# An embedding from the $C(X) \rtimes_{\alpha,r}\Gamma$ into $L^{\infty}(X) \ltimes \Gamma$.

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!!

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, $C(X) \rtimes_{\alpha,r} \Gamma \subset L^{\infty}(X) \ltimes \Gamma$.