Checking Ruan axioms

I need to check only one axiom for matrix bimodule made from certain bimodule. Here some preparatory definitions.

Matricial approach. (See here for details) For arbitrary linear space $V$ denote by $M_n(V)$ the the linear space of matriсes which elements are vectors from $V$. Consider $M_n(W)$ as normed $M_n(\mathbb{C})$-$M_n(\mathbb{C})$-bimodule with outer multiplication defined by equalities $$(\alpha v)_{i,j}=\sum\limits_{k=1}^n\alpha_{i,k}v_{k,j}\qquad\qquad (v \alpha)_{i,j}=\sum\limits_{k=1}^n\alpha_{k,j}v_{i,k}\qquad\qquad v\in M_n(V), \alpha\in M_n(\mathbb{C})$$ Assume that each $M_n(V)$ is endowed with norm $\Vert\cdot\Vert_{M_n(V)}$ such that

• For all $v\in M_n(V), \alpha\in M_n(\mathbb{C})$ $$\Vert \alpha v\Vert_{M_n(V)}\leq\Vert \alpha\Vert \Vert v \Vert_{M_n(V)}\qquad\qquad \Vert v \alpha\Vert_{M_n(V)}\leq\Vert \alpha\Vert \Vert v \Vert_{M_n(V)}$$
Where we consider $M_n(\mathbb{C})$ with the usual operator norm.
• For all $v_1\in M_n(V)$ $v_2\in M_m(V)$ $$\left\Vert\begin{pmatrix}v_1 & 0\\ 0&v_2\end{pmatrix}\right\Vert_{M_{n+m}(V)}=\max(\Vert v_1\Vert_{M_n(V)}, \Vert v_2\Vert_{M_m(V)})$$

then the pair $(V, \{\Vert\cdot\Vert_{M_n(V)}:n\in\mathbb{N}\})$ is called an operator space. Two properties mentioned above are called Ruan axioms.

Non-matricial approach. (See here for details) A unital $C^*$-algebra $\mathcal{A}$ is called properly infinite if there exist a family $\{S_n:n\in\mathbb{N}\}\subset\mathcal{A}$ such that $S_k^*S_l=\delta_k^l 1_\mathcal{A}$. If $\mathcal{A}$ is a $C^*$-algebra of bounded operators on Hilbert space then $\{S_n:n\in\mathbb{N}\}$ is a family of isometries with pairwise orthogonal images.

Let $\mathcal{A}$ and $\mathcal{C}$ be unital properly infinite $C^*$-algebras with families of "isometries with pairwise orthogonal images" $\{S_n:n\in\mathbb{N}\}\subset\mathcal{A}$ and $\{T_n:n\in\mathbb{N}\}\subset\mathcal{C}$.

Let $W$ be normed unital contractive $\mathcal{A}$-$\mathcal{C}$-bimodule such that for all $w_1,w_2\in W$ we have $$\Vert S_1 w_1 T_1^* + S_2 w_2 T_2^*\Vert=\max(\Vert w_1\Vert, \Vert w_2 \Vert)$$ Barry Johnson called this property as operator convexity.

Now we want to endow Ruan bimodule with operator space structure. For each $M_n(W)$ we define a norm $\Vert\cdot\Vert_{M_n(W)}$ by equality $$\Vert w\Vert_{M_n(W)}=\Vert\sum\limits_{i=1}^n\sum\limits_{j=1}^n S_i w_{i,j} T_j^*\Vert,\qquad w\in M_n(W)$$ I've checked that the second property is satisfied, but I don't know how to prove the first inequality. It must be an easy straightforward computation.

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It might be helpful to tell us where these definitions come from-- some work of Helemskii maybe? – Matthew Daws Jan 10 '12 at 9:32
Ok, but this is a long story – Norbert Jan 10 '12 at 9:40
I was just thinking of a reference! I mean-- did you just come up with this all on your own? Or is this a problem from a book? What have you already tried to do? – Matthew Daws Jan 10 '12 at 9:42

Okay, so the first point is that the family of operators $(S_i S_j^*)$ for "matrix units". Indeed, the usual matrix units $(e_{i,j})$ satisfy that $e_{i,j}^*=e_{j,i}$ and $e_{i,j} e_{k,l} = \delta^j_k e_{i,l}$. So the map $e_{i,j} \mapsto S_i S_j^*$ is a $*$-isomorphism. Hence, for $\alpha\in M_n$, we have that $$\|\alpha\| = \| \sum_{i,j} \alpha_{i,j} S_i S_j^*\|.$$
So then $$\|\alpha w\| = \| \sum_{i,j,k} S_i \alpha_{i,k} w_{k,j} T_j^* \| = \| \sum_{i,j,k,l} S_i \alpha_{i,k} S_k^* S_l w_{l,j} T_j^* \| = \| a v \|,$$ say where $$a = \sum_{i,k} S_i \alpha_{i,k} S_k^*, \quad v = \sum_{j,l} S_l w_{l,j} T_j^*.$$ So $$\|\alpha w\| \leq \|a\| \|v\| = \|\alpha\| \|w\|.$$