Roe algebra of a countably infinite set of points

Question

First let me state some definitions.

Let $X$ be a second countable, proper metric space. Let $H$ be a separable Hilbert space equipped with a nondegenerate $*$-representation $C_{0}(X)\rightarrow B(H)$ where no nonzero element of $C_{0}(X)$ acts as a compact operator.

Let $T\in B(H)$. $T$ is said to be locally compact if for any compact $K\subset X$, we have $\chi_{K}T$ and $T\chi_{K}$ are compact operators. $T$ is said to have finite propagation if $\sup\{d(y,x):(y,x)\in\mathrm{supp}(T)\}<\infty$, where $\mathrm{supp}(T)$ consists of all $(y,x)\in X\times X$ such that for all open neighborhoods $U$ of $x$ and $V$ of $y$ we have $\chi_{V}T\chi_{U}\neq 0$.

The Roe C*-algebra $C^{*}(X)$ is the operator norm closure of the set of all finite propagation, locally compact operators on $H$.

For example, if $X$ is a point, then $C^{*}(X)$ is the algebra of compact operators.

My question is: If $X$ is a countably infinite collection of points, what is $C^{*}(X)$? What is its K-theory?

Answer 1

You have to specify a metric on your infinite collection of points, since without this the question does not make any sense.

But besides that, the K-theory of the Roe algebra is coarsely invariant. So if we have a proper metric space X which has bounded geometry, then we may pass to a uniformly discrete subset $Y \subset X$ without changing the K-theory of the Roe algebra, i.e., $K_\ast(C^\ast(X)) \cong K_\ast(C^\ast(Y))$. So your question is to compute the K-theory of the Roe algebra of any bounded geometry proper metric space. This is exactly the coarse Baum--Connes conjecture: $K_\ast(C^\ast(Y)) \cong \lim_{d \to \infty} K_\ast(P_d(Y))$ for any discrete metric space Y of bounded geometry, where the $K_\ast(\cdot)$ on the right hand side denotes K-homology and $P_d(Y)$ is the Rips complex of $Y$.

Answer 2

Sorry for a late reply but perhaps this will be helpful.

Let $X$ be a locally compact metric space with metric $d$. Let $\rho \colon C_0(X) \to \mathcal{H}_X$ be a non-degenerate ample $*$-representation. Let $T$ be a bounded operator on $\mathcal{H}_X$.

-The support of $T$ is the complement of the open subset in $X \times X$ $$ \{ (x,y) \in X \times X \mbox{ such that there exists $f$ and $g$ in $C_0(X)$ satisfying $f(x) \neq 0$, $g(x) \neq 0$ and $fTg = 0$} \} $$ -If there exists $r > 0$ such that whenever $d(x,y) > r$ then $(x,y)$ is not in the support of $T$ then $T$ is said to have finite propagation (in particular propagation less than $r$).

-The operator $T$ is locally compact if $f T$ and $Tf$ is compact for any $f \in C_0(X)$.

We define $C\left[ X \right] \subset \mathcal{B}(\mathcal{H}_X)$ to be the $*$-subalgebra of locally compact, finite propagation operators. The Roe algebra $C^*(X)$ is the closure of $C \left[ X \right]$ inside $\mathcal{B}(\mathcal{H}_X)$.

It takes some time to check that $C^*(X)$ does not depend on the ample representation nor the Hilbert space (nor even the coarse equivalence class) up to non-canonical isomorphism. Considering this, it is routine to check that when $X$ is a proper countable discrete metric space then $C \left[ X \right]$ is equivalent to the following set of operators.

Let $T \in \mathcal{B}(\ell^2(X) \otimes \mathcal{H})$ be written as a family $(T_{x,y})_{x,y \in X}$ of operators on $\mathcal{H}$ where $\langle T (\delta_y \otimes \mathrm{id}), (\delta_x \otimes \mathrm{id}) \rangle = T_{x,y}$. Then $T$ is locally compact if and only if $T_{x,y}$ is a compact operator for all $x,y \in X$ and $T$ has finite propagation if there exists an $r > 0$ such that $T_{x,y} = 0$ whenever $d(x,y) > r$. The closure of the $*$-subalgebra of locally compact finite propagation operators inside $\mathcal{B}(\ell^2(X) \otimes \mathcal{H})$ is the Roe algebra $C^*(X)$.

Things become a bit more interesting when $X = G$ where $G$ is a discrete countable group. There always exists a proper metric on $G$ (when $G$ is finitely generated then we can choose the metric to be the word metric). Now I claim that the Roe algebra is none other than $$ \ell^{\infty}(G, \mathcal{K}) \rtimes_r G $$ where $\mathcal{K}$ are the compact operators on some infinite dimensional, separable Hilbert space. This is easy to show but we have to wade through a lot of definitions and actions. So lets fix some actions but first set $A = \ell^{\infty}(G, \mathcal{K})$. So $G$ acts on $A$ by $*$-automorphism: $$ \alpha \colon G \to \mathrm{Aut}(A) $$ where $\alpha_g(f) (h) = f(g^{-1}h)$ for all $f \in A$ and $g,h \in G$. Define a $*$-representation of $A$ on $\ell^2(G, \mathcal{H})$ by $$ \pi \colon A \to \mathcal{B}(\ell^2(G, \mathcal{H})) $$ where $\pi(f)(\xi)(g) = f(g) \left[ \xi(g) \right]$ for all $f \in A$, $\xi \in \ell^2(G, \mathcal{H})$ and $g \in G$. Let $\lambda \colon G \to \mathcal{U}(\ell^2(G, \mathcal{H}))$ be the unitary representation induced by the left action on $G$. That is $\lambda_g(\xi) (h) = \xi(g^{-1} h)$ for all $\xi \in \ell^2(G, \mathcal{H})$ and $g,h \in G$. Now notice that $(\lambda, \pi, \ell^2(G, \mathcal{H}))$ is a covariant representation. That is $$ \lambda_g \pi(f) \lambda_g^* = \pi(\alpha_g(f)) $$ for all $g \in G$ and $f \in A$. That is because $$ \lambda_g \left[\pi(f) (\xi)\right] (h) = \pi(f) (\xi)(g^{-1} h) = f(g^{-1} h) \left[ \xi(g^{-1} h) \right] = \alpha_g(f)(h) \left[ \lambda_g(\xi)(h) \right] = \pi(\alpha_g(f)) \lambda_g (\xi) (h) $$ for all $g, h \in G$, $f \in A$ and $\xi \in \ell^2(G, \mathcal{H})$. Hence we have a $*$-homomorphism $\sigma \colon C_c(G,A) \to \mathcal{B}(\ell^2(G, \mathcal{H})$ such that $$ \sigma \left( \sum_{g \in G} f_g \delta_g \right) = \sum_{g \in G} \pi(f_g) \lambda_g $$ for all $\sum_{g \in G} f_g \delta_g \in C_c(G, A)$, where $C_c(G,A)$ is a $*$-algebra under $\alpha$-twisted convolution and involution. Now it shouldn't take too much convincing that $\sigma (C_c(G, A)) = C \left[ G \right]$.

Typically $A \rtimes_r G$ is defined as follows. Start with a faithful representation $\pi \colon A \to \mathcal{B}(\mathcal{H}_0)$ and defined a new representation $$ \tilde \pi \colon A \to \mathcal{B}(\mathcal{H}_0 \otimes \ell^2(G)) $$ where $\tilde \pi (f) (v \otimes \delta_g) = (\alpha_{g^{-1}}(a)(v)) \otimes \delta_g$. It's not hard to show that $(1 \otimes \tilde \lambda, \tilde \pi, \mathcal{H}_0 \otimes \ell^2(G))$ is a covariant representation where $1 \otimes \tilde \lambda \colon G \to \mathcal{U}( \mathcal{H}_0 \otimes \ell^2(G))$ satisfies $(1 \otimes \tilde \lambda_g) (v \otimes \delta_h ) =v \otimes \delta_{gh} $ for all $v \in \mathcal{H}_0$ and $g,h \in G$. Then $A \rtimes_r G$ is the closure of the $*$-algebra $C_c(G, A)$ inside $\mathcal{B}(\mathcal{H}_0 \otimes \ell^2(G) )$.

In our case $A = \ell^{\infty}(G, \mathcal{K})$, $\mathcal{H}_0 = \ell^2(G, \mathcal{H})$ and $\pi \colon A \to \mathcal{B}(\ell^2(G)\otimes \mathcal{H})$ is the usual inclusion. So just to be clear $\ell^{\infty}(G, \mathcal{K}) \rtimes_r G \subset \mathcal{B}(\ell^2(G) \otimes \mathcal{H} \otimes \ell^2(G) )$. Now define a unitary \begin{align*} U \colon \ell^2(G) \otimes \mathcal{H} \otimes \ell^2(G) & \to \ell^2(G) \otimes \ell^2(G) \otimes \mathcal{H} \\ \delta_g \otimes v \otimes \delta_h & \mapsto \delta_{g} \otimes \delta_{hg} \otimes v \end{align*} Hence \begin{align*} U \tilde \pi(f) (\delta_g \otimes v \otimes \delta_h ) & = U \left[ \alpha_{h^{-1}}(f)(\delta_g \otimes v) \otimes \delta_h \right] \\ & =U( \delta_g \otimes f(hg) v \otimes \delta_h) \\ & = \delta_{g} \otimes \delta_{hg} \otimes f(hg) v \\ & = 1 \otimes \pi(f) (U( \delta_g \otimes v \otimes \delta_h )) \end{align*} It follows that $U \tilde \pi (f) U^* = 1 \otimes \pi(f)$. Likewise $U (1 \otimes \tilde \lambda_g) U^* = 1 \otimes \lambda_g \otimes 1 $. Hence it follows that $$ U (\ell^{\infty}(G, \mathcal{K}) \rtimes_r G )U^* \cong \mathbb{C}1 \otimes C^*(G) $$

So I end on some comments and questions. Much of this argument has been based on proposition 5.1.3. of http://www.ams.org/mathscinet-getitem?mr=2391387. A question that has plagued me a bit is what does the Roe algebra look like when $G$ is a locally compact second countable group? I can't imagine it is $L^{\infty}(G) \rtimes_r G$ because this space isn't always defined (functions in $L^{\infty}(G)$ are not necessarily continuous with respect to the action).

As for the $K$-theory of this space it would be really nice if you could compute it directly. I don't know how one would go about doing it but I would be really interested to see your methods.