Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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|cite|improve this question
1  
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 '14 at 12:06
    
Yes you are correct. I shall correct this. – Chris Cave Nov 5 '14 at 10:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.