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If any of the terminology is unclear then please don't hesitate to point it out. My question is: is it true that when $G$ is a locally compact second countable group then: \begin{equation*} C_0(G) \rtimes_r G \cong \mathcal{K}(L^2(G)) \end{equation*} as $C^*$-algebras? First we consider why this is true in the discrete countable case. Let $G$ be a discrete countable group, then there is a homomorphism $\alpha \colon G \to \mbox{Aut}(c_0(G))$ where \begin{equation*} \alpha_g( f) (x) = f(g^{-1}x). \end{equation*} There is a faithful representation $\pi \colon c_0(G) \to \mathcal{B}(\ell^2(G))$ where $(\pi(f) \xi)(s) = f(s) \cdot \xi(s)$ for all $\xi \in \ell^2(G)$ and $s \in G$. Define $C_c(G, c_0(G))$ to be the $*$-algebra of all compactly supported functions with values in $c_0(G)$ with $\alpha$-twisted convolution product and $*$-operation. Let $\lambda \colon G \to \mathcal{B}(\ell^2 (G))$ be the left regular representation, where $\lambda_g(\xi)(s) = \xi(g^{-1} s)$ for all $\xi \in \ell^2(G)$ and $g,s \in G$. Now $(\lambda, \pi, \ell^2(G))$ is a covariant representation of $C_c(G, c_0(G)) \to \mathcal{B}(\ell^2(G))$, call the representation $\sigma$.

First I want to show that $\overline{\sigma(C_c(G, c_0(G)))} \subset \mathcal{B}(\ell^2(G))$ are the compact operators. This is going to be done by showing that this coincides with the closure of the finite rank operators. Take a finite operator $F$ and write $F_{x,y} = \langle F \delta_y, \delta_x \rangle$ where $\delta_x$ is the characteristic function of the point $x \in G$. Now $F$ corresponds to the operator $$ \sum_{g \in G} \pi(f_g) \lambda_g \in \sigma(C_c(G,c_0(G))) $$ where $f_g(x) = F_{x,g^{-1}x}$. It now follows that anything in the closure of $\sigma(C_c(G,c_0(G)))$ can be approximated by operators of finite rank. Hence $\overline{\sigma(C_c(G, c_0(G)))} = \mathcal{K}(\ell^2(G))$.

Now we show the left regular covariant representation of $C_c(G,c_0(G))$ can be mapped $*$-isomorphically onto $\overline{\sigma(C_c(G, c_0(G)))}$.

Define a representation $\tilde \pi \colon c_0(G) \to \mathcal{B}(\ell^2(G) \otimes \ell^2(G))$ where \begin{equation*} \tilde \pi(f)(\delta_g \otimes \delta_h) = f(hg) \delta_g \otimes \delta_h \end{equation*} Then $(1 \otimes \lambda, \tilde \pi, \ell^2(G) \otimes \ell^2(G))$ is a covariant representation and $c_0(G) \rtimes_r G$ is the completion of $C_c(G, c_0(G))$ with respect to the norm induced by the covariant representation.

Define a unitary $U \colon \ell^2(G) \otimes \ell^2(G) \to \ell^2(G) \otimes \ell^2(G)$ by $U(\delta_x \otimes \delta_y) = \delta_x \otimes \delta_{yx}$. Then \begin{align*} U \tilde \pi(f)(\delta_s \otimes \delta_t) & = U((\alpha_{t^{-1}}(f) \delta_s) \otimes \delta_t) \\ & = U((f(ts) \delta_s) \otimes \delta_t)\\ & = f(ts) \delta_s \otimes \delta_{ts}\\ & = \delta_s \otimes (f(ts) \delta_{ts})\\ & = (1 \otimes f)(U(\delta_s \otimes \delta_t)). \end{align*} It follows that $U \tilde \pi(f) U^* = 1 \otimes f$ for all $f \in c_0(G)$. It also follows that $U (1\otimes \lambda_g) U^* = 1 \otimes \lambda_g$ for all $g \in G$. Hence \begin{equation*} U(c_0(G) \rtimes_r G) U^* = \mathbb{C}1 \otimes \overline{\sigma(C_c(G, c_0(G)))} \cong \mathcal{K}(\ell^2(G)). \end{equation*} This follows an argument in proposition 5.1.3. in "$C^*$-algebras and Finite-Dimensional Approximations" by Nathanial P. Brown and Narutaka Ozawa.

To define the reduced cross product $C_0(G) \rtimes_r G$, one takes the same space of compactly supported functions $C_c(G, C_0(G))$ but we now complete in the norm $\|f\|_1 = \int_G \|f(t)\| \, d \mu(t)$. We denote $L^1(G, C_0(G))$ to be the completion in this norm. Given a covariant representation $(U, \pi, \mathcal{H})$ this gives rise to a $*$-homomorphism $L^1(G,C_0(G)) \to \mathcal{B}(\mathcal{H})$.

Let $\tilde \pi \colon C_0(G) \to \mathcal{B}(L^2(G,L^2(G))$ where $\tilde \pi (f) (\xi)(s) = \pi(\alpha_{s^{-1}} (f)) \xi(s)$ and $\Lambda \colon G \to \mathcal{U}(G, L^2(G))$, where $\Lambda_g(\xi)(s) = \xi(g^{-1} s)$. Then $(\Lambda, \tilde \pi, \mathcal{B}(G, L^2(G))$ is a covariant representation and $C_0(G) \rtimes_r G $ is the completion of $L^1(G, C_0(G))$ with respect to the norm induced by this covariant representation.

In the locally compact case, it is not obvious to me how to follow. First to show that $C_0(G) \rtimes_r G \subset \mathcal{K}(L^2(G))$, one idea is that one can define a map $U \colon C_c(G,C_0(G)) \to \mathcal{K}(L^2(G))$ where \begin{equation*} (U(f) \xi)(s) = \int_G f_g(s) \cdot \xi(g^{-1}x) \, d\mu(g). \end{equation*} The reason this is true I think is because the kernel $f \colon G \times G \to \mathbb{C}$, $f(g,x) = f_g(x)$ is a Hilbert--Schmidt kernel hence $U(f)$ is a compact operator. However it is unclear whether for $f \in L^1(G, C_0(G))$, whether $f$ is a Hilbert--Schmidt kernel. I have no idea how to show the reverse inclusion either.

Thanks for reading!

share|improve this question
Why $\mathbb{C}1 \otimes C^*(\pi(c_0(G)), \{\lambda_g : g \in G\}) \cong \mathcal{K}(\ell^2(G))$? And a first feeling is that $\lambda_g$ is not a compact operator. –  ougao Aug 11 at 12:06
Yes you are correct. I shall correct this. –  Chris Cave Nov 5 at 10:16

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