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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 –  userNaN Jan 10 '12 at 9:40
2  
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

1 Answer 1

up vote 3 down vote accepted

I believe these ideas are probably from Wittstock's paper: "Injectivity of the module tensor product of semi-Ruan modules." see or this journal See section 4 in particular. Indeed, the required part is Proposition 4.4, helpfully without proof. It would have been good to have put all this in your question!

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 $$ \|\alpha w\| \leq \|a\| \|v\| = \|\alpha\| \|w\|. $$

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I will add this reference to my question. And great thanks for your answer! –  userNaN Jan 10 '12 at 11:37

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