The reduced crossed product $A\rtimes_{\alpha,r}\Gamma$ doesn't depend on the choice of the faithful representation $A \subset \mathbb{B}(\mathcal{H})$.

[I am following the book by "Brown and Ozawa". I have few questions about the proof. I am giving a detailed explanation of the proof and my questions below]

Idea: Rather than computing the norm of $x \in \mathbb{B}(\mathcal{H} \otimes l^2(\Gamma))$, we will cut by the (finte-rank) projections $(1 \otimes P)$ and show that the norm of $(1\otimes P)x(1\otimes P)$ is independent of the representation $A \subset \mathbb{B}(\mathcal{H})$ (where we use the fact that there is a unique $C^*$-norm on $M_n(A)$, when $A$ is a $C^*$-algebra).

Let $F \subset \Gamma$ be a finite set. Let $P$ be the projection onto $\text{span}\{\delta_g: g \in F\}$. Let $\{\delta_s \otimes \delta_t\}_{s,t \in F}$ be the canonical matrix units of $P\mathbb{B}(l^2(\Gamma))P \cong M_{F}(\mathbb{C})$ and fix some arbitrary elements $a \in A$ and $s \in \Gamma$. Let $\pi:A \to \mathbb{B}(\mathcal{H}\otimes l^2(\Gamma))$ be a regular representation. $(1 \otimes P)\pi(a)=(1 \otimes P)(\pi(a))(1\otimes P)=\sum_{q \in F}\alpha_{q^{-1}}(a) \otimes (\delta_q \otimes \delta_q)$ $$(1 \otimes P)\pi(a)\left(x \otimes \left(\sum_{t \in \Gamma} f(t)\delta_t\right)\right)=(1 \otimes P)\left(\sum_{t \in \Gamma}\pi(a)(x \otimes f(t)\delta_t)\right)$$ $$=(1 \otimes P)\left(\sum_{t \in \Gamma}\alpha_{t^{-1}}(a)(x) \otimes f(t)\delta_t)\right)=\sum_{t \in F}\alpha_{t^{-1}}(a)(x) \otimes f(t)\delta_t$$ Also, $$(1 \otimes P)(\pi(a))(1\otimes P)\left(x \otimes \left(\sum_{t \in \Gamma} f(t)\delta_t\right)\right)=(1 \otimes P)(\pi(a))\left(\sum_{t \in F}x \otimes f(t)\delta_t\right)$$ $$=(1\otimes P)\left(\sum_{t \in F}\alpha_{t^{-1}}(a)(x) \otimes f(t)\delta_t\right)=\sum_{t \in F}\alpha_{t^{-1}}(a)(x) \otimes f(t)\delta_t$$ Moreover $(\delta_t \otimes \delta_t)(f)=\langle f,\delta_t \rangle \delta_t=f(t)\delta_t$ gives us the claim mentioned in the beginning of the step.

We have $$\pi(a)=\sum_{t \in \Gamma}\alpha_{t^{-1}}(a) \otimes (\delta_t \otimes \delta_t)$$ where, convergence is in the strong operator topology. (I am not quite sure, what I should conclude from this line)

We see that $$(1 \otimes P)(\pi(a))(1\otimes \lambda_s)(1\otimes P)=(1 \otimes P)(\pi(a))(1\otimes P)(1\otimes \lambda_s)(1 \otimes P)$$ $$=\left(\sum_{q \in F}\alpha_{q^{-1}}(a) \otimes (\delta_q \otimes \delta_q)\right)\left(1\otimes P\lambda_sP \right)$$ $$=\left(\sum_{q \in F}\alpha_{q^{-1}}(a) \otimes (\delta_q \otimes \delta_q)\right)\left(\sum_{p \in F \cap sF}1 \otimes \left(\delta_{s^{-1}p}\otimes \delta_p\right)\right)$$ $$=\sum_{p \in F \cap sF}\alpha_{p^{-1}}(a) \otimes \left(\delta_{s^{-1}p}\otimes \delta_p\right) \in A \otimes M_\mathbb{F}(\mathbb{C})$$ Observe that when $q=s^{-1}p$ $$\left(\delta_q \otimes \delta_q\right)\left(\delta_{s^{-1}p}\otimes \delta_p\right)(f)=\left(\delta_q \otimes \delta_q\right)\langle f,\delta_p \rangle \delta_{s^{-1}p}=\langle f,\delta_p \rangle\langle \delta_{s^{-1}p},\delta_q \rangle\delta_q=\delta_q \otimes \delta_p (f)$$

Now if $x=\sum_s\pi(a_s)(1 \otimes \lambda_s) \in C_c(\Gamma,A) \subset \mathbb{B}(\mathcal{H} \otimes l^2(\Gamma))$, then we have $$ (1 \otimes P)x(1 \otimes P)=\sum_{s \in \Gamma}\sum_{p \in F \cap sF}\alpha_{p^{-1}}(a_s) \otimes \left(\delta_{s^{-1}p}\otimes \delta_p\right) \in A \otimes M_{\mathbb{F}}(\mathbb{C})\cong \mathbb{M}_{\mathbb{F}}(A)$$ Thus, the norm of $(1 \otimes P)x(1\otimes P)$ doesn't depend on the embedding $A \subset \mathbb{B}(\mathcal{H})$.

Now, if $\Gamma$ were countable, $l^2(\Gamma)$ would be separable. Hence there exists a sequence of finite rank projections $\{P_n\}$ such that $P_1 \le P_2 \le P_3 \le \ldots \le P_n \le \ldots$ such that $P_n \to \text{Id}$ in strong Operator topology. Then $$\sup_n\|\left(1\otimes P_n\right)x\left(1\otimes P_n\right)\|=\|x\|$$ Since, the left hand side doesn't depend on the choice of the faithful representation, the right hand side is also independent of the choice. I think when $\Gamma$ is uncountable, netifying this arguement may give the result.

Please let me know, if the above ideas are correct or not. Thanks for the help!!

  • $\begingroup$ can we view$\delta_s $ as $(0\dots,1,0 \dots,0)$ ? $\endgroup$ – math112358 Nov 19 '17 at 12:46
  • $\begingroup$ You can as long as $\Gamma$ is discrete. $\endgroup$ – tattwamasi amrutam Nov 19 '17 at 14:56
  • $\begingroup$ I think your idea is correct! You said if Γ were countable, l2(Γ) would be separable. Hence there exists a sequence of finite rank projections {Pn} such that P1≤P2≤P3≤…≤Pn≤…such that Pn→Id in strong Operator topology.How can we get this conclusion?Would you mind asking this question,thank you! $\endgroup$ – math112358 Nov 19 '17 at 15:14
  • $\begingroup$ Since $l^2(\Gamma)$ is separable, let $\{e_i\}$ be the orthonormal basis for the space. Define $E_n=\text{span }\{e_j: 1 \le j \le n\}$. Then $E_n$ is closed. Let $P_n$ be the projection onto $E_n$. You can show that $P_n$ so defined satisifes the above mentioned condition. $\endgroup$ – tattwamasi amrutam Nov 19 '17 at 15:18

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