Characterizations

In connection with the column Hilbert space , it is enough to calculate the row norm

$$\Vert T\Vert _{\mathrm{row}} :=\sup_{n \in {\mathbb{N}}} \sup_{\... ...}(X)}\leq 1} \left\Vert\left[Tx_1 \ldots Tx_n\right]\right\Vert _{M_{1,n}(Y)}$$

or the column norm

$$\Vert T\Vert _{\mathrm{col}} :=\sup_{n \in {\mathbb{N}}} \sup_{\... ...rray}{c}Tx_1\\ \vdots \\ Tx_n \end{array} \right]\right\Vert _{M_{n,1}(X)}$$

of an operator $T$, instead of the $\mathrm{cb}$-norm, to ascertain the complete boundedness.

Let $X$ be an operator space and $S:{\mathcal{C}}_\H\rightarrow X$ bzw. $T:X\rightarrow {\mathcal{C}}_\H$. Then we have $\Vert S\Vert _\mathrm{cb}= \Vert S\Vert _{\mathrm{row}}$ resp. $\Vert T\Vert _\mathrm{cb}= \Vert T\Vert _{\mathrm{col}}$ ([Mat94, Prop. 4] resp. [Mat94, Prop. 2]).

The column Hilbert space is characterized as follows,

(A)

as a hilbertian operator space [Mat94, Thm. 8]:

For an operator space $X$ on an Hilbert space $\H$, we have the following equivalences:

1. $X$ is completely isometric to ${\mathcal{C}}_\H$.
2. For all operator spaces $Y$ and all $T:X\rightarrow Y$ we have $\Vert T\Vert _{\mathrm{cb}}=\Vert T\Vert _{\mathrm{row}}$, and for all $S:Y\rightarrow X$ we have $\Vert S\Vert = \Vert S\Vert _{\mathrm{row}}$. For all operator spaces $Y$ and all $T:Y\rightarrow X$ we have $\Vert T\Vert _\mathrm{cb}= \Vert T\Vert _{\mathrm{col}}$, and for all $S:X\rightarrow Y$ we have $\Vert S\Vert = \Vert S\Vert _{\mathrm{col}}$.
3. $X$ coincides with the maximal hilbertian operator space on columns and with the minimal hilbertian operator space on rows. That means isometrically
   

   $$M_{n,1}(X) = M_{n,1}(\mathit{MAX}(\H))$$

   $$M_{1,n}(X) = M_{1,n}(\mathit{MIN}(\H))$$

   
   

(B)

as an operator space: For an operator space $X$ TFAE:

1. There is a Hilbert space $\H$, such that $X \stackrel{\mathrm{cb}}{=}{\mathcal{C}}_\H$ completely isometrically.
2. We have
   $$M_{n,1}(X) = \oplus_2 M_1(X)$$
   and
   $$M_{1,n}(X) = M_{1,n}(\mathit{MIN}(M_1(X)))$$
   isometrically. [Mat94, Thm. 10].
3. $\mathit{CB}(X)$ with the composition as multiplication is an operator algebra [Ble95, Thm. 3.4].