2
$\begingroup$

Suppose that $\Gamma$ is a discrete group acting on a compact Hausdorff space $X$. Suppose that there exists a quasi-invariant measure $\nu$ on $X$ which has full support. Show that there is an embedding from the reduced crossed product $C(X) \rtimes_{\alpha,r}\Gamma$ into the von Neumann crossed product $L^{\infty}(X) \ltimes \Gamma$.

Define a map $T:C(X) \to \mathbb{BL}^2(X, \nu)$ by $T(f)=M_f$, where $M_f(g)=fg$. Then $T$ is an isometry. (The proof is here.) The map $\tilde{T}:L^{\infty}(X,\nu) \to \mathbb{B}(L^2(X,\nu))$ defined by $\tilde{T}(f)=M_f$, where $M_f(g)=fg$ is an isometric $*$-homomorphism. Moreover the map $j: C(X) \to L^{\infty}(X,\nu)$ is defined by $j(f)=f$ is injective (look here.) The reduced crossed product $C(X) \rtimes_{\alpha,r}\Gamma$ is the norm closure of the elements of the form $\left\{\sum_s\pi(f_s)(1 \otimes \lambda_s): \text{finite sum}\right\}$ inside $\mathbb{B}\left(L^2(X,\nu) \otimes l^2(\Gamma)\right)$, where $$\pi: C(X) \to\mathbb{B}\left(L^2(X,\nu) \otimes l^2(\Gamma)\right)$$ is defined by $\pi(f)(\xi_t \otimes \delta_t)=T(\alpha_{t^{-1}}(f))(\xi_t)\otimes \delta_t$ and $\alpha_{t^{-1}}:C(X) \to C(X)$ is defined by $\alpha_{t^{-1}}(f)(x)=f(t.x)$. The crossed product von Neumann algebra $ L^{\infty}(X)\ltimes \Gamma$ is the weak$^*$-closure of the elements of the form $\left\{\sum_s\tilde{\pi}(f_s)(1 \otimes \lambda_s): \text{finite sum}\right\}$ inside $\mathbb{B}\left(L^2(X,\nu) \otimes l^2(\Gamma)\right)$, where $$\tilde{\pi}: L^{\infty}(X) \to\mathbb{B}\left(L^2(X,\nu) \otimes l^2(\Gamma)\right)$$ is defined by $\tilde{\pi}(f)(\xi_t \otimes \delta_t)=\tilde{T} (\alpha_{t^{-1}}(f))(\xi_t)\otimes \delta_t$ and $\alpha_{t^{-1}}:L^{\infty}(X) \to L^{\infty}(X)$ is defined by $\alpha_{t^{-1}}(f)(x)=f(t.x)$.

Now I want to show that $j$, defined from $C(X) \rtimes_{\alpha,r}\Gamma \to L^{\infty}(X) \ltimes \Gamma$ by $j\left(\sum_s\pi(f_s)(1\otimes \lambda_s)\right)=\sum_s\tilde{\pi}(f_s)(1\otimes \lambda_s)$ is an embedding. Whenever, $f \in C(X)$, $\pi(f)=\tilde{\pi}(f)$. Thus, $j\left(\sum_s\pi(f_s)(1\otimes \lambda_s)\right)=\sum_s\tilde{\pi}(f_s)(1\otimes \lambda_s)=\sum_s\pi(f_s)(1\otimes \lambda_s)$.

Now I am a bit confused on what to show. What exactly do I show now??

Thanks for the help!!

$\endgroup$
0
$\begingroup$

The norm closure of the set $\left\{\sum_{s}\pi(f_s)(1\otimes \lambda_s)\right\}$ inside $\mathbb{B}(L^2(X,\nu) \otimes l^2(\Gamma))$ is a subset of the weak$^*$-closure of the set he set $\left\{\sum_{s}\pi(f_s)(1\otimes \lambda_s)\right\}$ inside $\mathbb{B}(L^2(X,\nu)\otimes l^2(\Gamma))$. Thus, $C(X) \rtimes_{\alpha,r} \Gamma \subset L^{\infty}(X) \ltimes \Gamma$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.