Abstract versus concrete operator spaces

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Definition: An (abstract) operator space $X$ is a linear space together with a sequence $\{\|\cdot\|_n\}_{n=1}^\infty$ of complete norms such that

  • $\|x\oplus y\|_{m+n}= \max\{\|x\|_m, \|y\|_n\}$ for $x\in M_m(X), y \in M_n(Y)$.
  • $\|\alpha x \beta\|_n \le \|\alpha\|\|x\|_m\|\beta\|$ for $\alpha \in M_{n,m}(\mathbb{C}), x \in M_m(\mathbb{C}), \beta \in M_{m,n}(\mathbb{C})$.

Example: Let $X$ be a closed subspace of a unital $C^*$-algebra $A$. Then the inclusion $M_n(X)\subseteq M_n(A)$ induces complete matrix norms on $M_n(X)$ for all $n$, and in this way $X$ becomes an operator space. We refer to such an operator space as a concrete operator space, and Ruan showed that every abstract operator space is completely isometric to a concrete operator space.

To get some feeling with how useful this result is, consider the following:

If $X=[x_{ij}]\in M_n(X)$, consider the statement $$\|x_{kl}\|\le \|X\| \le \sum_{i,j=1}^n \|x_{ij}\|.$$

(1) If we view $X$ as a concrete operator space, this follows immediately from the result as known for $C^*$-algebras. Is this correct?

(2) Is it possible to deduce this result directly from the definition of operator space? Comments like "I don't think so" are also greatly appreciated!

Thanks for your help!

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For (1), you are correct. These inequalities are readily verified for a $C^*$-algebra, as $C^*$-algebras are completely isometric to concrete $C^*$-algebras, i.e. norm-closed $*$-subalgebras of the algebra $\mathbb{B}(\mathcal{H})$ bounded linear operators on some Hilbert space $\mathcal{H}$, and here we are dealing with an operator norm.

For (2), both inequalities follow from the second condition in your definition. For $m\le n$, let $e_m\in M_{n,1}(\mathbb{C})$ be the matrix whose $(m,1)$-entry is $1$, and all other entries are $0$. Then $\|e_m\|=1=\|e_m^*\|$ for all $m$. Now, to see the first inequality, for fixed $k,l\le n$ we have $$\|x_{k,l}\|=\|e_k^*Xe_l\|\le\|X\|,$$ and taking a sup over $k,l$ gets the result. For the second inequality, we can write $$X=\sum_{i,j=1}^ne_i[x_{i,j}]e_j^*.$$ Now apply the triangle inequality and your second condition.