Completion of operator space

Question

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

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

I now want to understand what the completion $Y$ of the operator space $X$ is, in a way that $Y$ becomes an operator space.

The first step should be to begin to define $Y$ to be the Banach space completion of $X$. To define an operator space structure on $Y$, we still require norms on $$M_2(Y), M_3(Y), M_4(Y), \dots$$ and it is not clear to me how to do this with the definition of abstract operator spaces.

If we use the concrete operator space approach (Ruan's theorem), then we can choose a Hilbert space $H$ and a complete isometry $$\sigma: X \to B(H).$$ We can then define $Y = \overline{\sigma(X)}\subseteq B(H)$ which shows that the Banach space completion of $X$ carries the canonical structure of an operator space (of course, it should be shown that this does not depend on the choice of representation and stuff like that). I would like to know if there is a way to avoid Ruan's theorem?

To illustrate where such a situation as in this question may happen: if $X,Y$ are operator spaces, then one can define norms on $M_n(X \odot Y)$ (where $X\odot Y$ is the algebraic tensor product) which satisfy the two requirements above. Thus, $X\odot Y$ becomes an abstract operator space. But then one defines the projective tensor product $X\hat{\otimes} Y$ to be the completion of this space (whatever this means).

Answer 1

The crucial ingredient is that the uniform structure on $M_n(X)$ is already determined by the norm on $X$, there is only a choice in the isometric structure. More precisely, if $[a_{ij}]\in M_n(X)$, then $$ \max_{i,j}\lVert a_{ij}\rVert\leq \lVert [a_{ij}]\rVert\leq \sum_{i,j}\lVert a_{ij}\rVert. $$ This can be seen from the abstract definition of an operator space as follows:

For one, $\|a_{kl}\|=\|E_{1k}[a_{ij}] E_{l1}\|\leq \lVert E_{1i}\rVert\lVert [a_{ij}]\rVert\lVert E_{j1}\rVert=\lVert[a_{ij}]\rVert$. For the second inequality, we have $$ \lVert [a_{ij}]\rVert=\left\lVert \sum_{i,j}a_{ij}\otimes E_{ij}\right\rVert\leq \sum_{i,j}\lVert E_{i1}(a_{ij}\otimes E_{11})E_{1j}\rVert\leq \sum_{i,j}\lVert a_{ij}\oplus 0_{n-1}\rVert\leq \sum_{i,j}\lVert a_{ij}\rVert. $$ Thus $M_n(Y)$ contains $M_n(X)$ as a dense subspace (with respect to this uniform structure, no norm is chosen yet). Then we can define the norm on $M_n(Y)$ simply by identifying it with the normed completion of $M_n(X)$.

Answer 2

Theorem: If $X$ is an operator space, there exists a unique operator space structure on the Banach space completion $Y$ of $X$ such that the norms of $M_n(Y)$ restrict to the norms of $M_n(X)$ (i.e. such that $X$ becomes an operator subsystem of $Y$).

Proof: Uniqueness (as suggested by MaoWao's answer):

Convergence in $M_n(Y)$ is equivalent with convergence of all entries in $Y$ (see the other answer). From this, it easily follows that $M_n(X)$ is norm-dense in $M_n(Y)$. Thus, there is only one possible extension of the norm of $M_n(X)$ to a norm on $M_n(Y)$.

Existence:

• Abstract operator space approach (as suggested in the comments above): We can embed $X$ completely isometrically into its bidual $X^{**}$ using the canonical map. The closure $Y$ of $X$ inside $X^{**}$ (which is a model for the Banach space completion of $X$) inherits a desired operator space structure.

• Concrete operator space approach: Choose a Hilbert space $H$ and a complete isometry $X \subseteq B(H)$. Then $Y:= \overline{X}\subseteq B(H)$ does what we want.