(x)$ (resp., $\sigma(x)$, $d_{G}(x)$, etc.) for $d(x,e_{X})$ (resp., $\sigma(x,e_{X})$, $d_{G}(x_{i},e)$, etc.). For $n\in\mathbb{N}$, consider the ball $B^{\sigma}_{n}(e_{X})$. As $\sigma$ is of bounded geometry, this ball contains finitely many points, and as each point in $X$ has finitely many coordinates different from $e$, $x\in B^{\sigma}_{n}(e_{X})$ implies that there is some $k_{n}\in\mathbb{N}$ such that $x_{i}=e$ for all $i>k_{n}$. Let $n-1<\sigma(x)\leq n$. As $\sigma(x)\geq d(x)$, we have $d(x)=\sum_{i=1}^{k_{n}}d_{G}(x_{i})\leq n$. Then

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$$\rho_{c}(x)=\sum_{i=1}^{k_{n}}c_{i}d_{G}(x_{i})\leq\max(c_{1},\ldots,c_{k_{n}})\sum_{i=1}^{k_{n}}d_{G}(x_{i})=c_{k_{n}}d(x)\leq c_{k_{n}}n.$$

Set $c_{k_{n}}=n$. Then $\rho_{c}(x)\leq n^{2}\leq(\sigma(x)+1)^{2}$ for any $x\in X$. ∎

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Theorem 5.3.

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Let $G$ be a finite group. Then $X=G^{\infty}$ and $Y=\sqcup_{i\in\mathbb{N}}G^{i}$ have property $\mathcal{B}\mathcal{G}$-A.

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Proof.

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Once again we restrict ourselves to the case of $X$. Let $\sigma\in\mathcal{B}\mathcal{G}(X,d)$. Combining Lemmas 5.1 and 5.2, there exists a sequence $c\in\mathcal{C}$ such that the metric $\rho_{c}\in\mathcal{B}\mathcal{G}(X,d)$ satisfies $\rho_{c}(x,y)\leq(\sigma(x,y)+1)^{2}$ for any $x,y\in X$.

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Let $R=n$, $\varepsilon$ arbitrary. Let $i(n)$ satisfy $c_{i(n)}>n$. As $\lim_{i\to\infty}c_{i}=\infty$, such $i(n)$ exists. Set

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$$\xi_{x}(z)=\left\{\begin{array}[]{cl}\alpha,&\mbox{\ if\ }x_{i}=z_{i}\mbox{\ for\ }i>i(n);\\ 0,&\mbox{otherwise},\end{array}\right.$$

where $\alpha$ is chosen to satisfy $\|\xi_{x}\|=1$. If $\rho_{c}(x,y)=\sum_{i\in\mathbb{N}}c_{i}d_{G}(x_{i},y_{i})<n$ then $x_{i}=y_{i}$ for any $i>i(n)$. Then $\xi_{x}-\xi_{y}=0$, hence $\xi$ has $(n,\varepsilon)$ variation with respect to the metric $\rho_{c}$. If $z\in\operatorname{supp}\xi_{x}$ then $x_{i}=z_{i}$ for any $i>i(n)$, hence $\rho_{c}(x,y)\leq\max(c_{1},\ldots,c_{i(n)})i(n)\operatorname{diam}(G)=ni(n)\operatorname{diam}(G)$, and we can set $S=S(c,n,\varepsilon)=ni(n)\operatorname{diam}(G)$. ∎

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6. Example 2

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Let $Z$ be the countable metric space with the metric $d$ such that $d(x,y)=1$ for any distinct points $x,y\in Z$. It is obviously of unbounded geometry, coarsely equivalent to a single point, and satisfies the standard version of property (A). Nevertheless,

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Theorem 6.1.

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The space $(Z,d)$ does not have property $\mathcal{B}\mathcal{G}$-A.

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Proof.

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Assume the contrary. Then $C^{*}_{\mathcal{B}\mathcal{G}(Z,d)}$ is exact by Theorem 4.2. Let us show that there exists $\rho\in\mathcal{B}\mathcal{G}(Z,d)$ such that the space $(Z,\rho)$ does not have property A. Let $\{G_{n}\}_{n\in\mathbb{N}}$ be a sequence of finite graphs that form an expander, and let $\rho$ be a metric such that $(Z,\rho)$ is isometric to the coarse union $\sqcup_{n\in\mathbb{N}}G_{n}$ after identifying points of $Z$ with the vertices of $\sqcup_{n\in\mathbb{N}}G_{n}$. Being an expander, $(X,\rho)$ has bounded geometry and does not have property A (it is shown in [5, Corollary 1.2] that the uniform Roe algebra of an expander is not exact). This contradicts the assumed exactness of $C^{*}_{\mathcal{B}\mathcal{G}}(Z,d)$, as $C^{*}_{u}(Z,\rho)$ is its $C^{*}$-subalgebra. ∎

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The $C^{*}$-algebra $C^{*}_{\mathcal{B}\mathcal{G}}(Z,d)$ was described in [2].

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This example shows that property $\mathcal{B}\mathcal{G}$-A is not coarsely equivalent and hints to a more subtle equivalence relation, which we introduce in the next section.

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7. $\mathcal{B}\mathcal{G}$-coarse equivalence

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Recall that, given metric spaces $(X,d_{X})$ and $(Y,d_{Y})$, a map $f:X\to Y$ is coarse if there exists a homeomorphism $\psi$ of $(0,\infty]$ such that $d_{Y}(f(x),f(y))\leq\psi(d_{X}(x,y))$ for any $x,y\in X$. A self-map $h:X\to X$ is close to identity with respect to $d_{X}$ (we write $h\sim_{d_{X}}\operatorname{id}_{X}$) if there exists $C>0$ such that $d_{X}(x,h(x))\leq C$ for any $x\in X$. $(X,d_{X})$ and $(Y,d_{Y})$ are coarsely equivalent if there exist coarse maps $f:X\to Y$, $g:Y\to X$ such that $g\circ f\sim_{d_{X}}\operatorname{id}_{X}$ and $f\circ g\sim_{d_{Y}}\operatorname{id}_{Y}$.

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Theorem 7.1.

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Let $(X,d_{X})$ and $(Y,d_{Y})$ be uniformly discrete metric spaces, and let $f:X\to Y$ and $g:Y\to X$ be coarse maps such that $g\circ f\sim_{d_{X}}\operatorname{id}_{X}$ and $f\circ g\sim_{d_{Y}}\operatorname{id}_{y}$. Then the following are equivalent:

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1. there exists $N\in\mathbb{N}$ such that $|g^{-1}(x)|\leq N$ for any $x\in X$ and $|f^{-1}(y)|\leq N$ for any $y\in Y$;

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2. for any $\sigma_{X}\in\mathcal{B}\mathcal{G}(X,d_{X})$ there exists $\rho_{X}\in\mathcal{B}\mathcal{G}(X,d_{X})$ such that $\sigma_{X}\preceq\rho_{X}$, and $\rho_{Y}\in\mathcal{B}\mathcal{G}(Y,d_{Y})$ (and, symmetrically, for any $\sigma_{Y}\in\mathcal{B}\mathcal{G}(Y,d_{Y})$ there exists $\rho_{Y}\in\mathcal{B}\mathcal{G}(Y,d_{Y})$ such that $\sigma_{Y}\preceq\rho_{Y}$, and $\rho_{X}\in\mathcal{B}\mathcal{G}(X,d_{X})$) such that $f$ and $g$ determine coarse equivalence between $(X,\rho_{X})$ and $(Y,\rho_{Y})$.

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Proof.

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If (2) holds then there are bounded geometry metrics $\rho_{X}$ and $\rho_{Y}$ determining coarse equivalence between $(X,\rho_{X})$ and $(Y,\rho_{Y})$. As $f$ and $g$ determine coarse equivalence between $(X,\rho_{X})$ and $(Y,\rho_{Y})$, there exists $C>0$ such that $\rho_{X}(z,g\circ f(z))\leq C$ for any $z\in X$. Let $y\in Y$. If $f^{-1}(y)=\emptyset$ then there is nothing to prove, so suppose that there is $x\in X$ with $f(x)=y$. If $z\in f^{-1}(y)$, i.e. $f(z)=f(x)=y$, then $\rho_{X}(z,g(y))=\rho_{X}(z,g\circ f(z))\leq C$, which implies that $z\in B_{C}^{\rho_{X}}(g(y))$, and, using bounded geometry of $\rho_{X}$, we may set $N=\max_{u\in X}|B_{C}^{\rho_{X}}(u)|$. Then $|f^{-1}(y)|\leq N$ for any $y\in Y$. Symmetrically, we can bound $|g^{-1}(x)|$.

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Now suppose that (1) holds. We shall prove the first half of (2) (the second half is proved symmetrically).

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Let $\sigma_{X}\in\mathcal{B}\mathcal{G}(X,d_{X})$. By Lemma 2.3, there exists $\tau_{X}\in\mathcal{B}\mathcal{G}(X,d_{X})$ such that $\tau_{X}(x,g\circ f(x))=d_{X}(x,g\circ f(x))$ for any $x\in X$. Then there exists $\rho_{X}\in\mathcal{B}\mathcal{G}(X,d_{X})$ such that $\sigma_{X},\tau_{X}\preceq\rho_{X}$.

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Set

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$$\rho_{Y}(y_{1},y_{2})=\rho_{X}(g(y_{1}),g(y_{2}))+d_{Y}(y_{1},y_{2}).$$ (2)

Being a sum of a metric and a pseudometric, it is a metric. Clearly, $\rho_{Y}\preceq d_{Y}$. To check bounded geometry condition, let $z\in B_{R}^{\rho_{Y}}(y)$. Then $\rho_{Y}(y,z)\leq R$, hence $\rho_{X}(g(y),g(z))\leq R$, which implies that $g(z)\in B_{R}^{\rho_{X}}(g(y))$. There is $C>0$ such that $|B_{R}^{\rho_{X}}(x)|\leq C$ for any $x\in X$, and for each $u=g(z)\in B_{R}^{\rho_{X}}(g(y))$ there are not more than $N$ points $z$ with $g(z)=u$, so $|B_{R}^{\rho_{Y}}(y)|\leq NC$.

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Next we check that $g\circ f\sim_{\rho_{X}}\operatorname{id}_{X}$ and $f\circ g\sim_{\rho_{Y}}\operatorname{id}_{Y}$. The first one obviously follows from the construction of $\rho_{X}$: $\rho_{X}(x,g\circ f(x))=d_{X}(x,g\circ f(x))$, and from $g\circ f\sim_{d_{X}}\operatorname{id}_{X}$. Let $C>0$ satisfy $d_{Y}(y,f\circ g(y))\leq C$ and $d_{X}(x,g\circ f(x))\leq C$ for any $x\in X$, $y\in Y$. Then

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$$\rho_{Y}(y,f\circ g(y)) = \rho_{X}(g(y),g\circ f\circ g(y))+d_{Y}(y,f\circ g(y))$$

$$= \rho_{X}(x,g\circ f(x))+d_{Y}(y,f\circ g(y))\ \leq\ 2C,$$

where $x=g(y)$.

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Finally, we have to check that $f$ and $g$ are coarse with respect to the metrics $\rho_{X}$ and $\rho_{Y}$. For $g$, the estimate

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$$\rho_{X}(g(y_{1}),g(y_{2}))\leq\rho_{Y}(y_{1},y_{2})$$

for any $y_{1},y_{2}\in Y$ follows directly from (2). For $f$ we have

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$$\rho_{Y}(f(x_{1}),f(x_{2})) = \rho_{X}(g\circ f(x_{1}),g\circ f(x_{2}))+d_{Y}(f(x_{1}),f(x_{2}))$$

$$\leq \rho_{X}(g\circ f(x_{1}),x_{1})+\rho_{X}(x_{1},x_{2})+\rho_{X}(x_{2},g\circ f(x_{2}))$$

$$+\ d_{Y}(f(x_{1}),f(x_{2})))$$

$$\leq \rho_{X}(x_{1},x_{2})+2C+\phi(d_{X}(x_{1},x_{2}))$$

$$\leq \rho_{X}(x_{1},x_{2})+2C+\phi(\rho_{X}(x_{1},x_{2}))$$

$$= \psi(\rho_{X}(x_{1},x_{2})),$$

where $C$ satisfies $\rho_{Y}(y,f\circ g(y))\leq C$ and $\rho_{X}(x,g\circ f(x))\leq C$ for any $x\in X$, $y\in Y$, and $\phi$ is a homeomorphism of $(0,\infty]$ such that $d_{Y}(f(x_{1}),f(x_{2}))\leq\phi(d_{X}(x_{1},x_{2}))$ for any $x_{1},x_{2}\in X$. ∎

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Definition 7.2.

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Let $(X,d_{X})$ and $(Y,d_{Y})$ be uniformly discrete metric spaces. We say that they are $\mathcal{B}\mathcal{G}$-coarsely equivalent if there exist coarse maps $f:X\to Y$ and $g:Y\to X$ such that they determine coarse equivalence between $(X,d_{X})$ and $(Y,d_{Y})$ and either of the two equivalent conditions of Theorem 7.1 holds.

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Clearly, the space $Z$ from the example 2 and a single point space are not $\mathcal{B}\mathcal{G}$-coarsely equivalent.

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Corollary 7.3.

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If $(X,d_{X})$ and $(Y,d_{Y})$ are of bounded geometry then coarse equivalence and $\mathcal{B}\mathcal{G}$-coarse equivalence coincide.

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Recall that $C^{*}$-algebras $A$ and $B$ are called stably $*$-isomorphic if $A\otimes\mathbb{K}$ and $B\otimes\mathbb{K}$ are $*$-isomorphic, where $\mathbb{K}$ denotes the algebra of compact operators on a separable Hilbert space.

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Proposition 7.4.

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Let $(X,d_{X})$ and $(Y,d_{Y})$ be $\mathcal{B}\mathcal{G}$-coarsely equivalent spaces. Then $C^{*}_{\mathcal{B}\mathcal{G}}(X,d_{X})$ and $C^{*}_{\mathcal{B}\mathcal{G}}(Y,d_{Y})$ are stably $*$-isomorphic.

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Proof.

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A reduction to the case of surjective $f$ implementing coarse equivalence and an explicit bijection between the underlying spaces $X\times\mathbb{N}$ and $Y\times\mathbb{N}$ were constructed in [1, Theorem 4] for metrics of bounded geometry, and this bijection depends only on the cardinality of $f^{-1}(y)$, $y\in Y$, which is uniformly bounded and does not depend on the metric, so passing to the direct limit agrees with the corresponding $*$-isomorphism between $C^{*}_{u}(X,\rho_{X})\otimes\mathbb{K}$ and $C^{*}_{u}(Y,\rho_{Y})\otimes\mathbb{K}$. To finish the proof, note that $\operatorname{dir}\lim_{\rho_{X}\in\mathcal{B}\mathcal{G}(X,d_{X})}(C^{*}_{u}(X,\rho_{X})\otimes\mathbb{K})\cong(\operatorname{dir}\lim_{\rho_{X}\in\mathcal{B}\mathcal{G}(X,d_{X})}C^{*}_{u}(X,\rho_{X}))\otimes\mathbb{K}$. ∎

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Proposition 7.5.

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Let $(X,d_{X})$ and $(Y,d_{Y})$ be $\mathcal{B}\mathcal{G}$-coarsely equivalent discrete metric spaces, and let $(X,d_{X})$ satisfy property $\mathcal{B}\mathcal{G}$-A. Then $(Y,d_{Y})$ satifies property $\mathcal{B}\mathcal{G}$-A too.

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Proof.

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Let $f$, $g$ be the maps from Definition 7.2, and let $\sigma_{X}\in\mathcal{B}\mathcal{G}(X,d_{X})$. Then there exists a homeomorphism $\psi_{X}$ of $(0,\infty]$ and $\rho_{X}\in\mathcal{B}\mathcal{G}(X,d_{X})$ such that $\rho_{X}(x_{1},x_{2})\leq\psi_{X}(\sigma_{X}(x_{1},x_{2}))$ for any $x_{1},x_{2}\in X$ and $(X,\rho_{X})$ has property A. Let $\sigma_{Y}\in\mathcal{B}\mathcal{G}(Y,d_{Y})$. By definition, there exists $\rho_{Y}\in\mathcal{B}\mathcal{G}(Y,d_{Y})$ such that $(X,\rho_{X})$ and $(Y,\rho_{Y})$ are coarsely equivalent, hence $(Y,\rho_{Y})$ has property A. It remains to show that there exists a homeomorphism $\psi_{Y}$ of $(0,\infty]$ such that $\rho_{Y}(y_{1},y_{2})\leq\psi_{Y}(\sigma_{Y}(y_{1},y_{2}))$ for any $y_{1},y_{2}\in Y$.

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As $g:(Y,\sigma_{Y})\to(X,\sigma_{X})$ is a coarse map, there exists a homeomorphism $\alpha$ of $(0,\infty]$ such that

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$$\sigma_{Y}(y_{1},y_{2})\geq\alpha(\sigma_{X}(g(y_{1}),g(y_{2})).$$

Then, we have

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$$\psi_{X}(\sigma_{X}(g(y_{1}),g(y_{2})))\geq\rho_{X}(g(y_{1}),g(y_{2})).$$

As $g:(Y,\rho_{Y})\to(X,\rho_{X})$ is also a coarse map, there exists a homeomorphism $\beta$ of $(0,\infty]$ such that

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$$\beta(\rho_{X}(g(y_{1}),g(y_{2})))\geq\rho_{Y}(y_{1},y_{2}).$$

Combining these three inequalitites, we get $\psi_{Y}(\sigma_{Y}(y_{1},y_{2}))\geq\rho_{Y}(y_{1},y_{2})$ for any $y_{1},y_{2}\in Y$, where $\psi_{Y}=\beta\circ\psi_{X}\circ\alpha$. ∎

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