A completely positive equivariant map $\varphi: A \to B$ induces a map $A \rtimes_r \Gamma \to B \rtimes_r \Gamma.$

Question

Recall the construction of the reduced crossed product:

Let $\Gamma$ be a discrete group and $A$ be a $C^*$-algebra with an action $\alpha: \Gamma\to \operatorname{Aut}(A)$. Consider the $*$-algebra $C_c(\Gamma,A)$ of finitely supported functions $\Gamma \to A$ with the $\alpha$-twisted multiplication and involution. Then we can build a canonical faithful representation of $C_c(\Gamma,A)$ as follows: start with a faithful representation $A \subseteq B(H)$. This induces a new faithful representation $\pi: A \to B(H \otimes \ell^2(\Gamma))$ by $\pi(a)(\xi \otimes \delta_g) = \alpha_{g}^{-1}(a)\xi \otimes \delta_g$. Considering the left regular representation $\lambda: \Gamma \to U(\ell^2(H)): g \mapsto (\delta_h \mapsto \delta_{gh})$, we obtain an induced faithful representation $$C_c(\Gamma,A) \to B(H \otimes \ell^2(\Gamma)): \sum_{s \in \Gamma} a_s s \mapsto \sum_{s \in \Gamma} \pi(a_s)(1 \otimes \lambda_s)$$ which induces a $C^*$-norm on $C_c(\Gamma,A)$. The reduced crossed product $A \rtimes_r \Gamma$ is the $C^*$-completion of $C_c(\Gamma,A)$ with respect to this norm, and does not depend on the choice of faithful representation $A\subseteq B(H)$.

Let $\Gamma$ be a discrete group and let $\varphi: A \to B$ be a $\Gamma$-equivariant completely positive contraction between the $\Gamma$-$C^*$-algebras $A$ and $B$. I want to show the following:

The induced map $C_c(\Gamma,A) \to C_c(\Gamma,B): \sum_{s \in \Gamma} a_s s \mapsto \sum_{s \in \Gamma}\varphi(a_s)s$ is bounded, hence extends uniquely to a map $\varphi \rtimes_r \Gamma: A \rtimes_r \Gamma \to B \rtimes_r \Gamma.$

Attempt: Let $\pi_A: A \to B(H_A \otimes \ell^2(\Gamma))$ and $\pi_B: B \to B(H_B \otimes \ell^2(\Gamma))$ be faithful representations as above. Then by the $C^*$-identity. $$\|\sum_s \varphi(a_s)s\|^2 = \|\sum_s \pi_B(\varphi(a_s))(1 \otimes \lambda_s)\|^2$$ $$=\|\sum_{s,t} (1 \otimes \lambda_{s^{-1}}) \pi_B(\varphi(a_s^*)\varphi(a_t)) (1\otimes \lambda_t)\|.$$

This looks like something we could apply Cauchy-Schwarz for completely positive maps on, but the surrounding factors $1 \otimes \lambda_{s^{-1}}$ and $1 \otimes \lambda_t$ complicate this. Maybe I need to apply Cauchy-Schwarz on some carefully crafted matrix. Does anybody see how I can continue?

Of course, other approaches are also welcome! I am also interested in the following: if $\varphi$ is injective, then is the extension $A \rtimes_r \Gamma \to B \rtimes_r \Gamma$ also injective?

Answer 1

This is Exercise 4.1.4 in the book by Brown+Ozawa (is that where this question ultimately comes from?) The "Hint" is "The reduced case is easy." Hmm.

Well, it might perhaps help to look at this section of the book. Indeed, Proposition 4.1.5, and its proof, shows that if $F\subseteq\Gamma$ is a finite-set, and $P:\ell^2(\Gamma)\rightarrow \ell^2(F)$ the projection, then $$ (1\otimes P)x(1\otimes P) = \sum_{s\in\Gamma} \sum_{p\in F\cap sF} \alpha_{p^{-1}}(a_s)\otimes e_{p, s^{-1}p} \in A\otimes M_F(\mathbb C) \cong M_F(A). $$ Here $x = \sum a_s\lambda_s \in C_c(\Gamma, A) \subseteq A \rtimes_r \Gamma \subseteq \mathcal B(H\otimes\ell^2(\Gamma))$ and $e_{p,q}$ are the matrix units of $M_F(\mathbb C)$. The point of this is to show that the norm of $x$, acting on $H\otimes\ell^2(\Gamma)$, does not depend upon the particular representation $A\subseteq\mathcal B(H)$ (because $M_F(A)$ has a unique norm).

However, this formula also gives us a convenient way to compute the norm of $\varphi\rtimes 1$. For, if $y=(\varphi\rtimes 1)x$, then $y=\sum_s \varphi(a_s)\lambda_s$, and so $$ (1\otimes P)y(1\otimes P) = \sum_{s\in\Gamma} \sum_{p\in F\cap sF} \beta_{p^{-1}}(\varphi(a_s))\otimes e_{p, s^{-1}p} = \sum_{s\in\Gamma} \sum_{p\in F\cap sF} \varphi(\alpha_{p^{-1}}(a_s))\otimes e_{p, s^{-1}p}, $$ the first equality by applying to above to $B\rtimes_r\Gamma$, and the second equality using that $\varphi$ is equivariant. However, this then equals $$ (\varphi\otimes 1_F)\big( (1\otimes P)x(1\otimes P) \big), $$ where $(\varphi\otimes 1_F)$ is the dilated map $M_F(A)\rightarrow M_F(B)$. As $\varphi$ is CCP by assumption, $\varphi\otimes 1_F$ is contractive, and so $$ \| (1\otimes P)y (1\otimes P) \| \leq \|(1\otimes P)x(1\otimes P)\| \leq \|x\|. $$ Taking the SOT limit as $F$ increases, so $P\rightarrow 1$, shows that $\|y\| \leq \|x\|$ as required.

Answer 2

Here is a proof free of Hilbert space representations. I will use the notation $G$ instead of $\Gamma$ for the group, which could be also locally compact, and $\alpha$ will denote the action of $G$ on $A$, which is assumed to be strongly continuous.

The main point is to observe that $A\rtimes_r G$ can be represented as a $C^*$-subalgebra of $M(A\otimes K(L^2G))$. This is done via the integrated form of the covariant pair $((1\otimes M)\circ\tilde\alpha, 1\otimes\lambda)\colon (A,G)\to M(A\otimes K)$, where $M\colon C_0(G)\to B(L^2(G))$ denotes the multiplication representation and $\tilde\alpha\colon A\to C_b(G,A)\subseteq M(A\otimes C_0(G))$ is defined by $\tilde\alpha(a)(t):=\alpha_{t^{-1}}(a)$.

In this way we can view $A\rtimes_r G$ as a $C^*$-subalgebra of $M(A\otimes K)$. Indeed, it is always contained in the relative multiplier algebra:

$$\tilde M(A\otimes K):=\{m\in M(A\otimes K): m(1\otimes K), (1\otimes K)m\subseteq A\otimes K\}.$$

Now, if $\rho\colon A\to B$ is a cp (completely positive) map, then we can built the cp map $\rho\otimes id\colon A\otimes K\to B\otimes K$ as the minimal tensor product is functorial for cp maps. Moreover, we can extend $\rho\otimes id$ to a cp map (still denoted by) $\rho\otimes id\colon \tilde M(A\otimes K)\to \tilde M(B\otimes K)$. And if $\rho$ is injective or faithful, then so is $\rho\otimes id$.

Now, to get the desired extension $\rho\rtimes_r G\colon A\rtimes_r G\to B\rtimes_r G$ we just need to observe that the restriction of $\rho\otimes id$ maps $A\rtimes_r G\subseteq \tilde M(A\otimes K)$ into $B\rtimes_r G\subseteq \tilde M(B\otimes K)$.

This works for all locally compact groups $G$, not only discrete groups, and the argument also allows for (equiivariant) cp maps $\rho\colon A\to M(B)$, in which case $\rho\rtimes_r G$ will be a cp map $A\rtimes_r G\to M(B\rtimes_r G)$.

Summarizing, we can say that the reduced crossed product is functorial for equivariant cp maps.