# Bounded operators with infinite matrix representations

I asked this question on StackExchange originally, but I'm giving it a go here as well.

Suppose that $A$ is a unital $C^*$-algebra, $\varphi\colon A\to B(H)$ is a unital, completely positive map and that $\Gamma$ is a non-empty set. If $A\subseteq B(K)$ for some Hilbert space $K$, we can consider the space $A'$ of bounded operators acting on $K\otimes\ell^2(\Gamma)$ such that each entry of its matrix representation is an element of $A$. By "matrix representation", I mean that for each $x\in A'$ we can write $x=[x_{r,s}]_{r,s\in \Gamma}$ where $x_{r,s}=\pi_rx\iota_s\in A$ with $\pi_r\colon K\otimes\ell^2(\Gamma)\to K$ given by $\pi_r(\sum\xi_t\otimes\delta_t)=\xi_r$ (each $\delta_r$ being the point mass at $r\in\Gamma$) and $\iota_s\colon K\to K\otimes\ell^2(\Gamma)$ given by $\iota_s(\xi)=\xi\otimes\delta_s$.

If we define $\tilde{\varphi}(x)$ for each $x\in A'$ to be the bounded operator on $H\otimes\ell^2(\Gamma)$ such that $$\tilde{\varphi}(x)_{r,s}=\varphi(x_{r,s}),$$ we clearly get a unital linear map $\tilde{\varphi}\colon A'\to B(H\otimes\ell^2(\Gamma))$. However, I find it hard to prove that this map is also completely positive. It occurred to me that I could maybe use the Stinespring dilation theorem, i.e., that there exists a Hilbert space $H'$, an isometry $V\colon H\to H'$ and a representation $\pi\colon A\to B(H)$ such that $\varphi(x)=V^*\pi(x)V$ for all $x\in A$. However, I run into some problems, namely that I need $\pi$ to be strongly continuous. If $x=a^*a\in M_n(A')$ is positive and we write $a=[a_{kl}]_{k,l=1}^n$, where $a_{kl}=[(a_{kl})_{r,s}]_{r,s\in \Gamma}$ for each $k,l=1,\ldots,n$, I then find that $$(x_{i,j})_{r,s}=\sum_{k=1}^n\sum_{t\in\Gamma}((a_{ki})_{t,r})^*(a_{kj})_{t,s}\in A,$$ where the series converges in the strong operator topology, but a bit of nudging around with the expression $\langle\tilde{\varphi}(x)\xi,\xi\rangle$ for some $\xi\in H\otimes\ell^2(\Gamma)$ doesn't lead to anything (for me at least).

There must be some easier way, but I haven't been able to find one yet...

• Here is the MSE post in case someone wants to see: math.stackexchange.com/q/944799/166535 – Joonas Ilmavirta Sep 29 '14 at 19:33
• If we identify $B(H\otimes \ell^2(\Gamma))$ with the minimal (i.e., spatial) tensor product $B(H)\otimes B(\ell^2(\Gamma))$, (and the same for $K$), don't we just have $\tilde{\varphi}=\varphi\otimes id_{\ell^2(\Gamma)}$? If this is so then complete positivity is immediate, I think. – Mike Jury Sep 29 '14 at 21:37
• But it isn't necessarily true that we can make that identification? For instance, by [Brown-Ozawa, Ex. 3.3.6] $B(\ell^2)\odot B(\ell^2)$ is not norm-dense in $B(\ell^2\otimes\ell^2)$... – Bryder Sep 30 '14 at 11:25